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A357512
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a(n) = Sum_{k = 0..n} k^5 * binomial(n,k)^2 * binomial(n+k,k)^2
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7
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0, 4, 1188, 126144, 10040000, 682492500, 41503541940, 2325305113600, 122429236976640, 6140504039242500, 296222848665342500, 13841644170257145792, 629814531655430506944, 28019919084086921883892, 1222770835880665252492500, 52476371578141941012480000, 2219374467089388085650636800
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OFFSET
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0,2
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COMMENTS
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Let m >= 2 and set u(n) = Sum_{k = 0..n} k^(2*m+1) * binomial(n,k)^2 * binomial(n+k,k)^2. We conjecture that there is a finite set of primes P(m) such that u(n-1) == 0 (mod n^4) for all odd numbers n not divisible by an element of P(m). For example, calculation suggests that P(2) = {3} (the present sequence), while conjecturally P(3) = {3, 5}, P(4) = {3, 7}, P(5) = {3, 5}, P(6) = {3, 11}, P(7) = {3, 5, 7, 13}, P(8) = {3}, P(9) = {3} and P(10) = {3, 7, 19}.
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LINKS
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FORMULA
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Conjecture: a(n-1) == 0 (mod n^4) for all n congruent to 1 or 5 mod 6 (checked up to n = 499).
Note: the Apery numbers A(n) = A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2 satisfy the supercongruence A(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Introduction).
Recurrence: a(1) = 4, a(2) = 1188, a(3) = 126144 and thereafter
(n - 3)*(2*n - 3)*(3*n^4 - 18*n^3 + 40*n^2 - 39*n + 13)*(n - 2)^3*(n - 1)^4*a(n) = (n - 3)*(2*n - 1)*(n + 1)*(105*n^7 - 879*n^6 + 2981*n^5 - 5124*n^4 + 4376*n^3 - 1236*n^2 - 442*n + 195)*(n - 2)^3*a(n-1) - n^3*(n - 3)*(2*n - 3)*(n + 1)*(105*n^7 - 591*n^6 + 1253*n^5 - 1346*n^4 + 784*n^3 - 124*n^2 - 90*n + 33)*a(n-2) + n^3*(2*n - 1)*(n + 1)*(3*n^4 - 6*n^3 + 4*n^2 - n - 1)*(n - 1)^4*a(n-3).
a(n) ~ n^(7/2) * (1 + sqrt(2))^(4*n + 2) / (2^(19/4) * Pi^(3/2)). - Vaclav Kotesovec, Oct 04 2022
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MAPLE
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seq( add( k^5 * binomial(n, k)^2 * binomial(n+k, k)^2, k = 0..n ), n = 0..20 );
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PROG
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(PARI) a(n) = sum(k = 0, n, k^5 * binomial(n, k)^2 * binomial(n+k, k)^2); \\ Michel Marcus, Oct 04 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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