A168561 -id:A168561 - cp's OEIS frontend

A164581 a(n) = 5*a(n - 1) + a(n - 2), with a(0)=1, a(1)=2.

1, 2, 11, 57, 296, 1537, 7981, 41442, 215191, 1117397, 5802176, 30128277, 156443561, 812346082, 4218173971, 21903215937, 113734253656, 590574484217, 3066606674741, 15923607857922, 82684645964351, 429346837679677, 2229418834362736, 11576441009493357

Offset: 0

Vincenzo Librandi, Aug 17 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,1).

Cf. A000032, A001077, A052924, A078343, A179237, A180148, A329723.

[ n le 2 select (n) else 5*Self(n-1)+Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Sep 12 2013

LinearRecurrence[{5, 1}, {1, 2}, 40] (* or *) Rest[CoefficientList[Series [x (1 - 3 x) / (1 - 5 x - x^2), {x, 0, 40}], x]] (* Harvey P. Dale, May 02 2011 *)

Vec((1-3*x)/(1-5*x-x^2) + O(x^40)) \\ Colin Barker, Oct 13 2015

a(n) = 5*a(n-1)+a(n-2) = A052918(n)-3*A052918(n-1).
G.f.: (1-3*x)/(1-5*x-x^2).
a(n) = A052918(n) + A015449(n). - R. J. Mathar, Jul 06 2012
a(n) = (2^(-1-n)*((5-sqrt(29))^n*(1+sqrt(29))+(-1+sqrt(29))*(5+sqrt(29))^n))/sqrt(29). - Colin Barker, Oct 13 2015
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*5^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*5^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = A052918(n) -3*A052918(n-1). - R. J. Mathar, Feb 14 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) - 7*(-1)^n/a(n)) = 3/10, since 1/(a(n) - 7*(-1)^n/a(n)) = b(n) - b(n+1), where b(n) = (1/5) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 7*(-1)^n/a(n)) = 1/10, since 1/(a(n) - 7*(-1)^n/a(n)) = c(n) + c(n+1), where c(n) = (1/5) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

A134513 Triangle read by rows: T(n, k) = binomial(ceiling((n+k)/2), floor((n-k)/2)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 3, 3, 4, 4, 1, 1, 1, 6, 6, 5, 5, 1, 1, 4, 4, 10, 10, 6, 6, 1, 1, 1, 10, 10, 15, 15, 7, 7, 1, 1, 5, 5, 20, 20, 21, 21, 8, 8, 1, 1, 1, 15, 15, 35, 35, 28, 28, 9, 9, 1, 1, 6, 6, 35, 35, 56, 56, 36, 36, 10, 10, 1, 1

Offset: 0

Gary W. Adamson, Oct 28 2007

Old name: abs(A049310 * A097806).
Equivalently, T(n,k) = A168561(n,k) + A168561(n,k+1).
Row sums = A062114: (1, 2, 3, 6, 9, 16, 25, 42, 67, ...).
Triangle A046854 = abs(A097806 * A049310).

			First few rows of the triangle:
  1;
  1,  1;
  1,  1,  1;
  2,  2,  1,  1;
  1,  3,  3,  1,  1;
  3,  3,  4,  4,  1,  1;
  1,  6,  6,  5,  5,  1,  1;
  4,  4, 10, 10,  6,  6,  1,  1;
  1, 10, 10, 15, 15,  7,  7,  1,  1;
  ...

Cf. A046854, A049310, A062114, A097806.

abs(A049310 * A097806) as infinite lower triangular matrices.

