A109466 -id:A109466 - cp's OEIS frontend

A057083 Scaled Chebyshev U-polynomials evaluated at sqrt(3)/2; expansion of 1/(1 - 3x + 3x^2).

1, 3, 6, 9, 9, 0, -27, -81, -162, -243, -243, 0, 729, 2187, 4374, 6561, 6561, 0, -19683, -59049, -118098, -177147, -177147, 0, 531441, 1594323, 3188646, 4782969, 4782969, 0, -14348907, -43046721, -86093442, -129140163, -129140163, 0

Offset: 0

Wolfdieter Lang, Aug 11 2000

With different sign pattern, see A000748.

Conjecture: Let M be any endomorphism on any vector space, such that M^3 = 1 (identity). Then (1-M)^n = A057681(n) - A057682(n)*M + z(n)*M^2, where z(0) = z(1) = 0 and, apparently, z(n+2) = a(n). - Stanislav Sykora, Jun 10 2012

Robert Israel, Table of n, a(n) for n = 0..4170
T. Alden Gassert, Discriminants of simplest 3^n-tic extensions, arXiv preprint arXiv:1409.7829 [math.NT], 2014.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=3, q=-3.
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs. (38) and (45),lhs, m=3.
Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-3).

Cf. A000748, A010892, A049310, A057078, A057681, A057682, A129339.

I:=[1,3]; [n le 2 select I[n] else 3*Self(n-1) - 3*Self(n-2): n in [1..30]]; // G. C. Greubel, Oct 23 2018

seq(3^(n/2)*orthopoly[U](n,sqrt(3)/2),n=0..100); # Robert Israel, Nov 21 2016

Join[{a=1,b=3},Table[c=3*b-3*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 17 2011 *)
CoefficientList[Series[1/(1 - 3 x + 3 x^2), {x, 0, 35}], x] (* Michael De Vlieger, Jul 30 2017 *)

a(n)=([0,1; -3,3]^n*[1;3])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,3,3) for n in range(1, 37)] # Zerinvary Lajos, Apr 23 2009

a(n) = S(n, sqrt(3))*(sqrt(3))^n with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310.
a(2*n) = A057078(n)*3^n; a(2*n+1)= A010892(n)*3^(n+1).
G.f.: 1/(1-3*x+3*x^2).
Binomial transform of A057079. a(n) = Sum_{k=0..n} 2*binomial(n, k)*cos((k-1)Pi/3). - Paul Barry, Aug 19 2003
For n > 5, a(n) = -27*a(n-6) - Gerald McGarvey, Apr 21 2005
a(n) = Sum_{k=0..n} A109466(n,k)*3^k. - Philippe Deléham, Nov 12 2008
a(n) = Sum_{k=1..n} binomial(k,n-k) * 3^k *(-1)^(n-k) for n>0; a(0)=1. - Vladimir Kruchinin, Feb 07 2011
By the conjecture: Start with x(0)=1, y(0)=0, z(0)=0 and set x(n+1) = x(n) - z(n), y(n+1) = y(n) - x(n), z(n+1) = z(n) - y(n). Then a(n) = z(n+2). This recurrence indeed ends up in a repetitive cycle of length 6 and multiplicative factor -27, confirming G. McGarvey's observation. - Stanislav Sykora, Jun 10 2012
G.f.: Q(0) where Q(k) =  1 + k*(3*x+1) + 9*x - 3*x*(k+1)*(k+4)/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Mar 15 2013
G.f.: G(0)/(2-3*x), where G(k)= 1 + 1/(1 - x*(k+3)/(x*(k+4) + 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013
a(n) = Sum_{k = 0..floor(n/3)} (-1)^k*binomial(n+2,3*k+2). Sykora's conjecture in the Comments section follows easily from this. - Peter Bala, Nov 21 2016
From Vladimir Shevelev, Jul 30 2017: (Start)
a(n) = 2*3^(n/2)*cos(Pi*(n-2)/6);
a(n) = K_2(n+2) - K_1(n+2);
For m,n>=1, a(n+m) = a(n-1)*K_1(m+1) + K_2(n+1)*K_2(m+1) + K_1(n+1)*a(m-1) where K_1 = A057681, K_2 = A057682. (End)

A030191 Scaled Chebyshev U-polynomial evaluated at sqrt(5)/2.

Original entry on oeis.org

1, 5, 20, 75, 275, 1000, 3625, 13125, 47500, 171875, 621875, 2250000, 8140625, 29453125, 106562500, 385546875, 1394921875, 5046875000, 18259765625, 66064453125, 239023437500, 864794921875, 3128857421875, 11320312500000, 40957275390625, 148184814453125

Offset: 0

Wolfdieter Lang

Number of (s(0), s(1), ..., s(2n+4)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+4, s(0) = 1, s(2n+4) = 5. - Herbert Kociemba, Jun 14 2004
Binomial transform of A002878. - Philippe Deléham, Oct 04 2005
Diagonal of square array A216219. - Philippe Deléham, Mar 15 2013
Lim_{n->oo} a(n+1)/a(n) = 2 + phi = A296184, where phi = A001622. - Wolfdieter Lang, Nov 16 2023~

			G.f. = 1 + 5*x + 20*x^2 + 75*x^3 + 275*x^4 + 1000*x^5 + 3625*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Santiago Alzate, Oscar Correa, and Rigoberto Flórez, Fibonacci identities from Jordan Identities, arXiv:2009.02639 [math.NT], 2020.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0 ,b=1; p=5, q=-5.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs. (38) and (45), lhs, m=5.
László Németh and László Szalay, Sequences Involving Square Zig-Zag Shapes, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.
Index entries for linear recurrences with constant coefficients, signature (5,-5).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000032, A000045, A001622.

