A092143 - OEIS

%I #54 Nov 06 2024 10:19:43

%S 1,2,6,48,240,8640,60480,3870720,104509440,10450944000,114960384000,

%T 198651543552000,2582470066176000,506164132970496000,

%U 113886929918361600000,116620216236402278400000,1982543676018838732800000,11562194718541867489689600000,219681699652295482304102400000

%N Cumulative product of all divisors of 1..n.

%C 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

%H G. C. Greubel, <a href="/A092143/b092143.txt">Table of n, a(n) for n = 1..190</a>

%H Angelo B. Mingarelli, <a href="https://arxiv.org/abs/0705.4299">Abstract factorials</a>, arXiv:0705.4299 [math.NT], 2007-2012.

%F 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

%F Sum_{n>=1} 1/a(n) = A117871. - _Amiram Eldar_, Nov 17 2020

%F log(a(n)) ~ n * log(n)^2 / 2. - _Vaclav Kotesovec_, Jun 20 2021

%F a(n) = Product_{k=1..n} k^floor(n/k). - _Vaclav Kotesovec_, Jun 24 2021

%F From _Ridouane Oudra_, Oct 31 2024: (Start)

%F a(n) = Product_{k=1..n} A007955(k).

%F a(n) = Product_{k=1..n} k^(tau(k)/2).

%F a(n) = sqrt(A175493(n)). (End)

%e a(6) = 1*2*3*2*4*5*2*3*6 = 8640.

%p seq(sqrt(mul(k^numtheory[tau](k), k=1..n)), n=1..40); # _Ridouane Oudra_, Oct 31 2024

%t 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 *)

%t Table[Product[k^Floor[n/k], {k, 1, n}], {n, 1, 25}] (* _Vaclav Kotesovec_, Jun 24 2021 *)

%t FoldList[Times, Times @@@ Divisors[Range[25]]] (* _Paolo Xausa_, Nov 06 2024 *)

%o (PARI) my(z=1); for(i=1,25, fordiv(i,j,z*=j); print1(z, ", "))

%o (Magma) [(&*[j^Floor(n/j): j in [1..n]]): n in [1..30]]; // _G. C. Greubel_, Feb 05 2024

%o (SageMath) [product(j^(n//j) for j in range(1,n+1)) for n in range(1,31)] # _G. C. Greubel_, Feb 05 2024

%Y Cf. A006218, A010786, A092144, A117871.

%Y Cf. A129364, A129365, A129439, A345466.

%Y Cf. A000005, A007955, A175493.

%K nonn

%O 1,2

%A _Jon Perry_, Mar 31 2004