Better definition, offset changed to 0, and more terms from Jinyuan Wang, Jan 25 2025

A309213 A007318 + A065941 - A049310.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 2, 6, 5, 1, 1, 5, 12, 6, 1, 2, 3, 14, 17, 8, 1, 3, 7, 14, 24, 26, 9, 1, 2, 12, 27, 30, 45, 33, 11, 1, 1, 9, 45, 62, 70, 66, 45, 12, 1, 2, 5, 44, 111, 147, 120, 104, 54, 14, 1, 3, 11, 39, 128, 273, 273, 217, 140, 69, 15, 1, 2, 18, 65, 139, 366, 546, 518, 329, 200, 80, 17, 1

Offset: 0

Gary W. Adamson, Jul 04 2007

Row sums = 1, 3, 7, 14, 25, 45, 85, ... (This is probably a new sequence and should be added to the OEIS.) - N. J. A. Sloane, Aug 09 2019

			First few rows of the triangle are:
1,
2, 1,
3, 3, 1,
2, 6, 5, 1,
1, 5, 12, 6, 1,
2, 3, 14, 17, 8, 1,
3, 7, 14, 24, 26, 9, 1,
...

Jinyuan Wang, Table of n, a(n) for n = 0..5150

Cf. A007318, A065941, A049310, A117591, A131375, A168561.

T007318(n, k) = binomial(n, k);
T065941(n, k) = binomial(n - (k+1)\2, k\2);
T049310(n, k) = if ((n+k)%2, 0, (-1)^((n+k)/2 + k) * binomial((n+k)/2, k));
T(n, k) = T007318(n, k) + T065941(n, k) - T049310(n, k); \\ Michel Marcus, Apr 28 2014

A007318 + A065941 - A168561 as infinite lower triangular matrices.

The old definition of A131376 did not match the data, as Michel Marcus pointed out. The definition there has been corrected, keeping the old data. The present sequence uses the old definition with corrected data from Michel Marcus. - N. J. A. Sloane, Aug 09 2019

More terms from Jinyuan Wang, Aug 29 2019

A123185 Triangular array from a zero coefficient sum of recursive polynomials: p(k, x) = x*p(k - 1, x) + p(k - 2, x).

Original entry on oeis.org

1, -1, 1, 2, -3, 1, -1, 3, -3, 1, 2, -4, 4, -3, 1, -1, 5, -7, 5, -3, 1, 2, -5, 9, -10, 6, -3, 1, -1, 7, -12, 14, -13, 7, -3, 1, 2, -6, 16, -22, 20, -16, 8, -3, 1, -1, 9, -18, 30, -35, 27, -19, 9, -3, 1, 2, -7, 25, -40, 50, -51, 35, -22, 10, -3, 1

Offset: 0

Roger L. Bagula, Oct 02 2006

Except for p(0,x)=1 all sum to zero: Table[Sum[CoefficientList[p[n, x], x][[m]], {m, 1, n + 1}], {n, 0, 12}] {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

			1
-1, 1
2, -3, 1
-1, 3, -3, 1
2, -4, 4, -3, 1
-1, 5, -7, 5, -3, 1
2, -5, 9, -10, 6, -3, 1

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)

Cf. A114525, A168561.

S:= series((1 - t + (1-2*x)*t^2)/(1 - t*x - t^2), t, 21):
R:= [seq(coeff(S,t,n),n=0..19)]:
seq(seq(coeff(R[n],x,j),j=0..n-1),n=1..20); # Robert Israel, Jul 12 2016

p[0, x] = 1; p[1, x] = x - 1; p[2, x] = x^2 - 3x + 2; p[k_, x_] := p[k, x] = x*p[k - 1, x] + p[k - 2, x] ; w = Table[CoefficientList[p[n, x], x], {n, 0, 10}]; Flatten[w]

p(k, x) = x*p(k - 1, x) + p(k - 2, x)  for k >= 3 with p(0,x) = 1, p(1,x) = x-1 and p(2,x) = x^2-3x+2. (Corrected by Robert Israel, Jul 12 2016)
G.f.: g(t,x) = (1 - t + (1-2x) t^2)/(1 - t x - t^2). - Robert Israel, Jul 12 2016
For n >= 1, T(n,k) = A114525(n,k) + 2*A114525(n-1,k) - 5*A168561(n-1,k), where we take A114525(n-1,n) = A168561(n-1,n) = 0. - Robert Israel, Jul 13 2016

Edited by Robert Israel, Jul 13 2016

A290864 Numbers k such that the k-th Fibonacci polynomial evaluated at k is prime.

