A031138 -id:A031138 - cp's OEIS frontend

A072256 a(n) = 10*a(n-1) - a(n-2) for n > 1, a(0) = a(1) = 1.

1, 1, 9, 89, 881, 8721, 86329, 854569, 8459361, 83739041, 828931049, 8205571449, 81226783441, 804062262961, 7959395846169, 78789896198729, 779939566141121, 7720605765212481, 76426118085983689, 756540575094624409, 7488979632860260401, 74133255753507979601, 733843577902219535609

Offset: 0

Lekraj Beedassy, Jul 08 2002

Any k in the sequence is followed by 5*k + 2*sqrt(2*(3*k^2 - 1)).
Gives solutions for x in 3*x^2 - 2*y^2 = 1. Corresponding y is given by A054320(n-1). [corrected by Jon E. Schoenfield, Jun 08 2018]
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7,8,9} which do not end in 0. - Tanya Khovanova, Jan 10 2007
For n >= 2, a(n) equals the permanent of the (2n-2) X (2n-2) tridiagonal matrix with sqrt(8)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 8 = 0. - Colin Barker, Feb 09 2014
The aerated sequence [b(n)]n>=1 = [1, 0, 9, 0, 89, 0, 881, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -12, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. - Peter Bala, May 12 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Christian Aebi and Grant Cairns, Equable Parallelograms on the Eisenstein Lattice, arXiv:2401.08827 [math.CO], 2024. See p. 14.
Jean-Paul Allouche, Jeffrey Shallit, and Manon Stipulanti, Combinatorics on words and generating Dirichlet series of automatic sequences, arXiv:2401.13524 [math.CO], 2024.
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 283).
Bruno Deschamps, Sur les bonnes valeurs initiales de la suite de Lucas-Lehmer, Journal of Number Theory, Volume 130, Issue 12, December 2010, Pages 2658-2670.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Cf. A031138, A046172, A054320, A108299.
Row 10 of array A094954.
First differences of A004189.
Essentially the same as A138288.

a:=[1,1];; for n in [3..20] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

[n le 2 select 1 else 10*Self(n-1)-Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 10 2014

seq( simplify(ChebyshevU(n,5) -9*ChebyshevU(n-1,5)), n=0..20); # G. C. Greubel, Jan 14 2020

a[n_]:= a[n]= 10a[n-1] -a[n-2]; a[0]=a[1]=1; Table[ a[n], {n, 0, 20}]
CoefficientList[Series[(1-9x)/(1-10x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2014 *)
Table[ChebyshevU[n, 5] -9*ChebyshevU[n-1, 5], {n,0,20}] (* G. C. Greubel, Jan 14 2020 *)
LinearRecurrence[{10,-1},{1,1},20] (* Harvey P. Dale, Jun 17 2022 *)

a(n)=([0,1; -1,10]^n*[1;1])[1,1] \\ Charles R Greathouse IV, May 10 2016

vector(21, n, polchebyshev(n-1,2,5) -9*polchebyshev(n-2,2,5) ) \\ G. C. Greubel, Jan 14 2020

