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a(n) = 7*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 7.
+10
36
2, 7, 51, 364, 2599, 18557, 132498, 946043, 6754799, 48229636, 344362251, 2458765393, 17555720002, 125348805407, 894997357851, 6390330310364, 45627309530399, 325781497023157, 2326097788692498, 16608466017870643, 118585359913786999, 846705985414379636
COMMENTS
a(n+1)/a(n) converges to (7+sqrt(53))/2 = 7.14005... = A176439. Lim a(n)/a(n+1) as n approaches infinity = 0.1400549... = 2/(7+sqrt(53)) = (sqrt(53)-7)/2 = 1/ A176439 = A176439 - 7. In general sequences with recurrence a(n) = k*a(n-1)+a(n-2) with a(0)=2 and a(1)=k [and a(-1)=0] have generating function (2-k*x)/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n+1) = b(5n), a(3n+2)=b(5*n+3), a(3n+3)=2*b(5n+4) where b(n) is the sequence of numerators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n)/2, for n>=1, is the sequence of numerators of continued fraction convergents to sqrt(k^2/4+1).]
For the sequence given above k=7 which implies that it is associated with A041090. For a similar statement about sequences with recurrence a(n) = k*a(n-1)+a(n-2) but with a(0)=1 [and a(-1)=0] see A054413; a sequence that is associated with A041091. (End)
FORMULA
a(n) = ((7+sqrt(53))/2)^n + ((7-sqrt(53))/2)^n.
E.g.f. : 2exp(7x/2)cosh(sqrt(53)x/2); a(n)=2^(1-n)sum{k=0..floor(n/2), C(n, 2k)53^k7^(n-2k)}. a(n)=2T(n, 7i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003 Limit(a(n+k)/a(k), k=infinity) = ( A086902(n) + A054413(n-1)*sqrt(53))/2.
EXAMPLE
a(4) = 7*a(3) + a(2) = 7*364 + 51 = 2599.
MATHEMATICA
RecurrenceTable[{a[0] == 2, a[1] == 7, a[n] == 7 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *) LinearRecurrence[{7, 1}, {2, 7}, 30] (* Harvey P. Dale, May 25 2023 *)
PROG
(Magma) I:=[2, 7]; [n le 2 select I[n] else 7*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016
AUTHOR
Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Sep 18 2003
Chebyshev polynomials S(n,83) + S(n-1,83) with Diophantine property.
+10
6
1, 84, 6971, 578509, 48009276, 3984191399, 330639876841, 27439125586404, 2277116783794691, 188973253929372949, 15682502959354160076, 1301458772372465913359, 108005395603955316648721
COMMENTS
(9*a(n))^2 - 85*b(n)^2 = -4 with b(n)= A097841(n) give all positive solutions of this Pell equation.
FORMULA
a(n) = S(n, 83) + S(n-1, 83) = S(2*n, sqrt(85)), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 83) = A097839(n). a(n) = (-2/9)*i*((-1)^n)*T(2*n+1, 9*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120. G.f.: (1+x)/(1 - 83*x + x^2).
a(n) = 83*a(n-1) - a(n-2) for n > 1; a(0)=1, a(1)=84. - Philippe Deléham, Nov 18 2008 a(n) = (2/9)*(9/2 o 9/2 o ... o 9/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 84, 0, 6971, 0, 578509, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 83*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -81, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. b(n) = 1/2*( (-1)^n - 1 )*F(n,9) + 1/9*( 1 + (-1)^(n+1) )*F(n+1,9), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1). Exp( Sum_{n >= 1} 18*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 18* A099371(n)*x^n. Exp( Sum_{n >= 1} (-18)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 18* A099371(n)*(-x)^n. (End)
EXAMPLE
All positive solutions of Pell equation x^2 - 85*y^2 = -4 are (9=9*1,1), (756=9*84,82), (62739=9*6971,6805), (5206581=9*578509,564733), ...
MATHEMATICA
CoefficientList[Series[(1+x)/(1-83x+x^2), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)
PROG
(PARI) my(x='x+O('x^20)); Vec((1+x)/(1-83*x+x^2)) \\ G. C. Greubel, Jan 13 2019 (Magma) m:=20; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-83*x+x^2) )); // G. C. Greubel, Jan 13 2019 (Sage) ((1+x)/(1-83*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019 (GAP) a:=[1, 84];; for n in [3..20] do a[n]:=83*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2019
Chebyshev polynomials S(n,27) + S(n-1,27) with Diophantine property.
+10
5
1, 28, 755, 20357, 548884, 14799511, 399037913, 10759224140, 290100013867, 7821941150269, 210902311043396, 5686540457021423, 153325690028535025, 4134107090313424252, 111467565748433919779, 3005490168117402409781
COMMENTS
(5*a(n))^2 - 29*b(n)^2 = -4 with b(n) = A097835(n) give all positive solutions of this Pell equation.
