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Search: id:A073751
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| A073751 |
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Primes numbers that when multiplied in order yield the sequence of colossally abundant numbers A004490. |
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7
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| 2, 3, 2, 5, 2, 3, 7, 2, 11, 13, 2, 3, 5, 17, 19, 23, 2, 29, 31, 7, 3, 37, 41, 43, 2, 47, 53, 59, 5, 61, 67, 71, 73, 11, 79, 2, 83, 3, 89, 97, 13, 101, 103, 107, 109, 113, 127, 131, 137, 139, 2, 149, 151, 7, 157, 163, 167, 17, 173, 179, 181, 191, 193, 197, 199, 19, 211, 3
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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The Mathematica program presents a very fast method of computing the factors of colossally abundant numbers. The 100th number has a sigma[n]/n ratio of 10.5681.
This calculation assumes that the ratio of consecutive colossally abundant numbers is always prime, which is implied by a conjecture mentioned in Lagarias' paper.
The ratio of consecutive colossally abundant numbers is prime for at least the first 10^7 terms. The 10^7-th term is a 77908696-digit number which has a sigma(n)/n value of 33.849.
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..10000
J. C. Lagarias, An elementary problem equivalent to the Riemann hypothesis, Am. Math. Monthly 109 (#6, 2002), 534-543.
Eric Weisstein's World of Mathematics, Colossally Abundant Number
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MATHEMATICA
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pFactor[f_List] := Module[{p=f[[1]], k=f[[2]]}, N[Log[(p^(k+2)-1)/(p^(k+1)-1)]/Log[p]]-1]; maxN=100; f={{2, 1}, {3, 0}}; primes=1; lst={2}; x=Table[pFactor[f[[i]]], {i, primes+1}]; For[n=2, n<=maxN, n++, i=Position[x, Max[x]][[1, 1]]; AppendTo[lst, f[[i, 1]]]; f[[i, 2]]++; If[i>primes, primes++; AppendTo[f, {Prime[i+1], 0}]; AppendTo[x, pFactor[f[[ -1]]]]]; x[[i]]=pFactor[f[[i]]]]; lst
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CROSSREFS
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Cf. A004490.
Sequence in context: A100761 A027748 A000705 this_sequence A108501 A166226 A088167
Adjacent sequences: A073748 A073749 A073750 this_sequence A073752 A073753 A073754
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KEYWORD
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nonn
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AUTHOR
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T. D. Noe (noe(AT)sspectra.com), Aug 07 2002
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