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A327934
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Numbers k such that there is no prime p such that p^p divides k, but for its arithmetic derivative a positive finite number of such primes exist.
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11
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15, 35, 39, 51, 55, 87, 91, 95, 111, 115, 119, 123, 143, 155, 158, 159, 183, 187, 203, 215, 219, 225, 235, 247, 259, 267, 275, 287, 291, 295, 299, 303, 319, 323, 327, 329, 335, 339, 355, 371, 374, 391, 395, 403, 407, 410, 411, 415, 427, 441, 447, 451, 471, 473, 482, 511, 515, 519, 525, 527, 533, 535, 543, 551, 559, 579
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OFFSET
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1,1
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COMMENTS
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a(15) = 158, being even, is the first term not in A080774 or A202237, while 275 (= 5^2 * 11) is the first odd term not in the latter.
After 1, 3375 (= 3^3 * 5^3) is the first term in A202237 that is not present in this sequence, even though the overwhelming majority of the terms of A202237 are also included here. On the other hand, A080774 is a genuine subsequence of this sequence, as the sum of prime factors of such semiprimes is always a multiple of 4.
In contrast to A360110, this is not a multiplicative semigroup; For example, although 15 = 3*5 and 275 = 5^2 * 11 are both present, their product 15*275 = 4125 = 3 * 5^3 * 11 is not. - Antti Karttunen, Jan 31 2023
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LINKS
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EXAMPLE
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1 has arithmetic derivative 1' = A003415(1) = 0. There are an infinite number of primes p such that p^p | 0, but because infinity is not a finite number, 1 is excluded from this sequence.
275 = 5^2 * 11 has no divisor of the form p^p, but its arithmetic derivative 275' = 135 = 3^3 * 5 has one divisor of the form p^p, therefore 275 is included in this sequence.
4125 = 3 * 5^3 * 11 has arithmetic derivative 4125' = A003415(4125) = 4225 = 5^2 * 13^2, that has no divisor of the form p^p, therefore 4125 is not included in this sequence.
(End)
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PROG
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(PARI)
A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
A129251(n) = { my(f = factor(n)); sum(k=1, #f~, (f[k, 2]>=f[k, 1])); };
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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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