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A092143 Cumulative product of all divisors of 1..n. 13
1, 2, 6, 48, 240, 8640, 60480, 3870720, 104509440, 10450944000, 114960384000, 198651543552000, 2582470066176000, 506164132970496000, 113886929918361600000, 116620216236402278400000 (list; graph; refs; listen; history; text; internal format)
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

Table of n, a(n) for n=1..16.

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 <= 16, k++, Do[n = n*d, {d, Divisors[k]}]; Sow[n]]][[2, 1]] (* Jean-Fran├žois Alcover, Oct 30 2012 *)

Table[Product[k^Floor[n/k], {k, 1, n}], {n, 1, 20}] (* Vaclav Kotesovec, Jun 24 2021 *)

PROG

(PARI) z=1; for(i=1, 20, fordiv(i, j, z*=j); print1(", "z))

CROSSREFS

Cf. A010786, A117871, A129364, A129365, A129439, A345466.

Sequence in context: A052688 A052657 A230714 * A281027 A271961 A239836

Adjacent sequences:  A092140 A092141 A092142 * A092144 A092145 A092146

KEYWORD

nonn

AUTHOR

Jon Perry, Mar 31 2004

STATUS

approved

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Last modified July 25 23:39 EDT 2021. Contains 346294 sequences. (Running on oeis4.)