Search: a204518 -id:a204518
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A001541
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a(0) = 1, a(1) = 3; for n > 1, a(n) = 6*a(n-1) - a(n-2).
(Formerly M3037 N1231)
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+10
116
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1, 3, 17, 99, 577, 3363, 19601, 114243, 665857, 3880899, 22619537, 131836323, 768398401, 4478554083, 26102926097, 152139002499, 886731088897, 5168247530883, 30122754096401, 175568277047523, 1023286908188737, 5964153172084899, 34761632124320657
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OFFSET
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0,2
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COMMENTS
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Chebyshev polynomials of the first kind evaluated at 3.
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 8*y^2 = 1, the corresponding values of y are in A001109. For n > 0, the ratios a(n)/A001090(n) may be obtained as convergents to sqrt(8): either successive convergents of [3; -6] or odd convergents of [2; 1, 4]. - Lekraj Beedassy, Sep 09 2003 [edited by Jon E. Schoenfield, May 04 2014]
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r = sqrt(8). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions greater than 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r = sqrt(2). - Benoit Cloitre, Feb 24 2004
This sequence give numbers n such that (n-1)*(n+1)/2 is a perfect square. Remark: (i-1)*(i+1)/2 = (i^2-1)/2 = -1 = i^2 with i = sqrt(-1) so i is also in the sequence. - Pierre CAMI, Apr 20 2005
a(n) is prime for n = {1, 2, 4, 8}. Prime a(n) are {3, 17, 577, 665857}, which belong to A001601(n). a(2k-1) is divisible by a(1) = 3. a(4k-2) is divisible by a(2) = 17. a(8k-4) is divisible by a(4) = 577. a(16k-8) is divisible by a(8) = 665857. - Alexander Adamchuk, Nov 24 2006
The upper principal convergents to 2^(1/2), beginning with 3/2, 17/12, 99/70, 577/408, comprise a strictly decreasing sequence; essentially, numerators=A001541 and denominators=A001542. - Clark Kimberling, Aug 26 2008
Numbers such that 2n^2 - 2 is a square. Also integer square roots of the expression 2*n^2 + 1, at values of n given by A001542. Also see A228405 regarding 2n^2 -+ 2^k generally for k >= 0. - Richard R. Forberg, Aug 20 2013
Values of x (or y) in the solutions to x^2 - 6xy + y^2 + 8 = 0. - Colin Barker, Feb 04 2014
Panda and Ray call the numbers in this sequence the Lucas-balancing numbers C_n (see references and links).
Partial sums of X or X+1 of Pythagorean triples (X,X+1,Z). - Peter M. Chema, Feb 03 2017
a(n)/A001542(n) is the closest rational approximation to sqrt(2) with a numerator not larger than a(n), and 2*A001542(n)/a(n) is the closest rational approximation to sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001653 and A002315 give a complete set of closest rational approximations to sqrt(2) with restricted numerator or denominator. a(n)/A001542(n) > sqrt(2) > 2*A001542(n)/a(n). - A.H.M. Smeets, May 28 2017
x = a(n), y = A001542(n) are solutions of the Diophantine equation x^2 - 2y^2 = 1 (Pell equation). x = 2*A001542(n), y = a(n) are solutions of the Diophantine equation x^2 - 2y^2 = -2. Both together give the set of fractional approximations for sqrt(2) obtained from limited fractions obtained from continued fraction representation to sqrt(2). - A.H.M. Smeets, Jun 22 2017
a(n) is the radius of the n-th circle among the sequence of circles generated as follows: Starting with a unit circle centered at the origin, every subsequent circle touches the previous circle as well as the two limbs of hyperbola x^2 - y^2 = 1, and lies in the region y > 0. - Kaushal Agrawal, Nov 10 2018
All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0<a<b<c are given by a=A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0<n. - Michael Somos, Jun 26 2022
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REFERENCES
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Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009)
J. W. L. Glaisher, On Eulerian numbers (formulas, residues, end-figures), with the values of the first twenty-seven, Quarterly Journal of Mathematics, vol. 45, 1914, pp. 1-51.
