Search: a093131 -id:a093131
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A020876
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a(n) = ((5+sqrt(5))/2)^n + ((5-sqrt(5))/2)^n.
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+10
15
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2, 5, 15, 50, 175, 625, 2250, 8125, 29375, 106250, 384375, 1390625, 5031250, 18203125, 65859375, 238281250, 862109375, 3119140625, 11285156250, 40830078125, 147724609375, 534472656250, 1933740234375, 6996337890625, 25312988281250, 91583251953125
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OFFSET
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0,1
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COMMENTS
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Number of no-leaf edge-subgraphs in Moebius ladder M_n.
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LINKS
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FORMULA
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Also, a(n) = (sqrt(5)*phi)^n + (sqrt(5)/phi)^n, where phi = golden ratio. - N. J. A. Sloane, Aug 08 2014
Let S(n, m)=sum(k=0, n, binomial(n, k)*fibonacci(m*k)), then for n>0 a(n)= S(2*n, 2)/S(n, 2). - Benoit Cloitre, Oct 22 2003
a(n)= 5*a(n-1) - 5*a(n-2).
G.f.: (2-5*x)/(1-5*x+5*x^2). (End)
a((2*m+1)*k)/a(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*a(2*(m-i)*k) + 5^(m*k).
a(m+r)*a(n+s) + a(m+s)*a(n+r) = 2*a(m+n+r+s) + 5^(n+s)*a(m-n)*a(r-s).
a(n)^2 - a(n+1)*a(n-1) = -5^n.
a(n)^2 - a(n+r)*a(n-r) = -5^(n-r+1)*A093131(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = -5^(n+1)*A093131(m-n).
a(m+n) + 5^(n)*a(m-n) = a(m)*a(n).
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EXAMPLE
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G.f. = 2 + 5*x + 15*x^2 + 50*x^3 + 175*x^4 + 625*x^5 + 2250*x^6 + ...
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MAPLE
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G:=(x, n)-> cos(x)^n+cos(3*x)^n:
seq(simplify(4^n*G(Pi/10, 2*n)), n=0..22); # Gary Detlefs, Dec 05 2010
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MATHEMATICA
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Table[Sum[LucasL[2*i] Binomial[n, i], {i, 0, n}], {n, 0, 50}] (* T. D. Noe, Sep 10 2011 *)
CoefficientList[Series[(2 - 5 x)/(1 - 5 x + 5 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 08 2014 *)
LinearRecurrence[{5, -5}, {2, 5}, 30] (* Harvey P. Dale, Mar 13 2016 *)
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PROG
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(Sage) [lucas_number2(n, 5, 5) for n in range(0, 24)] # Zerinvary Lajos, Jul 08 2008
(Magma) [Floor(((5+Sqrt(5))/2)^n+((5-Sqrt(5))/2)^n): n in [0..30]]; // Vincenzo Librandi, Aug 08 2014
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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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EXTENSIONS
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STATUS
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approved
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1, 4, 15, 55, 200, 725, 2625, 9500, 34375, 124375, 450000, 1628125, 5890625, 21312500, 77109375, 278984375, 1009375000, 3651953125, 13212890625, 47804687500, 172958984375, 625771484375, 2264062500000, 8191455078125
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OFFSET
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1,2
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COMMENTS
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Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 3, s(2n) = 5.
The array belongs to a family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the o.g.f. (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse x*(1-x)/(1 + (t-1)*x*(1-x)). See A091867 for more info on this family. Here t = -4 (mod signs in the results).
Let C(x) = (1 - sqrt(1-4x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t).
O.g.f.: G(x) = x*(1-x)/(1 - 5x*(1-x)) = P(Cinv(x),-5).
Inverse O.g.f.: Ginv(x) = (1 - sqrt(1 - 4*x/(1+5x)))/2 = C(P(x,5)) (signed A026378). Cf. A030528. (End)
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LINKS
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FORMULA
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G.f.: x*(1-x)/(1-5*x+5*x^2) = g1(3, x)/(1-g1(3, x)), g1(3, x) := x*(1-x)/(1-2*x)^2 (g.f. first column of A030523).
