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.

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A333564 a(n) = [x^n] ( c(x)/c(-x) )^n, where c(x) = (1 - sqrt( 1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

Original entry on oeis.org

2, 8, 56, 384, 2752, 20096, 148864, 1114112, 8403968, 63787008, 486584320, 3727196160, 28649455616, 220869853184, 1707123245056, 13223868760064, 102636144295936, 797982357192704, 6213784327684096, 48452953790480384, 378291752487878656, 2956824500391378944
Offset: 1

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Author

Peter Bala, Apr 07 2020

Keywords

Comments

It can be shown that a(n) satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p. We conjecture that a(n) satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 3 and positive integers n and k. Some examples are given below.
More generally, for integer r and positive integer s, we conjecture that the sequence a(r,s;n) := 2^(r*n) * Sum_{k = 0..s*n-1} (-1)^(s*n-1+k)*( binomial(n+k,k) )^r satisfies the same congruences. This is the case r = s = 1.

Examples

			Examples of congruences:
a(11) - a(1) = 486584320 - 2 = 2*(11^3)*182789 == 0 ( mod 11^3 ).
a(2*11) - a(2) = 2956824500391378944 - 8 = (2^3)*(3^2)*(11^3)*
107*288357478039 == 0 ( mod 11^3 ).
a(5^2) - a(5) = 1420245922851693002752 - 2752 = (2^6)*(3^3)*(5^6)* 7*19*1123*352183001 == 0 ( mod 5^6 ).
		

Crossrefs

Programs

  • Maple
    seq( (2^n)*add( (-1)^(n-1+k)*binomial(n+k,k), k = 0..n-1), n = 1..25);
    # alternative program
    a := proc (n) option remember; `if`(n = 1, 2, `if`(n = 2, 8, `if`(n = 3, 56, ((3*n+4)*a(n-1)+(36*n-76)*a(n-2)+(32*n-80)*a(n-3))/n)))
    end proc:
    seq(a(n), n = 1..25);
  • Mathematica
    a[n_] := -2^(n) Binomial[2n, n-1] Hypergeometric2F1[1, 2n +1, n + 2, 2];
    Table[Simplify[a[n]], {n, 1, 22}] (* Peter Luschny, Apr 13 2020 *)
    c[x_] := (1-Sqrt[1-4x])/(2x); ser[n_] := Series[(c[x]/c[-x])^n, {x, 0, 22}];
    Table[SeriesCoefficient[ser[n], n], {n, 1, 22}] (* Peter Luschny, Apr 14 2020 *)
  • Python
    from itertools import count, islice
    def A333564_gen(): # generator of terms
        yield (a:=2)
        c = 1
        for n in count(1):
            yield a<>1
    A333564_list = list(islice(A333564_gen(),20)) # Chai Wah Wu, Apr 26 2023

Formula

a(n) = 2^n * Sum_{k = 0..n-1} (-1)^(n-1+k)*binomial(n+k,k) = 2^n * A014300(n).
a(n) = (-1)^(n+1) + Sum_{k = 0..n-1} n^2/((n-k)*(2*n-k))*C(n-k,k)*C(3*n-2*k-1,n-k).
Congruences: a(p) == 2 ( mod p^3 ) for all prime p >= 3, follows from previous formula.
a(n) = Sum_{k = 1..n} (-1)^(n+k)*(3*k-1)*2^(k-1)*A000108(k-1).
a(n) = (1/2)*(A119259(n) - (-1)^n).
a(n) ~ 8^n / (3*sqrt(Pi*n)).
P-recursive with recurrence n*(3*n - 4)*a(n) = (21*n^2 - 40*n + 12)*a(n-1) + 4*(3*n - 1)*(2*n - 3)*a(n-2) with a(1) = 2 and a(2) = 8. Cf. A333565.
Alternative form: (a(n) + a(n-1))/(a(n) - a(n-2)) = P(n)/Q(n), where P(n) = 4*(3*n - 1)*(2*n - 3) and Q(n) = (21*n^2 - 40*n + 12).
Also, n*a(n) = (3*n + 4)*a(n-1) + 4*(9*n - 19)*a(n-2) + 16*(2*n - 5)*a(n-3) with a(1) = 2, a(2) = 8 and a(3) = 56.
O.g.f.: A(x) = 4*x/(1 - 8*x + (1 + 4*x)*sqrt(1 - 8*x)), which satisfies the differential equation (x + 1)*(4*x + 1)*(8*x - 1)*A'(x) + (16*x^2 - 4*x + 7)*A(x) + 2*(1 - 2*x) = 0.
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