Cf. A002878, A039717, A081567, A109466, A216219.

a:=[1,5];; for n in [3..30] do a[n]:=5*(a[n-1]-a[n-2]); od; a; # G. C. Greubel, Dec 28 2019

I:=[1,5]; [n le 2 select I[n] else 5*(Self(n-1) - Self(n-2)): n in [1..30]]; // G. C. Greubel, Dec 28 2019

seq(coeff(series(1/(1-5*x+5*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Dec 28 2019

Table[MatrixPower[{{2,1},{1,3}},n][[1]][[2]],{n,0,44}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
a[ n_]:= (((5 + Sqrt[5])/2)^(n + 1) - ((5 - Sqrt[5])/2)^(n + 1)) / Sqrt[5] // Expand; (* Michael Somos, Aug 27 2015 *)
Table[If[EvenQ[n], 5^(n/2)*LucasL[n+1], 5^((n+1)/2)*Fibonacci[n+1]], {n,0,30}] (* G. C. Greubel, Dec 28 2019 *)

{a(n) = imag((quadgen(5) + 2)^(n+1))}; /* Michael Somos, Aug 27 2015 */

[lucas_number1(n,5,5) for n in range(1, 22)] # Zerinvary Lajos, Apr 22 2009

a(n) = (sqrt(5))^n*U(n, sqrt(5)/2).
G.f.: 1/(1 - 5*x + 5*x^2).
a(2*k+1) = 5^(k+1)*Fibonacci(2*k+2).
a(2*k) = 5^k*Lucas(2*k+1).
a(n-1) = Sum_{k=0..n} C(n, k)*Fibonacci(2*k). - Benoit Cloitre, Jun 21 2003
a(n) = 5*a(n-1) - 5*a(n-2). - Benoit Cloitre, Oct 23 2003
a(n-1) = (((5+sqrt(5))/2)^n - ((5-sqrt(5))/2)^n)/sqrt(5) is the 2nd binomial transform of Fibonacci(n), the first binomial transform of Fibonacci(2n) and its n-th term is the n-th term of the third binomial transform of Fibonacci(3n) divided by 2^n. - Paul Barry, Mar 23 2004
a(n) = Sum_{k-0..n} 5^k*A109466(n,k). - Philippe Deléham, Nov 28 2006
a(n) = 5*A039717(n), n>0. - Philippe Deléham, Mar 12 2013
a(n) = A216219(n,n+3) = A216219(n,n+4) = A216219(n+3,n) = A216219(n+4,n). - Philippe Deléham, Mar 15 2013
G.f.: 1/(1-5*x/(1+x/(1-x))). - Philippe Deléham, Mar 15 2013
a(n) = -a(-2-n) * 5^(n+1) for all n in Z. - Michael Somos, Aug 27 2015
E.g.f.: exp((5-sqrt(5))*x/2)*((5 + sqrt(5))*exp(sqrt(5)*x) - 5 + sqrt(5))/(2*sqrt(5)). - Stefano Spezia, Dec 29 2019
a(n) = Sum_{k=0..n} A081567(n-k)*2^k. - Philippe Deléham, Mar 10 2023

A026729 Square array of binomial coefficients T(n,k) = binomial(n,k), n >= 0, k >= 0, read by downward antidiagonals.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 0, 2, 1, 0, 0, 1, 3, 1, 0, 0, 0, 3, 4, 1, 0, 0, 0, 1, 6, 5, 1, 0, 0, 0, 0, 4, 10, 6, 1, 0, 0, 0, 0, 1, 10, 15, 7, 1, 0, 0, 0, 0, 0, 5, 20, 21, 8, 1, 0, 0, 0, 0, 0, 1, 15, 35, 28, 9, 1, 0, 0, 0, 0, 0, 0, 6, 35, 56, 36, 10, 1, 0, 0, 0, 0, 0, 0, 1, 21, 70, 84, 45, 11, 1, 0, 0, 0, 0

Offset: 0

N. J. A. Sloane, Jan 19 2003

The signed triangular matrix T(n,k)*(-1)^(n-k) is the inverse matrix of the triangular Catalan convolution matrix A106566(n,k), n=k>=0, with A106566(n,k) = 0 if nPhilippe Deléham, Aug 01 2005
As a number triangle: unsigned version of A109466. - Philippe Deléham, Oct 26 2008
A063967*A130595 as infinite lower triangular matrices. - Philippe Deléham, Dec 11 2008
Modulo 2, this sequence becomes A106344. - Philippe Deléham, Dec 18 2008
Let {a_(k,i)}, k>=1, i=0,...,k, be the k-th antidiagonal of the array. Then s_k(n) = Sum_{i=0..k}a_(k,i)* binomial(n,k) is the n-th element of the k-th column of A111808. For example, s_1(n) = binomial(n,1) = n is the first column of A111808 for n>1, s_2(n) = binomial(n,1) + binomial(n,2) is the second column of A111808 for n>1, etc. Therefore, in cases k=3,4,5,6,7,8, s_k(n) is A005581(n), A005712(n), A000574(n), A005714(n), A005715(n), A005716(n), respectively. Besides, s_k(n+5) = A064054(n). - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012
As a triangle, T(n,k) = binomial(k,n-k). - Peter Bala, Nov 27 2015
For all n >= 0, k >= 0, the k-th homology group of the n-torus H_k(T^n) is the free abelian group of rank T(n,k) = binomial(n,k). See the Math Stack Exchange link below. - Jianing Song, Mar 13 2023

			Array begins
  1 0 0 0 0 0 ...
  1 1 0 0 0 0 ...
  1 2 1 0 0 0 ...
  1 3 3 1 0 0 ...
  1 4 6 4 1 0 ...
As a triangle, this begins
  1
  0 1
  0 1 1
  0 0 2 1
  0 0 1 3 1
  0 0 0 3 4 1
  0 0 0 1 6 5 1
  ...
Production array is
  0    1
  0    1   1
  0   -1   1   1
  0    2  -1   1  1
  0   -5   2  -1  1  1
  0   14  -5   2 -1  1  1
  0  -42  14  -5  2 -1  1  1
  0  132 -42  14 -5  2 -1  1  1
  0 -429 132 -42 14 -5  2 -1  1  1
  ... (Cf. A000108)

Muniru A Asiru, Rows n=0..50 of triangle, flattened
Tom Copeland, Addendum to Elliptic Lie Triad
Math Stack Exchange, Homology of the n-torus using the Künneth Formula
Lili Mu and Sai-nan Zheng, On the Total Positivity of Delannoy-Like Triangles, Journal of Integer Sequences, Vol. 20 (2017), Article 17.1.6.
L. W. Shapiro, S. Getu, Wen-Jin Woan and L. C. Woodson, The Riordan Group, Discrete Appl. Maths. 34 (1991) 229-239.

