cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A271469 G.f. satisfies A(x) = 1 + x*(A(x)^3 - A(x)^4 + A(x)^5).

Original entry on oeis.org

1, 1, 4, 23, 155, 1142, 8910, 72350, 605056, 5175866, 45077560, 398348733, 3562916317, 32192775763, 293410452560, 2694283228653, 24902681767987, 231496130358758, 2162985033344112, 20301976721356134, 191336242071696514, 1809916398759630481, 17178063381786563194, 163536967014934201972, 1561247114394683682834, 14943175106109268856975
Offset: 0

Views

Author

Paul D. Hanna, Apr 08 2016

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 23*x^3 + 155*x^4 + 1142*x^5 + 8910*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 9*x^2 + 54*x^3 + 372*x^4 + 2778*x^5 + 21873*x^6 +...
A(x)^3 = 1 + 3*x + 15*x^2 + 94*x^3 + 663*x^4 + 5025*x^5 + 39970*x^6 +...
A(x)^4 = 1 + 4*x + 22*x^2 + 144*x^3 + 1041*x^4 + 8016*x^5 + 64470*x^6 +...
A(x)^5 = 1 + 5*x + 30*x^2 + 205*x^3 + 1520*x^4 + 11901*x^5 + 96850*x^6 +...
A(x)^6 = 1 + 6*x + 39*x^2 + 278*x^3 + 2115*x^4 + 16848*x^5 + 138816*x^6 +...
A(x)^7 = 1 + 7*x + 49*x^2 + 364*x^3 + 2842*x^4 + 23044*x^5 + 192325*x^6 +...
where A(x) = 1 + x*(A(x)^3 - A(x)^4 + A(x)^5),
and A(x)^2 = 1 + x*(A(x)^3 + A(x)^6),
and A(x)^3 = 1 + x*(A(x)^3 + A(x)^5 + A(x)^7),
and A(x)^4 = 1 + x*(A(x)^3 + A(x)^5 + A(x)^6 + A(x)^8),
and A(x)^5 = 1 + x*(A(x)^3 + A(x)^5 + A(x)^6 + A(x)^7 + A(x)^9), etc.
The g.f. satisfies A(x) = F(x*A(x)^3) and F(x) = A(x/F(x)^3) where
F(x) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 9*x^5 + 21*x^6 + 51*x^7 +...+ A001006(n-1)*x^n +...
is a g.f. of the Motzkin numbers (A001006, shifted right 1 place).
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + x + 3*x^2 + 13*x^3 + 66*x^4 + 366*x^5 + 2148*x^6 +...+ A219537(n)*x^n +...
satisfies G(x) = 1 + x*(G(x)^2 - G(x)^3 + G(x)^4).
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[(1/x*InverseSeries[Series[8*x^4/(1 + x - Sqrt[1 - 2*x - 3*x^2])^3, {x, 0, 20}], x])^(1/3), x] (* Vaclav Kotesovec, Apr 16 2016 *)
  • PARI
    /* Formula A(x) = 1 + x*(A(x)^3 - A(x)^4 + A(x)^5): */
    {a(n)=local(A=1); for(i=1, n, A=1+x*(A^3-A^4+A^5) +x*O(x^n)); polcoeff(A, n)}
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    /* Formula using Series Reversion involving Motzkin numbers: */
    {a(n)=local(A=1); A=(1+x-sqrt(1-2*x-3*x^2+x^3*O(x^n)))/(2*x); polcoeff( (1/x*serreverse(x/A^3))^(1/3), n)}
    for(n=0, 25, print1(a(n), ", "))

Formula

G.f. A(x) satisfies:
(1) A(x)^2 = 1 + x*(A(x)^3 + A(x)^6).
(2) A(x)^3 = 1 + x*(A(x)^3 + A(x)^5 + A(x)^7).
Let F(x) = (1+x - sqrt(1 - 2*x - 3*x^2)) / (2*x), then g.f. A(x) satisfies:
(3) A(x) = ( (1/x)*Series_Reversion(x/F(x)^3) )^(1/3),
(4) A(x) = F(x*A(x)^3) and F(x) = A(x/F(x)^3),
where F(x) = 1 + x*M(x) such that M(x) = 1 + x*M(x) + x^2*M(x)^2 is the g.f. of the Motzkin numbers (A001006).
Let G(x) = 1 + x*(G(x)^2 - G(x)^3 + G(x)^4), then g.f. A(x) satisfies:
(5) A(x) = (1/x)*Series_Reversion(x/G(x)),
(6) A(x) = G(x*A(x)) and G(x) = A(x/G(x)),
where G(x) is the g.f. of A219537.
a(n) ~ sqrt((34 + (34102 - 8262*sqrt(17))^(1/3) + (34102 + 8262*sqrt(17))^(1/3)) / 1632) * ((28 + (513243 - 4131*sqrt(17))^(1/3)/3 + (19009 + 153*sqrt(17))^(1/3)) / 8)^n / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Apr 16 2016
D-finite recurrence: 8*n*(2*n-1)*(4*n-1)*(4*n+1)*(204*n^4 - 1341*n^3 + 3191*n^2 - 3286*n + 1242)*a(n) = 12*(45696*n^8 - 391776*n^7 + 1376164*n^6 - 2580579*n^5 + 2808064*n^4 - 1797694*n^3 + 651566*n^2 - 119476*n + 8160)*a(n-1) - 6*(n-2)*(29376*n^7 - 237168*n^6 + 760044*n^5 - 1236774*n^4 + 1082233*n^3 - 496791*n^2 + 108530*n - 8400)*a(n-2) + 9*(n-3)*(n-2)*(3*n-8)*(3*n-4)*(204*n^4 - 525*n^3 + 392*n^2 - 111*n + 10)*a(n-3). - Vaclav Kotesovec, Apr 16 2016
From Seiichi Manyama, Aug 06 2023: (Start)
a(n) = (1/n) * Sum_{k=0..n-1} binomial(n,k) * binomial(3*n+k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * binomial(n,k) * binomial(5*n-2*k,n-1-k) for n > 0. (End)
G.f.: A(x) = sqrt(B(x)) where B(x) is the g.f. of A370474. - Seiichi Manyama, Mar 31 2024
a(n) = (1/n) * Sum_{k=0..floor((n-1)/2)} binomial(n,k) * binomial(4*n-k,n-1-2*k) for n > 0. - Seiichi Manyama, Apr 01 2024
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(3*n/2+3*k/2+1/2,n)/(3*n+3*k+1). - Seiichi Manyama, Apr 04 2024