A110272 -id:A110272 - cp's OEIS frontend

A255494 Triangle read by rows: coefficients of numerator of generating functions for powers of Pell numbers.

1, 1, 1, 1, 4, 1, 1, 13, 13, 1, 1, 38, 130, 38, 1, 1, 105, 1106, 1106, 105, 1, 1, 280, 8575, 26544, 8575, 280, 1, 1, 729, 62475, 567203, 567203, 62475, 729, 1, 1, 1866, 435576, 11179686, 32897774, 11179686, 435576, 1866, 1, 1, 4717, 2939208, 207768576, 1736613466, 1736613466, 207768576, 2939208, 4717, 1

Offset: 0

N. J. A. Sloane, Mar 06 2015

Note that Table 8 by Falcon should be labeled with the powers n (not r) and that the labels are off by 1. - R. J. Mathar, Jun 14 2015

			Triangle begins:
  1;
  1,    1; # see A079291
  1,    4,      1; # see A110272
  1,   13,     13,        1;
  1,   38,    130,       38,        1;
  1,  105,   1106,     1106,      105,        1;
  1,  280,   8575,    26544,     8575,      280,      1;
  1,  729,  62475,   567203,   567203,    62475,    729,    1;
  1, 1866, 435576, 11179686, 32897774, 11179686, 435576, 1866, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
S. Falcon, On The Generating Functions of the Powers of the K-Fibonacci Numbers, Scholars Journal of Engineering and Technology (SJET), 2014; 2 (4C):669-675.

Cf. A000129, A079291, A094706, A110272.

Diagonals: A094706, A255495, A255496, A255497, A255498.

P:= func< n | Round(((1 + Sqrt(2))^n - (1 - Sqrt(2))^n)/(2*Sqrt(2))) >;
function T(n,k)
  if k eq 0 or k eq n then return 1;
  else return P(n-k+1)*T(n-1,k-1) + P(k+1)*T(n-1,k);
  end if; return T;
end function;
[T(n,k): k in [0..n], n in [0..12]];

T[n_, k_]:= T[n,k]= If[k==0 || k==n, 1, Fibonacci[n-k+1, 2]*T[n-1, k-1] + Fibonacci[k+1, 2]*T[n-1, k]]; Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 19 2021 *)

@CachedFunction
def P(n): return lucas_number1(n, 2, -1)
def T(n,k): return 1 if (k==0 or k==n) else P(n-k+1)*T(n-1, k-1) + P(k+1)*T(n-1, k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 19 2021

From G. C. Greubel, Sep 19 2021: (Start)
T(n, k) = P(n-k+1)*T(n-1, k-1) + P(k+1)*T(n-1, k), where T(n, 0) = T(n, n) = 1 and P(n) = A000129(n).
T(n, k) = T(n, n-k).
T(n, 1) = A094706(n).
T(n, 2) = A255495(n-2).
T(n, 3) = A255496(n-3).
T(n, 4) = A255497(n-4).
T(n, 5) = A255498(n-5). (End)

A099930 a(n) = Pell(n) * Pell(n-1) * Pell(n-2) / 10.

Original entry on oeis.org

1, 12, 174, 2436, 34307, 482664, 6791772, 95567064, 1344731653, 18921807828, 266250046986, 3746422451772, 52716164405255, 741772724044560, 10437534301224120, 146867252940711408, 2066579075472320521, 29078974309550454492, 409172219409185308518, 5757490046038128779316

Offset: 3

Ralf Stephan, Nov 03 2004

Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 13.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 10.
Index entries for linear recurrences with constant coefficients, signature (12,30,-12,-1).

Cf. A000129. Third column of triangle A099927. Cf. A001656, A084175.

Drop[CoefficientList[Series[x^3/((1+2x-x^2)(1-14x-x^2)),{x,0,20}],x],3] (* or *) LinearRecurrence[{12,30,-12,-1},{1,12,174,2436},20] (* Harvey P. Dale, Feb 26 2012 *)

G.f.: x^3 / ((1+2*x-x^2)*(1-14*x-x^2)).
a(n) = 12*a(n-1) + 30*a(n-2) - 12*a(n-3) - a(n-4); a(3)=1, a(4)=12, a(5)=174, a(6)=2436. - Harvey P. Dale, Feb 26 2012
From Peter Bala, Mar 30 2015: (Start)
The following remarks assume an offset of 0.
The o.g.f. A(x) = 1/( (1 + 2*x - x^2)*(1 - 14*x - x^2) ). Hence A(x) (mod 4) = 1/(1 + 2*x - x^2)^2 (mod 4). It follows by Theorem 1 of Heninger et al. that sqrt(A(x)) = 1 + 6*x + 69*x^2 + 804*x^3 + ... has integral coefficients.
Sum_{n >= 0} a(n+3)*x^n = exp( Sum_{n >= 1} Pell(4*n)/Pell(n)*x^n/n ). Cf. A001656, A084175. (End)
a(n+1) = (1/2)*Sum_{k=1..n-1} ( A014445(k)*A110272(n-k) ) for n > 1. - Michael A. Allen, Jan 25 2023

A213688 a(n) = Sum_{i=0..n} A000129(i)^3.