The official entry for Pascal's triangle is A007318. See also A052553 (the same array read by upward antidiagonals).
Cf. A030528 (subtriangle for 1<=k<=n).
Cf. A109466, A102426.

nmax:=15;; T:=List([0..nmax],n->List([0..nmax],k->Binomial(n,k)));;
b:=List([2..nmax],n->OrderedPartitions(n,2));;
a:=Flat(List([1..Length(b)],i->List([1..Length(b[i])],j->T[b[i][j][1]][b[i][j][2]]))); # Muniru A Asiru, Jul 17 2018

/* As triangle: */ [[Binomial(k, n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Nov 29 2015

seq(seq(binomial(k,n-k),k=0..n),n=0..12); # Peter Luschny, May 31 2014

Table[Binomial[k, n - k], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 28 2015 *)

As a number triangle, this is defined by T(n,0) = 0^n, T(0,k) = 0^k, T(n,k) = T(n-1,k-1) + Sum_{j, j>=0} (-1)^j*T(n-1,k+j)*A000108(j) for n>0 and k>0. - Philippe Deléham, Nov 07 2005
As a triangle read by rows, it is [0, 1, -1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 22 2006
As a number triangle, this is defined by T(n, k) = Sum_{i=0..n} (-1)^(n+i)*binomial(n, i)*binomial(i+k, i-k) and is the Riordan array ( 1, x*(1+x) ). The row sums of this triangle are F(n+1). - Paul Barry, Jun 21 2004
Sum_{k=0..n} x^k*T(n,k) = A000007(n), A000045(n+1), A002605(n), A030195(n+1), A057087(n), A057088(n), A057089(n), A057090(n), A057091(n), A057092(n), A057093(n) for n=0,1,2,3,4,5,6,7,8,9,10. - Philippe Deléham, Oct 16 2006
T(n,k) = A109466(n,k)*(-1)^(n-k). - Philippe Deléham, Dec 11 2008
G.f. for the triangular interpretation: -1/(-1+x*y+x^2*y). - R. J. Mathar, Aug 11 2015
For T(0,0) = 0, the triangle below has the o.g.f. G(x,t) = [t*x(1+x)]/[1-t*x(1+x)]. See A109466 for a signed version and inverse, A030528 for reverse and A102426 for a shifted version. - Tom Copeland, Jan 19 2016

A015443 Generalized Fibonacci numbers: a(n) = a(n-1) + 8*a(n-2).

Original entry on oeis.org

1, 1, 9, 17, 89, 225, 937, 2737, 10233, 32129, 113993, 371025, 1282969, 4251169, 14514921, 48524273, 164643641, 552837825, 1869986953, 6292689553, 21252585177, 71594101601, 241614783017, 814367595825, 2747285859961

Offset: 0

Olivier Gérard

Construct a graph as follows: form the graph whose adjacency matrix is the tensor product of that of P_3 and [1,1;1,1], then add a loop at each of the extremity nodes. a(n-1) counts walks of length n between adjacent nodes. - Paul Barry, Nov 12 2004
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 9*a(n-2) equals the number of 9-colored compositions of n with all parts >= 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 1, 6, 1, 24, 6, 16, 1, 6, 24, 110, 6, 56, 16, 24, 2, 16, 6, 60, 24, ... - R. J. Mathar, Aug 10 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Joerg Arndt, Matters Computational (The Fxtbook), p. 318
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Index entries for linear recurrences with constant coefficients, signature (1,8)

Cf. A015442, A015441, A100302, A100303.

[ n eq 1 select 1 else n eq 2 select 1 else Self(n-1)+8*Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 23 2011

CoefficientList[Series[1/(1-x-8*x^2), {x,0,50}], x] (* G. C. Greubel, Apr 30 2017 *)

Vec(1/(1-x-8*x^2)+O(x^99)) \\ Charles R Greathouse IV, Feb 03 2014

[lucas_number1(n,1,-8) for n in range(1, 27)] # Zerinvary Lajos, Apr 22 2009

a(n) = (((1+sqrt(33))/2)^(n+1) - ((1-sqrt(33))/2)^(n+1))/sqrt(33).
a(n) = Sum_{k=0..n} A109466(n,k)*(-8)^(n-k). - Philippe Deléham, Oct 26 2008
G.f.: 1/(1-x-8*x^2). - R. J. Mathar, Apr 07 2011
a(n) = (Sum_{1<=k<=n+1, k odd} C(n+1,k)*33^((k-1)/2))/2^n. - Vladimir Shevelev, Feb 05 2014

A107920 Lucas and Lehmer numbers with parameters (1 +- sqrt(-7))/2.

