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A357568
a(n) = 9*binomial(2*n,n)^2 - 8*binomial(3*n,n).
9
1, 12, 204, 2928, 40140, 547512, 7535472, 105077376, 1484848332, 21237645000, 306972655704, 4477160465856, 65802123629424, 973487343836448, 14483651478207360, 216550246159148928, 3251660678391659724, 49011343741651501800, 741221951008966181160, 11243583961952559386400
OFFSET
0,2
COMMENTS
Conjectures:
1) a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 3.
These are stronger supercongruences than those satisfied separately by the sequences {binomial(2*n,n)} = A000984 and {binomial(3*n,n)} = A005809.
2) More generally, for k >= 1, the sequence {9*binomial(2*n,n)^k - k*(2^k)*binomial(3*n,n): n >= 0} may satisfy the same supercongruences. This is the case k = 2. See A357509 for the case k = 1.
LINKS
C. Helou and G. Terjanian, On Wolstenholme’s theorem and its converse, J. Number Theory 128 (2008), 475-499.
FORMULA
a(n) = 9*A002894(n) - 8*A005809(n) = 9*A000984(n)^2 - 8*A005809(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^5) for all primes p >= 7 (apply Helou and Terjanian, Section 3, Proposition 2).
From Stefano Spezia, Jul 17 2024: (Start)
G.f.: 16*cos(arccos(1-27*x/2)/6)/sqrt(4-27*x) + 18*EllipticK(16*x)/Pi.
E.g.f.: 9*hypergeom([1/2, 1/2], [1, 1], 16*x) - 8*hypergeom([1/3, 2/3], [1/2, 1], 27*x/4).
a(n) ~ 9*2^(4*n)/(n*Pi). (End)
EXAMPLE
Examples of supercongruences:
a(11) - a(1) = 4477160465856 - 12 = (2^2)*3*(11^5)*101*22937 == 0 (mod 11^5).
a(5^2) - a(5) = 143816772358933669354266172512 - 547512 = (2^3)*3*(5^9)*167191* 194659*94271599039 == 0 (mod 5^9).
MAPLE
seq(9*binomial(2*n, n)^2 - 8*binomial(3*n, n), n = 0..20);
MATHEMATICA
A357568[n_] := 9*Binomial[2*n, n]^2 - 8*Binomial[3*n, n];
Array[A357568, 25, 0] (* Paolo Xausa, Jul 17 2024 *)
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Oct 21 2022
STATUS
approved