0,2

The thirteen listed terms satisfy the linear recurrence a(n) = 13a(n - 1) + 104a(n - 2) - 260a(n - 3) - 260a(n - 4) + 104a(n - 5) + 13a(n - 6) - a(n - 7) for n>6 - John W. Layman (layman(AT)math.vt.edu), Apr 14 2000

A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.

A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 74.

G.f.: 1/(1-13*x-104*x^2+260*x^3+260*x^4-104*x^5-13*x^6+x^7) = 1/((1+x)*(1-3*x+x^2)*(1+7*x+x^2)*(1-18*x+x^2)) (see Comments to A055870).

a(n) = 5*a(n-1)+F(n-5)*Fibonomial(n+5, 5), n >= 1, a(0) = 1; F(n) = A000045(n) (Fibonacci). a(n) = 18*a(n-1)-a(n-2)+((-1)^n)*Fibonomial(n+4, 4), n >= 2; a(0) = 1, a(1) = 13; Fibonomial(n+4, 4) = A001656(n).

with(combinat):a:=n->1/240*fibonacci(n)*fibonacci(n+1)*fibonacci(n+2)*fibonacci(n+3)*fibonacci(n+4)*fibonacci(n+5): seq(a(n), n=1..17); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 07 2007

Adjacent sequences: A001655 A001656 A001657 this_sequence A001659 A001660 A001661

Sequence in context: A034833 A013518 A101004 this_sequence A034911 A133284 A012570

nonn

njas

More terms and formulae from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Jul 13 200, who also observes that Layman's recurrence is indeed true for all n >= 7.