Original entry on oeis.org

2, 5, 71, 8419

Offset: 1

Bobby Jacobs, Aug 12 2017

Numbers k such that A084844(k) = A117715(k,k) is prime.
a(5) > 9200. - Giovanni Resta, Aug 13 2017
Except for a(1), all terms == 1 or 5 (mod 6). - Robert Israel, Aug 13 2017

			5 is in the sequence because A117715(5,5) = 701 is prime.

Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Wikipedia, Fibonacci polynomials

Cf. A000045, A011973, A084844, A117715, A162515, A168561.

select(t -> isprime(combinat:-fibonacci(t,t)), [2,seq(seq(6*i+j,j=[1,5]),i=0..100)]); # Robert Israel, Aug 13 2017

Select[Range[100], PrimeQ@ Fibonacci[#, #] &] (* Giovanni Resta, Aug 13 2017 *)

a(4) from Giovanni Resta, Aug 13 2017

A317403 a(n)=(-1)^((n-2)(n-1)/2)2^(n-1)*n^(n-3).

Original entry on oeis.org

1, 1, -4, -32, 400, 6912, -153664, -4194304, 136048896, 5120000000, -219503494144, -10567230160896, 564668382613504, 33174037869887488, -2125764000000000000, -147573952589676412928, 11034809241396899282944, 884295678882933431599104, -75613185918270483380568064

Offset: 1

Rigoberto Florez, Aug 26 2018

sign

Discriminant of Fibonacci polynomials.

Fibonacci polynomials are defined as F(0)=0, F(1)=1 and F(n)=x*F(n-1)+F(n-2) for n>1. Coefficients are given in triangle A168561 with offset 1.

Rigoberto Flórez, Robinson Higuita, and Alexander Ramírez, The resultant, the discriminant, and the derivative of generalized Fibonacci polynomials, arXiv:1808.01264 [math.NT], 2018.
Rigoberto Flórez, Robinson Higuita, and Antara Mukherjee, Star of David and other patterns in the Hosoya-like polynomials triangles, 2018.
R. Flórez, R. Higuita and A. Mukherjees, Characterization of the strong divisibility property for generalized Fibonacci polynomials, Integers, 18 (2018), Paper No. A14.
R. Flórez, N. McAnally, and A. Mukherjees, Identities for the generalized Fibonacci polynomial, Integers, 18B (2018), Paper No. A2.
Eric Weisstein's World of Mathematics, Discriminant
Eric Weisstein's World of Mathematics, Fibonacci Polynomial

Cf. A006645, A001629, A001871, A006645, A007701, A045618, A045925, A093967, A168561, A193678, A317404, A317405, A317408, A317451, A318184, A318197.

Essentially the same as A127670.

[(-1)^((n-2)*(n-1) div 2)*2^(n-1)*n^(n-3): n in [1..20]]; // Vincenzo Librandi, Aug 27 2018

Array[(-1)^((#-2)*(#-1)/2)*2^(#-1)*#^(#-3)&,20]

concat([1], [poldisc(p) | p<-Vec(x/(1-x^2-y*x) - x + O(x^20))]) \\ Andrew Howroyd, Aug 26 2018

A320508 T(n,k) = binomial(n - k - 1, k), 0 <= k < n, and T(n,n) = (-1)^n, triangle read by rows.

