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A364508
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a(n) = (7*n)!*(6*n)!*(2*n)! / ((4*n)!^2 * (3*n)!^2 * n!).
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4
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1, 350, 594594, 1299170600, 3164045050530, 8188909171581600, 22035578229399735000, 60924423899585957558848, 171839010049825493742617250, 492149504510899056782561257748, 1426695143534668869395862598229344, 4176678405144148418744441910948978000
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OFFSET
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0,2
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COMMENTS
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LINKS
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FORMULA
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a(n) = Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k) * binomial(6*n, 3*n + k)^2 (showing a(n) to be integral). Compare with Dixon's identity Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k)^3 = (3*n)!/n!^3 = A006480(n).
P-recursive: a(n) = (7/4)*(6*n-1)*(6*n-5)*(7*n-1)*(7*n-2)*(7*n-3)*(7*n-4)*(7*n-5)*(7*n-6)/((3*n-1)*(3*n-2)*(4*n-1)^2*(4*n-3)^2*n^2) * a(n-1) with a(0) = 1.
a(n) ~ c^n * sqrt(21)/(12*Pi*n), where c = (7^7)/(2^8).
a(n) = [x^n] G(x)^(14*n), where the power series G(x) = 1 + 25*x + 12798*x^2 + 13543850*x^3 + 18933663145*x^4 + 30733263922830*x^5 + 54771428143877503*x^6 + 104061045049532102971*x^7 + 207134582792235253663131*x^8 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^14, where the power series F(x) = 1 + 25*x + 21548*x^2 + 31466125*x^3 + 57506245907*x^4 + 119069165444705*x^5 + 266966985031172547*x^6 + 632553825380957995891*x^7 + 1560815989686060202098169*x^8 + ... appears to have integer coefficients.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.
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EXAMPLE
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Examples of supercongruences:
a(7) - a(1) = 60924423899585957558848 - 350 = 2*(7^4)*12687301936606821649 == 0 (mod 7^4).
a(11) - a(1) = 4176678405144148418744441910948978000 - 350 = 2*(5^2)*7*(11^3)* 29*139*1181*1883311859633620981885519 == 0 (mod 11^3).
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MAPLE
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seq( (7*n)!*(6*n)!*(2*n)! / ((4*n)!^2 * (3*n)!^2 * n!), n = 0..15);
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MATHEMATICA
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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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