OFFSET
1,2
COMMENTS
Let p be a prime and let ordp(n,p) denote the exponent of the largest power of p which divides n. For example, ordp(48,2)=4 since 48 = 3*(2^4). Let b(n) = A006218(n) = Sum_{k=1..n} floor(n/k). The prime factorization of a(n) appears to be given by the following conjectural formula: ordp(a(n),p) = b(floor(n/p)) + b(floor(n/p^2)) + b(floor(n/p^3)) + ... . Compare with the comments in A129365. - Peter Bala, Apr 15 2007
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..190
Angelo B. Mingarelli, Abstract factorials, arXiv:0705.4299 [math.NT], 2007-2012.
FORMULA
a(n) = Product_{k=1..n} {floor(n/k)}!. This formula is due to Sebastian Martin Ruiz. - Peter Bala, Apr 15 2007; Formula corrected by R. J. Mathar, May 06 2008
Sum_{n>=1} 1/a(n) = A117871. - Amiram Eldar, Nov 17 2020
log(a(n)) ~ n * log(n)^2 / 2. - Vaclav Kotesovec, Jun 20 2021
a(n) = Product_{k=1..n} k^floor(n/k). - Vaclav Kotesovec, Jun 24 2021
EXAMPLE
a(6) = 1*2*3*2*4*5*2*3*6 = 8640.
MATHEMATICA
Reap[For[n = k = 1, k <= 25, k++, Do[n = n*d, {d, Divisors[k]}]; Sow[n]]][[2, 1]] (* Jean-François Alcover, Oct 30 2012 *)
Table[Product[k^Floor[n/k], {k, 1, n}], {n, 1, 25}] (* Vaclav Kotesovec, Jun 24 2021 *)
PROG
(PARI) my(z=1); for(i=1, 25, fordiv(i, j, z*=j); print1(z, ", "))
(Magma) [(&*[j^Floor(n/j): j in [1..n]]): n in [1..30]]; // G. C. Greubel, Feb 05 2024
(SageMath) [product(j^(n//j) for j in range(1, n+1)) for n in range(1, 31)] # G. C. Greubel, Feb 05 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Jon Perry, Mar 31 2004
STATUS
approved