OFFSET
0,3
COMMENTS
In general, the sequence with g.f. Product_{k>=1} 1/(1-x^k)^Fibonacci(k+z), where z is nonnegative integer, is asymptotic to phi^(n + z/4) / (2 * sqrt(Pi) * 5^(1/8) * n^(3/4)) * exp((phi/10 - 1/2) * Fibonacci(z) - Fibonacci(z+1)/10 + 2 * 5^(-1/4) * phi^(z/2) * sqrt(n) + s), where s = Sum_{k>=2} (Fibonacci(z) + Fibonacci(z+1) * phi^k) / ((phi^(2*k) - phi^k - 1)*k) and phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Aug 06 2015
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..4550
Loic Foissy, The Hopf algebra of Fliess operators and its dual pre-Lie algebra, 2013.
W. S. Gray, K. Ebrahimi-Fard, Affine SISO Feedback Transformation Group and Its Faa di Bruno Hopf Algebra, arXiv:1411.0222 [math.OC], 2014.
Vaclav Kotesovec, Asymptotics of the Euler transform of Fibonacci numbers, arXiv:1508.01796 [math.CO], Aug 07 2015
Vaclav Kotesovec, Asymptotics of sequence A034691
FORMULA
G.f.: Product_{k>0} 1/(1 - x^k)^Fibonacci(k).
a(n) ~ phi^n / (2 * sqrt(Pi) * 5^(1/8) * n^(3/4)) * exp(-1/10 + 2*5^(-1/4)*sqrt(n) + s), where s = Sum_{k>=2} phi^k / ((phi^(2*k) - phi^k - 1)*k) = 0.600476601392575912969719494850393576083765123939643511355547131467... and phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Aug 06 2015
G.f.: exp(Sum_{k>=1} x^k/(k*(1 - x^k - x^(2*k)))). - Ilya Gutkovskiy, May 29 2018
EXAMPLE
G.f. = 1 + x + 2*x^2 + 4*x^3 + 8*x^4 + 15*x^5 + 30*x^6 + 56*x^7 + 108*x^8 + 203*x^9 + ...
MAPLE
F:= proc(n) option remember; (<<1|1>, <1|0>>^n)[1, 2] end:
a:= proc(n) option remember; `if`(n=0, 1, add(add(d*
F(d), d=numtheory[divisors](j))*a(n-j), j=1..n)/n)
end:
seq(a(n), n=0..40); # Alois P. Heinz, Jan 12 2017
MATHEMATICA
CoefficientList[Series[Product[1/(1-x^k)^Fibonacci[k], {k, 1, 40}], {x, 0, 40}], x] (* Vaclav Kotesovec, Aug 05 2015 *)
PROG
(PARI) ET(v)=Vec(prod(k=1, #v, 1/(1-x^k+x*O(x^#v))^v[k]))
ET(vector(40, n, fibonacci(n)))
(SageMath)
def EulerTransform(a):
@cached_function
def b(n):
if n == 0: return 1
s = sum(sum(d * a(d) for d in divisors(j)) * b(n-j) for j in (1..n))
return s//n
return b
a = BinaryRecurrenceSequence(1, 1)
b = EulerTransform(a)
print([b(n) for n in range(36)]) # Peter Luschny, Nov 11 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Franklin T. Adams-Watters, Oct 21 2009
EXTENSIONS
First formula corrected by Vaclav Kotesovec, Aug 05 2015
STATUS
approved