OFFSET
0,2
COMMENTS
Entry number six in the "Big Table" of Almkvist et al. (see links). The period T(x) = Sum_{n>=0} a(n)*x^(2*n) is also the first x-derivative of the 6-volume associated to the algebraic variety V6 = P1 & P2 & P3, with P1 : X1^2 + Y1^2 = X2^2 + Y2^2, P2 : X2^2 + Y2^2 = X3^2 + Y3^2, P3 : x=(X1^2 + X2^2 + X3^2 + Y1^2 + Y2^2 + Y3^2)^3*(1 - X1*X2*X3*Y1*Y2*Y3). The small x limit reduces V6 to a 6-ball with 6-volume proportional to x. Similar constructions are known to exist for a few other geometries on Almkvist's list, most notably #3: A186420, and #16: A039699.
LINKS
Gert Almkvist, Christian van Enckevort, Duco van Straten, and Wadim Zudilin, Tables of Calabi-Yau Equations, arXiv:math/0507430 [math.AG], 2005-2010, p. 10.
G. Almkvist et al., Raw Calabi-Yau Period Data, Johannes Gutenberg Universität Mainz, 2019, case 1.6.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
FORMULA
G.f.: 4F3({1/4, 3/4, 1/2, 1/2}, {1, 1, 1}, 1024*x).
Define the period integral: dt(x) = dz1*dz2*dz3/sqrt(1-32*x*cos(z1)*cos(z2)*cos(z3)).
T(x) = (1/(2*Pi)^3)*Integral_{0..2*Pi,0..2*Pi,0..2*Pi} dt(x),
the Picard-Fuchs coefficients: (c0,c1,c2,c3,c4) = (768*x, 14592*x^2-1, x*(25344*x^2-7), 2*x^2*(5120*x^2-3), x^3*(32*x-1)*(32*x+1)),
and the certificate function: G(z1,z2,z3) = (16*sin(z1)*(
48*x*cos(z1)
+ cos(z2)*cos(z3)
+ 48*x*cos(z1)*(cos(z3)^2 + cos(z2)^2)
+ 2304*x^2*cos(z1)^2*cos(z2)*cos(z3)
+ 80*x*cos(z1)*cos(z2)^2*cos(z3)^2
+ 384*x^2*cos(z1)^2*(cos(z2)*cos(z3)^3 + cos(z2)^3*cos(z3))
+ 256*x^2*cos(z1)^2*cos(z2)^3*cos(z3)^3)
)/(3*(1 - 32*x*cos(z1)*cos(z2)*cos(z3))^(7/2)),
Then: 0 = Sum_{n=0..4} cn*d^n/dx^n dt(x) + d/dz1 G(z1,z2,z3) + d/dz2 G(z2,z3,z1) + d/dz3 G(z3,z1,z2), thus: 0 = Sum_{n=0..4} cn*d^n/dx^n T(x).
Furthermore, let (a1,a2,a3) = (c1,c2,c3)/c0, then also: 0 = (1/2)*a2*a3 - (1/8)*a3^3 + d/dx(a2) - (3/4)*a3*d/dx(a3) - (1/2)*d^2/dx^2(a3) - a1.
D-finite with recurrence: n^4*a(n) - 16*(4*n-1)*(4*n-3)*(-1+2*n)^2*a(n-1) = 0. - R. J. Mathar, Jan 27 2020
a(n) ~ 2^(10*n-1/2) / (Pi*n)^2. - Amiram Eldar, Oct 19 2025
From Peter Bala, Oct 19 2025: (Start)
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 Mestrovic, equation 39).
a(n) = binomial(4*n, n)*binomial(3*n, n)*binomial(2*n, n)^2 = [x^n] (1 + x)^(4*n) * [x^n] (1 + x)^(3*n) * [x^n] (1 + x)^(2*n) * [x^n] (1 + x)^(2*n).
a(n) = [x^n] F(x)^(48*n), where F(x) = 1 + x + 110*x^2 + 32219*x^3 + 133400606*x^4 + 6735458262*x^5 + 3813588923865*x^6 + ... appears to have integer coefficients.
We conjecture that for each integer k, the sequence defined by f_k(n) = [x^n] F(x)^(k*n) satisfies the above supercongruences.
E(x) = exp( (1/48)*Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 158*x^2 + 51491*x^3 + 23055534*x^4 + 12350497090*x^5 + 7424497087209*x^6 + ... appears to have integer coefficients.
We conjecture that for each integer k, the sequence defined by e_k(n) = [x^n] E(x)^(k*n) also satisfies the above supercongruences. (End)
MATHEMATICA
Binomial[4*#, 2*#]*Binomial[2*#, #]^3&/@Range[0, 10]
PROG
(SageMath)
print([binomial(4*n, 2*n)*binomial(2*n, n)^3 for n in range(10)]) # F. Chapoton, Sep 14 2025
(Python)
from math import comb
def A307618(n): return comb(4*n, n)*comb(3*n, n)*comb(2*n, n)**2 # Chai Wah Wu, Feb 17 2026
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bradley Klee, Jun 04 2019
STATUS
approved