G. K. Panda, Some fascinating properties of balancing numbers, In Proc. of Eleventh Internat. Conference on Fibonacci Numbers and Their Applications, Cong. Numerantium 194 (2009), 185-189.
A. Patra, G. K. Panda, and T. Khemaratchatakumthorn. "Exact divisibility by powers of the balancing and Lucas-balancing numbers." Fibonacci Quart., 59:1 (2021), 57-64; see C(n).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022
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LINKS
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J. M. Katri and D. R. Byrkit, Problem E1976, Amer. Math. Monthly, 75 (1968), 683-684.
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FORMULA
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E.g.f.: exp(3*x)*cosh(2*sqrt(2)*x). Binomial transform of A084128. - Paul Barry, May 16 2003
a(n) = sqrt(8*((A001109(n))^2) + 1).
a(n) = T(n, 3), with Chebyshev's T-polynomials A053120. (End)
a(n) = ((3+2*sqrt(2))^n + (3-2*sqrt(2))^n)/2.
a(n) ~ (1/2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all elements x of the sequence, 2*x^2 - 2 is a square. Limit_{n -> infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 10 2002 [corrected by Peter Pein, Mar 09 2009]
a(n) = ((-1+sqrt(2))^n + (1+sqrt(2))^n + (1-sqrt(2))^n + (-1-sqrt(2))^n)/4 (with interpolated zeros).
E.g.f.: cosh(x)*cosh(sqrt(2)x) (with interpolated zeros). (End)
a(n+1) - A001542(n+1) = A090390(n+1) - A046729(n) = A001653(n); a(n+1) - 4*A079291(n+1) = (-1)^(n+1). Formula generated by the floretion - .5'i + .5'j - .5i' + .5j' - 'ii' + 'jj' - 2'kk' + 'ij' + .5'ik' + 'ji' + .5'jk' + .5'ki' + .5'kj' + e. - Creighton Dement, Nov 16 2004
a(n) = 3*a(n-1) + 4*A001542(n-1); e.g., a(4) = 99 = 3*17 + 4*12. - Zak Seidov, Dec 19 2013
a(n) = cos(n * arccos(3)) = cosh(n * log(3 + 2*sqrt(2))). - Daniel Suteu, Jul 28 2016
Inverse binomial transform of A084130.
Sum_{n>=0} (-1)^n*a(n)/n! = cosh(2*sqrt(2))/exp(3) = 0.4226407909842764637... (End)
a(n) = S(n, 6) - 3*S(n-1, 6), for n >= 0, with S(n, 6) = A001109(n+1), (Chebyshev S of A049310). See the first comment and the formula a(n) = T(n, 3). - Wolfdieter Lang, Nov 22 2020
a(n) = [x^n] (3*x + sqrt(1 + 8*x^2))^n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) hold for all prime p and positive integers n and k.
O.g.f. A(x) = 1 + x*d/dx(log(B(x))), where B(x) = 1/sqrt(1 - 6*x + x^2) is the o.g.f. of A001850. (End)
Sum_{n >= 1} 1/(a(n) - 2/a(n)) = 1/2.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 1/a(n)) = 1/4.
Sum_{n >= 1} 1/(a(n)^2 - 2) = 1/2 - 1/sqrt(8). (End)
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EXAMPLE
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G.f. = 1 + 3*x + 17*x^2 + 99*x^3 + 577*x^4 + 3363*x^5 + 19601*x^6 + 114243*x^7 + ...
