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%I #20 Jan 31 2021 20:28:09
%S 1,2,21,4,5,6,21,16,81,10,341,12,1365,42,285,16,85,18,87381,20,21,22,
%T 1398101,24,125,26,81,84,89478485,90,341,256,1815,102,1365,36,
%U 22906492245,38,117,80,349525,42,5461,44,4545,598,23456248059221,48,1029,50,1479,52
%N a(n) is the smallest positive multiple k of n such that every length of the runs of 0's and 1's in the binary representation of k divides n.
%C By "run" of 0's or 1's, it is meant: Think of binary k as a string of 0's and 1's. A single run of the digit b (0 or 1) is made up completely of consecutive digits all equal to b, and is bounded on its ends by either the digit 1-b or the end of the string.
%H Giovanni Resta, <a href="/A162536/b162536.txt">Table of n, a(n) for n = 1..400</a>
%e For n = 9, we check: 9 in binary is 1001, which has a run of two 0's, and 2 does not divide 9. Checking further: 2*9 = 18 = 10010, which still doesn't work. 3*9 = 27 = 11011 in binary, which has two runs of two 1's. 4*9 = 36 = 100100 in binary, 5*9 = 45 = 101101 in binary, 6*9 = 54 = 110110 in binary, 7*9 = 63 = 111111 in binary, 8*9 = 72 = 1001000 in binary, none of which work. But 9*9 = 81 = 1010001 in binary, which has three runs of one 1 each, a run of one 0, and a run of three 0's. Since 9 is divisible by both of these lengths (1 and 3), a(9) = 81.
%t a[n_] := Block[{m}, If[n>2 && PrimeQ[n], m=1; While[Mod[m,n] > 0, m=4*m+1], m=n; While[! AllTrue[ Union[ Length /@ Split[ IntegerDigits[m, 2]]], Mod[n, #] == 0 &], m += n]]; m]; Array[a, 60] (* _Giovanni Resta_, Aug 11 2019 *)
%Y Cf. A162534, A162537.
%K nonn,base
%O 1,2
%A _Leroy Quet_, Jul 05 2009
%E More terms from _Sean A. Irvine_, Jan 26 2011
%E More terms from _Giovanni Resta_, Aug 11 2019
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