Binomial transform of Fibonacci(2n+2).
a(n) = (sqrt(5)/2 + 5/2)^n*(3*sqrt(5)/10 + 1/2) - (5/2 - sqrt(5)/2)^n*(3*sqrt(5)/10 - 1/2). (End)
a(n) = (1/5)*Sum_{r=1..9} sin(3*r*Pi/10)*sin(r*Pi/2)*(2*cos(r*Pi/10))^(2n)).
a(n) = 5*a(n-1) - 5*a(n-2).
a(n) = Sum_{k=0..n} Sum_{i=0..n} binomial(n, i)*binomial(k+i+1, 2k+1). - Paul Barry, Jun 22 2004
Limit_{k->infinity} a(n+k)/a(k) = (A020876(n) + A093131(n)*sqrt(5))/2.
(End)
Limit_{k->infinity} a(k+1)/a(k) = 1 + phi^2 = (5 + sqrt(5)) / 2.
a(n) = a(n-1) * 3 + A081567(n-2) (not proved).
(End)
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MATHEMATICA
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CoefficientList[Series[(1 - x) / (1 - 5 x + 5 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *)
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PROG
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(PARI) Vec(x*(1-x)/(1-5*x+5*x^2) + O(x^40)) \\ Altug Alkan, Nov 20 2015
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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A316269
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Array T(n,k) = n*T(n,k-1) - T(n,k-2) read by upward antidiagonals, with T(n,0) = 0, T(n,1) = 1, n >= 2.
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+10
9
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0, 0, 1, 0, 1, 2, 0, 1, 3, 3, 0, 1, 4, 8, 4, 0, 1, 5, 15, 21, 5, 0, 1, 6, 24, 56, 55, 6, 0, 1, 7, 35, 115, 209, 144, 7, 0, 1, 8, 48, 204, 551, 780, 377, 8, 0, 1, 9, 63, 329, 1189, 2640, 2911, 987, 9, 0, 1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10
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OFFSET
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2,6
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COMMENTS
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Define {x(k)} to be an integer sequence satisfying all the following conditions:
(i) {x(k)} satisfies second-order linear recursion, that is, there exists two integers P, Q such that x(k+2) = P*x(k+1) + Q*x(k) holds for all k >= 0.
(ii) {x(k)} is (not necessarily strictly) increasing. ({A000035(k)} satisfies condition (i), but it doesn't satisfy this.)
(iii) All terms in {x(k)} do not share a common factor. ({A024023(k)} satisfies both conditions (i) and (ii), but all terms share a common factor 2.)
(iv) {x(k)} satisfies strong divisibility, that is, gcd(x(m),x(n)) = x(gcd(m,n)) holds for all m, n >= 0. ({A093131(k)} satisfies all conditions (i) to (iii), but 5 = gcd(A093131(2),A093131(3)) != A093131(gcd(2,3)) = 1.)
(v) For all positive integers n, there eventually exists some m > 0 such that n divides x(m). ({A002275(k)} satisfies all conditions (i) to (iv), but 2, 5 and 10 never divide any term.)
Then it's easy to show that the only solutions to {x(k)} are x(k) = A172236(n,k) or x(k) = T(n,k), i.e., x(0) = 0, x(1) = 1, P >= 1, Q = 1 or P >= 2, Q = -1.
The case n = 0 is not included since it gives the period-4 signed sequence 0, 1, 0, -1, 0, 1, 0, -1, ..., the g.f. of which is the inverse of the 4th cyclotomic polynomial.
The case n = 1 is not included since it gives the period-6 signed sequence 0, 1, 1, 0, -1, -1, ..., the g.f. of which is the inverse of the 6th cyclotomic polynomial.
