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A357509
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a(n) = 2*binomial(3*n,n) - 9*binomial(2*n,n).
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9
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-7, -12, -24, -12, 360, 3738, 28812, 201672, 1355112, 8936070, 58427226, 380724552, 2479017996, 16151245488, 105359408760, 688338793488, 4504288103784, 29521135717470, 193771020939510, 1273649831269200, 8382448392851610, 55234026483856110, 364347399072847320
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OFFSET
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0,1
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COMMENTS
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For integers j and k, not necessarily positive, define u(n) = k^2*(k - 1)*binomial(j*n,n) - j^2*(j - 1)*binomial(k*n,n). We conjecture that u(p) == u(1) (mod p^5) for all primes p >= 7. This is essentially the case (j, k) = (3, 2). [Follows from Helou and Terjanian (2008), Section 3, Proposition 2.]
Conjecture: for r >= 2, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5. - Peter Bala, Oct 13 2022
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LINKS
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FORMULA
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a(p) == a(1) (mod p^4) for all primes p >= 5 by Meštrović, Section 3, equation 15.
Conjecture: the stronger supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7.
The conjecture is true: apply Helou and Terjanian, Section 3, Proposition 2. - Peter Bala, Oct 22 2022
The conjecture was proved by Aidagulov and Alekseyev; see the remarks following Corollary 2. - Peter Bala, Oct 29 2022
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MAPLE
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seq(2*binomial(3*n, n) - 9*binomial(2*n, n), n = 0..20);
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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