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MAPLE
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a[0]:=1: a[1]:=3: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006
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MATHEMATICA
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Table[Simplify[(1/2) (3 + 2 Sqrt[2])^n + (1/2) (3 - 2 Sqrt[2])^n], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
a[ n_] := If[n == 0, 1, With[{m = Abs @ n}, m Sum[4^i Binomial[m + i, 2 i]/(m + i), {i, 0, m}]]]; (* Michael Somos, Jul 11 2011 *)
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PROG
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(PARI) {a(n) = real((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */
(PARI) {a(n) = subst( poltchebi( abs(n)), x, 3)}; /* Michael Somos, Apr 07 2003 */
(PARI) {a(n) = if( n<0, a(-n), polsym(1 - 6*x + x^2, n) [n+1] / 2)}; /* Michael Somos, Apr 07 2003 */
(PARI) {a(n) = polchebyshev( n, 1, 3)}; /* Michael Somos, Jul 11 2011 */
(PARI) a(n)=([1, 2, 2; 2, 1, 2; 2, 2, 3]^n)[3, 3] \\ Vim Wenders, Mar 28 2007
(Magma) [n: n in [1..10000000] |IsSquare(8*(n^2-1))] // Vincenzo Librandi, Nov 18 2010
(Haskell)
a001541 n = a001541_list !! (n-1)
a001541_list =
1 : 3 : zipWith (-) (map (* 6) $ tail a001541_list) a001541_list
(Scheme, with memoization-macro definec)
(definec (A001541 n) (cond ((zero? n) 1) ((= 1 n) 3) (else (- (* 6 (A001541 (- n 1))) (A001541 (- n 2))))))
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CROSSREFS
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Cf. A055997 = numbers n such that n(n-1)/2 is a square.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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A001075
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a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).
(Formerly M1769 N0700)
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+10
104
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1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647
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OFFSET
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0,2
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COMMENTS
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Chebyshev's T(n,x) polynomials evaluated at x=2.
x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).
Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002
For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]
a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003
a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008
The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008
Pisano period lengths: 1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 48 = 0. - Colin Barker, Feb 10 2014
A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.
M =
1, 1, 0, 0, 0, 0, ...
2, 0, 3, 0, 0, 0, ...
2, 0, 0, 3, 0, 0, ...
2, 0, 0, 0, 3, 0, ...
2, 0, 0, 0, 0, 3, ...
...
The triangle generated from M is:
1,
1, 1,
3, 1, 3,
11, 3, 3, 9,
41, 11, 9, 9, 27,
...
Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016
For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016
If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.
a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.
a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)
(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018
Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020
a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022
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REFERENCES
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Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
Eugene McDonnell, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.
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LINKS
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FORMULA
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G.f.: (1 - 2*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(2*x)*cosh(sqrt(3)*x).
a(n) = 4*a(n-1) - a(n-2) = a(-n).
a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n >= 0. See A049310 and A053120.
a(n) = sqrt(1 + 3*A001353(n)) (cf. Richardson comment, Oct 10 2002).
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n, 2*k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01, 2004
a(n) = ((2 + sqrt(3))^n + (2 - sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2 + sqrt(3))^(n)).
a(n) = cosh(n * log(2 + sqrt(3))).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k)*3^k. - Paul Barry, May 08 2003
a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006
Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k - 4)/(x*(3*k - 1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = (tan(Pi/12)^n + tan(5*Pi/12)^n)/2. - Greg Dresden, Oct 01 2020
a(n) = (1/2)^n * [x^n] ( 4*x + sqrt(1 + 12*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - (3/2)/a(n)) = 1.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (1/2)/a(n)) = 1/3.
Sum_{n >= 1} 1/(a(n)^2 - 3/2) = 1 - 1/sqrt(3). (End)
a(n) = binomial(2*n, n) + 2*Sum_{k > 0} binomial(2*n, n+2*k)*cos(k*Pi/3). - Greg Dresden, Oct 11 2022
2*a(n) + 2^n = 3*Sum_{k=-n..n} (-1)^k*binomial(2*n, n+6*k). - Greg Dresden, Feb 07 2023
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EXAMPLE
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2^6 - 1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 - 1 = 31 divides a(2^3) = 18817, therefore 31 is prime.