a(n) = (3-sqrt(6))/6 * (5+2*sqrt(6))^n + (3+sqrt(6))/6 * (5-2*sqrt(6))^n.
a(n) = (2*A031138(n) + 1)/3 = sqrt((2*A054320(n-1)^2 + 1)/3), n >= 1.
a(n) = U(n-1, 5)-U(n-2, 5) = T(2*n-1, sqrt(3))/sqrt(3) with Chebyshev's U- and T- polynomials and U(-1, x) := 0, U(-2, x) := -1, T(-1, x) := x.
G.f.: (1-9*x)/(1-10*x+x^2).
6*a(n)^2 - 2 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 10 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 8) = a(n+1). - Benoit Cloitre, Nov 10 2002
a(n)*a(n+3) = 80 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = L(n-1,10), where L is defined as in A108299; see also A054320 for L(n,-10). - Reinhard Zumkeller, Jun 01 2005
a(n) = A138288(n-1) for n > 0. - Reinhard Zumkeller, Mar 12 2008
a(n) = sqrt(A046172(n)). - Paul Weisenhorn, May 15 2009
a(n) = ceiling(((3-sqrt(6))*(5+2*sqrt(6))^n)/6). - Paul Weisenhorn, May 23 2020
E.g.f.: exp(5*x)*(3*cosh(2*sqrt(6)*x) - sqrt(6)*sinh(2*sqrt(6)*x))/3. - Stefano Spezia, Oct 25 2023
From Peter Bala, May 08 2025: (Start)
a(n) = (-1)^n * Dir(n-1, -5), where Dir(n, x) denotes the n-th row polynomial of A244419.
For arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = -8 with a(n) := (3-sqrt(6))/6 * (5+2*sqrt(6))^n + (3+sqrt(6))/6 * (5-2*sqrt(6))^n as given above (the particular case x = 0 is noted in the Comments section).
a(n+1/2) = 1/sqrt(3) * A001079(n).
a(n+3/4) + a(n+1/4) = sqrt(2/3) * sqrt(1 + sqrt(3)) * A001079(n).
a(n+3/4) - a(n+1/4) = 4 * sqrt(sqrt(3) - 1) * A004189(n).
a(n) divides a(3*n-1); a(n) divides a(5*n-2); in general, for k >= 0, a(n) divides a((2*k+1)*n - k).
Sum_{n >= 2} 1/(a(n) - 1/a(n)) = 1/8 (telescoping series: for n >= 2, 1/(a(n) - 1/a(n)) = 1/A291181(n-2) - 1/A291181(n-1).)
Product_{n >= 2} ((a(n) + 1)/(a(n) - 1))^(-1)^n = sqrt(3/2) (telescoping product: Product_{n = 2..k} (((a(n) + 1)/(a(n) - 1))^(-1)^n)^2 = 3/2 * (1 - (-1)^(k+1)/(3*A098308(k))).) (End)

Edited by Robert G. Wilson v, Jul 17 2002

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

Original entry on oeis.org

1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, 102558961, 1015229051, 10049731549, 99482086439, 984771132841, 9748229241971, 96497521286869, 955226983626719, 9455772314980321, 93602496166176491, 926569189346784589, 9172089397301669399, 90794324783669909401

Offset: 0

Ignacio Larrosa Cañestro, Feb 27 2000

Chebyshev's even-indexed U-polynomials evaluated at sqrt(3).
a(n)^2 is a star number (A003154).
Any k in the sequence has the successor 5*k + 2*sqrt(3(2*k^2 + 1)). - Lekraj Beedassy, Jul 08 2002
{a(n)} give the values of x solving: 3*y^2 - 2*x^2 = 1. Corresponding values of y are given by A072256(n+1). x + y = A001078(n+1). - Richard R. Forberg, Nov 21 2013
The aerated sequence (b(n))n>=1 = [1, 0, 11, 0, 109, 0, 1079, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

			a(1)^2 = 121 is the 5th star number (A003154).

G. C. Greubel, Table of n, a(n) for n = 0..1000
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Star Number
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (10,-1).
Index entries for sequences related to Chebyshev polynomials.

A member of the family A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, which are the expansions of (1+x) / (1-kx+x^2) with k = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, May 04 2004
Cf. A003154, A006061, A031138, A007667, A004189.
Cf. A138281. Cf. A100047.
Cf. A142238.

a:=[1,11];; for n in [3..30] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jul 22 2019

I:=[1,11]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

CoefficientList[Series[(1+x)/(1-10x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Mar 22 2015 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Numerator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A142238 *)
a[3/2, 20] (* Gerry Martens, Jun 07 2015 *)

a(n)=subst(poltchebi(n+1)-poltchebi(n),x,5)/4;

