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a(n) = Sum_{k = 0..n} ( binomial(n+k-1,k) + binomial(n+k-1,k)^2 ).
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%I #26 May 09 2023 15:36:44

%S 2,4,20,166,1812,22504,297362,4067298,56897300,809019580,11649254520,

%T 169444978124,2485270719570,36707044807996,545386321069862,

%U 8144809732228666,122177690210103060,1839933274439787940,27804610626798500372,421476329345312885304,6406685025104178888312

%N a(n) = Sum_{k = 0..n} ( binomial(n+k-1,k) + binomial(n+k-1,k)^2 ).

%C Conjectures:

%C 1) a(p) == 4 (mod p^5) for all primes p >= 7 (checked up to p = 499). Note that A000984(p) == 2 (mod p^3) and A333592(p) == 2 (mod p^3) for all primes p >= 5.

%C 2) For r >= 2, and all primes p >= 5, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).

%C 3) More generally, let m be a positive integer and set u(n) = 2*m*Sum_{k = 0..m*n} binomial(n+k-1,k) + (m + 1)*Sum_{k = 0..m*n} binomial(n+k-1,k)^2. Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. This is the case m = 1. See A357673 for the case m = 2.

%C 4) For r >= 2, and all primes p >= 5, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ).

%F a(n) = A000984(n) + A333592(n).

%F a(p) == 4 (mod p^3) for all primes p >= 5.

%F a(n) ~ 2^(4*n) / (3*Pi*n).

%e Examples of supercongruences:

%e a(19) - a(1) = 421476329345312885304 - 4 = (2^2)*(5^2)*(19^5)*1913*2383*373393 == 0 (mod 19^5).

%e a(25) - a(5) = 5375188503768910125546897504 - 22504 = (2^3)*(5^10)*1858537* 37019662696111 == 0 (mod 5^10).

%p seq(add( binomial(n+k-1,k) + binomial(n+k-1,k)^2, k = 0..n ), n = 0..20);

%o (PARI) a(n) = sum(k = 0, n, binomial(n+k-1,k) + binomial(n+k-1,k)^2); \\ _Michel Marcus_, Oct 24 2022

%o (Python)

%o from math import comb

%o def A357671(n): return comb(n<<1,n)+sum(comb(n+k-1,k)**2 for k in range(n+1)) if n else 2 # _Chai Wah Wu_, Oct 28 2022

%Y Cf. A000984, A333592, A357509, A357565, A357566, A357672, A357673, A357674.

%K nonn,easy

%O 0,1

%A _Peter Bala_, Oct 10 2022