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...
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MAPLE
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orthopoly[T](n, 2) ;
end proc:
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MATHEMATICA
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Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]
LinearRecurrence[{4, -1}, {1, 2}, 30] (* Harvey P. Dale, Aug 22 2015 *)
a[ n_] := LucasL[2*n, x]/2 /. x->Sqrt[2]; (* Michael Somos, Sep 05 2022 *)
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PROG
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(PARI) {a(n) = subst(poltchebi(abs(n)), x, 2)};
(PARI) {a(n) = real((2 + quadgen(12))^abs(n))};
(PARI) {a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};
(PARI) my(x='x+O('x^30)); Vec((1-2*x)/(1-4*x+x^2)) \\ G. C. Greubel, Dec 19 2017
(SageMath) [lucas_number2(n, 4, 1)/2 for n in range(0, 25)] # Zerinvary Lajos, May 14 2009
(Haskell)
a001075 n = a001075_list !! n
a001075_list =
1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list
(SageMath)
def a(n):
Q = QuadraticField(3, 't')
u = Q.units()[0]
(Magma) I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A023110
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Squares which remain squares when the last digit is removed.
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+10
30
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0, 1, 4, 9, 16, 49, 169, 256, 361, 1444, 3249, 18496, 64009, 237169, 364816, 519841, 2079364, 4678569, 26666896, 92294449, 341991049, 526060096, 749609641, 2998438564, 6746486769, 38453641216, 133088524969, 493150849009, 758578289296, 1080936581761
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OFFSET
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1,3
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COMMENTS
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For the first 4 terms the square has only one digit. It is understood that deleting this digit yields 0. - Colin Barker, Dec 31 2017
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REFERENCES
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R. K. Guy, Neg and Reg, preprint, Jan 2012.
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LINKS
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Jon E. Schoenfield, Table of n, a(n) for n = 1..70 (terms 1..38 from David W. Wilson, terms 39..40 from Robert G. Wilson v, terms 41..67 from Dmitry Petukhov)
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FORMULA
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Appears to satisfy a(n)=1444*a(n-7)+a(n-14)-76*sqrt(a(n-7)*a(n-14)) for n >= 16. For n = 15, 14, 13, ... this would require a(1) = 16, a(0) = 49, a(-1) = 169, ... - Henry Bottomley, May 08 2001; edited by Robert Israel, Sep 28 2014
G.f.: x^2*(1 + 4*x + 9*x^2 + 16*x^3 + 49*x^4 + 169*x^5 + 256*x^6 - 1082*x^7 - 4328*x^8 - 9738*x^9 - 4592*x^10 - 6698*x^11 - 6698*x^12 - 4592*x^13 + 361*x^14 + 1444*x^15 + 3249*x^16 + 256*x^17 + 169*x^18 + 49*x^19 + 16*x^20) / ((1 - x)*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)*(1 - 1442*x^7 + x^14)).
a(n) = 1443*a(n-7) - 1443*a(n-14) + a(n-21) for n>22.
(End)
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MAPLE
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count:= 1: A[1]:= 0:
for n from 0 while count < 35 do
for t in [1, 4, 6, 9] do
if issqr(10*n^2+t) then
count:= count+1;
A[count]:= 10*n^2+t;
fi
od
od:
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MATHEMATICA
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fQ[n_] := IntegerQ@ Sqrt@ Quotient[n^2, 10]; Select[ Range@ 1000000, fQ]^2 (* Robert G. Wilson v, Jan 15 2011 *)
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PROG
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(PARI) for(n=0, 1e7, issquare(n^2\10) & print1(n^2", ")) \\ M. F. Hasler, Jan 16 2012
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A031149
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Numbers whose square with its last digit deleted is also a square.
