OFFSET
0,2
COMMENTS
Let L(n) = A000032(n) denote the n-th Lucas number.
For a fixed positive integer k, the power series expansion of exp( Sum_{n >= 1} L(k*n)x^n/n ) has integer coefficients given by the formula F(k*n)/F(k), where F(n) = A000045(n) [Johnson, 2.22].
The power series expansion of exp( Sum_{n >= 1} L(k*n)*L(2*k*n) *x^n/n ) has integer coefficients given by ( F(k*(n + 1))*F(k*(n + 2))*F(k*(n + 3)) )/( F(k)*F(2*k)*F(3*k) )
The present sequence is the particular case k = 2. See A001655 for the case k = 1.
LINKS
Eric W. Weisstein, Fibonacci Number
Index entries for linear recurrences with constant coefficients, signature (21,-56,21,-1).
FORMULA
a(n) = ( F(2*n + 2)*F(2*n + 4)*F(2*n + 6) )/( F(2)*F(4)*F(6) ).
a(n) = (1/8) * Sum_{k = 0..n} F(2*k + 2)*F(6*n - 6*k + 6).
O.g.f.: 1/( (1 - 3*x + x^2)*(1 - 18*x + x^2) ) = 1/8 * Sum_{n >= 0} F(2*n + 2)*x^n * Sum_{n >= 0} F(6*n + 6)*x^n.
O.g.f. also equals exp( Sum_{n >= 1} ( trace( M^(2*n) + M^(6*n) )*x^n/n ), where M is the 2X2 matrix [ 1, 1; 1, 0 ].
Recurrences: a(n) = 21*a(n-1) - 56*a(n-2) + 21*a(n-3) - a(n-4).
Also a(0) = 1 and for n >= 1, a(n) = (1/n)*Sum_{k = 1..n} L(2*k)*L(4*k)*a(n-k).
From Peter Bala, Aug 19 2022: (Start)
Sum_{n >= 0} 1/a(n) = 40/3 - 8*Sum_{n >= 1} 1/F(2*n) = 40/3 - 8*A153386.
Sum_{n >= 0} (-1)^n/a(n) = - 88/3 + 40*Sum_{n >= 1} (-1)^(n+1)/F(2*n). Cf. A265288. (End)
MAPLE
seq((1/24)*fibonacci(2*n+2)*fibonacci(2*n+4)*fibonacci(2*n+6), n = 0 .. 16);
MATHEMATICA
Table[1/8 * Sum[Fibonacci[2*k + 2]*Fibonacci[6*n - 6*k + 6], {k, 0, n}], {n, 0, 17}] (* or *) RecurrenceTable[{a[n] == 21*a[n - 1] - 56*a[n - 2] + 21*a[n - 3] - a[n - 4], a[1] == 1, a[2] == 21, a[3] == 385, a[4] == 6930}, a, {n, 17}] (* Michael De Vlieger, Mar 18 2015 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Mar 18 2015
STATUS
approved