A191797 a(n) = binomial(F(n), 2) where F(n) = A000045(n).
0, 0, 0, 1, 3, 10, 28, 78, 210, 561, 1485, 3916, 10296, 27028, 70876, 185745, 486591, 1274406, 3337236, 8738290, 22879230, 59901985, 156830905, 410597496, 1074972528, 2814337800, 7368069528, 19289917153, 50501756955, 132215475106, 346144864780, 906219437046
Offset: 0
Examples
a(7) = binomial(13,2) = 78.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,1,-5,-1,1).
Programs
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Maple
with(combinat): seq(binomial(fibonacci(n), 2), n = 0 .. 30);
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Mathematica
Table[Binomial[Fibonacci[n], 2], {n, 0, 39}] (* Alonso del Arte, Apr 04 2013 *) -
PARI
a(n) = binomial(fibonacci(n), 2); \\ Michel Marcus, Sep 07 2015
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PARI
concat(vector(3), Vec(x^3 / ((1+x)*(1-x-x^2)*(1-3*x+x^2)) + O(x^40))) \\ Colin Barker, Mar 26 2017
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Python
from sympy import binomial, fibonacci def a(n): return binomial(fibonacci(n), 2) # Indranil Ghosh, Mar 26 2017
Formula
a(n) = 3*a(n-1) + 1*a(n-2) - 5*a(n-3) - 1*a(n-4) + 1*a(n-5).
G.f.: x^3/(1-3*x-x^2+5*x^3+x^4-x^5) = x^3/((1+x)*(1-x-x^2)*(1-3*x+x^2)).
a(n) + a(n+1) = A056014(n+1). - R. J. Mathar, Jun 24 2011
a(n) = (2*F(n)^2 - F(n+4) + 3*F(n+1))/4, F(n) = A000045(n). - Gary Detlefs, Jan 05 2013
a(n) = Sum_{k=1..n-2} A122931(k). - J. M. Bergot, Apr 05 2013
a(n) = (2^(-1-n)*(-(-1)^n*2^(1+n) + sqrt(5)*(1-sqrt(5))^n + (3-sqrt(5))^n - sqrt(5)*(1+sqrt(5))^n + (3+sqrt(5))^n)) / 5. - Colin Barker, Mar 26 2017
a(n) ~ phi^(2*n) / 10, where phi is the golden ratio (A001622). - Amiram Eldar, Aug 26 2025