The congruence property: let p be an odd prime which is not divisible by n^2 - 4, then T(n,(p-1)/2) == 1/2(((n-2)/p) - ((n+2)/p)) (mod p), T(n,(p+1)/2) == 1/2(((n-2)/p) + ((n+2)/p)) (mod p). Here ((n-2)/p) is the Legendre symbol. Or equivalently:
((n-2)/p)...((n+2)/p)...T(n,(p-1)/2) mod p...T(n,(p+1)/2) mod p
.....1...........1...............0....................1
....-1..........-1...............0...................-1
.....1..........-1...............1....................0
....-1...........1..............-1....................0
To prove this, rewrite (n +- sqrt(n^2-4))/2 as ((sqrt(n+2) +- sqrt(n-2))/2)^2.
Let E(n,m) be the smallest number l such that m divides T(n,l), we have: E(n,p) divides (p - ((n^2-4)/p))/2 for odd primes p that are not divisible by n^2 - 4. E(n,p) = p for odd primes p that are divisible by n^2 - 4. E(n,2) = 2 for even n and 3 for odd n.
E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide T(n,E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 - 4, p^2 is never divisible by T(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1.
Given n, the largest possible value of E(n,m)/m is 1 for even n and 3/2 for odd n. It can be obtained by the value of E(n,2) described above.
Let pi(n,m) be the Pisano period of T(n,k) modulo m, i.e, the smallest number l such that T(n,k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p + ((n-2)/p) if ((n+2)/p) = -1 and (p - ((n-2)/p))/2 if ((n+2)/p) = 1. pi(n,p) = p for odd primes p that are divisible by n - 2 and 2p for odd primes p that are divisible by n + 2. pi(n,2) = 2 even n and 3 for odd n.
pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide T(n,E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 - 4, p^2 is never divisible by T(n,p), so pi(n,p^e) = p^e or 2p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1),pi(n,m_2)) if gcd(m_1,m_2) = 1.
Given n, the largest possible value of pi(n,m)/m is:
Parity of n...n + 2 is a power of 2 or 3...max{pi(n,m)/m}.....obtained by
....even..........yes (even exponent)............1...........pi(n,2^e) = 2^e
....even...........yes (odd exponent)...........4/3............pi(n,3) = 4
....even...................no....................2.............pi(n,p) = 2p (p >= 3 is any prime factor of n + 2)
.....odd..................yes....................2..........pi(n,3^e) = 2*3^e
.....odd...................no....................3..........pi(n,2p^e) = 6p^e (p >= 5 is any prime factor of n + 2)
The largest possible value of pi(n,m)/m is obtained by infinitely many m except for the case n = 10, in which we have pi(10,3) = 6, pi(10,7) = 8, pi(10,21) = 24 and pi(10,m)/m <= 14/13 for all other m. [Corrected by Jianing Song, Nov 04 2018]
Let z(n,m) be the number of zeros in a period of T(n,k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: for odd primes p that are not divisible by n^2 - 4, z(n,p) = 2 if ((n+2)/p) = -1; 1 or 2 if ((n+2)/p) = 1. z(n,p) = 1 for odd primes p that are divisible by n - 2 and 2 for odd primes p that are divisible by n + 2. z(n,2) = 1.
For all odd primes p, z(n,p) = 2 if and only if pi(n,p) is even, z(n,p) = 1 if and only if pi(n,p) is odd. For all odd primes p, if E(n,p) is even then z(n,p) = 2 (the converse is not necessarily true). [Comment revised by Jianing Song, Jul 06 2019]
z(n,p^e) = z(n,p) for all odd primes p. z(n,4) = 1 if n == 2, 3 (mod 4) and 2 if n == 0, 1 (mod 4). z(n,2^e) = 1 for even n and 2 for odd n, e >= 3.
By induction it is easy to show that T(n,k) = T(n,m+1)*T(n,k-m) - T(n,m)*T(k-m-1). Let k = 2m we have T(n,2m) = T(n,m)*(T(n,m+1)-T(m-1)); let k = 2m+1 we have T(n,2m+1) = T(n,m+1)^2 - T(n,m)^2 = (T(n,m+1)+T(n,m))*(T(n,m+1)-T(n,m)). So T(n,k) is composite if n >= 3, k >= 3. - Jianing Song, Jul 06 2019
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LINKS
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FORMULA
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T(2,k) = k; T(n,k) = (((n+sqrt(n^2 - 4))/2)^k - ((n - sqrt(n^2 - 4))/2)^k)/sqrt(n^2 - 4), n >= 3, k >= 0.