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+10
22
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0, 1, 2, 3, 4, 7, 13, 16, 19, 38, 57, 136, 253, 487, 604, 721, 1442, 2163, 5164, 9607, 18493, 22936, 27379, 54758, 82137, 196096, 364813, 702247, 870964, 1039681, 2079362, 3119043, 7446484, 13853287, 26666893, 33073696, 39480499, 78960998, 118441497, 282770296
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OFFSET
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1,3
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COMMENTS
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For the first 4 terms, the square has only one digit, but in analogy to the sequences in other bases (A204502, A204512, A204514, A204516, A204518, A204520, A004275, A055793, A055792), it is understood that deleting this digit yields 0.
Solutions x to x^2 = 10*y^2 + j, j in {0,1,4,6,9}, in increasing order of x.
j=0 occurs only for x=0.
Let M be the 2 X 2 matrix [19, 60; 6, 19].
Solutions of x^2 = 10*y^2 + 1 are (x,y)^T = M^k (1,0)^T for k >= 0.
Solutions of x^2 = 10*y^2 + 4 are (x,y)^T = M^k (2,0)^T for k >= 0.
Solutions of x^2 = 10*y^2 + 6 are (x,y)^T = M^k (4,1)^T and M^k (16,5)^T for k >= 0.
Solutions of x^2 = 10*y^2 + 9 are (x,y)^T = M^k (3,0)^T, M^k (7,2)^T, M^k (13,4)^T for k >= 0.
Since (1,0)^T <= (2,0)^T <= (3,0)^T <= (4,1)^T <= (7,2)^T <= (13,4)^T <= (16,5)^T <= (19,6)^T = M (1,0)^T (element-wise) and M has positive entries, we see that the terms always occur in this order, for successive k.
The eigenvalues of M are 19 + 6*sqrt(10) and 19 - 6*sqrt(10).
From this follow my formulas below and the G.f. (End)
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REFERENCES
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R. K. Guy, Neg and Reg, preprint, Jan 2012. [From N. J. A. Sloane, Jan 12 2012]
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LINKS
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FORMULA
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Appears to satisfy: a(n)=38a(n-7)-a(n-14) which would require a(-k) to look like 3, 2, 1, 4, 7, 13, 16, 57, 38, 19, 136, ... for k>0. - Henry Bottomley, May 08 2001
Empirical g.f.: x^2*(1 + 2*x + 3*x^2 + 4*x^3 + 7*x^4 + 13*x^5 + 16*x^6 - 19*x^7 - 38*x^8 - 57*x^9 - 16*x^10 - 13*x^11 - 7*x^12 - 4*x^13) / ((1 - 38*x^7 + x^14)). - Colin Barker, Jan 17 2014
a(n) = 38*a(n-7) - a(n-14) for n>15 (conjectured). - Colin Barker, Dec 31 2017
With e1 = 19 + 6*sqrt(10) and e2 = 19 - 6*sqrt(10),
a(2+7k) = (e1^k + e2^k)/2,
a(3+7k) = e1^k + e2^k,
a(4+7k) = (3/2) (e1^k + e2^k),
a(5+7k) = (2+sqrt(10)/2) e1^k + (2-sqrt(10)/2) e2^k,
a(6+7k) = (7/2+sqrt(10)) e1^k + (7/2-sqrt(10)) e2^k,
a(7+7k) = (13/2+2 sqrt(10)) e1^k + (13/2-2 sqrt(10)) e2^k,
a(8+7k) = (8+5 sqrt(10)/2) e1^k + (8-5 sqrt(10)/2) e2^k. - Robert Israel, Feb 16 2016
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EXAMPLE
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364813^2 = 133088524969, 115364^2 = 13308852496.
