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A357955
a(n) = 3*binomial(4*n,n) - 20*binomial(3*n,n) + 54*binomial(2*n,n).
3
37, 60, 108, 60, -660, 60, 82404, 1411848, 17540460, 191318820, 1952058108, 19175376324, 184118073828, 1743153802320, 16359157606200, 152693295412560, 1420516291306860, 13190159377278324, 122358232382484420, 1134645084249344400, 10522118980232969340
OFFSET
0,1
COMMENTS
Conjectures:
1) a(p) == a(1) (mod p^7) for all primes p >= 3 except p = 7.
2) a(p^r) == a(p^(r-1)) ( mod p^(3*r+5) ) for r >= 2 and all primes p >= 3.
These are stronger supercongruences than those satisfied separately by the sequences {binomial(2*n,n)} = A000984, {binomial(3*n,n)} = A005809 and {binomial(4*n,n)} = A005810.
Conjecture 1) was proved by Aidagulov and Alekseyev; see the remarks following Corollary 2. - Peter Bala, Oct 29 2022
LINKS
R. R. Aidagulov and M. A. Alekseyev, On p-adic approximation of sums of binomial coefficients, Journal of Mathematical Sciences 233:5 (2018), 626-634; arXiv:1602.02632 [math.NT], 2018.
C. Helou and G. Terjanian, On Wolstenholme’s theorem and its converse, J. Number Theory 128 (2008), 475-499.
FORMULA
a(n) = 3*A005810(n) - 20*A005809(n) + 54*A000984(n).
a(k*p^r) == a(k*p^(r-1)) ( mod p^(3*r) ) for positive integers k and r and for all primes p >= 5 (see Meštrović, Section 6, equation 39).
a(p) == a(1) (mod p^6) for all primes p >= 7 (apply Helou and Terjanian, Section 3, Proposition 2).
From Stefano Spezia, Jul 17 2024: (Start)
G.f.: 54/sqrt(1-4*x) - 40*cos(arccos(1-27*x/2)/6)/sqrt(4-27*x) + 3*hypergeom([1/4, 1/2, 3/4}, {1/3, 2/3}, 4^4*x/3^3).
E.g.f.: 54*exp(2*x)*BesselI(0, 2*x) - 20*hypergeom([1/3, 2/3], [1/2, 1], 27*x/4) + 3*hypergeom([1/4, 1/2, 3/4], [1/3, 2/3, 1], 4^4*x/3^3).
a(n) ~ exp(2*n*arctanh(229/283))*sqrt(6/(n*Pi)). (End)
EXAMPLE
Examples of supercongruences:
a(11) - a(1) = 19175376324 - 60 = (2^3)*3*(11^7)*41 == 0 (mod 11^7).
a(5^2) - a(5) = 726506045044361132812560 - 60 = (2^2)*3*(5^11)*41*30241552444123 == 0 (mod 5^11).
MAPLE
seq(3*binomial(4*n, n) - 20*binomial(3*n, n) + 54*binomial(2*n, n), n = 0..20);
MATHEMATICA
A357955[n_] := 3*Binomial[4*n, n] - 20*Binomial[3*n, n] + 54*Binomial[2*n, n];
Array[A357955, 25, 0] (* Paolo Xausa, Jul 17 2024 *)
PROG
(Python)
from math import comb
def A357955(n): return 54*comb(m:=n<<1, n)+3*comb(m<<1, n)-20*comb(m+n, n) # Chai Wah Wu, Oct 24 2022
KEYWORD
sign,easy
AUTHOR
Peter Bala, Oct 22 2022
STATUS
approved