Original entry on oeis.org

1, 1, -1, 1, 0, 1, 1, 1, 0, -1, 1, 2, 0, 0, 1, 1, 3, 1, 0, 0, -1, 1, 4, 3, 0, 0, 0, 1, 1, 5, 6, 1, 0, 0, 0, -1, 1, 6, 10, 4, 0, 0, 0, 0, 1, 1, 7, 15, 10, 1, 0, 0, 0, 0, -1, 1, 8, 21, 20, 5, 0, 0, 0, 0, 0, 1, 1, 9, 28, 35, 15, 1, 0, 0, 0, 0, 0, -1, 1, 10, 36

Offset: 0

Franck Maminirina Ramaharo, Oct 14 2018

Differs from A164925 in signs.
The n-th row consists of the coefficients in the expansion of (-x)^n + (((1 + sqrt(1 + 4*x))/2)^n -((1 - sqrt(1 + 4*x))/2)^n )/sqrt(1 + 4*x).
The coefficients in the expansion of Sum_{j=0..floor((n - 1)/2)} T(n,k)*x^(n - 2*j - 1) yield the n-th row in A168561, the coefficients of the n-th Fibonacci polynomial.
Row n sums up to Fibonacci(n) + (-1)^n (A008346).

			Triangle begins:
    1;
    1, -1;
    1,  0,  1;
    1,  1,  0, -1;
    1,  2,  0,  0,  1;
    1,  3,  1,  0,  0, -1;
    1,  4,  3,  0,  0,  0, 1;
    1,  5,  6,  1,  0,  0, 0, -1;
    1,  6, 10,  4,  0,  0, 0,  0, 1;
    1,  7, 15, 10,  1,  0, 0,  0, 0, -1;
    1,  8, 21, 20,  5,  0, 0,  0, 0,  0, 1;
    1,  9, 28, 35, 15,  1, 0,  0, 0,  0, 0, -1;
    ...

Wikipedia, Fibonacci polynomials

Inspired by A123018.

Cf. A007318, A026729, A049310, A052553, A164925, A168561.

Table[Table[Binomial[n - k - 1, k], {k, 0, n}], {n, 0, 12}]//Flatten

create_list(binomial(n - k - 1, k), n, 0, 12, k, 0, n);

G.f.: 1/((1 + x*y)*(1 - y - x*y^2)).
E.g.f.: exp(-x*y) + (exp(y*(1 + sqrt(1 + 4*x))/2) - exp(y*(1 - sqrt(1 + 4*x))/2))/sqrt(1 + 4*x).
T(n,1) = A023443(n).

A346038 Triangle read by rows T(n, k) such that Fib(n, x+1) = Sum_{k=1..n} T(n, k)*Fib(k, x) where Fib(n, x) is the n-th Fibonacci polynomial.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 0, 3, 3, 1, -2, 2, 6, 4, 1, -4, -3, 7, 10, 5, 1, -3, -12, 0, 16, 15, 6, 1, 5, -18, -21, 11, 30, 21, 7, 1, 20, -4, -50, -24, 35, 50, 28, 8, 1, 29, 48, -51, -98, -9, 78, 77, 36, 9, 1, 1, 124, 45, -164, -150, 42, 147, 112, 45, 10, 1, -94, 128, 282, -67, -365, -177, 154, 250, 156, 55, 11, 1

Offset: 1

Michel Marcus, Jul 02 2021

			Triangle begins:
   1;
   1,  1;
   1,  2, 1;
   0,  3, 3,  1;
  -2,  2, 6,  4, 1;
  -4, -3, 7, 10, 5, 1;
  ...
The first 3 Fibonacci polynomials are 1, x, x^2 + 1. So F3(n, x+1) = x^2 + 2*x + 2  = 1*1 + 2*x + 1*(x^2+1) = 1*F(1,x) + 2*F(2, x) + 1*F(3,x), so the 3rd row is [1, 2, 1].

Michel Marcus, Rows n=1..100 of triangle, flattened
Math StackExchange, Variable transformation x->x+1 in Fibonacci-polynomial Fn(x)
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.

Cf. A000012, A000027, A000217, A005581: diagonals.

