A001658 Fibonomial coefficients.

1, 13, 273, 4641, 85085, 1514513, 27261234, 488605194, 8771626578, 157373300370, 2824135408458, 50675778059634, 909348684070099, 16317540120588343, 292806787575013635, 5254201798026392211, 94282845030238533383, 1691836875411111866723, 30358781826262552258596

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

It appears that a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7) for n > 6. - John W. Layman, Apr 14 2000

Layman's formula is correct. - Wolfdieter Lang, Jul 13 2000

Layman's formula is a consequence of formula 2.8 (p. 116) of Lind (1971). - Dale Gerdemann, May 08 2016

References

• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• T. D. Noe, Table of n, a(n) for n = 0..200
• 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.
• D. A. Lind, A Determinant Involving Binomial Coefficients, Part 1, Part 2, Fibonacci Quarterly 9.2, 1971.
• Index entries for linear recurrences with constant coefficients, signature (13, 104, -260, -260, 104, 13, -1).

Crossrefs

Cf. A001654, A001656, A001657.

Programs

Mathematica

f[n_] := Times @@ Fibonacci[Range[n+1, n+6]]/240; Table[f[n], {n, 0, 20}] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
LinearRecurrence[{13,104,-260,-260,104,13,-1},{1,13,273,4641,85085,1514513,27261234},20] (* Harvey P. Dale, Aug 24 2014 *)

PARI

b(n,k)=prod(j=1,k,fibonacci(n+j)/fibonacci(j));
vector(20,n,b(n-1,6))  \\ Joerg Arndt, May 08 2016

Formula

From Wolfdieter Lang, Jul 13 2000: (Start)

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). (End)

From Gary Detlefs, Dec 03 2012: (Start)

a(n) = F(n+1)*F(n+2)*F(n+3)*F(n+4)*F(n+5)*F(n+6)/240.

a(n) = (F(n+5)^2 - F(n+4)^2)*(F(n+3)^4 - 1)/240, where F(n) = A000045(n). (End)

Conjecture: a(n) = F(7)^(n-6) + Sum_{i=3..n-5} F(i-2)F(6)^{i-1}F(7)^{n-i-5} + Sum_{j=3..i} F(i-2)F(j-2)F(5)^{j-1}F(6)^{i-j}F(7)^{n-i-5} + Sum_{k=3..j} F(i-2)F(j-2)F(k-2)F(4)^{k-1}F(5)^{j-k}F(6)^{i-j}F(7)^{n-i-5} + Sum_{l=3..k} F(i-2)F(j-2)F(k-2)F(l-2)F(3)^{l-1}F(4)^{k-l}F(5)^{j-k}F(6)^{i-j}F(7)^{n-i-5} + Sum_{m=3..l} F(i-2)F(j-2)F(k-2)F(l-2)F(m-2)F(m)F(3)^{l-m}F(4)^{k-l}F(5)^{j-k}F(6)^{i-j}F(7)^{n-i-5}, where F(n)=A000045(n). - Dale Gerdemann, May 08 2016

G.f.: exp( Sum_{k>=1} F(7*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extensions

More terms from Wolfdieter Lang, Jul 13 2000