cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A364174 a(n) = (9*n)!*(5*n/2)!*(3*n/2)!/((5*n)!*(9*n/2)!*(3*n)!*(n/2)!).

Original entry on oeis.org

1, 48, 4862, 549120, 65132550, 7945986048, 987291797996, 124259864002560, 15789207515217990, 2021092963752345600, 260227401685879140612, 33665720694993527504896, 4372592850984736084611996, 569819472537519480058675200, 74468439316740019538310543000
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295442, defined by A295442(n) = (18*n)!*(5*n)!*(3*n)!/((10*n)!*(9*n)!*(6*n)!*n!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 12). Here we are essentially considering the sequence {A295442(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (3*n/2)! := Gamma(1 + 3*n/2).
This sequence is only conjecturally an integer sequence.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

Crossrefs

Programs

  • Maple
    seq( simplify((9*n)!*(5*n/2)!*(3*n/2)!/((5*n)!*(9*n/2)!*(3*n)!*(n/2)!)), n = 0..15)

Formula

a(n) ~ c^n * 1/sqrt(2*Pi*n), where c = 2*(3^7)/(5^3) * sqrt(15) = 135.5234332504899....
a(n) = 108*(9*n - 1)*(9*n - 5)*(9*n - 7)*(9*n - 11)*(9*n - 13)*(9*n - 17)/(5*n*(n - 1)*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9))*a(n-2) for n >= 2 with a(0) = 1 and a(1) = 48.