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A191242
Reversion of x-x^2-x^3-2*x^4
1
1, 1, 3, 12, 50, 224, 1054, 5121, 25509, 129591, 668811, 3496740, 18481512, 98585788, 530068840, 2869725800, 15630429306, 85589391884, 470905310206, 2601941245750, 14432082902820, 80328808797750, 448527122885700, 2511672193514250
OFFSET
1,3
COMMENTS
For the reversion of x - a*x^2 - b*x^3 - c*x^4 (a!=0, b!=0, c!=0) we have
a(n) = sum(k=1,n-1, (sum(j=0..k, a^(-n+3*k-j+1)*b^(n-3*k+2*j-1)*c^(k-j)*binomial(j,n-3*k+2*j-1)*binomial(k,j)))*binomial(n+k-1,n-1))/n, n>1, a(1)=1.
LINKS
FORMULA
a(n) = sum(k=1..n-1, (sum(j=0..k, binomial(j,n-3*k+2*j-1)*2^(k-j)*binomial(k,j)))*binomial(n+k-1,n-1))/n, n>1, a(1)=1.
MATHEMATICA
a[1] = 1; a[n_] := Sum[Sum[Binomial[j, n - 3k + 2j - 1]*2^(k - j)* Binomial[k, j], {j, 0, k}]*Binomial[n + k - 1, n - 1], {k, 1, n - 1}]/n;
Array[a, 24] (* Jean-François Alcover, Jul 23 2018 *)
PROG
(Maxima)
a(n):=sum((sum(binomial(j, n-3*k+2*j-1)*2^(k-j)*binomial(k, j), j, 0, k))*binomial(n+k-1, n-1), k, 1, n-1)/n;
(PARI) x='x+O('x^66); /* that many terms */
Vec(serreverse(x-x^2-x^3-2*x^4)) /* show terms */ /* Joerg Arndt, May 28 2011 */
(Magma) [&+[Binomial(i, n-3*k+2*i-1)*2^(k-i)*Binomial(k, i)*Binomial(n+k-1, n-1)/n: k in [0..25], i in [0..n]]: n in [1..25]]; // Vincenzo Librandi, Jul 23 2018
CROSSREFS
Sequence in context: A151178 A378078 A151179 * A105479 A151180 A268650
KEYWORD
nonn
AUTHOR
Vladimir Kruchinin, May 28 2011
STATUS
approved