A081079 -id:A081079 - cp's OEIS frontend

A092521 a(n) = 8a(n-1) - 8a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.

1, 8, 56, 385, 2640, 18096, 124033, 850136, 5826920, 39938305, 273741216, 1876250208, 12860010241, 88143821480, 604146740120, 4140883359361, 28382036775408, 194533374068496, 1333351581704065, 9138927697859960

Offset: 1

K. S. Bhanu (bhanu_105(AT)yahoo.com) and M. N. Deshpande, Apr 06 2004

a(n) such that 9*(T(a(n)-1) + T(a(n+1)-1)) = 7*(T(a(n) + a(n+1) - 1)), where T(i) denotes the i-th triangular number.
Partial sums of Chebyshev sequence S(n,7) = U(n,7/2) = A004187(n+1). - Wolfdieter Lang, Aug 31 2004
From Klaus Purath, Aug 06 2025: (Start)
Numbers k such that both 3*k + 1 and 15*k + 1 are perfect squares. Also the sum of two consecutive terms is a square.
Take any recurrence (r) of the form (3,-1) with initial value 0 followed by an arbitrary positive integer i. Then the product of two consecutive terms of r divided by 3*i^2 gives the current sequence. (End)

			G.f. = x + 8*x^2 + 56*x^3 + 385*x^4 + 2640*x^5 + 18096*x^6 + ... - _Michael Somos_, Jan 23 2025

Michael De Vlieger, Table of n, a(n) for n = 1..1197
Francesca Arici and Jens Kaad, Gysin sequences and SU(2)-symmetries of C*-algebras, arXiv:2012.11186 [math.OA], 2020.
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A212336 for more sequences with g.f. of the type 1/(1 - k*x + k*x^2 - x^3).

Cf. A000032, A004187, A058038, A081079.

A092521:= func< n | (Lucas(4*n+2) -3)/15 >; // G. C. Greubel, Jun 12 2025

a[1] = 1; a[2] = 8; a[3] = 56; a[n_] := a[n] = 8 a[n - 1] - 8 a[n - 2] + a[n - 3]; Table[ a[n], {n, 20}] (* Robert G. Wilson v, Apr 08 2004 *)
Table[(LucasL[4n+2]-3)/15, {n, 1, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)
LinearRecurrence[{8,-8,1},{1,8,56},30] (* Harvey P. Dale, Dec 27 2015 *)

Vec(x/((1-x)*(1-7*x+x^2)) + O(x^100)) \\ Altug Alkan, Oct 29 2015

def A092521(n): return (lucas_number2(4*n+2,1,-1) -3)//15 # G. C. Greubel, Jun 12 2025

G.f.: x/(1 - 8*x + 8*x^2 - x^3) = x/((1 - x)*(1 - 7*x + x^2)).
a(n) = 7*a(n-1) - a(n-2) + 1, n>=2, a(0):=0, a(1)=1.
a(n) = (S(n, 7)-S(n-1, 7) -1)/5, n>=1, with S(n, 7) = U(n, 7/2) = A004187(n+1).
a(n) = A058038(n)/3.
a(n) = (1/3)*Sum_{k=0..n} Fibonacci(4*k). - Gary Detlefs, Dec 07 2010
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jan 23 2025
From G. C. Greubel, Jun 12 2025: (Start)
a(n) = A081079(n)/15.
E.g.f.: (1/15)*( exp(7*x/2)*( 3*cosh(p*x) + sqrt(5)*sinh(p*x) ) - 3*exp(x) ), where p = 3*sqrt(5)/2. (End)

Edited and extended by Robert G. Wilson v, Apr 08 2004

A156088 Alternating sum of the squares of the first n even-indexed Fibonacci numbers.

Original entry on oeis.org

0, -1, 8, -56, 385, -2640, 18096, -124033, 850136, -5826920, 39938305, -273741216, 1876250208, -12860010241, 88143821480, -604146740120, 4140883359361, -28382036775408, 194533374068496, -1333351581704065, 9138927697859960

Offset: 0

Stuart Clary, Feb 04 2009

Apart from signs, same as A092521.

Natural bilateral extension (brackets mark index 0): ..., 2640, -385, 56, -8, 1, 0, [0], -1, 8, -56, 385, -2640, 18096, ... This is (-a(n))-reversed followed by a(n). That is, a(-n) = -a(n-1).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-8,-8,-1).

Cf. A000032, A000045, A081079, A092521, A103434, A103433, A156089.

A156088:= func< n | (-1)^n*(Lucas(4*n+2)-3)/15 >; // G. C. Greubel, Jun 12 2025

a[n_]:= If[n >= 0, Sum[(-1)^k Fibonacci[2k]^2, {k,n}], Sum[ -(-1)^k Fibonacci[-2k]^2, {k,-n-1}]];
LinearRecurrence[{-8,-8,-1}, {0,-1,8}, 41] (* G. C. Greubel, Jun 12 2025 *)

def A156088(n): return (-1)^n*(lucas_number2(4*n+2,1,-1) -3)//15 # G. C. Greubel, Jun 12 2025

Let F(n) be the n-th Fibonacci number, A000045(n), and L(n) be the n-th Lucas number, A000032(n), then: (Start)
a(n) = Sum_{k=1..n} (-1)^k F(2*k)^2.
Closed form: a(n) = (-1)^n * (L(4*n+2) - 3)/15.
Factored closed form: a(n) = (1/3) * (-1)^n * F(n)*L(n)*F(n+1)*L(n+1) = (1/3)*(-1)^n * F(2*n)*F(2*n+2).
Recurrence: a(n) + 8*a(n-1) + 8*a(n-2) + a(n-3) = 0.
G.f.: -x/(1 + 8*x + 8*x^2 + x^3) = -x/((1 + x)(1 + 7*x + x^2)). (End)
From G. C. Greubel, Jun 12 2025: (Start)
a(n) = (-1)^n*A081079(n)/15.
E.g.f.: (1/15)*( exp(-7*x/2)*( 3*cosh(p*x) - sqrt(5)*sinh(p*x) ) - 3*exp(-x) ), where p = 3*sqrt(5)/2. (End)

cp's OEIS Frontend

A092521 a(n) = 8a(n-1) - 8a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

Extensions

A156088 Alternating sum of the squares of the first n even-indexed Fibonacci numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

cp's OEIS Frontend

A092521 a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

Extensions

A156088 Alternating sum of the squares of the first n even-indexed Fibonacci numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

A092521 a(n) = 8a(n-1) - 8a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.