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.

A385641 Partial sums of A097893.

Original entry on oeis.org

1, 3, 8, 20, 51, 133, 356, 972, 2695, 7557, 21372, 60840, 174097, 500295, 1442720, 4172752, 12099411, 35161001, 102375400, 298586652, 872177273, 2551118623, 7471195500, 21904500500, 64286141881, 188844619563, 555216323396, 1633658183432, 4810340397375, 14173698242137
Offset: 0

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Author

Mélika Tebni, Aug 03 2025

Keywords

Comments

Second partial sums of the central trinomial coefficients (A002426).
Third partial sums of A025178 (sequence starting 1, 0, 2, 4, 12, 32, 90 .... with offset 0).
For p prime of the form 4*k + 3 (A002145), a(p) + 1 == 0 (mod p).
For p Pythagorean prime (A002144), a(p) - 3 == 0 (mod p).
Sequences with g.f. (1-x)^k / sqrt(1-2*x-3*x^2): this sequence (k=-2), A097893 (k=-1), A002426 (k=0), A025178 (k=1), A024997 (k=2), A026083 (k=3). - Mélika Tebni, Aug 25 2025

Crossrefs

Programs

  • Maple
    a := series(exp(x)*(BesselI(0, 2*x) + 2*int(BesselI(0, 2*x), x) + int(int(BesselI(0, 2*x), x), x)), x = 0, 30): seq(n!*coeff(a, x, n), n = 0 .. 29);
  • PARI
    a(n) = sum(k=0, n, sum(i=0, k, sum(j=0, i, binomial(i, i-j)*binomial(j, i-j)))); \\ Michel Marcus, Aug 06 2025
  • Python
    from math import comb as C
    def a(n):
        return sum(C(n+1, k+1)*C(2*(k//2), k//2) for k in range(n + 1))
    print([a(n) for n in range(30)])
    

Formula

G.f.: (1 / sqrt((1 + x)*(1 - 3*x))) / (1 - x)^2.
E.g.f.: exp(x)*(BesselI(0, 2*x) + 2*g(x) + Integral_{x=-oo..oo} g(x) dx) where g(x) = Integral_{x=-oo..oo} BesselI(0, 2*x) dx.
D-finite with recurrence n*a(n) = (4*n-1)*a(n-1) - (2*n+1)*a(n-2) - (4*n-5)*a(n-3) + 3*(n-1)*a(n-4).
a(0) = 1, a(1) = 3 and a(n) = a(n-2) - 1 + 2*A383527(n) for n >= 2.
a(n) = Sum_{k=0..n} binomial(n+1, k+1)*A128014(k).
a(n) = Sum_{k=0..n} (2*A247287(k) + k+1).
a(n) ~ 3^(n + 5/2) / (8*sqrt(Pi*n)). - Vaclav Kotesovec, Aug 03 2025