OFFSET
0,2
COMMENTS
The integrality of this sequence can be used to prove Chebyshev's estimate C(1)*x/log(x) <= #{primes <= x} <= C(2)*x/log(x), for x sufficiently large; the constant C(1) = 0.921292... and C(2) = 1.105550.... Chebyshev's approach used the related step function floor(x) -floor(x/2) -floor(x/3) -floor(x/5) +floor(x/30). See A182067.
This sequence is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin.
The o.g.f. Sum_{n >= 0} a(n)*z^n is a generalized hypergeometric series of type 8F7 (see Bober, Table 2, Entry 31) and is an algebraic function of degree 483840 over the field of rational functions Q(z) (see Rodriguez-Villegas). Bober remarks that the monodromy group of the differential equation satisfied by the o.g.f. is W(E_8), the Weyl group of the E_8 root system.
See the Bala link for the proof that a(n), n = 0,1,2..., is an integer.
Congruences: a(p^k) == a(p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integer k (write a(n) as C(30*n,15*n)*C(15*n,5*n)/C(6*n,n) and use equation 39 in Mestrovic, p. 12). More generally, the congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) may hold for any prime p >= 5 and any positive integers n and k. Cf. A295431. - Peter Bala, Jan 24 2020
From Peter Bala, Aug 28 2025: (Start)
Conjectures: 7*a(n)/(2*n + 1), a(n)/(3*n + 1), a(n)/(5*n + 1) and 42*a(n)/((2*n + 1)*(3*n + 1)*(5*n + 1)) are integers for all n (checked up to n = 1000).
More generally, calculation suggests that for k = 2, 3 or 5 and r >= 1, there exists a constant C(k, r) such that C(k, r)*a(n)/Product_{i = 1..r, i coprime to k} (k*n + i) is an integer for all n.
It appears that a(n)/(30*n - 1) is integral for all n (checked up to n = 1000). More generally, for r >= 1, we conjecture that there exists a constant D(r) such that D(r)*a(n)/Product_{i = 1..r, i coprime to 30} (30*n - i) is integral for all n.
Similar results may hold for all the 52 sporadic integral factorial ratio sequences listed in A295431. (End)
LINKS
N. J. A. Sloane, Table of n, a(n) for n = 0..50
Peter Bala, Proof of the integrality of A211417 and A211418
Frits Beukers, Hypergeometric functions, how special are they?, Notices Amer. Math. Soc. 61 (2014), no. 1, 48--56. MR3137256
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.
Florian Fürnsinn and Sergey Yurkevich, Algebraicity of hypergeometric functions with arbitrary parameters, arXiv:2308.12855 [math.CA], 2023.
Romeo Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Fernando Rodriguez Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math.NT/0701362, 2007.
Fernando Rodriguez Villegas, Mixed Hodge numbers and factorial ratios, arXiv:1907.02722 [math.NT], 2019.
K. Soundararajan, Integral Factorial Ratios, arXiv:1901.05133 [math.NT], 2019.
Wadim Zudilin, Integer-valued factorial ratios, MathOverflow question 26336, 2010.
FORMULA
a(n) ~ 2^(14*n-1) * 3^(9*n-1/2) * 5^(5*n-1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Aug 30 2016
a(n) = binomial(30*n,15*n)*binomial(15*n,5*n)/binomial(6*n,n) = binomial(30*n,15*n)*binomial(16*n,6*n)/binomial(16*n,n). - Chai Wah Wu, Feb 15 2026
MATHEMATICA
Table[(30 n)!*n!/((15 n)!*(10 n)!*(6 n)!), {n, 0, 5}] (* Michael De Vlieger, Oct 02 2015 *)
PROG
(PARI) a(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!);
vector(10, n, a(n-1)) \\ Altug Alkan, Oct 02 2015
(Magma) [Factorial(30*n)*Factorial(n)/(Factorial(15*n)*Factorial(10*n)*Factorial(6*n)): n in [0..10]]; // Vincenzo Librandi, Oct 03 2015
(Python)
from math import comb
def A211417(n): return comb(30*n, 15*n)*comb(15*n, 5*n)//comb(6*n, n) # Chai Wah Wu, Feb 15 2026
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Apr 11 2012
STATUS
approved