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.

A085364 a(0)=1, for n>0: a(n) = 6*13^(n-1) - (1/2)*Sum_{i=1..n-1} a(i)*a(n-i).

Original entry on oeis.org

1, 6, 60, 654, 7458, 87378, 1042152, 12587730, 153479508, 1885010946, 23285957604, 289018502682, 3601315495050, 45023019250398, 564465885846216, 7094214579174558, 89351097367355826, 1127492973620753010
Offset: 0

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Author

Mario Catalani (mario.catalani(AT)unito.it), Jun 25 2003

Keywords

Comments

From G. C. Greubel, May 23 2020: (Start)
This sequence is part of a class of sequences, for m >= 0, with the properties:
a(n) = 2*m*(4*m+1)^(n-1) - (1/2)*Sum_{k=1..n-1} a(k)*a(n-k).
a(n) = Sum_{k=0..n} m^k * binomial(n-1, n-k) * binomial(2*k, k).
n*a(n) = 2*((2*m+1)*n - (m+1))*a(n-1) - (4*m+1)*(n-2)*a(n-2).
a(n) = (2*m) * Hypergeometric2F1(-n+1, 3/2; 2; -4*m), for n > 0.
(4*m + 1)^n = Sum_{k=0..n} Sum_{j=0..k} a(j)*a(k-j).
G.f.: sqrt( (1 - t)/(1 - (4*m+1)*t) ).
This sequence is the case of m=3. (End)

Crossrefs

Cf. A001022 (13^n), A085362, A085363.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( Sqrt((1-x)/(1-13*x)) )); // G. C. Greubel, May 23 2020
    
  • Maple
    seq(coeff(series( sqrt((1-x)/(1-13*x)) , x, n+1), x, n), n = 0..30); # G. C. Greubel, May 23 2020
  • Mathematica
    CoefficientList[Series[Sqrt[(1-x)/(1-13x)], {x, 0, 25}], x]
  • PARI
    my(x='x+O('x^66)); Vec(sqrt((1-x)/(1-13*x))) \\ Joerg Arndt, May 10 2013
    
  • Sage
    def A085362_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( sqrt((1-x)/(1-13*x)) ).list()
    A085362_list(30) # G. C. Greubel, May 23 2020

Formula

G.f.: sqrt((1-x)/(1-13*x))
Sum_{i=0..n} Sum_{j=0..i} a(j)*a(i-j) = 13^n.
D-finite with recurrence: n*a(n) = 2*(7*n-4)*a(n-1) - 13*(n-2)*a(n-2). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ 2*sqrt(3)*13^(n-1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 14 2012
a(0) = 1; a(n) = (6/n) * Sum_{k=0..n-1} (n+k) * a(k). - Seiichi Manyama, Mar 28 2023
From Seiichi Manyama, Aug 22 2025: (Start)
a(n) = (1/4)^n * Sum_{k=0..n} 13^k * binomial(2*k,k) * binomial(2*(n-k),n-k)/(1-2*(n-k)).
a(n) = Sum_{k=0..n} (-3)^k * 13^(n-k) * binomial(2*k,k)/(1-2*k) * binomial(n-1,n-k). (End)