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A189712
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Numbers m such that for each prime p that divides m, there is a k(p) such that k(p) == 1 (mod p), k(p) divides m evenly, and m divides k(p)!/2 evenly.
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0
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30, 56, 60, 90, 105, 120, 132, 144, 168, 180, 210, 240, 252, 264, 280, 288, 306, 315, 336, 351, 360, 380, 396, 400, 420, 432, 480, 495, 504, 520, 525, 528, 540, 546, 552, 560, 576, 612, 616, 630, 660, 672, 702, 720, 735, 756, 760, 792, 800, 810, 840, 858, 864, 900, 918, 924, 960, 990, 992, 1008, 1040, 1050, 1053, 1056, 1080, 1092, 1100, 1104
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OFFSET
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1,1
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COMMENTS
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The order of every simple group must be a term of this sequence, so A001034 is a subsequence of this sequence. The Dahlke link shows the sequence with 72 and 112 (which are not in this sequence) added.
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LINKS
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PROG
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(PARI) findk(p, n) = {for (k = 1, n, if (((k % p) == 1) && ((n % k) == 0) && ((k! % 2) == 0) && (((k!/2) % n) == 0), return (1)); ); return (0); }
isok(n) = {vp = factor(n)[, 1]~; for (i = 1, #vp, if (! findk(vp[i], n), return (0)); ); return (1); } \\ Michel Marcus, Aug 22 2013
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CROSSREFS
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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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