A218482 First differences of the binomial transform of the partition numbers (A000041).
1, 1, 3, 8, 21, 54, 137, 344, 856, 2113, 5179, 12614, 30548, 73595, 176455, 421215, 1001388, 2371678, 5597245, 13166069, 30873728, 72185937, 168313391, 391428622, 908058205, 2101629502, 4853215947, 11183551059, 25718677187, 59030344851, 135237134812, 309274516740
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 3*x^2 + 8*x^3 + 21*x^4 + 54*x^5 + 137*x^6 + 344*x^7 +... The g.f. equals the product: A(x) = (1-x)/((1-x)-x) * (1-x)^2/((1-x)^2-x^2) * (1-x)^3/((1-x)^3-x^3) * (1-x)^4/((1-x)^4-x^4) * (1-x)^5/((1-x)^5-x^5) * (1-x)^6/((1-x)^6-x^6) * (1-x)^7/((1-x)^7-x^7) *... and also equals the series: A(x) = 1 + x*(1-x)/((1-x)-x)^2 + x^4*(1-x)^2/(((1-x)-x)*((1-x)^2-x^2))^2 + x^9*(1-x)^3/(((1-x)-x)*((1-x)^2-x^2)*((1-x)^3-x^3))^2 + x^16*(1-x)^4/(((1-x)-x)*((1-x)^2-x^2)*((1-x)^3-x^3)*((1-x)^4-x^4))^2 +...
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
Crossrefs
Programs
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Maple
b:= proc(n) option remember; add(combinat[numbpart](k)*binomial(n,k), k=0..n) end: a:= n-> b(n)-b(n-1): seq(a(n), n=0..50); # Alois P. Heinz, Aug 19 2014
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Mathematica
Flatten[{1, Table[Sum[Binomial[n-1,k]*PartitionsP[k+1],{k,0,n-1}],{n,1,30}]}] (* Vaclav Kotesovec, Jun 25 2015 *)
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PARI
{a(n)=sum(k=0,n,(binomial(n,k)-if(n>0,binomial(n-1,k)))*numbpart(k))} for(n=0,40,print1(a(n),", "))
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PARI
{a(n)=local(X=x+x*O(x^n));polcoeff(prod(k=1,n,(1-x)^k/((1-x)^k-X^k)),n)}
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PARI
{a(n)=local(X=x+x*O(x^n));polcoeff(sum(m=0,n,x^m*(1-x)^(m*(m-1)/2)/prod(k=1,m,((1-x)^k - X^k))),n)}
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PARI
{a(n)=local(X=x+x*O(x^n));polcoeff(sum(m=0,n,x^(m^2)*(1-X)^m/prod(k=1,m,((1-x)^k - x^k)^2)),n)}
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PARI
{a(n)=local(X=x+x*O(x^n));polcoeff(exp(sum(m=1,n+1,x^m/((1-x)^m-X^m)/m)),n)}
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PARI
{a(n)=local(X=x+x*O(x^n));polcoeff(exp(sum(m=1,n+1,sigma(m)*x^m/(1-X)^m/m)),n)}
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PARI
{a(n)=local(X=x+x*O(x^n));polcoeff(prod(k=1,n,(1 + x^k/(1-X)^k)^valuation(2*k,2)),n)}
Formula
G.f.: Product_{n>=1} (1-x)^n / ((1-x)^n - x^n).
G.f.: Sum_{n>=0} x^n * (1-x)^(n*(n-1)/2) / Product_{k=1..n} ((1-x)^k - x^k).
G.f.: Sum_{n>=0} x^(n^2) * (1-x)^n / Product_{k=1..n} ((1-x)^k - x^k)^2.
G.f.: exp( Sum_{n>=1} x^n/((1-x)^n - x^n) / n ).
G.f.: exp( Sum_{n>=1} sigma(n) * x^n/(1-x)^n / n ), where sigma(n) is the sum of divisors of n (A000203).
G.f.: Product_{n>=1} (1 + x^n/(1-x)^n)^A001511(n), where 2^A001511(n) is the highest power of 2 that divides 2*n.
a(n) ~ exp(Pi*sqrt(n/3) + Pi^2/24) * 2^(n-2) / (n*sqrt(3)). - Vaclav Kotesovec, Jun 25 2015
Comments