A266964 Expansion of Product_{k>=1} (1 - k*x^k)^k.
1, -1, -4, -5, -3, 23, 44, 104, 70, -93, -465, -1155, -1882, -1904, 804, 6195, 18755, 33296, 47327, 35198, -28493, -176199, -453792, -805453, -1126396, -1028297, -18994, 2946491, 8248080, 16444480, 25436984, 30736635, 22263981, -16098311, -102681575
Offset: 0
Keywords
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..5000 from Vaclav Kotesovec)
Programs
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Magma
m:=50; R
:=PowerSeriesRing(Rationals(), m); Coefficients(R! ( (&*[(1-k*q^k)^k: k in [1..m]]) )); // G. C. Greubel, Oct 30 2018
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Maple
seq(coeff(series(mul((1-k*x^k)^k,k=1..n),x,n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Oct 31 2018
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Mathematica
nmax = 40; CoefficientList[Series[Product[(1-k*x^k)^k, {k, 1, nmax}], {x, 0, nmax}], x] (* More efficient program: *) nmax = 40; s = 1-x; Do[s*=Sum[Binomial[k, j]*(-1)^j*k^j*x^(j*k), {j, 0, nmax/k}]; s = Expand[s]; s = Take[s, Min[nmax + 1, Exponent[s, x] + 1, Length[s]]];, {k, 2, nmax}]; Take[CoefficientList[s, x], nmax] -
PARI
N=66; x='x+O('x^N); Vec(prod(k=1, N, (1-k*x^k)^k)) \\ Seiichi Manyama, Nov 18 2017 -
Ruby
def s(f_ary, g_ary, n) s = 0 (1..n).each{|i| s += i * f_ary[i] * g_ary[i] ** (n / i) if n % i == 0} s end def A(f_ary, g_ary, n) ary = [1] a = [0] + (1..n).map{|i| s(f_ary, g_ary, i)} (1..n).each{|i| ary << (1..i).inject(0){|s, j| s + a[j] * ary[-j]} / i} ary end def A266964(n) A((0..n).map{|i| -i}, (0..n).to_a, n) end p A266964(50) # Seiichi Manyama, Nov 18 2017
Formula
a(0) = 1 and a(n) = -(1/n) * Sum_{k=1..n} (Sum_{d|k} d^(2+k/d)) * a(n-k) for n > 0. - Seiichi Manyama, Nov 02 2017
From Seiichi Manyama, Nov 14 2017: (Start)
A generalized Euler transform.
Suppose given two sequences f(n) and g(n), n>0, we define a new sequence a(n), n>=0, by Product_{n>0} (1 - g(n)*x^n)^(-f(n)) = a(0) + a(1)*x + a(2)*x^2 + ...
Since Product_{n>0} (1 - g(n)*x^n)^(-f(n)) = exp(Sum_{n>0} (Sum_{d|n} d*f(d)*g(d)^(n/d))*x^n/n), we see that a(n) is given explicitly by a(n) = (1/n) * Sum_{k=1..n} b(k)*a(n-k) where b(n) = Sum_{d|n} d*f(d)*g(d)^(n/d).
Examples:
1. If we set g(n) = 1, we get the usual Euler transform.
2. If we set f(n) = -h(n) and g(n) = -1, we get the weighout transform (cf. A026007).
3. If we set f(n) = -n and g(n) = n, we get this sequence.
(End)