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.

Showing 1-2 of 2 results.

A317133 G.f.: Sum_{n>=0} binomial(4*(n+1), n)/(n+1) * x^n / (1+x)^(n+1).

Original entry on oeis.org

1, 3, 15, 85, 526, 3438, 23358, 163306, 1167235, 8490513, 62648451, 467769217, 3527692298, 26832220834, 205601792340, 1585604105312, 12297768490441, 95861469636203, 750611119223931, 5901214027721577, 46564408929573723, 368644188180241449, 2927350250765841801, 23310167641788680947, 186089697960587977233, 1489085453187335910243
Offset: 0

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Author

Paul D. Hanna, Jul 21 2018

Keywords

Comments

Note that: binomial(4*(n+1), n)/(n+1) = A002293(n+1) for n >= 0, where F(x) = Sum_{n>=0} A002293(n)*x^n satisfies F(x) = 1 + x*F(x)^4.
Compare the g.f. to:
(C1) M(x) = Sum_{n>=0} binomial(2*(n+1), n)/(n+1) * x^n / (1+x)^(n+1) where M(x) = 1 + M(x) + M(x)^2 is the g.f. of Motzkin numbers (A001006).
(C2) 1 = Sum_{n>=0} binomial(m*(n+1), n)/(n+1) * x^n / (1+x)^(m*(n+1)) holds for fixed m.
(C3) If S(x,p,q) = Sum_{n>=0} binomial(p*(n+1),n)/(n+1) * x^n/(1+x)^(q*(n+1)), then Series_Reversion ( x*S(x,p,q) ) = x*S(x,q,p) holds for fixed p and q.

Examples

			G.f.: A(x) = 1 + 3*x + 15*x^2 + 85*x^3 + 526*x^4 + 3438*x^5 + 23358*x^6 + 163306*x^7 + 1167235*x^8 + 8490513*x^9 + 62648451*x^10 + ...
such that
A(x) = 1/(1+x) + 4*x/(1+x)^2 + 22*x^2/(1+x)^3 + 140*x^3/(1+x)^4 + 969*x^4/(1+x)^5 + 7084*x^5/(1+x)^6 + ... + A002293(n+1)*x^n/(1+x)^(n+1) + ...
RELATED SERIES.
Series_Reversion( x*A(x) )  =  x/((1+x)^4 - x)  =  x - 3*x^2 + 3*x^3 + 5*x^4 - 22*x^5 + 27*x^6 + 28*x^7 - 163*x^8 + 235*x^9 + 134*x^10 + ...
which equals the sum:
Sum_{n>=0} binomial(n+1, n)/(n+1) * x^(n+1)/(1+x)^(4*(n+1)).

Crossrefs

Cf. A316371, A127897, A317134, A349361, A349362, A349363, A349364.

Programs

  • Mathematica
    Rest[CoefficientList[InverseSeries[Series[x/((1 + x)^4 - x), {x, 0, 20}], x], x]] (* Vaclav Kotesovec, Jul 22 2018 *)
  • PARI
    {a(n) = my(A = sum(m=0, n, binomial(4*(m+1), m)/(m+1) * x^m / (1+x +x*O(x^n))^(1*(m+1)))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    {a(n) = my(A = (1/x) * serreverse( x/((1+x)^4 - x +x*O(x^n)) ) ); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) satisfies:
(1) A(x) = (1 + x*A(x))^4 / (1+x).
(2) A(x) = (1/x) * Series_Reversion( x/((1+x)^4 - x) ).
(3) A(x) = Sum_{n>=0} binomial(4*(n+1), n)/(n+1) * x^n / (1+x)^(n+1).
a(n) ~ 229^(n + 3/2) / (sqrt(Pi) * 2^(7/2) * n^(3/2) * 3^(3*n + 9/2)). - Vaclav Kotesovec, Jul 22 2018
a(n) = (1/(n+1)) * Sum_{k=0..n} (-1)^k * binomial(n+1,k) * binomial(4*n-4*k+4,n-k). - Seiichi Manyama, Mar 23 2024

A317135 G.f.: Sum_{n>=0} binomial(4*n+2, n)/(2*n+1) * x^(n+1)/(1+x)^(2*n+1).

Original entry on oeis.org

1, 1, 4, 18, 92, 504, 2897, 17235, 105233, 655687, 4152461, 26650675, 172961053, 1133152365, 7484233381, 49780558057, 333155274455, 2241793462963, 15158132783457, 102938343190333, 701783155862751, 4801340686195787, 32954688422181493, 226853921031126233, 1565828056187052419, 10834714749540715871, 75142241780769151970, 522240807596491906516
Offset: 0

Views

Author

Paul D. Hanna, Jul 22 2018

Keywords

Comments

Equals the self-convolution square-root of A317134.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 18*x^3 + 92*x^4 + 504*x^5 + 2897*x^6 + 17235*x^7 + 105233*x^8 + 655687*x^9 + 4152461*x^10 + ...
such that
A(x) = 1/(1+x) + 2*x/(1+x)^3 + 9*x^2/(1+x)^5 + 52*x^3/(1+x)^7 + 340*x^4/(1+x)^9 + 2394*x^5/(1+x)^11 + ... + A069271(n)*x^n/(1+x)^(2*n+1) + ...

Crossrefs

Cf. A317134.

Programs

  • Mathematica
    CoefficientList[Sqrt[1/x * InverseSeries[Series[4*x/((1 + x)^2 + Sqrt[(1 + x)^4 - 4*x])^2, {x, 0, 30}], x]], x] (* Vaclav Kotesovec, Jul 22 2018 *)
  • PARI
    {a(n) = my(A = sum(m=0, n, binomial(4*m+2, m)/(2*m+1) * x^m / (1+x +x*O(x^n))^(2*m+1))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    {a(n) = my(A = sqrt( (1/x) * serreverse( 4*x/((1+x)^2 + sqrt( (1+x)^4 - 4*x  + x*O(x^n)))^2 ))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) satisfies:
(1) A(x) = ( (1 + x*A(x)^2)^2 + sqrt( (1 + x*A(x)^2)^4 - 4*x*A(x)^2 ) ) / 2.
(2) A(x) = sqrt( (1/x) * Series_Reversion( 4*x/((1+x)^2 + sqrt( (1+x)^4 - 4*x ))^2 ) ).
(3) A(x) = Sum_{n>=0} binomial(4*n+2, n)/(2*n+1) * x^(n+1)/(1+x)^(2*n+1).
a(n) ~ 37^(1/4) * (101 + 16*sqrt(37))^(n + 1/2) / (4*sqrt(Pi) * n^(3/2) * 3^(3*n + 5/2)). - Vaclav Kotesovec, Jul 22 2018
Showing 1-2 of 2 results.

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