

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
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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 edgeweight 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 edgeweight is assigned as F(2). For k>1, T(k) is obtained from T(k1) by attaching F(k) vertices to pendents in T(k1) except r. In T(k), r is retained as in T(k1) and all the new edgeweights 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, for k>1 D(k)D(k1) is this sequence.
Retaining the notation of D(k) above it follows, for k>1 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) = prod(k=1..n, F(k)) * (F(n+2)1).  Franklin T. AdamsWatters, 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. AdamsWatters, Jun 23 2011. */


CROSSREFS

Sequence in context: A320343 A002856 A093461 * A153524 A153552 A295199
Adjacent sequences: A191991 A191992 A191993 * A191995 A191996 A191997


KEYWORD

nonn


AUTHOR

K.V.Iyer, Venkata Subba Reddy P., Charles R Greathouse IV, Jun 21 2011


STATUS

approved