For n >= 2, Sum_{i=1..k} 1/T(n,2^i) = 2/n - ((u^(2^k-1) + v^(2^k-1))/(u + v)) * (1/T(n,2^k)), where u = (n + sqrt(n^2 - 4))/2, v = (n - sqrt(n^2 - 4))/2 are the two roots of the polynomial x^2 - n*x + 1. As a result, Sum_{i=>1} 1/T(n,2^i) = (n - sqrt(n^2 - 4))/2. - Jianing Song, Apr 21 2019
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EXAMPLE
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The array starts in row n = 2 with columns k >= 0 as follows:
0 1 2 3 4 5 6
0 1 3 8 21 55 144
0 1 4 15 56 209 780
0 1 5 24 115 551 2640
0 1 6 35 204 1189 6930
0 1 7 48 329 2255 15456
0 1 8 63 496 3905 30744
0 1 9 80 711 6319 56160
0 1 10 99 980 9701 96030
0 1 11 120 1309 14279 155760
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MATHEMATICA
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Table[If[# == 2, k, Simplify[(((# + Sqrt[#^2 - 4])/2)^k - ((# - Sqrt[#^2 - 4])/2)^k)/Sqrt[#^2 - 4]]] &[n - k + 2], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Jul 19 2018 *)
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PROG
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(PARI) T(n, k) = if (k==0, 0, if (k==1, 1, n*T(n, k-1) - T(n, k-2)));
tabl(nn) = for(n=2, nn, for (k=0, nn, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 03 2018
(PARI) T(n, k) = ([n, -1; 1, 0]^k)[2, 1] \\ Jianing Song, Nov 10 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A083861
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Square array T(n,k) of second binomial transforms of generalized Fibonacci numbers, read by ascending antidiagonals, with n, k >= 0.
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+10
4
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0, 0, 1, 0, 1, 5, 0, 1, 5, 19, 0, 1, 5, 20, 65, 0, 1, 5, 21, 75, 211, 0, 1, 5, 22, 85, 275, 665, 0, 1, 5, 23, 95, 341, 1000, 2059, 0, 1, 5, 24, 105, 409, 1365, 3625, 6305, 0, 1, 5, 25, 115, 479, 1760, 5461, 13125, 19171, 0, 1, 5, 26, 125, 551, 2185, 7573, 21845, 47500, 58025
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OFFSET
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0,6
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COMMENTS
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Row n >= 0 of the array gives the solution to the recurrence b(k) = 5*b(k-1) + (n - 6)*b(k-2) for k >= 2 with b(0) = 0 and b(1) = 1. The rows are the binomial transforms of the rows of array A083857. The rows are the second binomial transforms of the generalized Fibonacci numbers in array A083856.
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LINKS
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FORMULA
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T(n, k) = (((5 + sqrt(4*n + 1))/2)^k - ((5 - sqrt(4*n + 1))/2)^k)/sqrt(4*n + 1).
O.g.f. for row n >= 0: -x/(-1 + 5*x + (n-6)*x^2) . - R. J. Mathar, Dec 02 2007
T(n,k) = 5*T(n,k-1) + (n - 6)*T(n,k-2) for k >= 2 with T(n,0) = 0 and T(n,1) = 1 for all n >= 0.
T(n,k) = Sum_{i = 0..k} binomial(k,i) * A083857(n,i).
T(n,k) = Sum_{i = 0..k} Sum_{j = 0..i} binomial(k,i) * binomial(i,j) * A083856(n,j). (End)
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EXAMPLE
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Array T(n,k) (with rows n >= 0 and columns k >= 0) begins as follows:
0, 1, 5, 19, 65, 211, 665, 2059, 6305, 19171, ...
