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A211337
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Numbers k for which the number of divisors, tau(k), is congruent to 1 modulo 3.
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9
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1, 6, 8, 10, 14, 15, 21, 22, 26, 27, 33, 34, 35, 38, 39, 46, 48, 51, 55, 57, 58, 62, 64, 65, 69, 74, 77, 80, 82, 85, 86, 87, 91, 93, 94, 95, 106, 111, 112, 115, 118, 119, 120, 122, 123, 125, 129, 133, 134, 141, 142, 143, 145, 146, 155, 158, 159, 161, 162
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OFFSET
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1,2
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COMMENTS
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The numbers of terms not exceeding 10^k, for k = 1, 2, ..., are 4, 36, 366, 3635, 36499, 365456, 3654240, 36538501, 365382167, 3653804173, ... . Conjecture: the asymptotic density of this sequence exists and equals 3*zeta(3)/Pi^2 = 0.3653814847007... (A346602), so, a(n) ~ k*n with k = Pi^2/(3*zeta(3)) = 2.73686555524... . This conjecture is true if this sequence and A211338 have the same density (see A059269). - Amiram Eldar, Jan 06 2024
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LINKS
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FORMULA
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Conjecture: a(n) ~ k*n where k = 2/prod(1 - (p-1)/(p^(3*k))) = 2.7290077... where p ranges over the primes and k ranges over the positive integers. - Charles R Greathouse IV, Apr 13 2012
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EXAMPLE
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The divisors of 10 are: 1, 2, 5, 10 (4 divisors). 4 is congruent to 1 modulo 3. Thus 10 is a member of this sequence.
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MATHEMATICA
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Select[Range[162], Mod[DivisorSigma[0, #], 3] == 1 &] (* T. D. Noe, Apr 21 2012 *)
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PROG
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(PARI) {plnt=1 ; mxind=100 ; for(k=1, 10^6,
if(numdiv(k) % 3 == 1, print(k); plnt++; if(mxind+1 == plnt, break() )))}
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CROSSREFS
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This is an extension of A030513 (numbers with 4 divisors).
The union of A059269 and A211338 is the complementary sequence to this one.
The definition of this sequence uses A000005 (the number of divisors of n).
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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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