Original entry on oeis.org

0, 1, 1, -1, -3, -1, 5, 7, -3, -17, -11, 23, 45, -1, -91, -89, 93, 271, 85, -457, -627, 287, 1541, 967, -2115, -4049, 181, 8279, 7917, -8641, -24475, -7193, 41757, 56143, -27371, -139657, -84915, 194399, 364229, -24569, -753027, -703889, 802165, 2209943, 605613, -3814273

Offset: 0

Michael Somos, May 28 2005

The sequences A001607, A077020, A107920, A167433, A169998 are all essentially the same except for signs.
This is an example of a sequence of Lehmer numbers. In this case, the two parameters, alpha and beta, are (1 +- i*sqrt(7))/2. Bilu, Hanrot, Voutier and Mignotte show that all terms of a Lehmer sequence a(n) have a primitive factor for n > 30. Note that for this sequence, a(30) = 24475 = 5*5*11*89 has no primitive factors. - T. D. Noe, Oct 29 2003
Row sums of Riordan array (1/(1+2*x^2), x/(1+2*x^2)). - Paul Barry, Sep 10 2005
Pisano period lengths: 1, 1, 8, 2, 24, 8, 21, 2, 24, 24, 10, 8, 168, 21, 24, 4, 144, 24, 360, 24, ... - R. J. Mathar, Aug 10 2012
This is the Lucas Sequence U_n(P, Q) = U_n(1, 2). V_n(1, 2) = A002249(n). - Raphie Frank, Dec 25 2013
Note that (A002249(n)/2)^2 + 7*(a(n)/2)^2 = 2^n for all n in N. This is a specific case of the Lucas sequence identity (V_n/2)^2 - D*(U_n/2)^2 = Q^n where V_n = (a^n + b^n), U_n = (a^n - b^n)/(a - b), Q = (a*b) = 2 and D = (a - b)^2 = -7; a = (1 + sqrt(-7))/2 and b = (1 - sqrt(-7))/2. - Raphie Frank, Nov 26 2015
For the special case where |a(n)| = 1, true for n if and only if n is in {1, 2, 3, 5, 13} = {A215795(n) + 1} = {A060728(n) - 2}, then, additionally, by the Lucas sequence identity (U_2n = U_n*V_n), we have (a(2n)/2)^2 + 7*(a(n)/2)^2 = 2^n. - Raphie Frank, Nov 26 2015

			G.f. = x + x^2 - x^3 - 3*x^4 - x^5 + 5*x^6 + 7*x^7 - 3*x^8 - 17*x^9 - 11*x^10 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Christian Ballot, Lucasnomial Fuss-Catalan Numbers and Related Divisibility Questions, J. Int. Seq., Vol. 21 (2018), Article 18.6.5.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
F. Beukers, The multiplicity of binary recurrences, Compositio Mathematica, Tome 40 (1980) no. 2 , p. 251-267. See Theorem 2 p. 259.
Y. Bilu, G. Hanrot, P. M. Voutier and M. Mignotte, Existence of primitive divisors of Lucas and Lehmer numbers, [Research Report] RR-3792, INRIA. 1999, pp.41, HAL Id : inria-00072867.
M. Mignotte, Propriétés arithmétiques des suites récurrentes, Besançon, 1988-1989, see p. 14. In French.
Ronald Orozco López, Deformed Differential Calculus on Generalized Fibonacci Polynomials, arXiv:2211.04450 [math.CO], 2022.
Eric Weisstein's World of Mathematics, Lehmer Number
Wikipedia, Lucas Sequence
Index entries for linear recurrences with constant coefficients, signature (1,-2).
Index entries for sequences related to Chebyshev polynomials.

Similar sequences: A001607, A077020, A107920, A167433, A169998.

Cf. A002249, A049310, A060728, A109466, A215795, A227078.

[0] cat [n le 2 select 1 else Self(n-1)-2*Self(n-2): n in [1..45]]; // Vincenzo Librandi, Nov 27 2015

a:= n-> (Matrix([[1,1],[ -2,0]])^n)[1,2]: seq(a(n), n=0..45); # Alois P. Heinz, Sep 03 2008

LinearRecurrence[{1, -2}, {0, 1}, 50] (* Vincenzo Librandi, Nov 27 2015 *)
a[ n_] := Im[ ((1 + Sqrt[-7]) / 2)^n // FullSimplify] 2 / Sqrt[7]; (* Michael Somos, Jan 19 2017 *)
a[n_] := If[n < 2, n, Hypergeometric2F1[1 - n/2, (1 - n)/2, 1 - n, 8]];
Table[a[n], {n, 0, 45}] (* Peter Luschny, Feb 23 2018 *)

PARI
```
{a(n) = imag(quadgen(-7)^n)};
```

my(x='x+O('x^100)); concat(0, Vec(x/(1-x+2*x^2))) \\ Altug Alkan, Dec 04 2015

[lucas_number1(n,1,2) for n in range(0, 46)] # Zerinvary Lajos, Apr 22 2009

G.f.: x / (1 - x + 2*x^2).
a(n) = a(n-1) - 2*a(n-2).
a(n) = -(-1)^n*A001607(n).
From Paul Barry, Sep 10 2005: (Start)
a(n+1) = Sum_{k=0..n} C((n+k)/2, k)*(-2)^((n-k)/2)*(1+(-1)^(n-k))/2.
a(n+1) = Sum_{k=0..floor(n/2)} C(n-k, k)*(-2)^k. (End)
a(n+1) = Sum_{k=0..n} A109466(n,k)*2^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = ((1 - i*sqrt(7))^n - (1 + i*sqrt(7))^n)*i/(2^n*sqrt(7)), where i=sqrt(-1). - Bruno Berselli, Jul 01 2011
(a(2*(A060728(n)) - 4))^2 = (A002249(A060728(n) - 2))^2 = 2^(A060728(n)) - 7 = A227078(n), the Ramanujan-Nagell squares. - Raphie Frank, Dec 25 2013
a(n) = -a(-n) * 2^n for all n in Z. - Michael Somos, Jan 19 2017
G.f.: x / (1 - x / (1 + 2*x / (1 - 2*x))). - Michael Somos, Jan 19 2017
a(n) = S(n-1, 1/sqrt(2))*(sqrt(2))^(n-1), n >= 0, with the Chebyshev S polynomials (coefficients in A049310), and S(-1, x) = 0. - Wolfdieter Lang, Feb 22 2018
a(n) = hypergeom([1-n/2, (1-n)/2], [1-n], 8) for n >= 2. - Peter Luschny, Feb 23 2018

