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A357566
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a(n) = ( Sum_{k = 0..n} binomial(n+k-1,k)^2 )^3 * ( Sum_{k = 0..n} binomial(n+k-1,k)^3 )^2.
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5
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1, 32, 3556224, 4816142496896, 14260946236464636800, 62923492736113950202540032, 355372959542696519903013302282592, 2376354966106399942850054560101358877184, 17973185649572984869873798116070605084766512000, 149319509846904520286037745483655872001727895961600000
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OFFSET
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0,2
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COMMENTS
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Conjectures:
1) a(p) == a(1) (mod p^5) for all odd primes p except p = 5 (checked up to p = 271).
2) a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) for r >= 2 and all primes p >= 5.
3) More generally, let m be a positive integer and set u(n) = ( Sum_{k = 0..m*n} binomial(n+k-1,k)^2 )^(m+2) * ( Sum_{k = 0..m*n} binomial(n+k-1,k)^3 )^(2*m). Then the supercongruences u(p) == u(1) (mod p^5) hold for all primes p >= 5.
4) u(p^r) == u(p^(r-1)) (mod p^(3*r+3)) for r >= 2 and all primes p >= 5.
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LINKS
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EXAMPLE
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a(7) - a(1) = 2376354966106399942850054560101358877184 - 32 = (2^5)*(7^5)*19*31*317*339247*25170329*2771351868561767 == 0 (mod 7^5).
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MAPLE
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seq((add(binomial(n+k-1, k)^2, k = 0..n))^3 * (add( binomial(n+k-1, k)^3, k = 0..n))^2, n = 0..20);
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PROG
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(PARI) a(n) = sum(k = 0, n, binomial(n+k-1, k)^2)^3 * sum(k = 0, n, binomial(n+k-1, k)^3)^2; \\ Michel Marcus, Oct 25 2022
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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