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A157162
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1/product(1 - a(n)*x^n, n=1..infinity) = 1+ sum(F(k+1)*x^k, k=1..infinity) = 1/(1-x-x^2), where F(n) = A000045(n) (Fibonacci numbers).
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3
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1, 1, 1, 1, 2, 2, 4, 5, 8, 10, 18, 24, 40, 52, 88, 125, 210, 286, 492, 702, 1144, 1638, 2786, 3986, 6704, 9640, 16096, 23964, 39650, 57794, 97108, 144245, 236880, 353010, 589298, 880828, 1459960, 2179068, 3604880, 5471094, 9030450, 13561742, 22542396, 34277634
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,5
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COMMENTS
| A formal infinite product representation for the o.g.f. series of the Fibonacci numbers (A000045).
In the context of Witt rings the o.g.f. is called associated unital series for the (infinite dimensional) Witt vector (a(1),a(2),...). Sometimes also called inverse Somos transform, here for the Fibonacci numbers.
1-x-x^2 = product(1 - a(n)*x^n, n=1..infinity).
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LINKS
| W. Lang: Recurrences for the general problem.
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FORMULA
| Recurrence I. With P(n,m) the set of partitions of n with m parts:
a(n)= F(n+1) - sum(sum(product(a(j)^e(j),j=1..m), p from P(n,m)), m=2..n), n>=2, with sum(j*e(j),j=1..n)=n, sum(e(j),j=1..n)=m for the partition p of n with m parts. F(n) = A000045(n) (Fibonacci numbers). Input a(1)=F(2)=1. See the array A008284(n,m) for the cardinalities of the sets P(n,m).
Recurrence II (simplified version). With the Lucas numbers L(n)=A000035(n), n>=1, as input (found by V. Jovovic, Mar 10 2009):
a(n) = (- sum(d*a(d)^(n/d), d|n with 1<=d<n) + L(n))/n, n>=2, a(1)=1.
Recurrence II. With the number array M0(n,vec(e)) given for any partition in A048996.
a(n) = - sum((d/n)*(a(d))^(n/d),d|n with 1<=d<n) + sum(((-1)^(m-1))*(1/m)*sum(M0(p)*F(2)^e(1)*...*F(n+1)^e(n),p=(1^e(1),...,n^e(n)) from P(n,m)), m=1..n) for n>=2; a(1)=F(2)=1. See recurrence 1 for the set P(n,m). The M0 numbers are m!/product(e(j)!,j=1..n).
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EXAMPLE
| Recurrence I: a(2) = F(3) - a(1)^2 = 1; a(4) = F(5) - (a(1)*a(3) + a(2)^2 +a(1)^2*a(2) + a(1)^4) = 5 - 4 = 1.
Recurrence II (simplified): a(4) = (-(a(1)^4 + 2*a(2)^2) + L(4))/4 = (-3 + 7)/4 = 1.
Recurrence II: a(4)= (-(a(1)^4 + 2*a(2)^2)/4 + 1*1*F(5) - (1/2)*(2*F(2)*F(4)+ 1*F(3)^2) +(1/3)*3*F(2)^2*F(3)-(1/4)*1*F(2)^4 = -3/4 +7/4 = 1.
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CROSSREFS
| A147542 (with the product instead of the reciprocal one).
Sequence in context: A027193 A126796 A206138 * A109434 A089299 A017910
Adjacent sequences: A157159 A157160 A157161 * A157163 A157164 A157165
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KEYWORD
| nonn,easy
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AUTHOR
| Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de) Aug 10 2009
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