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%I #14 Oct 29 2020 03:23:59
%S 1,1,2,5,12,28,65,151,350,807,1850,4221,9597,21760,49215,111032,
%T 249856,560835,1255854,2805969,6256784,13925698,30941050,68634679,
%U 152009239,336152787,742276931,1636747349,3604206106,7926412320,17410413153
%N First differences of the binomial transform of the distinct partition numbers (A000009).
%H Seiichi Manyama, <a href="/A129519/b129519.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: A(x) = Product_{n>=1} [1 + x^n/(1-x)^n].
%F a(n) = A266232(n) - A266232(n-1), for n>0. - _Vaclav Kotesovec_, Oct 30 2017
%F a(n) ~ exp(Pi*sqrt(n/6) + Pi^2/48) * 2^(n - 9/4) / (3^(1/4) * n^(3/4)). - _Vaclav Kotesovec_, Oct 30 2017
%e Product formula is illustrated by:
%e A(x) = [1 + x + x^2 + x^3 + x^4 + x^5 +...]*
%e [1 + x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 +...]*
%e [1 + x^3 + 3x^4 + 6x^5 + 10x^6 + 15x^7 +...]*
%e [1 + x^4 + 4x^5 + 10x^6 + 20x^7 + 35x^8 +...]*
%e [1 + x^5 + 5x^6 + 15x^7 + 35x^8 + 70x^9 +...]*...*
%e [1 + Sum_{k>=n+1} C(k-1,n)*x^k ]*...
%t Flatten[{1, Differences[Table[Sum[Binomial[n, k]*PartitionsQ[k], {k, 0, n}], {n, 0, 40}]]}] (* _Vaclav Kotesovec_, Oct 30 2017 *)
%o (PARI) {a(n)=polcoeff(prod(k=0,n,1+sum(i=k+1,n,binomial(i-1,k)*x^i +x*O(x^n))),n)}
%Y Cf. A000009, A218482, A266232, A307501.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Apr 18 2007
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