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A191994
(Sum of first n Fibonacci numbers) times (product of first n Fibonacci numbers).
2
1, 2, 8, 42, 360, 4800, 102960, 3538080, 196035840, 17520703200, 2529842515200, 590412901478400, 222813349683724800, 136001024583142118400, 134285149587387262464000, 214504624277084224347264000, 554361997358383529330695680000
OFFSET
1,2
COMMENTS
Let F(1), F(2), F(3), ... be the Fibonacci numbers 1, 1, 2, .... For k=1, we define the tree T(1) the path on two vertices with one identified as the root r. We assign the edge-weight F(1). T(2) is obtained from T(1) by attaching F(2) vertex to the pendents in T(1) except r. In T(2), r is retained as in T(1) and the new edge-weight is assigned as F(2). For k>1, T(k) is obtained from T(k-1) by attaching F(k) vertices to pendents in T(k-1) except r. In T(k), r is retained as in T(k-1) and all the new edge-weights are assigned F(k). With D(1)=1, for k>1 let D(k)=Sum of all distances d(r,x) taken across all vertices x in T(k). By induction it follows that for k>1, D(k)-D(k-1) is this sequence.
Retaining the notation of D(k) above, it follows, for k>1, that if D(k)=a(1)F(1)+ - - - +a(k)F(k) then D(k+1)=b(1)F(1)+ - - - +b(k)F(k)+b(k+1)F(k+1) where b(k+1) is the number of leaf nodes in T(k+1).
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..97
Eric Weisstein's World of Mathematics, Fibonacci Factorial Constant
FORMULA
a(n) ~ C*sqrt(phi^(n^2 + 3*n + 4)/5^(n+1)) where C = A062073 and phi = (1+sqrt(5))/2.
a(n) = (F(n+2)-1) * Product_{k=1..n} F(k). - Franklin T. Adams-Watters, Jun 23 2011
PROG
(PARI) s=0; p=1; for(n=1, 40, f=fibonacci(n); s+=f; p*=f; print1(s*p", ")) \\ Charles R Greathouse IV, Jun 21 2011
(PARI) a(n)=prod(k=1, n, fibonacci(k))*(fibonacci(n+2)-1) /* Franklin T. Adams-Watters, Jun 23 2011 */
CROSSREFS
Cf. A000071 (sum of Fibonacci numbers), A003266 (product of Fibonacci numbers).
Cf. A062073 (Fibonacci factorial constant).
Sequence in context: A320343 A002856 A093461 * A153524 A153552 A295199
KEYWORD
easy,nonn
STATUS
approved