A091137 The Hirzebruch numbers. a(n) = Product_{2 <= p <= n+1, p prime} p^floor(n / (p - 1)).

1, 2, 12, 24, 720, 1440, 60480, 120960, 3628800, 7257600, 479001600, 958003200, 2615348736000, 5230697472000, 31384184832000, 62768369664000, 32011868528640000, 64023737057280000, 51090942171709440000, 102181884343418880000, 33720021833328230400000, 67440043666656460800000

Offset: 0

Author

Henry Bottomley, Dec 19 2003

Keywords

• nonn
• changed

Comments

Largest number m such that number of times m divides k! is almost k/n for large k, i.e., largest m with A090624(m) = n.

This is always a relatively small multiple of n!, since the multiplicity with which a prime p divides n! is always <= n/(p-1); it is equal to floor(n/(p-1)) at least when n is a power of p. - Franklin T. Adams-Watters, May 31 2010

At least for most small n, a(n) = A002790(n) * n!; the first difference is n=15. It appears that A002790(n) * n! always divides a(n).

Conjecture: The denominators of the series reversion of the sequence with e.g.f. Polylog(2,x). - Benedict W. J. Irwin, Jan 05 2017

Not only is a(n) divisible by n!; a(n) is divisible by (n + 1)! as has been observed by Bedhouche and Bakir (see links and A363596). - Hal M. Switkay, Aug 15 2025

Examples

			Let n = 4. The partitions of 4 are [[4], [3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]]. Thus a(4) = lcm([5, 4*2, 3*3, 3*2*2, 2*2*2*2]) = 720.
		

References

• P. Curtz, Integration numérique ..., Note 12, C.C.S.A., Arcueil, 1969; see pp. 36, 56.
• F. Hirzebruch, Topological Methods in Algebraic Geometry, Springer, 3rd. ed., 1966; Lemma 1.7.3, p. 14. [From N. J. A. Sloane, Sep 06 2010]

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..443
• Abdelmalek Bedhouche and Bakir Farhi, On some products taken over the prime numbers, arXiv:2207.07957 [math.NT], 2022. See sigma_n p. 3.
• Victor M. Buchstaber and Alexander P. Veselov, Todd polynomials and Hirzebruch numbers, arXiv:2310.07383 [math.AT], Oct. 2023.
• Donghyun Kim and Jaeseong Oh, Extending the science fiction and the Loehr--Warrington formula, arXiv:2409.01041 [math.CO], 2024. See p. 32.

Crossrefs

Starts similarly to A002207 especially for even n and all values of A002207 seen so far seem to divide a(n).

Cf. A002790, A000142, A090622, A090624, A091136, A171080, A238963, A275314, A368093, A368116, A363596.

Programs

Maple

A091137 := proc(n) local a,i,p ; a := 1 ; for i from 1 do p := ithprime(i) ; if p > n+1 then break; fi; a := a*p^floor(n/(p-1)) ; od: a ; end:
seq(A091137(n), n = 0..47); # R. J. Mathar, Feb 23 2009

Mathematica

A027760[n_] := Product[d, {d, Select[ Divisors[n] + 1, PrimeQ]}]; a[n_] := a[n] = A027760[n]*a[n-1]; a[0] = 1; Table[ a[n], {n, 0, 18}] (* Jean-François Alcover, Oct 04 2011 *)

PARI

a(n) = local(r); r=1; forprime(p=2, n+1, r*=p^(n\(p-1))); r
\\ Franklin T. Adams-Watters, May 31 2010

Python

from math import prod
from sympy import primerange
def A091137(n): return prod(p**(n//(p-1)) for p in primerange(n+2))
# Chai Wah Wu, Apr 28 2023

SageMath

def a(n): return lcm(product(r + 1 for r in p) for p in Partitions(n))
# Or, more efficient:
from functools import cache
@cache
def a_rec(n):
    if n == 0: return 1
    p = mul(s for s in map(lambda i: i + 1, divisors(n)) if is_prime(s))
    return p * a_rec(n - 1)
print([a_rec(n) for n in range(22)]) # Peter Luschny, Dec 12 2023

Formula

a(n) = Product_p {p prime} p^floor(n/(p-1)).

a(2n+1) = 2*a(2n).

a(n+1) = A027760(n+1)*a(n). - Paul Curtz, Aug 01 2008

From Peter Luschny, Dec 11 2023: (Start)

a(n) = lcm_{p in P(n)} Product_{r in p}(r + 1), where P(n) are the partitions of n.

a(n) = lcm(A238963row(n)).

a(n) = A368116(1, n), seen as the lcm of the product of the 1-shifted partitions.

a(n) = A368093(1, n), seen as the cumulative product of the Clausen numbers A160014(1, n). (End)

a(n) = lcm({k: A275314(k) = n+1}). - Hal M. Switkay, Aug 13 2025

a(n) = (n + 1)! * A363596(n). - Hal M. Switkay, Aug 15 2025

Extensions

New name using a formula of the author by Peter Luschny, Dec 11 2023