A015440 a(n) = a(n-1) + 5*a(n-2), with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 6, 11, 41, 96, 301, 781, 2286, 6191, 17621, 48576, 136681, 379561, 1062966, 2960771, 8275601, 23079456, 64457461, 179854741, 502142046, 1401415751, 3912125981, 10919204736, 30479834641, 85075858321, 237475031526, 662854323131, 1850229480761, 5164501096416

Offset: 0

Olivier Gérard

Original name: Generalized Fibonacci numbers.
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 6*a(n-2) equals the number of 6-colored compositions of n with all parts >= 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 3, 6, 6, 1, 6, 21, 12, 18, 3, 40, 6, 56, 21, 6, 24, 16, 18, 360, 6, .... - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Jan 02 2024: (Start)
This sequence {a(n-1)}, with a(-1) = 0, appears in the formula for powers of phi21 := (1 + sqrt(21))/2 = A222134 = 2.791287..., together with A(n) = A365824(n), as phi21^n = A(n) + a(n-1)*phi21(n), for n >= 0.
Limit_{n->oo} a(n)/a(n-1) = phi21. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Joerg Arndt, Matters Computational (The Fxtbook), section 14.8, pp. 317-318
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
A. G. Shannon and J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, Vol. 21, No. 2, (2015), 35-42.
Paul Thomas Young, p-adic congruences for generalized Fibonacci sequences, The Fibonacci Quarterly, Vol. 32, No. 1, 1994.
Index entries for linear recurrences with constant coefficients, signature (1,5).

Cf. A006130, A006131, A015441, A049310, A222134, A365824.

[n le 2 select 1 else Self(n-1)+5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 06 2012

A015440 := proc(n)
    if n <= 1 then
        1;
    else
        procname(n-1)+5*procname(n-2) ;
    end if;
end proc: # R. J. Mathar, May 15 2016

a[n_]:=(MatrixPower[{{1,3},{1,-2}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{1, 5}, {1, 1}, 100] (* Vincenzo Librandi, Nov 06 2012 *)

a(n)=abs([1,3;1,-2]^n*[1;1])[2,1] \\ Charles R Greathouse IV, Feb 03 2014

[lucas_number1(n,1,-5) for n in range(1, 28)] # Zerinvary Lajos, Apr 22 2009

a(n) = a(n-1) + 5*a(n-2).
a(n) = (( (1+sqrt(21))/2 )^(n+1) - ( (1-sqrt(21))/2 )^(n+1))/sqrt(21).
a(n) = Sum_{k=0..ceiling(n/2)} 5^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
G.f.: 1/(1 - x - 5x^2). - R. J. Mathar, Sep 03 2008
a(n) = Sum_{k=0..n} A109466(n,k)*(-5)^(n-k). - Philippe Deléham, Oct 26 2008
From Jeffrey R. Goodwin, May 28 2011: (Start)
A special case of a more general class of Lucas sequences given by
U(n) = U(n-1) + (4^(m-1)-1)/3 U(n-2).
U(n) = (( (1+sqrt((4^(m)-1)/3))/2 )^(n+1) - ( (1-sqrt((4^(m)-1)/3))/2 )^(n+1))/sqrt((4^(m)-1)/3). Fix m = 2 to get the formula for the Fibonacci sequence, fix m = 3 to get the formula for a(n). (End)
G.f.: G(0)/(2-x), where G(k)= 1 + 1/(1 - x*(21*k-1)/(x*(21*k+20) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 20 2013
G.f.: Q(0)/x -1/x, where Q(k) = 1 + 5*x^2 + (k+2)*x - x*(k+1 + 5*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n) = (Sum_{k=1..n+1, k odd} binomial(n+1,k)*21^((k-1)/2))/2^n. - Vladimir Shevelev, Feb 05 2014
With an initial 0 prepended, the sequence [0, 1, 1, 6, 11, 41, 96, ...] satisfies the congruences a(n*p^k) == (3|p)*(7|p)*a(n*p^(k-1)) (mod p^k) for positive integers k and n and all primes p, where (n|p) denotes the Legendre symbol. See Young, Theorem 1, Corollary 1(i). - Peter Bala, Dec 28 2022
a(n) = sqrt(-5)^(n-1)*S(n-1,1/sqrt(-5)), for n >= 0, with the Chebyshev polynomial S(n, x) (see A049310). - Wolfdieter Lang, Nov 17 2023
From Peter Bala, Jun 27 2025: (Start)
The following products telescope:
Product_{k >= 0} (1 + 5^k/a(2*k+1)) = 1 + sqrt(21).
Product_{k >= 1} (1 - 5^k/a(2*k+1)) = 1/22 * (1 + sqrt(21)).
Product_{k >= 0} (1 + (-5)^k/a(2*k+1)) = (1/21) * (21 + sqrt(21)).
Product_{k >= 1} (1 - (-5)^k/a(2*k+1)) = (1/22) * (21 + sqrt(21)). (End)
E.g.f.: exp(x/2)*(sqrt(21)*cosh(sqrt(21)*x/2) + sinh(sqrt(21)*x/2))/sqrt(21). - Stefano Spezia, Jul 04 2025

A034867 Triangle of odd-numbered terms in rows of Pascal's triangle.

