A063019 Reversion of y - y^2 + y^3 - y^4.
0, 1, 1, 1, 1, 2, 7, 22, 57, 132, 308, 793, 2223, 6328, 17578, 47804, 130169, 360924, 1019084, 2900484, 8252860, 23445510, 66717135, 190750110, 548178735, 1580970612, 4568275692, 13217653582, 38306172442, 111248832992, 323794993924, 944208641464, 2757606964409
Offset: 0
Examples
G.f. A(x) = x + x^2 + x^3 + x^4 + 2*x^5 + 7*x^6 + 22*x^7 + 57*x^8 + 132*x^9 + 308*x^10 + 793*x^11 + 2223*x^12 + 6328*x^13 + 17578*x^14 + ... such that A(x - x^2 + x^3 - x^4) = x.
Links
- R. J. Mathar, Table of n, a(n) for n = 0..104
- John Engbers, David Galvin, Clifford Smyth, Restricted Stirling and Lah numbers and their inverses, arXiv:1610.05803 [math.CO], 2016. See p. 8.
- Vladimir Kruchinin, The method for obtaining expressions for coefficients of reverse generating functions, arXiv:1211.3244 [math.CO], 2012.
- Elżbieta Liszewska, Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
- Index entries for reversions of series
Programs
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Maple
F:= RootOf(y-y^2+y^3-y^4=x,y): S:= series(F, x, 40): seq(coeff(S,x,n),n=0..39); # Robert Israel, Jan 08 2019
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Mathematica
CoefficientList[InverseSeries[Series[y - y^2 + y^3 - y^4, {y, 0, 30}], x], x] g[d_] := g[d] = If[OddQ[d], 3, 1]; f[x_, y_, d_] := f[x, y, d] = If[x < 0 || y < x, 0, If[x == 0 && y == 0, 1, f[x - 1, y, 0] + f[x, y - If[d == 0, 1, g[d]], If[d == 0, 1, g[d] + d]]]]; Join[{0}, Table[f[n - 1, n - 1, 0], {n, 30}]] (* David Scambler, May 09 2012 *) -
Maxima
a(n):=if n<2 then n else (-1)^(n+1)*sum((sum(binomial(j,n-3*k+2*j-1)*(-1)^(2*j-k)*binomial(k,j),j,0,k))*binomial(n+k-1,n-1),k,1,n-1)/n; /* Vladimir Kruchinin, May 10 2011 */
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PARI
my(x='x+O('x^66)); Vec(serreverse(x-x^2+x^3-x^4)) /* Joerg Arndt, May 12 2011 */ -
Sage
# uses[Reversion from A063022] Reversion(x - x^2 + x^3 - x^4, 30) # Peter Luschny, Jan 08 2019
Formula
Conjecture: 32*n*(n-1)*(n-2)*a(n) -8*(n-1)*(n-2)*(16*n-9)*a(n-1) +2*(n-2)*(71*n^2+46*n-549)*a(n-2) +(97*n^3-2250*n^2+10859*n-14850)*a(n-3) -12*(4n-15)*(4*n-14)*(4*n-17)*a(n-4)=0. - R. J. Mathar, Oct 01 2012
Conjecture confirmed for n >= 5 using the fact that the g.f. satisfies
(24*x + 96)*g(x) + (-1104*x^2 - 1302*x + 456)*g'(x)
+ (-2688*x^3 - 1086*x^2 + 1086*x - 312)*g''(x)
+ (-768*x^4 + 97*x^3 + 142*x^2 - 128*x + 32)*g'''(x) = 6*x+24.
It is not true for n=4. - Robert Israel, Jan 08 2019
Recurrence: 16*(n-2)*(n-1)*n*(5*n-14)*a(n) = 4*(n-2)*(n-1)*(110*n^2 - 473*n + 468)*a(n-1) - (n-2)*(1015*n^3 - 6902*n^2 + 15391*n - 11232)*a(n-2) + 8*(2*n-5)*(4*n-13)*(4*n-11)*(5*n-9)*a(n-3). - Vaclav Kotesovec, Feb 12 2014
Conjecture: a(n+1) = (1/(n+1))*Sum_{k=0..floor(n/4)} binomial(n+1, n-4*k)*binomial(n+k, n) (compare to the formula from Peter Bala in A215340). - Joerg Arndt, Apr 01 2019
From Paul D. Hanna, Sep 01 2022: (Start)
G.f. A(x) satisfies:
A(x)^3 = A( x^3 + 3*x*(1-x)*A(x)^3 ), and
A(x)^3 = ( x^3 + 3*x*(1-x)*A(x)^3 ) * (1 + A(x)^3) / (1 - A(x)^12). (End)
Comments