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A367099
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Least positive integer such that the number of divisors having two distinct prime factors is n.
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2
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1, 6, 12, 24, 36, 60, 72, 120, 144, 216, 288, 360, 432, 960, 720, 864, 1296, 1440, 1728, 2160, 2592, 3456, 7560, 4320, 5184, 7776, 10800, 8640, 10368, 12960, 15552, 17280, 20736, 40320, 25920, 31104, 41472, 60480, 64800, 51840, 62208, 77760, 93312
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OFFSET
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0,2
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COMMENTS
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Does this contain every power of six, namely 1, 6, 36, 216, 1296, 7776, ...?
Yes, every power of six is a term, since 6^k = 2^k * 3^k is the least positive integer having n = tau(6^k) - (2k+1) divisors with two distinct prime factors. - Ivan N. Ianakiev, Nov 11 2023
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LINKS
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EXAMPLE
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The divisors of 60 having two distinct prime factors are: 6, 10, 12, 15, 20. Since 60 is the first number having five such divisors, we have a(5) = 60.
The terms together with their prime indices begin:
1: {}
6: {1,2}
12: {1,1,2}
24: {1,1,1,2}
36: {1,1,2,2}
60: {1,1,2,3}
72: {1,1,1,2,2}
120: {1,1,1,2,3}
144: {1,1,1,1,2,2}
216: {1,1,1,2,2,2}
288: {1,1,1,1,1,2,2}
360: {1,1,1,2,2,3}
432: {1,1,1,1,2,2,2}
960: {1,1,1,1,1,1,2,3}
720: {1,1,1,1,2,2,3}
864: {1,1,1,1,1,2,2,2}
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MATHEMATICA
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nn=1000;
w=Table[Length[Select[Divisors[n], PrimeNu[#]==2&]], {n, nn}];
spnm[y_]:=Max@@NestWhile[Most, y, Union[#]!=Range[0, Max@@#]&];
Table[Position[w, k][[1, 1]], {k, 0, spnm[w]}]
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PROG
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(PARI) a(n) = my(k=1); while (sumdiv(k, d, omega(d)==2) != n, k++); k; \\ Michel Marcus, Nov 11 2023
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CROSSREFS
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Positions of first appearances in A367098 (counts divisors in A007774).
A001221 counts distinct prime factors.
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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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