FORMULA
a(n) = S(n, 27) + S(n-1, 27) = S(2*n, sqrt(29)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 27)= A097781(n). a(n) = (-2/5)*i*((-1)^n)*T(2*n+1, 5*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120. G.f.: (1+x)/(1-27*x+x^2).
a(n) = (2/5)*(5/2 o 5/2 o ... o 5/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 28, 0, 755, 0, 20357, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 27*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -25, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. b(n) = (1/2)*( (-1)^n - 1 )*F(n,5) + (1/5)*( 1 + (-1)^(n+1) )*F(n+1,5), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1). Exp( Sum_{n >= 1} 10*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10* A052918(n)*x^n. Exp( Sum_{n >= 1} (-10)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10* A052918(n)*(-x)^n. (End)
EXAMPLE
All positive solutions of Pell equation x^2 - 29*y^2 = -4 are
(5=5*1,1), (140=5*28,26), (3775=5*755,701), (101785=5*20357,18901), ...
MATHEMATICA
a[n_] := -2/5*I*(-1)^n*ChebyshevT[2*n + 1, 5*I/2]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jun 21 2013, from 2nd formula *)
PROG
(PARI) {a(n) = (-1)^n * subst(2 * I / 5 * poltchebi(2*n + 1), 'x, -5/2 * I)}; /* Michael Somos, Nov 04 2008 */
Chebyshev polynomials S(n,51).
+10
5
1, 51, 2600, 132549, 6757399, 344494800, 17562477401, 895341852651, 45644872007800, 2326993130545149, 118631004785794799, 6047854250944989600, 308321935793408674801, 15718370871212897425251, 801328592496064360013000
COMMENTS
Used for all positive integer solutions of Pell equation x^2 - 53*y^2 = -4. See A097837 with A097838. a(n-1), with a(-1) := 0, and b(n) := A099368(n) give the proper and improper nonnegative solutions of the Pell equation b(n)^2 - 53*(7*a(n-1))^2 = +4, n >= 0. - Wolfdieter Lang, Jun 27 2013
FORMULA
a(n) = S(n, 51)=U(n, 51/2)= S(2*n+1, sqrt(53))/sqrt(53) with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x). a(n) = 51*a(n-1) - a(n-2), n >= 1, a(-1)=0, a(0)=1, a(1)=51.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap := (51+7*sqrt(53))/2 and am := (51-7*sqrt(53))/2 = 1/ap.
G.f.: 1/(1-51*x+x^2).
MATHEMATICA
LinearRecurrence[{51, -1}, {1, 51}, 30] (* G. C. Greubel, Jan 12 2019 *)
PROG
(PARI) my(x='x+O('x^30)); Vec(1/(1-51*x+x^2)) \\ G. C. Greubel, Jan 12 2019 (Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( 1/(1-51*x+x^2) )); // G. C. Greubel, Jan 12 2019 (Sage) (1/(1-51*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 12 2019 (GAP) a:=[1, 51];; for n in [2..30] do a[n]:=51*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2019
First differences of Chebyshev polynomials S(n,51) = A097836(n) with Diophantine property.
+10
5
1, 50, 2549, 129949, 6624850, 337737401, 17217982601, 877779375250, 44749530155149, 2281348258537349, 116304011655249650, 5929223246159194801, 302274081542463685201, 15410048935419488750450, 785610221624851462587749
COMMENTS
(7*b(n))^2 - 53*a(n)^2 = -4 with b(n)= A097837(n) give all positive solutions of this Pell equation.
FORMULA
a(n) = ((-1)^n)*S(2*n, 7*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.
G.f.: (1-x)/(1 - 51*x + x^2).
a(n) = S(n, 51) - S(n-1, 51) = T(2*n+1, sqrt(53)/2)/(sqrt(53)/2), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x) and T(n, x) Chebyshev's polynomials of the first kind, A053120.
EXAMPLE
All positive solutions of Pell equation x^2 - 53*y^2 = -4 are (7=7*1,1), (364=7*52,50), (18557=7*2651,2549), (946043=7*135149,129949), ...
MATHEMATICA
LinearRecurrence[{51, -1}, {1, 50}, 20] (* G. C. Greubel, Jan 13 2019 *)
PROG
(PARI) my(x='x+O('x^20)); Vec((1-x)/(1-51*x+x^2)) \\ G. C. Greubel, Jan 13 2019 (Magma) m:=20; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-51*x+x^2) )); // G. C. Greubel, Jan 13 2019 (Sage) ((1-x)/(1-51*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019 (GAP) a:=[1, 50];; for n in [3..20] do a[n]:=51*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2019
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