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%I #63 Sep 02 2022 01:54:35
%S 0,1,1,1,1,2,7,22,57,132,308,793,2223,6328,17578,47804,130169,360924,
%T 1019084,2900484,8252860,23445510,66717135,190750110,548178735,
%U 1580970612,4568275692,13217653582,38306172442,111248832992
%N Reversion of y - y^2 + y^3 - y^4.
%C Apparently: For n>0 number of Dyck (n-1)-paths with each ascent length equal to 0 or 1 modulo 4. - _David Scambler_, May 09 2012
%H R. J. Mathar, <a href="/A063019/b063019.txt">Table of n, a(n) for n = 0..104</a>
%H John Engbers, David Galvin, Clifford Smyth, <a href="https://arxiv.org/abs/1610.05803">Restricted Stirling and Lah numbers and their inverses</a>, arXiv:1610.05803 [math.CO], 2016. See p. 8.
%H Vladimir Kruchinin, <a href="http://arxiv.org/abs/1211.3244">The method for obtaining expressions for coefficients of reverse generating functions</a>, arXiv:1211.3244 [math.CO], 2012.
%H Elżbieta Liszewska, Wojciech Młotkowski, <a href="https://arxiv.org/abs/1907.10725">Some relatives of the Catalan sequence</a>, arXiv:1907.10725 [math.CO], 2019.
%H <a href="/index/Res#revert">Index entries for reversions of series</a>
%F 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
%F Conjecture confirmed for n >= 5 using the fact that the g.f. satisfies
%F (24*x + 96)*g(x) + (-1104*x^2 - 1302*x + 456)*g'(x)
%F + (-2688*x^3 - 1086*x^2 + 1086*x - 312)*g''(x)
%F + (-768*x^4 + 97*x^3 + 142*x^2 - 128*x + 32)*g'''(x) = 6*x+24. It
%F is not true for n=4. - _Robert Israel_, Jan 08 2019
%F 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
%F 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
%F From _Paul D. Hanna_, Sep 01 2022: (Start)
%F G.f. A(x) satisfies:
%F A(x)^3 = A( x^3 + 3*x*(1-x)*A(x)^3 ), and
%F A(x)^3 = ( x^3 + 3*x*(1-x)*A(x)^3 ) * (1 + A(x)^3) / (1 - A(x)^12). (End)
%e 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 + ...
%e such that A(x - x^2 + x^3 - x^4) = x.
%p F:= RootOf(y-y^2+y^3-y^4=x,y):
%p S:= series(F, x, 40):
%p seq(coeff(S,x,n),n=0..39); # _Robert Israel_, Jan 08 2019
%t CoefficientList[InverseSeries[Series[y - y^2 + y^3 - y^4, {y, 0, 30}], x], x]
%t 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 *)
%o (Maxima)
%o 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 */
%o (PARI)
%o x='x+O('x^66); Vec(serreverse(x-x^2+x^3-x^4)) /* _Joerg Arndt_, May 12 2011 */
%o (Sage) # uses[Reversion from A063022]
%o Reversion(x - x^2 + x^3 - x^4, 30) # _Peter Luschny_, Jan 08 2019
%Y Cf. A063022, A063023.
%K nonn,easy
%O 0,6
%A _Olivier Gérard_, Jul 05 2001
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