Original entry on oeis.org

1, 2, 3, 1, 4, 4, 5, 10, 1, 6, 20, 6, 7, 35, 21, 1, 8, 56, 56, 8, 9, 84, 126, 36, 1, 10, 120, 252, 120, 10, 11, 165, 462, 330, 55, 1, 12, 220, 792, 792, 220, 12, 13, 286, 1287, 1716, 715, 78, 1, 14, 364, 2002, 3432, 2002, 364, 14, 15, 455, 3003, 6435, 5005, 1365, 105, 1

Offset: 0

N. J. A. Sloane

Also triangle of numbers of n-sequences of 0,1 with k subsequences of consecutive 01 because this number is C(n+1,2*k+1). - Roger Cuculiere (cuculier(AT)imaginet.fr), Nov 16 2002
From Gary W. Adamson, Oct 17 2008: (Start)
Received from Herb Conn:
Let T = tan x, then
  tan x = T
  tan 2x = 2T / (1 - T^2)
  tan 3x = (3T - T^3) / (1 - 3T^2)
  tan 4x = (4T - 4T^3) / (1 - 6T^2 + T^4)
  tan 5x = (5T - 10T^3 + T^5) / (1 - 10T^2 + 5T^4)
  tan 6x = (6T - 20T^3 + 6T^5) / (1 - 15T^2 + 15T^4 - T^6)
  tan 7x = (7T - 35T^3 + 21T^5 - T^7) / (1 - 21T^2 + 35T^4 - 7T^6)
  tan 8x = (8T - 56T^3 + 56T^5 - 8T^7) / (1 - 28T^2 + 70T^4 - 28T^6 + T^8)
  tan 9x = (9T - 84T^3 + 126T^5 - 36T^7 + T^9) / (1 - 36 T^2 + 126T^4 - 84T^6 + 9T^8)
... To get the next one in the series, (tan 10x), for the numerator add:
  9....84....126....36....1 previous numerator +
  1....36....126....84....9 previous denominator =
  10..120....252...120...10 = new numerator
For the denominator add:
  ......9.....84...126...36...1 = previous numerator +
  1....36....126....84....9.... = previous denominator =
  1....45....210...210...45...1 = new denominator
...where numerators = A034867, denominators = A034839
(End)
Column k is the sum of columns 2k and 2k+1 of A007318. - Philippe Deléham, Nov 12 2008
Triangle, with zeros omitted, given by (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
The row polynomials N(n,x) = Sum_{k=0..floor((n-1)/2)} T(n-1,k)*x^k, and D(n,x) = Sum_{k=0..floor(n/2)} A034839(n,k)*x^k, n >= 1, satisfy the recurrences N(n,x) = D(n-1,x) + N(n-1,x), D(n,x) = D(n-1,x) + x*N(n-1,x), with inputs N(1,x) = 1 = D(1,x). This is due to the Pascal triangle A007318 recurrence. Q(n,x) := tan(n*x)/tan(x) satisfies the recurrence Q(n,x) = (1 + Q(n-1,x))/(1 - v(x)*Q(n-1,x)) with input Q(1,x) = 1 and v = v(x) := (tan(x))^2. This recurrence is obtained from the addition theorem for tan(n*x) using n = 1 + (n-1). Therefore Q(n,x) = N(n,-v(x))/D(n,-v(x)). This proves the Gary W. Adamson contribution from above. See also A220673. This calculation was motivated by an e-mail of Thomas Olsen. The Oliver/Prodinger and Ma references resort to HAKEM Al Memo 239, Item 16, for the tan(n*x) formula in terms of tan(x). - Wolfdieter Lang, Jan 17 2013
The infinitesimal generator (infinigen) for the Narayana polynomials A090181/A001263 can be formed from the row polynomials P(n,y) of this entry. The resulting matrix is an instance of a matrix representation of the analytic infinigens presented in A145271 for general sets of binomial Sheffer polynomials and in A001263 and A119900 specifically for the Narayana polynomials. Given the column vector of row polynomials V = (1, P(1,x) = 2x, P(2,y) = 3x + x^2, P(3,y) = 4x + 4x^2, ...), form the lower triangular matrix M(n,k) = V(n-k,n-k), i.e., diagonally multiply the matrix with all ones on the diagonal and below by the components of V. Form the matrix MD by multiplying A132440^Transpose = A218272 = D (representing derivation of o.g.f.s) by M, i.e., MD = M*D. The non-vanishing component of the first row of (MD)^n * V / (n+1)! is the n-th Narayana polynomial. - Tom Copeland, Dec 09 2015
The diagonals of this entry are A078812 (also shifted A128908 and unsigned A053122, which are embedded in A030528, A102426, A098925, A109466, A092865). Equivalently, the antidiagonals of A078812 are the rows of A034867. - Tom Copeland, Dec 12 2015
Binomial(n,2k+1) is also the number of permutations avoiding both 132 and 213 with k peaks, i.e., positions with w[i]w[i+2]. - Lara Pudwell, Dec 19 2018
Binomial(n,2k+1) is also the number of permutations avoiding both 123 and 132 with k peaks, i.e., positions with w[i]w[i+2]. - Lara Pudwell, Dec 19 2018
The row polynomial P(n, x) = Sum_{0..floor(n/2)} T(n, k)*x^k appears as numerator polynomial of the diagonal sequence m of triangle A104698 as follows. G(m, x) = P(m, x^2)/(1 - x)^(m+1), for m >= 0. - Wolfdieter Lang, May 14 2025
Number of acyclic orientations of the path graph on n+1 vertices, with k-1 sinks. - Per W. Alexandersson, Aug 15 2025

			Triangle T starts:
  n\k   0   1   2   3   4  5 ...   ----------------------------------------
0:    1
1:    2
2:    3   1
3:    4   4
4:    5  10   1
5:    6  20   6
6:    7  35  21   1
7:    8  56  56   8
8:    9  84 126  36   1
9:   10 120 252 120  10
 10:   11 165 462 330  55  1
 11:   12 220 792 792 220 12
... ... reformatted and extended by - _Wolfdieter Lang_, May 14 2025

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 136.

