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%I #67 Dec 14 2023 17:36:23
%S 1,2,12,24,720,1440,60480,120960,3628800,7257600,479001600,958003200,
%T 2615348736000,5230697472000,31384184832000,62768369664000,
%U 32011868528640000,64023737057280000,51090942171709440000,102181884343418880000,33720021833328230400000,67440043666656460800000
%N The Hirzebruch numbers. a(n) = Product_{2 <= p <= n+1, p prime} p^floor(n / (p - 1)).
%C 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.
%C 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
%C 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).
%C Conjecture: The denominators of the series reversion of the sequence with e.g.f. Polylog(2,x). - _Benedict W. J. Irwin_, Jan 05 2017
%D P. Curtz, Integration numérique ..., Note 12, C.C.S.A., Arcueil, 1969; see pp. 36, 56.
%D 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]
%H Michael De Vlieger, <a href="/A091137/b091137.txt">Table of n, a(n) for n = 0..443</a>
%H Abdelmalek Bedhouche and Bakir Farhi, <a href="https://arxiv.org/abs/2207.07957">On some products taken over the prime numbers</a>, arXiv:2207.07957 [math.NT], 2022. See sigma_n p. 3.
%H Victor M. Buchstaber and Alexander P. Veselov, <a href="https://doi.org/10.48550/arXiv.2310.07383">Todd polynomials and Hirzebruch numbers</a>, arXiv:2310.07383 [math.AT], Oct. 2023.
%F a(n) = Product_p {p prime} p^floor(n/(p-1)).
%F a(2n+1) = 2*a(2n).
%F a(n+1) = A027760(n+1)*a(n). - _Paul Curtz_, Aug 01 2008
%F From _Peter Luschny_, Dec 11 2023: (Start)
%F a(n) = lcm_{p in P(n)} Product_{r in p}(r + 1), where P(n) are the partitions of n.
%F a(n) = lcm(A238963row(n)).
%F a(n) = A368116(1, n), seen as the lcm of the product of the 1-shifted partitions.
%F a(n) = A368093(1, n), seen as the cumulative product of the Clausen numbers A160014(1, n). (End)
%e 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.
%p 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: for n from 0 to 20 do printf("%a,",A091137(n) ) ; od: # _R. J. Mathar_, Feb 23 2009
%t 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 *)
%o (PARI) a(n) = local(r); r=1; forprime(p=2, n+1, r*=p^(n\(p-1))); r
%o \\ _Franklin T. Adams-Watters_, May 31 2010
%o (Python)
%o from math import prod
%o from sympy import primerange
%o def A091137(n): return prod(p**(n//(p-1)) for p in primerange(n+2))
%o # _Chai Wah Wu_, Apr 28 2023
%o (SageMath)
%o def a(n): return lcm(product(r + 1 for r in p) for p in Partitions(n))
%o # Or, more efficient:
%o from functools import cache
%o @cache
%o def a_rec(n):
%o if n == 0: return 1
%o p = mul(s for s in map(lambda i: i + 1, divisors(n)) if is_prime(s))
%o return p * a_rec(n - 1)
%o print([a_rec(n) for n in range(22)]) # _Peter Luschny_, Dec 12 2023
%Y Starts similarly to A002207 especially for even n and all values of A002207 seen so far seem to divide a(n).
%Y Cf. A090622, A090624, A091136, A171080, A238963, A368093, A368116.
%Y Cf. A002790, A000142. - _Franklin T. Adams-Watters_, May 31 2010
%K nonn
%O 0,2
%A _Henry Bottomley_, Dec 19 2003
%E New name using a formula of the author by _Peter Luschny_, Dec 11 2023
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