Original entry on oeis.org

0, 1, 9, 134, 1862, 26251, 369251, 5196060, 73113372, 1028784997, 14476099149, 203694183170, 2866194639170, 40330419190351, 567492063162119, 7985219303802744, 112360562315573112, 1581033091723823881, 22246823846444284881, 313036566941955454910

Offset: 0

N. J. A. Sloane, Jun 18 2012

Bruno Berselli, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (13,18,-42,11,1).

Cf. A000129, A048739 (partial sums of A000129), A084158 (sum of squares of A000129).

Cf. A110272 (cubes of the Pell numbers).

A110272:=func; [&+[A110272(i): i in [0..n]]: n in [0..19]]; // Bruno Berselli, Jun 21 2012

LinearRecurrence[{13, 18, -42, 11, 1}, {0, 1, 9, 134, 1862}, 20] (* Bruno Berselli, Jun 21 2012 *)

G.f.: x*(1-4*x-x^2)/((1-x)*(1+2*x-x^2)*(1-14*x-x^2)). [Bruno Berselli, Jun 18 2012]

a(n) = ((3+sqrt(2))*(1+sqrt(2))^(3n+1)+(3-sqrt(2))*(1-sqrt(2))^(3n+1)-21*(-1)^n*((1+sqrt(2))^n+(1-sqrt(2))^n)+32)/224. [Bruno Berselli, Jun 18 2012]

A110273 a(n) = Pell(n)^3 + Pell(n+1)^3.

Original entry on oeis.org

1, 9, 133, 1853, 26117, 367389, 5169809, 72744121, 1023588937, 14402985777, 202665398173, 2851718540021, 40126725007181, 564625868522949, 7944888884612393, 111793070252410993, 1573047872420021137, 22134463284128711769, 311455533850231631029, 4382511937187348260781

Offset: 0

Paul Barry, Jul 18 2005

G. C. Greubel, Table of n, a(n) for n = 0..850
Index entries for linear recurrences with constant coefficients, signature (12,30,-12,-1).

Cf. A000129, A110272.

I:=[1,9,133,1853]; [n le 4 select I[n] else 12*Self(n-1) + 30*Self(n-2) - 12*Self(n-3) - Self(n-4): n in [1..31]]; // G. C. Greubel, Sep 17 2021

LinearRecurrence[{12,30,-12,-1},{1,9,133,1853},30] (* Harvey P. Dale, Jan 24 2018 *)
Sum[Fibonacci[Range[0, 30] +j, 2]^3, {j,0,1}] (* G. C. Greubel, Sep 17 2021 *)

[lucas_number1(n+1, 2, -1)^3 + lucas_number1(n, 2, -1)^3 for n in (0..30)] # G. C. Greubel, Sep 17 2021

G.f.: (1+x)*(1-4*x-x^2)/((1+2*x-x^2)*(1-14*x-x^2)).
a(n) = 12*a(n-1) + 30*a(n-2) - 12*a(n-3) - a(n-4).
a(n) = ( 3*(-1)^n*A001333(n) + Pell(3*n) + Pell(3*(n+1)) )/8.

A244352 a(n) = Pell(n)^3 - Pell(n)^2, where Pell(n) is the n-th Pell number (A000129).

Original entry on oeis.org

0, 0, 4, 100, 1584, 23548, 338100, 4798248, 67750848, 954701400, 13441659268, 189185124940, 2662308356400, 37463104912660, 527155118240244, 7417689205890000, 104375121328998144, 1468671237346368048, 20665783224031936900, 290789699203441908148

Offset: 0

Colin Barker, Jun 26 2014

			a(3) = Pell(3)^3 - Pell(3)^2 = 5^3 - 5^2 = 100.

Colin Barker, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (17,-25,-223,-79,95,-7,-1).

Cf. A000129, A079291, A110272.

Pell:= func< n | n eq 0 select 0 else Evaluate(DicksonSecond(n-1,-1),2) >;
[Pell(n)^3 - Pell(n)^2: n in [0..40]]; // G. C. Greubel, Aug 20 2022

CoefficientList[Series[4*x^2*(3*x^3-4*x^2+8*x+1) / ((x+1)*(x^2-6*x+1)*(x^2-2*x-1)*(x^2+14*x-1)), {x, 0, 20}], x] (* Vaclav Kotesovec, Jun 26 2014 *)

pell(n) = round(((1+sqrt(2))^n-(1-sqrt(2))^n)/(2*sqrt(2)))
vector(50, n, pell(n-1)^3-pell(n-1)^2)

def Pell(n): return lucas_number1(n,2,-1)
[Pell(n)^3 -Pell(n)^2 for n in (0..40)] # G. C. Greubel, Aug 20 2022

a(n) = A110272(n) - A079291(n).
G.f.: 4*x^2*(1+8*x-4*x^2+3*x^3) / ((1+x)*(1-6*x+x^2)*(1+2*x-x^2)*(1-14*x-x^2)).
a(n) = A045991(A000129(n)). - Michel Marcus, Jun 26 2014

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A255494 Triangle read by rows: coefficients of numerator of generating functions for powers of Pell numbers.

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A099930 a(n) = Pell(n) * Pell(n-1) * Pell(n-2) / 10.

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A213688 a(n) = Sum_{i=0..n} A000129(i)^3.

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A110273 a(n) = Pell(n)^3 + Pell(n+1)^3.

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A244352 a(n) = Pell(n)^3 - Pell(n)^2, where Pell(n) is the n-th Pell number (A000129).

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