(a(n)-1)^2 + a(n)^2 + (a(n)+1)^2 = b(n)^2 + (b(n)+1)^2 = c(n), where b(n) is A031138 and c(n) is A007667.
a(n) = 10*a(n-1) - a(n-2).
a(n) = (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1).
a(n) = U(2*(n-1), sqrt(3)) = S(n-1, 10) + S(n-2, 10) with Chebyshev's U(n, x) and S(n, x) := U(n, x/2) polynomials and S(-1, x) := 0. S(n, 10) = A004189(n+1), n >= 0.
6*a(n)^2 + 3 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then (-1)^n*q(n, -12) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-10)*(-1)^n, where L is defined as in A108299; see also A072256 for L(n,+10). - Reinhard Zumkeller, Jun 01 2005
From Reinhard Zumkeller, Mar 12 2008: (Start)
(sqrt(2) + sqrt(3))^(2*n+1) = a(n)*sqrt(2) + A138288(n)*sqrt(3);
a(n) = A138288(n) + A001078(n).
a(n) = A001079(n) + 3*A001078(n). (End)
a(n) = A142238(2n) = A041006(2n)/2 = A041038(2n)/4. - M. F. Hasler, Feb 14 2009
a(n) = sqrt(A006061(n)). - Zak Seidov, Oct 22 2012
a(n) = sqrt((3*A072256(n)^2 - 1)/2). - T. D. Noe, Oct 23 2012
(sqrt(3) + sqrt(2))^(2*n+1) - (sqrt(3) - sqrt(2))^(2*n+1) = a(n)*sqrt(8). - Bruno Berselli, Oct 29 2019
a(n) = A004189(n)+A004189(n+1). - R. J. Mathar, Oct 01 2021
E.g.f.: exp(5*x)*(2*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/2. - Stefano Spezia, May 16 2023
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 5), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 10*a(n)*a(n+1) + a(n+1)^2 = 12.
More generally, for arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 12 with a(n) := (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1) as given above.
a(n+1/2) = sqrt(3) * A001078(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6)*sqrt(sqrt(3) + 1) * A001078(n+1).
a(n+3/4) - a(n+1/4) = sqrt(sqrt(3) - 1) * A001079(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/12 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A004291(n) + 1/A004291(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3/2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3/2 * (1 - 1/A171640(k+2))). (End)

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A350923 a(0) = 2, a(1) = 2, and a(n) = 10*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

2, 2, 14, 134, 1322, 13082, 129494, 1281854, 12689042, 125608562, 1243396574, 12308357174, 121840175162, 1206093394442, 11939093769254, 118184844298094, 1169909349211682, 11580908647818722, 114639177128975534, 1134810862641936614, 11233469449290390602, 111199883630261969402

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Essentially the same as A157085. - R. J. Mathar, Feb 07 2022

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A031138, A054318, A097784, A253175, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350921, A350922, A350924, A350925, A350926.

LinearRecurrence[{11, -11, 1}, {2, 2, 14}, 25] (* Paolo Xausa, May 30 2025 *)

G.f.: 2*(1 - 10*x + 7*x^2)/((1 - x)*(1 - 10*x + x^2)). - Stefano Spezia, Jan 22 2022
From Hugo Pfoertner, Jan 22 2022: (Start)
a(n) = A031138(n) + 1.
a(n) = 3*A054318(n) - 1.
a(n) = 12*A097784(n-2) + 2 for n >= 2. (End)
a(n) = 2 * A253175(n) for n>=1. - Alois P. Heinz, Jan 22 2022

A007667 The sum of both two and three consecutive squares.

Original entry on oeis.org

5, 365, 35645, 3492725, 342251285, 33537133085, 3286296790925, 322023548377445, 31555021444198565, 3092070077983081805, 302991312620897818205, 29690056566770003102165

Offset: 1

N. J. A. Sloane, Robert G. Wilson v

			a(2) = 365 = 13^2+14^2 = 10^2+11^2+12^2.

M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 22.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..500
Index entries for sequences related to sums of squares
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A003154, A031138, A006061, A054320.

a:=[5, 365, 35645];; for n in [4..20] do a[n]:=99*a[n-1]-99*a[n-2] + a[n-3]; od; a; # G. C. Greubel, Jul 23 2019

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( 5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2)) )); // G. C. Greubel, Jul 23 2019

CoefficientList[Series[5*(1-26*x+x^2)/((1-x)*(1-98*x+x^2)),{x,0,20}],x] (* Vincenzo Librandi, Apr 16 2012 *)
LinearRecurrence[{99,-99,1},{5,365,35645},20] (* Harvey P. Dale, Dec 10 2024 *)

my(x='x+O('x^20)); Vec(5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2))) \\ G. C. Greubel, Jul 23 2019

(5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2))).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019

From Ignacio Larrosa Cañestro, Feb 27 2000: (Start)
a(n) = (b(n)-1)^2 + b(n)^2 + (b(n)+1)^2 = c(n)^2 + (c(n)+1)^2, where b(n) = A054320(n) and c(n) = A031138(n).
a(n) = 3*A006061(n) + 2.
a(n) = 99*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 3*(5 - 2*sqrt(6))/8*(sqrt(3) + sqrt(2))^(4*n) + 3*(5 + 2*sqrt(6))/8*(sqrt(3) - sqrt(2))^(4*n) + 5/4. (End)
G.f.: 5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2)). - Colin Barker, Apr 14 2012

Corrected by T. D. Noe, Nov 07 2006

A054318 a(n)-th star number (A003154) is a square.