0, 1, 5, 20, 75, 275, 1000, 3625, 13125, 47500, ...
0, 1, 5, 21, 85, 341, 1365, 5461, 21845, 87381, ...
0, 1, 5, 22, 95, 409, 1760, 7573, 32585, 140206, ...
0, 1, 5, 23, 105, 479, 2185, 9967, 45465, 207391, ...
0, 1, 5, 24, 115, 551, 2640, 12649, 60605, 290376, ...
0, 1, 5, 25, 125, 625, 3125, 15625, 78125, 390625, ...
...
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MAPLE
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seq(seq(round( (((5+sqrt(4*(n-k)+1))/2)^k - ((5-sqrt(4*(n-k)+1))/2)^k)/sqrt(4*(n-k)+1) ), k=0..n), n=0..10); # G. C. Greubel, Dec 27 2019
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MATHEMATICA
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T[n_, k_]:= Round[(((5 +Sqrt[4*n+1])/2)^k - ((5 -Sqrt[4*n+1])/2)^k)/Sqrt[4*n+1]]; Table[T[n-k, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 27 2019 *)
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PROG
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(PARI) T(n, k) = round( (((5+sqrt(4*n+1))/2)^k - ((5-sqrt(4*n+1))/2)^k)/sqrt(4*n + 1) );
for(n=0, 10, for(k=0, n, print1(T(n-k, k), ", "))) \\ G. C. Greubel, Dec 27 2019
(Magma)
T:= func< n, k | Round( (((5+Sqrt(4*n+1))/2)^k - ((5-Sqrt(4*n+1))/2)^k)/Sqrt(4*n + 1) ) >;
[T(n-k, k): k in [0..n], n in [0..10]]; // G. C. Greubel, Dec 27 2019
(Sage) [[round( (((5+sqrt(4*(n-k)+1))/2)^k - ((5-sqrt(4*(n-k)+1))/2)^k)/sqrt(4*(n-k)+1) ) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Dec 27 2019
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CROSSREFS
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Cf. A083856 (second inverse binomial transform), A083856 (first inverse binomial transform), A082297 (main diagonal).
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A217593
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Square array T, read by antidiagonals: T(n,k) = 0 if n-k >=1 or if k-n >= 9, T(0,k) = 1 for k = 0..8, T(n,k) = T(n-1,k) + T(n,k-1).
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+10
3
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1, 1, 0, 1, 1, 0, 1, 2, 0, 0, 1, 3, 2, 0, 0, 1, 4, 5, 0, 0, 0, 1, 5, 9, 5, 0, 0, 0, 1, 6, 14, 14, 0, 0, 0, 0, 1, 7, 20, 28, 14, 0, 0, 0, 0, 0, 8, 27, 48, 42, 0, 0, 0, 0, 0, 0, 8, 35, 75, 90, 42, 0, 0, 0, 0, 0, 0, 0, 43, 110, 165, 132, 0, 0, 0, 0, 0, 0, 0, 0, 43, 153, 275, 297, 132, 0, 0, 0, 0, 0, 0, 0, 0, 0, 196, 428, 572, 429, 0, 0, 0, 0, 0, 0, 0
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OFFSET
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0,8
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REFERENCES
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A hexagon arithmetic of E. Lucas.
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LINKS
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FORMULA
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T(n,n+7) = T(n,n+8) = A094865(n+3).
Sum_{k, 0<=k<=n} T(n-k,k) = A178381(n).
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EXAMPLE
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Square array begins :
1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, ...
0, 1, 2, 3, 4, 5, 6, 7, 8, 8, 0, 0, ...
0, 0, 2, 5, 9, 14, 20, 27, 35, 43, 43, 0, 0, ...
0, 0, 0, 5, 14, 28, 75, 110, 153, 196, 196, 0, 0, ....
0, 0, 0, 0, 14, 42, 90, 165, 275, 428, 624, 820, 820, 0, 0, ...
...
