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A068306
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Numbers n such that n*tau(n) < sigma(n)*bigomega(n) where tau(n)=A000005(n), sigma(n)=A000203(n), and bigomega(n)=A001222(n).
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1
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4, 8, 12, 16, 18, 20, 24, 27, 32, 36, 40, 48, 54, 56, 64, 72, 80, 81, 88, 96, 104, 108, 112, 128, 144, 160, 162, 176, 192, 208, 216, 224, 243, 256, 272, 288, 304, 320, 324, 352, 368, 384, 416, 432, 448, 464, 486, 512, 544, 576, 608, 640, 704, 729, 736, 768, 832
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OFFSET
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1,1
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COMMENTS
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This sequence contains numbers of the form p^u, u >= p, p prime; numbers of the form 2^k*p, p odd prime and k >= floor(log(p)/log(2)). Example: the first number of the form 2^k*67 in the sequence is 2^6*67=4288 because floor(log(67)/log(2))=6.
[Are we to understand that these are all the numbers in the sequence, or just some of them? - N. J. A. Sloane, Dec 27 2018]
If n=p^(p-1) with p prime, n*tau(n) - sigma(n)*bigomega(n) = 1.
If n > 1 then each of n, tau(n), sigma(n) and bigomega(n) are > 0. We can then write the inequality as tau(n)/bigomega(n) < sigma(n)/n.
Note that tau(n) and bigomega(n) only depend on the prime signature of n.
Suppose we choose some k from A025487. If k is not a term then no number with that prime signature is not a term. k is the number with the value for sigma(m)/m for numbers m with the same prime signature as k. (End)
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LINKS
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MAPLE
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filter:= proc(n) local L, t, j;
L:= ifactors(n)[2];
n * mul(t[2]+1, t=L) < mul(add(t[1]^j, j=0..t[2]), t=L)*add(t[2], t=L)
end proc:
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MATHEMATICA
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filterQ[n_] := n DivisorSigma[0, n] < DivisorSigma[1, n] PrimeOmega[n];
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PROG
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(PARI) isok(n) = n*numdiv(n) < sigma(n)*bigomega(n); \\ Michel Marcus, Dec 27 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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