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A351859
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a(n) = [x^n] (1 + x + x^2 + x^3)^(4*n)/(1 + x + x^2)^(3*n).
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2
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1, 1, 3, 19, 67, 251, 1137, 4803, 20035, 87013, 377753, 1634469, 7134385, 31261114, 137121113, 603206144, 2660097603, 11749336328, 51981371895, 230336544210, 1021976441817, 4539784391763, 20188837618799, 89871081815631, 400427435522737, 1785639575031501
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OFFSET
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0,3
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COMMENTS
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This sequence is the third in an infinite family of sequences defined as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k, so that G_1(x) = (1 + x)^2, and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. See A000984, the sequence of central binomial coefficients, for the case k = 1 and A351858 for the case k = 2. The present sequence is the case k = 3.
Given a power series G(x) with integer coefficients it is known that the sequence (g(n))n>=1 defined by g(n) := [x^n] G(x)^n satisfies the Gauss congruences g(n*p^r) == g(n*p^(r-1) (mod p^r) for all primes p and positive integers n and r.
Thus a(n) satisfies the Gauss congruences. Calculation suggests that, in fact, the stronger supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. These supercongruences are known to hold for the central binomial coefficients A000984(n) = [x^n] ((1 + x)^2)^n (Meštrović, equation 39).
More generally, if r is a positive integer and s an integer then the sequence defined by a(r,s;n) = [x^(r*n)] G_3(x)^(s*n) may satisfy the same supercongruences.
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REFERENCES
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R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
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LINKS
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FORMULA
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a(n) = Sum_{i = 0..n} Sum_{j = 0..n} Sum_{k = 0..j} (-1)^j* C(4n,n-2*i-j-k) *C(4n,i)*C(3n+j-1,j)*C(j,k).
The o.g.f. A(x) = 1 + x + 3*x^2 + 19*x^3 + 67*x^4 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2 + x^3)^4/(1 + x + x^2)^3) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 + x + x^2)^3/(1 + x + x^2 + x^3)^4 ) = 1 + x + 2*x^2 + 8*x^3 + 25*x^4 + 81*x^5 + 305*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).
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EXAMPLE
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Examples of supercongruences:
a(5) - a(1) = 251 - 1 = 2*(5^3) == 0 (mod 5^3)
a(2*7)- a(2) = 137121113 - 3 = 2*5*(7^4)*5711 == 0 (mod 7^4)
a(5^2) - a(5) = 1785639575031501 - 251 = 2*(5^6)*1373*3989*10433 == 0 (mod 5^6)
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MAPLE
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seq(add(add(add((-1)^j*binomial(4*n, n-2*i-j-k)*binomial(4*n, i)* binomial(3*n+j-1, j)*binomial(j, k), k = 0..j), j = 0..n), i = 0..n), n = 0..25);
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MATHEMATICA
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nterms=25; Table[Sum[Sum[Sum[(-1)^j*Binomial[4n, n-2i-j-k]Binomial[4n, i]Binomial[3n+j-1, j]Binomial[j, k], {k, 0, j}], {j, 0, n}], {i, 0, n}], {n, 0, nterms-1}] (* Paolo Xausa, Apr 11 2022 *)
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PROG
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(PARI)
a(n)=sum(i=0, n, sum(j=0, n, sum(k=0, j, (-1)^j*binomial(4*n, n-2*i-j-k)*binomial(4*n, i)*binomial(3*n+j-1, j)*binomial(j, k))));
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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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