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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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%I #18 Feb 09 2021 01:56:42

%S 30,56,60,90,105,120,132,144,168,180,210,240,252,264,280,288,306,315,

%T 336,351,360,380,396,400,420,432,480,495,504,520,525,528,540,546,552,

%U 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

%N 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.

%C 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.

%H Karl Dahlke, <a href="http://www.mathreference.com/grp-fin,loword.html">Finite Groups, Simple Groups of Low Order</a>

%o (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);}

%o isok(n) = {vp = factor(n)[,1]~; for (i = 1, #vp, if (! findk(vp[i], n), return (0));); return (1);} \\ _Michel Marcus_, Aug 22 2013

%Y Cf. A001034.

%K nonn

%O 1,1

%A _James V. Blowers_, Apr 25 2011