Square array, read by rows, with 0 omitted:
1, 1, 1, 1, 1, 1, 1, 1, 1
1, 2, 3, 4, 5, 6, 7, 8, 8
2, 5, 9, 14, 20, 27, 35, 43, 43
5, 14, 28, 48, 75, 110, 153, 196, 196
14, 42, 90, 165, 275, 428, 624, 820, 820
42, 132, 297, 572, 1000, 1624, 2444, 3264, 3264
132, 429, 1001, 2001, 3625, 6069, 9333, 12597, 12597
429, 1430, 3431, 7056, 13125, 22458, 35055, 47652, 47652
...
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KEYWORD
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AUTHOR
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STATUS
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approved
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A093130
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Third binomial transform of Fibonacci(3n).
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+10
2
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0, 2, 20, 160, 1200, 8800, 64000, 464000, 3360000, 24320000, 176000000, 1273600000, 9216000000, 66688000000, 482560000000, 3491840000000, 25267200000000, 182835200000000, 1323008000000000, 9573376000000000
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OFFSET
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0,2
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LINKS
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FORMULA
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G.f.: 2*x/(1-10*x+20*x^2).
a(n) = ((5+sqrt(5))^n - (5-sqrt(5))^n)/sqrt(5).
a(0)=0, a(1)=2, a(n) = 10*a(n-1) - 20*a(n-2). - Harvey P. Dale, Jun 24 2015
a(2*n) = 2^(2*n)*5^n*Fibonacci(2*n), a(2*n+1) = 2^(2*n+1)*5^n*Lucas(2*n+1). - G. C. Greubel, Dec 27 2019
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MAPLE
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seq(coeff(series(2*x/(1-10*x+20*x^2), x, n+1), x, n), n = 0..20); # G. C. Greubel, Dec 27 2019
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MATHEMATICA
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LinearRecurrence[{10, -20}, {0, 2}, 20] (* Harvey P. Dale, Jun 24 2015 *)
Table[If[EvenQ[n], 2^n*5^(n/2)*Fibonacci[n], 2^n*5^((n-1)/2)*LucasL[n]], {n, 0, 20}] (* G. C. Greubel, Dec 27 2019 *)
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PROG
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(PARI) my(x='x+O('x^20)); concat([0], Vec(2*x/(1-10*x+20*x^2))) \\ G. C. Greubel, Dec 27 2019
(Magma) I:=[0, 2]; [n le 2 select I[n] else 10*Self(n-1) - 20*Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 27 2019
(Sage)
P.<x> = PowerSeriesRing(ZZ, prec)
return P( 2*x/(1-10*x+20*x^2) ).list()
(GAP) a:=[0, 2];; for n in [3..20] do a[n]:=10*a[n-1]-20*a[n-2]; od; a; # G. C. Greubel, Dec 27 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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0, 1, 0, 1, 2, 0, 2, 3, 3, 0, 3, 8, 6, 4, 0, 5, 15, 20, 10, 5, 0, 8, 30, 45, 40, 15, 6, 0, 13, 56, 105, 105, 70, 21, 7, 0, 21, 104, 224, 280, 210, 112, 28, 8, 0, 34, 189, 468, 672, 630, 378, 168, 36, 9, 0, 55, 340, 945, 1560, 1680, 1260, 630, 240, 45, 10, 0
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OFFSET
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0,5
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COMMENTS
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LINKS
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FORMULA
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Sum_{k, 0<=k<=n} T(n,k)*x^k = A000045(n), A001906(n), A093131(n), A099453(n-1), A081574(n), A081575(n) for x = 0,1,2,3,4,5 respectively. Sum_{k, 0<=k<=n} T(n,k)*2^(n-k) = A014445(n).
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EXAMPLE
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Triangle begins :
0 ;
1,0 ;
1,2,0 ;
2,3,3,0 ;
3,8,6,4,0 ;
5,15,20,10,5,0 ;
...
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MATHEMATICA
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Flatten[Table[Binomial[n, k]Fibonacci[n-k], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Jan 16 2013 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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