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A092143
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Cumulative product of all divisors of 1..n.
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14
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1, 2, 6, 48, 240, 8640, 60480, 3870720, 104509440, 10450944000, 114960384000, 198651543552000, 2582470066176000, 506164132970496000, 113886929918361600000, 116620216236402278400000, 1982543676018838732800000, 11562194718541867489689600000
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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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
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EXAMPLE
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a(6) = 1*2*3*2*4*5*2*3*6 = 8640.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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