A268554 - OEIS

%I #33 Nov 06 2024 04:42:04

%S 1,36,6300,1552320,445945500,139815211536,46384755633216,

%T 16009450307136000,5689533506261190300,2067982222137781950000,

%U 765185639177176836418800,287266309673587605560908800,109149488451384203661831720000

%N Diagonal of the rational function 1/((1 - w - u v) * (1 - x y - x z - y z)).

%C Each second element (which is zero) is skipped. - _R. J. Mathar_, Mar 10 2016

%C Annihilating differential operator: (-x^2+432*x^4)*Dx^4 + (-5*x+4320*x^3)*Dx^3 + (-4+10644*x^2)*Dx^2 + 6012*x*Dx + 288.

%H Vaclav Kotesovec, <a href="/A268554/b268554.txt">Table of n, a(n) for n = 0..200</a>

%H A. Bostan, S. Boukraa, J.-M. Maillard, and J.-A. Weil, <a href="http://arxiv.org/abs/1507.03227">Diagonals of rational functions and selected differential Galois groups</a>, arXiv preprint arXiv:1507.03227 [math-ph], 2015.

%H R. Meštrović, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.

%H Jacques-Arthur Weil, <a href="http://www.unilim.fr/pages_perso/jacques-arthur.weil/diagonals/">Supplementary Material for the Paper "Diagonals of rational functions and selected differential Galois groups"</a>

%F Conjecture: n^3*(2*n-1)*a(n) -6*(4*n-1)*(3*n-1)*(3*n-2)*(4*n-3)*a(n-1)=0. - _R. J. Mathar_, Mar 10 2016

%F From _Vaclav Kotesovec_, Jul 01 2016: (Start)

%F a(n) = (4*n)! * (3*n)! / ((n!)^3 * (2*n)!^2).

%F a(n) ~ 2^(4*n - 3/2) * 3^(3*n + 1/2) / (Pi^(3/2) * n^(3/2)).

%F (End)

%F 0 = (-x^2+432*x^4)*y'''' + (-5*x+4320*x^3)*y''' + (-4+10644*x^2)*y'' + 6012*x*y' + 288*y, where y = 1 + 36*x^2 + 6300*x^4  + ... is the g.f. - _Gheorghe Coserea_, Jul 03 2016

%F From _Peter Bala_, Oct 16 2024: (Start)

%F a(n) = 4 * Sum_{k = 0..2*n-1} (-1)^(n+k) * binomial(2*n-1, k) * binomial(4*n+k-1, k) * A108625(2*n, 2*n-k) for n >= 1 (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). Cf. A002897.

%F a(n) = binomial(4*n, 2*n)*binomial(3*n, n)*binomial(2*n, n).

%F The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k (apply Meštrović, Section 6, equation 39).

%F a(n) = [x^(2*n)] (1 + x)^(4*n) * [x^n] (1 + x)^(3*n) * [x^n] (1 + x)^(2*n) = [x^n] F(x)^(36*n), where F(x) = 1 + x + 52*x^2 + 6919*x^3 + 1266837*x^4 + 275133604*x^5 + 66468858333*x^6 + 17272069128056*x^7 + 4732687104502730*x^8 + 1350192483617697301*x^9 + 397617338885817524186*x^10 + ... appears to have integer coefficients (checked up to O(x^500)).

%F Let E(x) = exp(Sum_{n >= 1} (1/36) *a(n)*x^n/n). Then E(x) = 1 + x + 88*x^2 + 14461*x^3 + 3115089*x^4 + 781116715*x^5 + 215898182457*x^6 + 63857605571783*x^7 + 19853845202113934*x^8 + 6413541401057933731*x^9 + 2135530251738770328084*x^10 + ... appears to have integer coefficients (checked up to O(x^500)).

%F a(n) = 36 * [x^n] ( x/series_reversion(E(x)) )^n.

%F For integer r and positive integer s, define sequences {u(n) : n >= 0} and {v(n) : n >= 0} by setting u(n) = [x^(s*n)] F(x)^(r*n) and v(n) = [x^(s*n)] E(x)^(r*n). We conjecture that both u(n) and v(n) satisfy the above supercongruences. (End)

%p A268554 := proc(n)

%p     1/(1-w-u*v)/(1-x*y-x*z-y*z) ;

%p     coeftayl(%,x=0,n) ;

%p     coeftayl(%,y=0,n) ;

%p     coeftayl(%,z=0,n) ;

%p     coeftayl(%,u=0,n) ;

%p     coeftayl(%,v=0,n) ;

%p     coeftayl(%,w=0,n) ;

%p end proc:

%p seq(A268554(2*n),n=0..40) ; # _R. J. Mathar_, Mar 10 2016

%t Table[(4*n)!*(3*n)!/((n!)^3*(2*n)!^2), {n, 0, 15}] (* _Vaclav Kotesovec_, Jul 01 2016 *)

%o (PARI)

%o my(x1='x1, x2='x2, x3='x3, y1='y1, y2='y2, y3='y3);

%o R = 1/((1 - y1 - y2*y3) * (1 - x1*x2 - x1*x3 - x2*x3));

%o diag(n, expr, var) = {

%o   my(a = vector(n));

%o   for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));

%o   for (k = 1, n, a[k] = expr;

%o        for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));

%o   return(a);

%o };

%o diag(11, R, [x1,x2,x3,y1,y2,y3]) \\ _Gheorghe Coserea_, Jun 30 2016

%Y Cf. A268545 - A268555.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Feb 29 2016