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A357569
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a(n) = binomial(3*n,n)^2 - 27*binomial(2*n,n).
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6
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-26, -45, 63, 6516, 243135, 9011205, 344597148, 13520945736, 540917244351, 21966327267885, 902702921361813, 37456461969311736, 1566697064604277788, 65973795093057780936, 2794203818388994498200, 118933541228931589568016, 5084343623375039833670079, 218184481964802862563857685
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OFFSET
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0,1
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COMMENTS
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Conjectures:
1) a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 3.
These are stronger supercongruences than those satisfied separately by the sequences {binomial(2*n,n)} = A000984 and {binomial(3*n,n)} = A05809.
2) More generally, for k >= 1, the sequence {2*binomial(3*n,n)^k - k*(3^(k+1))*binomial(2*n,n): n >= 0} may satisfy the same supercongruences. This is the case k = 2. See A357509 for the case k = 1.
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LINKS
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FORMULA
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a(k*p^r) == a(k*p^(r-1)) ( mod p^(3*r) ) for positive integers k and r and for all primes p >= 5 (see Meštrović, Section 6, equation 39).
a(p) == a(1) (mod p^5) for all primes p >= 7 (apply Helou and Terjanian, Section 3, Proposition 2).
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EXAMPLE
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Examples of supercongruences:
a(13) - a(1) = 65973795093057780936 + 45 = (3^2)*(13^5)*163*121122434651 == 0 (mod 13^5).
a(5^2) - a(5) = 2765555290416839473031163791322085183080 - 9011205 = (3^2)*(5^9)* 229*2333*6840413*74974087*574203805501 == 0 (mod 5^9).
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MAPLE
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seq(binomial(3*n, n)^2 - 27*binomial(2*n, n), n = 0..20);
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MATHEMATICA
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Table[Binomial[3n, n]^2-27*Binomial[2n, n], {n, 0, 30}] (* Harvey P. Dale, Jun 12 2023 *)
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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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