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A333472
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a(n) = [x^n] ( c (x/(1 + x)) )^n, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.
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1
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1, 1, 3, 13, 59, 276, 1317, 6371, 31131, 153292, 759428, 3780888, 18900389, 94805959, 476945913, 2405454213, 12158471195, 61574325840, 312365992620, 1587052145492, 8074474510884, 41131551386120, 209760563456920, 1070822078321520, 5471643738383781, 27982867986637151
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OFFSET
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0,3
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COMMENTS
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Let F(x) = 1 + f(1)*x + f(2)*x^2 + ... be a power series with integer coefficients. The associated sequence u(n) := [x^n] F(x)^n is known to satisfy the Gauss congruences: u(n*p^k) == u(n*p^(k-1)) ( mod p^k ) for any prime p and positive integers n and k. For certain power series F(x), stronger congruences may hold. Examples include F(x) = (1 + x)^2, F(x) = 1/(1 - x) and F(x) = c(x), where c(x) is the o.g.f. of the Catalan numbers A000108. The associated sequences (with some differences of offset) are A000984, A001700 and A025174, respectively.
Here we take F(x) = c(x/(1 + x)) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 9*x^5 + 21*x^6 + ... (cf. A001006 and A086246) and conjecture that the associated sequence a(n) = [x^n] ( c(x/(1 + x)) )^n satisfies the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and positive integers n and k. Cf. A333473.
More generally, we conjecture that for any positive integer r and any integer s the sequence a(r,s;n) := [x^(r*n)] ( c(x/(1 + x)) )^(s*n) also satisfies the above congruences.
Note that the sequence b(n) := [x^n] c(x)^n = A025174(n) satisfies the stronger congruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. The sequence d(n) := [x^n] ( (1 + x)*c(x/(1 + x)) )^n = A333093(n) appears to satisfy the same congruences.
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LINKS
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FORMULA
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a(n) = [x^n] ( (1 + x - sqrt(1 - 2*x - 3*x^2)) / (2*x) )^n.
a(n) ~ sqrt(((9386 + 1026*sqrt(57))^(1/3) + (9386 - 1026*sqrt(57))^(1/3) - 19)/228) * (((1261 + 57*sqrt(57))^(1/3) + (1261 - 57*sqrt(57))^(1/3) + 10)/6)^n / sqrt(Pi*n). - Vaclav Kotesovec, Mar 29 2020
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EXAMPLE
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Examples of congruences:
a(11) - a(1) = 3780888 - 1 = (11^2)*31247 == 0 ( mod 11^2 ).
a(3*7) - a(3) = 41131551386120 - 13 = (7^2)*13*23671*2727841 == 0 ( mod 7^2 ).
a(5^2) - a(5) = 27982867986637151 - 276 = (5^4)*13*74687*46113049 == 0 ( mod 5^4 ).
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MAPLE
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Cat := x -> (1/2)*(1-sqrt(1-4*x))/x:
G := x -> Cat(x/(1+x)):
H := (x, n) -> series(G(x)^n, x, 51):
seq(coeff(H(x, n), x, n), n = 0..25);
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MATHEMATICA
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Table[SeriesCoefficient[((1 + x - Sqrt[1 - 2*x - 3*x^2]) / (2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 29 2020 *)
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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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