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MAPLE
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for i from 1 to 150000 do if (floor(sqrt(10 * i^2 + 9)) > floor(sqrt(10 * i^2))) then print(floor(sqrt(10 * i^2 + 9))) end if end do;
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MATHEMATICA
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fQ[n_] := IntegerQ@ Sqrt@ Quotient[n^2, 10]; Select[ Range[ 0, 40000000], fQ] (* Harvey P. Dale, Jun 15 2011 *) (* modified by Robert G. Wilson v, Jan 16 2012 *)
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PROG
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(PARI) s=[]; for(n=0, 1e7, if(issquare(n^2\10), s=concat(s, n))); s \\ Colin Barker, Jan 17 2014; typo fixed by Zak Seidov, Jan 31 2014
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A204502
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Numbers such that floor[a(n)^2 / 9] is a square.
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+10
20
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0, 1, 2, 3, 4, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165, 168, 171, 174, 177
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OFFSET
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1,3
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COMMENTS
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Or, numbers n such that n^2, with its last base-9 digit dropped, is again a square. (Except maybe for the 3 initial terms whose square has only 1 digit in base 9.)
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LINKS
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FORMULA
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Conjecture: a(n) = 3*n-12 for n>5. G.f.: x^2*(x^2+x+1)*(x^3-x+1)/(x-1)^2. [Colin Barker, Nov 23 2012]
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MATHEMATICA
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Select[Range[0, 200], IntegerQ[Sqrt[Floor[#^2/9]]]&] (* Harvey P. Dale, May 05 2018 *)
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PROG
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(PARI) b=9; for(n=0, 200, issquare(n^2\b) & print1(n", "))
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CROSSREFS
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The squares are in A204503, the squares with last base-9 digit dropped in A204504, and the square roots of the latter in A028310.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A204514
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Numbers such that floor(a(n)^2 / 8) is again a square.
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+10
20
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0, 1, 2, 3, 6, 17, 34, 99, 198, 577, 1154, 3363, 6726, 19601, 39202, 114243, 228486, 665857, 1331714, 3880899, 7761798, 22619537, 45239074, 131836323, 263672646, 768398401, 1536796802, 4478554083, 8957108166, 26102926097, 52205852194, 152139002499, 304278004998, 886731088897
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OFFSET
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1,3
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COMMENTS
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Or: Numbers whose square, with its last base-8 digit dropped, is again a square. (Except maybe for the 3 initial terms whose square has only 1 digit in base 8.)
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LINKS
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FORMULA
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G.f. = (x^2 + 2*x^3 - 3*x^4 - 6*x^5)/(1 - 6*x^2 + x^4).
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MAPLE
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MATHEMATICA
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CoefficientList[Series[(x^2 + 2*x^3 - 3*x^4 - 6*x^5)/(x (1 - 6*x^2 + x^4)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Sep 28 2014 *)
LinearRecurrence[{0, 6, 0, -1}, {0, 1, 2, 3, 6}, 40] (* Harvey P. Dale, Nov 23 2022 *)
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PROG
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(PARI) b=8; for(n=0, 1e7, issquare(n^2\b) & print1(n", "))
(PARI) A204514(n)=polcoeff((x + 2*x^2 - 3*x^3 - 6*x^4)/(1 - 6*x^2 + x^4+O(x^(n+!n))), n-1, x)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A204516
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Numbers such that floor(a(n)^2 / 7) is a square.
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+10
20
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0, 1, 2, 3, 8, 16, 45, 127, 254, 717, 2024, 4048, 11427, 32257, 64514, 182115, 514088, 1028176, 2902413, 8193151, 16386302, 46256493, 130576328, 261152656, 737201475, 2081028097, 4162056194, 11748967107, 33165873224, 66331746448
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OFFSET
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1,3
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COMMENTS
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Or: Numbers whose square, with its last base-7 digit dropped, is again a square (where for the first 3 terms, dropping the digit is meant to yield zero).
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LINKS
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FORMULA
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G.f. = (x + 2*x^2 + 3*x^3 - 8*x^4 - 16*x^5 - 3*x^6 )/(1 - 16*x^3 + x^6).