Cf. A162515 and A168561 (Fibonacci polynomials coefficients).

rowV(n) = my(v= if (n==0, [0], n--; vector(n+1, k, k--; if (k%2==0, binomial(n-k/2, k/2))))); Pol(v); \\ A162515
rowT(n, vfp, vfp1) = {my(vp1 = vfp1[n], vc = vector(n), i=n); forstep (k = poldegree(vp1), 0, -1, vc[i] = polcoef(vp1, k)/polcoef(vfp[k+1], k); vp1 -= vfp[k+1]*vc[i]; i--;); vc;}
tabl(nn) = {my(vfp = vector(nn, k, rowV(k))); my(vfp1 = vector(nn, k, subst(vfp[k], x, x+1))); for(n=1, nn, print((rowT(n, vfp, vfp1))););}

A177717 A symmetrical triangle based on the Fibonacci Polynomials: p(x,n)=f(n,x)+x^(n-1)*f(n,1/x).

Original entry on oeis.org

2, 1, 1, 2, 0, 2, 1, 2, 2, 1, 2, 0, 6, 0, 2, 1, 3, 4, 4, 3, 1, 2, 0, 11, 0, 11, 0, 2, 1, 4, 6, 10, 10, 6, 4, 1, 2, 0, 17, 0, 30, 0, 17, 0, 2, 1, 5, 8, 20, 21, 21, 20, 8, 5, 1

Offset: 0

Roger L. Bagula, May 12 2010

Row sums are:

{2, 2, 4, 6, 10, 16, 26, 42, 68, 110,...}

			{2},
{1, 1},
{2, 0, 2},
{1, 2, 2, 1},
{2, 0, 6, 0, 2},
{1, 3, 4, 4, 3, 1},
{2, 0, 11, 0, 11, 0, 2},
{1, 4, 6, 10, 10, 6, 4, 1},
{2, 0, 17, 0, 30, 0, 17, 0, 2},
{1, 5, 8, 20, 21, 21, 20, 8, 5, 1}

Function form used from:http://functions.wolfram.com/HypergeometricFunctions/Fibonacci2General/26/01/02/0001/

Cf. A168561

f[n_, z_] := (1/(2 Sqrt[4 + z^2])) (-HypergeometricPFQ[{}, {}, n ((-I) Pi - \ Log[(1/2) (z + Sqrt[4 + z^2])])] - HypergeometricPFQ[{}, {}, n (I Pi - Log[( 1/2) (z + Sqrt[4 + z^2])])] + 2 HypergeometricPFQ[{}, {}, n Log[(1/ 2) (z + Sqrt[4 + z^2])]]);
Table[CoefficientList[FullSimplify[ExpandAll[f[n, x] + x^(n - 1)*f[n, 1/x]]], x], {n, 1, 10}];
Flatten[%]

f(x,n)=(1/(2 Sqrt[4 + z^2])) (-HypergeometricPFQ[{}, {}, n ((-I) Pi - Log[(1/2) (z + \ Sqrt[4 + z^2])])] - HypergeometricPFQ[{}, {}, n (I Pi - Log[(1/2) (z + Sqrt[ 4 + z^2])])] + 2 HypergeometricPFQ[{}, {}, n Log[(1/2) (z + Sqrt[4 + z^2])]]);
p(x,n)=f(x,n)+x^(n-1)*f(1/x,n);
t(n,m)=coefficients(p(x,n))

cp's OEIS Frontend

A164581 a(n) = 5*a(n - 1) + a(n - 2), with a(0)=1, a(1)=2.

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A134513 Triangle read by rows: T(n, k) = binomial(ceiling((n+k)/2), floor((n-k)/2)).

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A309213 A007318 + A065941 - A049310.

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A123185 Triangular array from a zero coefficient sum of recursive polynomials: p(k, x) = x*p(k - 1, x) + p(k - 2, x).

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A290864 Numbers k such that the k-th Fibonacci polynomial evaluated at k is prime.

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A317403 a(n)=(-1)^((n-2)*(n-1)/2)*2^(n-1)*n^(n-3).

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A320508 T(n,k) = binomial(n - k - 1, k), 0 <= k < n, and T(n,n) = (-1)^n, triangle read by rows.

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A317403 a(n)=(-1)^((n-2)(n-1)/2)2^(n-1)*n^(n-3).