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A127726
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Numbers that are 3-imperfect.
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7
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6, 120, 126, 2520, 2640, 30240, 32640, 37800, 37926, 55440, 685440, 758520, 831600, 2600640, 5533920, 6917400, 9102240, 10281600, 11377800, 16687440, 152182800, 206317440, 250311600, 475917120, 866829600, 1665709920, 1881532800, 2082137400, 2147450880
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OFFSET
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1,1
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COMMENTS
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The new terms come from the paper by Zhou and Zhu. This sequence also contains n = 9223372034707292160 = 2^31*3*5*17*257*65537, which has the product of five Fermat primes (A019434). For this n, n/3 is a 2-imperfect number (A127725). - T. D. Noe, Apr 03 2009
By definition, n is k-imperfect iff n = k*A206369(n).
So a k-imperfect number is always a multiple of k, and up to the first odd 3-imperfect number (larger than 10^49, if it exists, see Zhou & Zhu (2009)), all terms are a multiple of 6. (End)
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LINKS
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Weiyi Zhou and Long Zhu, On k-imperfect numbers, INTEGERS: Electronic Journal of Combinatorial Number Theory, 9 (2009), #A01.
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EXAMPLE
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6 = 2*3, so A206369(6) = (2 - 1)(3 - 1) = 2 = 6 / 3, so 6 is a term.
120 = 2^3 * 3 * 5, (8-4+2-1)*(3-1)*(5-1) = 40 = 120 / 3, so 120 is another term.
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MATHEMATICA
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okQ[n_] := 3 Sum[d*(-1)^PrimeOmega[n/d], {d, Divisors[n]}] == n;
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PROG
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(PARI) isok(n) = 3*sumdiv(n, d, d*(-1)^bigomega(n/d)) == n; \\ Michel Marcus, Oct 28 2017
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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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