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MATHEMATICA
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LinearRecurrence[{0, 0, 16, 0, 0, -1}, {0, 1, 2, 3, 8, 16, 45}, 30] (* or *) CoefficientList[Series[ (x+2x^2+3x^3-8x^4-16x^5-3x^6)/(1-16x^3+x^6), {x, 0, 30}], x] (* Harvey P. Dale, Apr 22 2023 *)
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PROG
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(PARI) b=7; for(n=0, 2e9, issquare(n^2\b) & print1(n", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A204520
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Numbers such that floor(a(n)^2 / 5) is a square.
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+10
19
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0, 1, 2, 3, 7, 9, 18, 47, 123, 161, 322, 843, 2207, 2889, 5778, 15127, 39603, 51841, 103682, 271443, 710647, 930249, 1860498, 4870847, 12752043, 16692641, 33385282
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OFFSET
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1,3
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COMMENTS
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Also: Numbers whose square, with its last base-5 digit dropped, is again a square. (For the three initial terms whose squares have only one digit in base 5, it is then understood that this yields zero.)
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LINKS
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FORMULA
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Empirical g.f.: -x^2*(x+1)*(3*x^6 + 4*x^5 + 14*x^4 - 5*x^3 - 2*x^2 - x-1) / ((x^4 - 4*x^2 - 1)*(x^4 + 4*x^2 - 1)). - Colin Barker, Sep 15 2014
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PROG
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(PARI) b=5; for(n=0, 2e9, issquare(n^2\b) && print1(n", "))
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CROSSREFS
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Cf. A031149, A055812, A204502, A204514, A204516, A204518 and A004275, A001075, A001541 for the analog in bases 10,...,6 and 4, 3, 2.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A055851
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a(n) and floor(a(n)/6) are both squares; i.e., squares that remain squares when written in base 6 and last digit is removed.
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+10
18
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0, 1, 4, 9, 25, 100, 729, 2401, 9604, 71289, 235225, 940900, 6985449, 23049601, 92198404, 684502569, 2258625625, 9034502500, 67074266169, 221322261601, 885289046404, 6572593581849, 21687323011225, 86749292044900
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OFFSET
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1,3
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COMMENTS
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For the first 3 terms, the above "base 6" interpretation is questionable, since they have only 1 digit in base 6. It is understood that dropping this digit yields 0. - M. F. Hasler, Jan 15 2012
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LINKS
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FORMULA
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Empirical g.f.: -x^2*(9*x^8+100*x^7+25*x^6-162*x^5-296*x^4-74*x^3+9*x^2+4*x+1) / ((x-1)*(x^2+x+1)*(x^6-98*x^3+1)). - Colin Barker, Sep 15 2014
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EXAMPLE
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a(5) = 100 because 100 = 10^2 = 244 base 6 and 24 base 6 = 16 = 4^2.
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PROG
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(PARI) b=6; for(n=1, 2e9, issquare(n^2\b) & print1(n^2, ", ")) \\ M. F. Hasler, Jan 15 2012
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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EXTENSIONS
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More terms added and offset changed to 1 by M. F. Hasler, Jan 16 2012
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STATUS
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approved
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0, 0, 0, 1, 3, 6, 17, 48, 96, 271, 765, 1530, 4319, 12192, 24384, 68833, 194307, 388614, 1097009, 3096720, 6193440, 17483311, 49353213, 98706426, 278635967, 786554688, 1573109376, 4440692161, 12535521795, 25071043590, 70772438609, 199781794032, 399563588064
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OFFSET
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1,5
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LINKS
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FORMULA
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G.f. = (x^4 + 3*x^5 + 6*x^6 + x^7)/(1 - 16*x^3 + x^6)
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PROG
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(PARI) b=7; for(n=1, 2e9, issquare(n^2\b) & print1(sqrtint(n^2\b), ", "))
(PARI) A204517(n)=polcoeff((x^4 + 3*x^5 + 6*x^6 + x^7)/(1 - 16*x^3 + x^6+O(x^n)), n)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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