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A213259
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The number of primorials that neither exceed nor divide the n-th colossally abundant number.
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1
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0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
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OFFSET
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1,13
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COMMENTS
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Also: the number of primorials p#<m such that p#/phi(p#)>m/phi(m), where m is the n-th colossally abundant (CA) number.
Note that only two CA numbers are primorials (2 and 6); all other CA numbers are not squarefree, while all primorials are squarefree.
The sequence is not monotonic but tends to grow, albeit very slowly.
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REFERENCES
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G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann. J. Math. Pures Appl. 63 (1984), 187-213.
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LINKS
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S. Ramanujan, Highly composite numbers, Annotated and with a foreword by J.-L. Nicolas and G. Robin, Ramanujan J., 1 (1997), 119-153.
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FORMULA
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Let x be the largest prime factor of the n-th CA number m, then:
(1) a(n) >= [sqrt(x)/log(x)] for all nonzero terms a(n).
(2) a(n) < 2 sqrt(x)/log(x) for all terms a(n).
(3) a(n+1) <= a(n)+1 (due to the factorization pattern of CA numbers).
Corollary: there are only finitely many zero terms a(n).
All formulas can be checked directly for small x (e.g., for x<100000).
For large x, (1) and (2) follow from Robin's Lemma (Robin, 1984, p.190).
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EXAMPLE
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For the first five CA numbers m, each smaller primorial divides m; therefore the initial five terms are zeros.
The 6th CA number, 360, is not divisible by one smaller primorial, 210; thus a(6)=1.
The 13th CA number, 21621600, is not divisible by two smaller primorials, 510510 and 9699690; thus a(13)=2.
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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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