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A049425
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Row sums of triangle A049404.
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9
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1, 1, 3, 9, 33, 141, 651, 3333, 18369, 108153, 678771, 4495041, 31324833, 228803589, 1744475643, 13852095741, 114235118721, 976176336753, 8627940414819, 78726234866553, 740440277799201, 7168107030092541, 71331617341611243, 728811735008913909, 7637128289949856833, 81995144342947130601
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OFFSET
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0,3
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LINKS
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FORMULA
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E.g.f.: exp(x+x^2+(x^3)/3).
a(n) = n! * sum(k=0..n, sum(j=0..k, binomial(3*j,n) * (-1)^(k-j)/(3^k * (k-j)!*j!))). [Vladimir Kruchinin, Feb 07 2011]
Conjecture: -a(n) +a(n-1) +(2*n-2)*a(n-2) + (2-3*n+n^2)*a(n-3)=0. - R. J. Mathar, Nov 14 2011
a(n) ~ exp(n^(2/3)+n^(1/3)/3-2*n/3-2/9)*n^(2*n/3)/sqrt(3)*(1+59/(162*n^(1/3))). - Vaclav Kotesovec, Oct 08 2012
Recurrence: a(n+3) = a(n+2)+2*(n+2)*a(n+1)+(n+2)*(n+1)*a(n).
It derives from the differential equation for the e.g.f.: A'(x) = (1+2*x+x^2)*A(x).
So, the above conjecture is true.
b(n) = a(n+1) = sum((n!/k!)*sum(bin(k,i)*bin(k-i+2,n-2*i-k)/3^i,i=0..k),k=0..n).
E.g.f. for b(n) = a(n+1): (1+t)^2*exp(t+t^2+t^3/3).
(End)
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MATHEMATICA
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Table[n!*SeriesCoefficient[E^(x+x^2+(x^3)/3), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 08 2012 *)
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PROG
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(PARI) x='x+O('x^66); Vec(serlaplace(exp(x+x^2+(x^3)/3))) \\ Joerg Arndt, May 04 2013
(Maxima) /* for b(n) = a(n+1) */
b(n) := sum((n!/k!)*sum(binomial(k, i)*binomial(k-i+2, n-2*i-k)/3^i, i, 0, k), k, 0, n);
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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