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2, 11, 83, 699, 6252, 58106, 554633, 5399099, 53356322, 533627511, 5388927513, 54859837434, 562267554552, 5796123147756, 60047675871333, 624801952898619, 6526036790730942, 68395815476047901, 718992874207884953, 7578808590187108199
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OFFSET
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1,1
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COMMENTS
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The Apéry numbers A005258 satisfy the supercongruences A005258(p) == 3 (mod p^3) and A005258(p-1) == 1 (mod p^3) for primes p >= 5. It easily follows that a(p) == 2 (mod p^3) for primes p >= 3. We conjecture that the stronger supercongruences a(p) == 2 (mod p^5) hold for primes p >= 5. See A212334 for the corresponding conjecture for the Apéry numbers A005259.
Conjecture: for r >= 2, and all primes p >= 5, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ). - Peter Bala, Oct 13 2022
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LINKS
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FORMULA
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a(n) = (1/2)*Sum_{k = 0..n} (2*n^2 - k*n + k^2)/(n*(n + k)) * binomial(n,k)^2 * binomial(n + k,k).
a(n) = (1/2)*Sum_{k = 0..n-1} (4*n + k)*(n - k)/(n*(n + k)) * binomial(n,k)^2* binomial(n + k,k) for n >= 1.
a(n) ~ 5^(3/4)*(13 + 5*sqrt(5))/(20*sqrt(22 + 10*sqrt(5))*Pi*n) * ((11 + 5*sqrt(5))/2)^n.
(11*n^2 - 31*n + 22)*n^2*a(n) = (121*n^4 - 462*n^3 + 607*n^2 - 322*n + 64)*a(n-1) + (11*n^2 - 9*n + 2)*(n - 2)^2*a(n-2) with a(1) = 2 and a(2) = 11.
The g.f. A(x) = 2*x + 11*x^2 + 83*x^3 + ... satisfies the differential equation
(x^5 + 13*x^4 + 22*x^3 + 9*x^2 - x)*A''(x) + (x^4 + 4*x^3 + 26*x^2 + 22*x - 1)*A'(x) + (2*x^2 - 16*x + 4)*A(x) + x^2 - 8*x + 2 = 0, with A(0) = 2 and A'(0) = 11.
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EXAMPLE
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Examples of superconguences:
a(5) - 2 = 6252 - 2 = 2*(5^5) == 0 (mod 5^5).
a(7) - 2 = 554633 - 2 = 3*(7^5)*11 == 0 (mod 7^5).
a(11) - 2 = 5388927513 - 2 = (11^5)*33461 == 0 (mod 11^5).
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MAPLE
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seq((1/2)*add((2*n^2 - k*n + k^2)/(n*(n + k)) * binomial(n, k)^2 * binomial(n + k, k), k = 0..n), n = 1..20);
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PROG
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(PARI) f(n) = sum(k=0, n, binomial(n, k)^2 * binomial(n+k, k)); \\ A005258
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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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