A016036 Row sums of triangle A000369.
1, 4, 31, 361, 5626, 109951, 2585269, 71066626, 2236441141, 79289379361, 3127129674736, 135802922499949, 6439320471558781, 331026965612789356, 18338413238239145731, 1089132347371148170381, 69033182553940825258594, 4651256393180943757676371
Offset: 1
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..200
- Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Crossrefs
Programs
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Magma
R
:=PowerSeriesRing(Rationals(), 30); Coefficients(R!(Laplace( Exp(1-(1-4*x)^(1/4)) -1 ))); // G. C. Greubel, Oct 02 2023 -
Mathematica
a[n_, m_] /; (n>= m>= 1):= a[n, m]= (4*(n-1)-m)*a[n-1,m] + a[n-1,m-1]; a[n_, m_] /; n
,0]= 0; a[1,1] = 1; a[n]:= Sum[a[n,m], {m, n}]; Table[a[n], {n,20}] (* Jean-François Alcover, Feb 28 2013 *) With[{nn=20},CoefficientList[Series[Exp[1-Surd[1-4x,4]]-1,{x,0,nn}],x] Range[0,nn]!]//Rest (* Harvey P. Dale, Apr 20 2016 *) -
Maxima
a(n):=((n-1)!*sum((sum(binomial(n+k-1,n-1)*sum(binomial(j,n-m-3*k+2*j)*binomial(k,j)*3^(-n+m+3*k-j)*2^(n-m-5*k+3*j)*(-1)^(n-m-k),j,0,k),k,1,n-m))/(m-1)!,m,1,n-1))+1; /* Vladimir Kruchinin, Oct 18 2011 */
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SageMath
def A016036_list(prec): P.
= PowerSeriesRing(QQ, prec) return P( exp(1-(1-4*x)^(1/4)) -1 ).egf_to_ogf().list() a=A016036_list(40); a[1:] # G. C. Greubel, Oct 02 2023
Formula
E.g.f.: exp(1 - (1-4*x)^(1/4)) - 1.
a(n) = 6*(2*n-5)*a(n-1) - 3*(16*n^2-96*n+145)*a(n-2) + 2*(4*n-15)*(2*n-7)*(4*n-13)*a(n-3) + a(n-4), n >= 4; a(0) = 1, a(1) = 1, a(2) = 4, a(3) = 31.
a(n) = 1 + (n-1)!*Sum_{m=1..n-1} ( Sum_{k=1..n-m} binomial(n+k-1,n-1) * ( Sum_{j=0..k} binomial(j,n-m-3*k+2*j)*binomial(k,j)*3^(-n+m+3*k-j)*2^(n-m-5*k+3*j)*(-1)^(n-m-k) ) )/(m-1)!. - Vladimir Kruchinin, Oct 18 2011
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator 1/(1-x)^3*d/dx. Cf. A001515, A015735 and A028575. - Peter Bala, Nov 25 2011
a(n) ~ 2^(2*n-3/2)*n^(n-3/4)*exp(1-n)*sqrt(Pi)/Gamma(3/4) * (1 - Gamma(3/4)/(n^(1/4)*sqrt(Pi)) + Gamma(3/4)^2/(4*sqrt(n/2)*Pi)). - Vaclav Kotesovec, Aug 10 2013
From Seiichi Manyama, Jan 20 2025: (Start)
a(n) = Sum_{k=0..n} (-1)^k * 4^(n-k) * |Stirling1(n,k)| * A000587(k).
a(n) = e * (-4)^n * n! * Sum_{k>=0} (-1)^k * binomial(k/4,n)/k!. (End)