Original entry on oeis.org

1, 5, 45, 441, 4361, 43165, 427285, 4229681, 41869521, 414465525, 4102785725, 40613391721, 402031131481, 3979697923085, 39394948099365, 389969783070561, 3860302882606241, 38213059042991845, 378270287547312205

Offset: 1

Ignacio Larrosa Cañestro, Feb 27 2000

A two-way infinite sequence which is palindromic.
Also indices of centered hexagonal numbers (A003215) which are also centered square numbers (A001844). - Colin Barker, Jan 02 2015
Also positive integers y in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0. - Colin Barker, Jan 02 2015

			a(2) = 5 because the 5th Star number (A003154) 121=11^2 is the 2nd that is a square.

Colin Barker, Table of n, a(n) for n = 1..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

A031138 is 3*a(n)-2. Cf. A003154, A006061, A182432, A211955.
Quintisection of column k=2 of A233427.
Cf. A001844, A003215, A253475.

a:=[1,5,45];; for n in [4..30] do a[n]:=11*a[n-1]-11*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jul 23 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) )); // G. C. Greubel, Jul 23 2019

CoefficientList[Series[x(1-6x+x^2)/((1-x)(1-10x+x^2)), {x,0,30}], x] (* Michael De Vlieger, Aug 11 2016 *)
LinearRecurrence[{11,-11,1},{1,5,45},30] (* Harvey P. Dale, Nov 05 2016 *)

a(n)=if(n<1,a(1-n),1/2+subst(poltchebi(n)+poltchebi(n-1),x,5)/12)

Vec(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) + O(x^30)) \\ Colin Barker, Jan 02 2015

(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019

a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 1/2 + (3 - sqrt(6))/12*(5 + 2*sqrt(6))^n + (3 + sqrt(6))/12*(5 - 2*sqrt(6))^n.
From Michael Somos, Mar 18 2003: (Start)
G.f.: x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)).
12*a(n)*a(n-1) + 4 = (a(n) + a(n-1) + 2)^2.
a(n) = a(1-n) = 10*a(n-1) - a(n-2) - 4.
a(n) = 12*a(n-1)^2/(a(n-1) + a(n-2)) - a(n-1).
a(n) = (a(n-1) + 4)*a(n-1)/a(n-2). (End)
From Peter Bala, May 01 2012: (Start)
a(n+1) = 1 + (1/2)*Sum_{k = 1..n} 8^k*binomial(n+k,2*k).
a(n+1) = R(n,4), where R(n,x) is the n-th row polynomial of A211955.
a(n+1) = (1/u)*T(n,u)*T(n+1,u) with u = sqrt(3) and T(n,x) the Chebyshev polynomial of the first kind.
Sum {k>=0} 1/a(k) = sqrt(3/2). (End)
A003154(a(n)) = A006061(n). - Zak Seidov, Oct 22 2012
a(n) = (4*a(n-1) + a(n-1)^2) / a(n-2), n >= 3. - Seiichi Manyama, Aug 11 2016
2*a(n) = 1+A072256(n). - R. J. Mathar, Feb 07 2022

More terms from James Sellers, Mar 01 2000

A261932 The first of two consecutive positive integers the sum of the squares of which is equal to the sum of the squares of ten consecutive positive integers.

Original entry on oeis.org

26, 48, 68, 126, 468, 866, 1226, 2268, 8406, 15548, 22008, 40706, 150848, 279006, 394926, 730448, 2706866, 5006568, 7086668, 13107366, 48572748, 89839226, 127165106, 235202148, 871602606, 1612099508, 2281885248, 4220531306, 15640274168, 28927951926

Offset: 1

Colin Barker, Sep 06 2015

For the first of the corresponding ten consecutive positive integers, see A261934.

