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Search: id:A092143
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| A092143 |
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Cumulative product of all divisors of 1..n. |
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+0
7
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| 1, 2, 6, 48, 240, 8640, 60480, 3870720, 104509440, 10450944000, 114960384000, 198651543552000, 2582470066176000, 506164132970496000, 113886929918361600000, 116620216236402278400000
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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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 (pbala(AT)toucansurf.com), Apr 15 2007
Mingarelli proved that the sum of the reciprocals of a(n) ~ 2.69179920 is irrational. - Jonathan Vos Post (jvospost3(AT)gmail.com), May 31 2007
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LINKS
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Angelo B. Mingarelli, Abstract factorials of arbitrary sets of integers, May 29 2007.
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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 (pbala(AT)toucansurf.com), 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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PROGRAM
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(PARI) z=1; for(i=1, 20, fordiv(i, j, z*=j); print1(", "z))
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CROSSREFS
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Cf. A129364, A129365, A129439.
Sequence in context: A052614 A052688 A052657 this_sequence A052593 A052586 A052554
Adjacent sequences: A092140 A092141 A092142 this_sequence A092144 A092145 A092146
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KEYWORD
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nonn
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AUTHOR
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Jon Perry (perry(AT)globalnet.co.uk), Mar 31 2004
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