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened
Jean-Luc Baril and José Luis Ramírez, Descent distribution on Catalan words avoiding ordered pairs of Relations, arXiv:2302.12741 [math.CO], 2023.
M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv preprint arXiv:1812.07112 [math.CO], 2018.
L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.
Sergi Elizalde, Johnny Rivera Jr., and Yan Zhuang, Counting pattern-avoiding permutations by big descents, arXiv:2408.15111 [math.CO], 2024. See p. 6.
S.-M. Ma, On some binomial coefficients related to the evaluation of tan(nx), arXiv preprint arXiv:1205.0735 [math.CO], 2012. - From _N. J. A. Sloane_, Oct 13 2012
K. Oliver and H. Prodinger, The continued fraction expansion of Gauss' hypergeometric function and a new application to the tangent function, Transactions of the Royal Society of South Africa, Vol. 76 (2012), 151-154, [DOI], [PDF]. - From _N. J. A. Sloane_, Jan 03 2013
Eric Weisstein's World of Mathematics, Tangent. [From _Eric W. Weisstein_, Oct 18 2008]

Cf. A000129, A007318, A034839, A034867, A084938, A131980, A220673.
Cf. A001263, A090181, A119900, A132440, A145271.
Cf. A030528, A053122, A078812, A092865, A098925, A102426, A109466, A128908, A218272, A104698.
From Wolfdieter Lang, May 14 2025:(Start)
Row length A008619. Row sums A000079. Alternating row sums  A009545(n+1).
Column sequences (with certain offsets): A000027, A000292, A000389, A000580, A000582, A001288, ... (End)

/* as a triangle */ [[Binomial(n+1,2*k+1): k in [0..Floor(n/2)]]: n in [0..20]]; // G. C. Greubel, Mar 06 2018

seq(seq(binomial(n+1,2*k+1), k=0..floor(n/2)), n=0..14); # Emeric Deutsch, Apr 01 2005

u[1, x_] := 1; v[1, x_] := 1; z = 12;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x]
v[n_, x_] := u[n - 1, x] + v[n - 1, x]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]  (* A034839 as a triangle *)
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]  (* A034867 as a triangle *)
(* Clark Kimberling, Feb 18 2012 *)
Table[Binomial[n+1, 2*k+1], {n,0,20}, {k,0,Floor[n/2]}]//Flatten (* G. C. Greubel, Mar 06 2018 *)

for(n=0,20, for(k=0,floor(n/2), print1(binomial(n+1,2*k+1), ", "))) \\ G. C. Greubel, Mar 06 2018

T(n,k) = C(n+1,2k+1) = Sum_{i=k..n-k} C(i,k) * C(n-i,k).
E.g.f.: 1+(exp(x)*sinh(x*sqrt(y)))/sqrt(y). - Vladeta Jovovic, Mar 20 2005
G.f.: 1/((1-z)^2-t*z^2). - Emeric Deutsch, Apr 01 2005
T(n,k) = Sum_{j = 0..n} A034839(j,k). - Philippe Deléham, May 18 2005
Pell(n+1) = A000129(n+1) = Sum_{k=0..n} T(n,k) * 2^k = (1/n!) Sum_{k=0..n} A131980(n,k) * 2^k. - Tom Copeland, Nov 30 2007
T(n,k) = A007318(n,2k) + A007318(n,2k+1). - Philippe Deléham, Nov 12 2008
O.g.f for column k, k>=0: (1/(1-x)^2)*(x/(1-x))^(2*k). See the G.f. of this array given above by Emeric Deutsch. - Wolfdieter Lang, Jan 18 2013
T(n,k) = (x^(2*k+1))*((1+x)^n-(1-x)^n)/2. - L. Edson Jeffery, Jan 15 2014

More terms from Emeric Deutsch, Apr 01 2005

A053404 Expansion of 1/((1+3x)(1-4*x)).

Original entry on oeis.org

1, 1, 13, 25, 181, 481, 2653, 8425, 40261, 141361, 624493, 2320825, 9814741, 37664641, 155441533, 607417225, 2472715621, 9761722321, 39434309773, 156574977625, 629786694901, 2508686426401, 10066126765213, 40170363882025

Offset: 0

Barry E. Williams, Jan 07 2000

Hankel transform is := 1,12,0,0,0,... - Philippe Deléham, Nov 02 2008

The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 13*a(n-2) equals the number of 13-colored compositions of n with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
A. K. Whitford, Binet's formula generalized, Fib. Quart., 15 (1977), pp. 21, 24, 29.
Index entries for linear recurrences with constant coefficients, signature (1,12).

Cf. A000045, A001045, A006130, A006131, A015440 - A015443, A015445 - A015447, A053404, A053428, A168579, A350468, A350469.

[((4^(n+1)) - (-3)^(n+1))/7: n in [0..30]]; // G. C. Greubel, Jan 16 2018

seq(simplify(hypergeom([1/2 - (1/2)*n, -(1/2)*n], [-n], -48)), n = 1..40); # Peter Bala, Jul 05 2025

CoefficientList[Series[1/((1 + 3 x) (1 - 4 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 06 2014 *)

a(n)=([0,1; 12,1]^n*[1;1])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

[lucas_number1(n,1,-12) for n in range(1, 25)] # Zerinvary Lajos, Apr 22 2009

a(n) = ((4^(n+1))-(-3)^(n+1))/7.
a(n) = a(n-1) + 12*a(n-2), n > 1; a(0)=1, a(1)=1.
From Paul Barry, Jul 30 2004: (Start)
Convolution of 4^n and (-3)^n.
G.f.: 1/((1+3x)(1-4x)); a(n) = Sum_{k=0..n, 4^k*(-3)^(n-k)} = Sum_{k=0..n, (-3)^k*4^(n-k)}. (End)
a(n) = Sum_{k, 0<=k<=n} A109466(n,k)*(-12)^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = (sum_{1<=k<=n+1, k odd} C(n+1,k)*7^(k-1))/2^n. - Vladimir Shevelev, Feb 05 2014
From Peter Bala, Jun 27 2025: (Start)
a(n) = hypergeom([1/2 - (1/2)*n, -(1/2)*n], [-n], -48) for n >= 1.
The following products telescope:
Product_{k >= 0} (1 + 12^k/a(2*k+1)) = 8.
Product_{k >= 1} (1 - 12^k/a(2*k+1)) = 4/25.
Product_{k >= 0} (1 + (-12)^k/a(2*k+1)) = 8/7.
Product_{k >= 1} (1 - (-12)^k/a(2*k+1)) = 28/25. (End)

More terms from James Sellers, Feb 02 2000

A030192 Scaled Chebyshev U-polynomial evaluated at sqrt(6)/2.