			26 is in the sequence because 26^2 + 27^2 = 7^2 + 8^2 + ... + 16^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,18,-18,0,0,-1,1).

Cf. A001652, A031138, A261933, A261934, A261935.

CoefficientList[Series[2 (4 x^8 - x^7 + x^5 - 63 x^4 + 29 x^3 + 10 x^2 + 11 x + 13)/((1 - x) (x^4 - 4 x^2 - 1) (x^4 + 4 x^2 - 1)), {x, 0, 45}], x] (* Vincenzo Librandi, Sep 07 2015 *)

Vec(-2*x*(4*x^8-x^7+x^5-63*x^4+29*x^3+10*x^2+11*x+13)/((x-1)*(x^4-4*x^2-1)*(x^4+4*x^2-1)) + O(x^40))

G.f.: -2*x*(4*x^8-x^7+x^5-63*x^4+29*x^3+10*x^2+11*x+13) / ((x-1)*(x^4-4*x^2-1)*(x^4+4*x^2-1)).

a(n) = a(n-1) + 18*a(n-4) - 18*a(n-5) - a(n-8) + a(n-9) for n>8. - Vincenzo Librandi, Sep 07 2015

A261933 The first of two consecutive positive integers the sum of the squares of which is equal to the sum of the squares of seventeen consecutive positive integers.

Original entry on oeis.org

40, 91, 2743, 6364, 192004, 445423, 13437571, 31173280, 940438000, 2181684211, 65817222463, 152686721524, 4606265134444, 10685888822503, 322372742188651, 747859530853720, 22561485688071160, 52339481270937931, 1578981625422792583, 3663015829434801484

Offset: 1

Colin Barker, Sep 06 2015

For the first of the corresponding seventeen consecutive positive integers, see A261935.

			40 is in the sequence because 40^2 + 41^2 = 5^2 + 6^2 + ... + 21^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,70,-70,-1,1).

Cf. A001652, A031138, A261932, A261934, A261935.

LinearRecurrence[{1,70,-70,-1,1},{40,91,2743,6364,192004},20] (* Harvey P. Dale, Oct 17 2015 *)

Vec(-x*(40*x^4+51*x^3-148*x^2+51*x+40)/((x-1)*(x^4-70*x^2+1)) + O(x^40))

G.f.: -x*(40*x^4+51*x^3-148*x^2+51*x+40) / ((x-1)*(x^4-70*x^2+1)).

A261934 The first of ten consecutive positive integers the sum of the squares of which is equal to the sum of the squares of two consecutive positive integers.

Original entry on oeis.org

7, 17, 26, 52, 205, 383, 544, 1010, 3755, 6949, 9838, 18200, 67457, 124771, 176612, 326662, 1210543, 2239001, 3169250, 5861788, 21722389, 40177319, 56869960, 105185594, 389792531, 720952813, 1020490102, 1887478976, 6994543241, 12936973387, 18311951948

Offset: 1

Colin Barker, Sep 06 2015

For the first of the corresponding two consecutive positive integers, see A261932.

			7 is in the sequence because 7^2 + 8^2 + ... + 16^2 = 26^2 + 27^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,18,-18,0,0,-1,1).

Cf. A001652, A031138, A261932, A261933, A261935.

LinearRecurrence[{1,0,0,18,-18,0,0,-1,1},{7,17,26,52,205,383,544,1010,3755},40] (* Harvey P. Dale, Mar 29 2018 *)

Vec(x*(2*x^8+2*x^7+x^6+2*x^5-27*x^4-26*x^3-9*x^2-10*x-7)/((x-1)*(x^4-4*x^2-1)*(x^4+4*x^2-1)) + O(x^40))

G.f.: x*(2*x^8+2*x^7+x^6+2*x^5-27*x^4-26*x^3-9*x^2-10*x-7) / ((x-1)*(x^4-4*x^2-1)*(x^4+4*x^2-1)).

A261935 The first of seventeen consecutive positive integers the sum of the squares of which is equal to the sum of the squares of two consecutive positive integers.

Original entry on oeis.org

5, 23, 933, 2175, 65849, 152771, 4609041, 10692339, 322567565, 748311503, 22575121053, 52371113415, 1579935906689, 3665229628091, 110572938347721, 256513702853499, 7738525748434325, 17952293970117383, 541586229452055573, 1256404064205363855

Offset: 1

Colin Barker, Sep 06 2015

For the first of the corresponding two consecutive positive integers, see A261933.

			5 is in the sequence because 5^2 + 6^2 + ... + 21^2 = 40^2 + 41^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,70,-70,-1,1).

Cf. A001652, A031138, A261932, A261933, A261934.

Vec(x*(21*x^4+18*x^3-560*x^2-18*x-5)/((x-1)*(x^4-70*x^2+1)) + O(x^40))

G.f.: x*(21*x^4+18*x^3-560*x^2-18*x-5) / ((x-1)*(x^4-70*x^2+1)).

A087125 Indices k of hex numbers H(k) that are also triangular.

Original entry on oeis.org

0, 5, 54, 539, 5340, 52865, 523314, 5180279, 51279480, 507614525, 5024865774, 49741043219, 492385566420, 4874114620985, 48248760643434, 477613491813359, 4727886157490160, 46801248083088245, 463284594673392294, 4586044698650834699, 45397162391834954700

Offset: 0

Eric W. Weisstein, Aug 14 2003

From the law of cosines, the non-Pythagorean triple {a(n), a(n)+1=A253475(n+1), A072256(n+1)} forms a near-isosceles triangle with the angle bounded by the consecutive sides equal to the regular tetrahedron's central angle (see A156546 and A247412). This implies also that a(n) are those numbers k such that (16/3)*A000217(k)+1 is a perfect square. - Federico Provvedi, Apr 04 2023

Colin Barker, Table of n, a(n) for n = 0..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A006244, A031138, A072256, A054318.

[Round((-4-(5-2*Sqrt(6))^n*(-2+Sqrt(6)) + (2+Sqrt(6))*(5 + 2*Sqrt(6))^n)/8): n in [0..25]]; // G. C. Greubel, Nov 04 2017

CoefficientList[Series[(-x^2+5*x)/((1-x)*(1-10*x+x^2)), {x, 0, 25}], x] (* G. C. Greubel, Nov 04 2017 *)
LinearRecurrence[{11,-11,1},{0,5,54},30] (* Harvey P. Dale, Jun 14 2022 *)
Table[(x Sqrt[z^(2 n + 1) + z^-(2 n + 1) - 2] - 4) / 8 //. {x -> Sqrt[2], y -> Sqrt[3], z -> (5 + 2 x y)}, {n, 0, 100}] // Round (* Federico Provvedi, Apr 16 2023 *)

concat(0, Vec(x*(x-5)/((x-1)*(x^2-10*x+1)) + O(x^50))) \\ Colin Barker, Jun 23 2015

G.f.: (-x^2+5*x)/((1-x)*(1-10*x+x^2)).
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) for n > 2. - Colin Barker, Jun 23 2015
a(n) = (-4 - (5-2*sqrt(6))^n*(-2 + sqrt(6)) + (2+sqrt(6))*(5+2*sqrt(6))^n)/8. - Colin Barker, Mar 05 2016
a(n) = 10*a(n-1) - a(n-2) + 4 for n > 1. - Charlie Marion, Feb 14 2023
a(n) = ((x^(n+1)+1)*(x^n-1))/(2*x^n*(x-1)), with x=5+2*sqrt(6). - Federico Provvedi, Apr 04 2023
a(n) = sqrt(3*A161680(A054318(n+1)) + 1/4) - 1/2 = floor(sqrt(3*A000217(A054318(n+1)-1) + 1/4)). - Federico Provvedi, Apr 16 2023

cp's OEIS Frontend

A072256 a(n) = 10*a(n-1) - a(n-2) for n > 1, a(0) = a(1) = 1.

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A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

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A350923 a(0) = 2, a(1) = 2, and a(n) = 10*a(n-1) - a(n-2) - 4 for n >= 2.

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A007667 The sum of both two and three consecutive squares.

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A054318 a(n)-th star number (A003154) is a square.

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A261932 The first of two consecutive positive integers the sum of the squares of which is equal to the sum of the squares of ten consecutive positive integers.

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