Original entry on oeis.org

1, 6, 30, 144, 684, 3240, 15336, 72576, 343440, 1625184, 7690464, 36391680, 172207296, 814893696, 3856118400, 18247348224, 86347378944, 408600184320, 1933516832256, 9149499887616, 43295898332160, 204878390667264, 969494954010624, 4587699380060160

Offset: 0

Wolfdieter Lang

Binomial transform of A001834. - Philippe Deléham, Nov 19 2009

Colin Barker, Table of n, a(n) for n = 0..1000
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=6, q=-6.
W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs. (38) and (45), lhs, m=6.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-6).

Cf. A083881.

Join[{a=1,b=6},Table[c=6*b-6*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)

a(n)=([0,1;-6,6]^n*[1;6])[1,1] \\ Charles R Greathouse IV, Jun 12 2015

Vec(1/(6*x^2-6*x+1) + O(x^100)) \\ Colin Barker, Jun 15 2015

[lucas_number1(n,6,6) for n in range(1, 21)] # Zerinvary Lajos, Apr 22 2009

a(n) = center term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 1 1 / 1 4 1 / 1 1 1]. M^n * [1 1 1] = [A083881(n) a(n) A083881(n)]. E.g., a(3) = 144 since M^3 * [1 1 1] = [54 144 54] = [A083881(3) a(3) A083881(3)]. - Gary W. Adamson, Dec 18 2004
a(n) = (sqrt(6))^n*U(n, sqrt(6)/2).
G.f.: 1/(6*(x^2-x+1/6)).
a(2*k+1) = 6^(k+1)*A001353(k), a(2*k) = 6^k*A001834(k).
Preceded by 0, this is the binomial transform of A001353. Its e.g.f. is then exp(3x)*sinh(sqrt(3)x)/sqrt(3). - Paul Barry, May 09 2003
a(n) = Sum_{k=0..n} A109466(n,k)*6^k. - Philippe Deléham, Oct 28 2008
a(n) = ((3+sqrt(3))^n - (3-sqrt(3))^n)/sqrt(12). - Al Hakanson (hawkuu(AT)gmail.com), Dec 29 2008
G.f.: A(x)= 1/(1-6*x+6*x^2) = G(0)/(1-3*x) where G(k) = 1 + 3*x/((1-3*x) - x*(1-3*x)/(x + (1-3*x)/G(k+1))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 28 2012

A015447 Generalized Fibonacci numbers: a(n) = a(n-1) + 11*a(n-2).

Original entry on oeis.org

1, 1, 12, 23, 155, 408, 2113, 6601, 29844, 102455, 430739, 1557744, 6295873, 23431057, 92685660, 350427287, 1369969547, 5224669704, 20294334721, 77765701465, 301003383396, 1156426099511, 4467463316867, 17188150411488

Offset: 0

Olivier Gérard

The compositions of n in which each positive integer is colored by one of p different colors are called p-colored compositions of n. For n>=2, 12*a(n-2) equals the number of 12-colored compositions of n, with all parts >= 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011

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

Cf. A015446, A015443.

[n le 2 select 1 else Self(n-1) + 11*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 06 2012

Join[{a=1,b=1},Table[c=b+11*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
LinearRecurrence[{1,11},{1,1},30] (* or *) CoefficientList[Series[ 1/(1-x-11 x^2),{x,0,50}],x] (* Harvey P. Dale, May 08 2011 *)

Vec(1/(1-x-11*x^2)+O(x^99)) \\ Charles R Greathouse IV, Feb 03 2014

[lucas_number1(n,1,-11) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009

a(n) = ( ( (1+3*sqrt(5))/2 )^(n+1) - ( (1-3*sqrt(5))/2 )^(n+1) )/(3*sqrt(5)).
a(n-1) = (1/3)*(-1)^n*Sum_{i=0..n} (-3)^i*Fibonacci(i)*C(n, i). - Benoit Cloitre, Mar 06 2004
a(n) = Sum_{k=0..n} A109466(n,k)*(-11)^(n-k). - Philippe Deléham, Oct 26 2008
G.f.: 1/(1 - x - 11*x^2). - Harvey P. Dale, May 08 2011
a(n) = ( Sum_{1<=k<=n+1, k odd} C(n+1,k)*45^((k-1)/2) )/2^n. - Vladimir Shevelev, Feb 05 2014

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A057083 Scaled Chebyshev U-polynomials evaluated at sqrt(3)/2; expansion of 1/(1 - 3*x + 3*x^2).

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A030191 Scaled Chebyshev U-polynomial evaluated at sqrt(5)/2.

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A026729 Square array of binomial coefficients T(n,k) = binomial(n,k), n >= 0, k >= 0, read by downward antidiagonals.

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GAP

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Maple

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A015443 Generalized Fibonacci numbers: a(n) = a(n-1) + 8*a(n-2).

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A107920 Lucas and Lehmer numbers with parameters (1 +- sqrt(-7))/2.

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Sage

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A015440 a(n) = a(n-1) + 5*a(n-2), with a(0) = a(1) = 1.

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A057083 Scaled Chebyshev U-polynomials evaluated at sqrt(3)/2; expansion of 1/(1 - 3x + 3x^2).

A053404 Expansion of 1/((1+3x)(1-4*x)).