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 A162536 a(n) = the smallest positive multiple of n where every length of the runs of 0's and 1's in the binary representation of a(n) divides n. 3
 1, 2, 21, 4, 5, 6, 21, 16, 81, 10, 341, 12, 1365, 42, 285, 16, 85, 18, 87381, 20, 21, 22, 1398101, 24, 125, 26, 81, 84, 89478485, 90, 341, 256, 1815, 102, 1365, 36, 22906492245, 38, 117, 80, 349525, 42, 5461, 44, 4545, 598, 23456248059221, 48, 1029, 50, 1479, 52 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS By "run" of 0's or 1's, it is meant: Think of binary n 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 by the edge of the string. LINKS Giovanni Resta, Table of n, a(n) for n = 1..400 EXAMPLE 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), then a(9) = 81. MATHEMATICA 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 *) CROSSREFS Cf. A162534, A162537 Sequence in context: A143247 A303216 A303218 * A100980 A122509 A024230 Adjacent sequences:  A162533 A162534 A162535 * A162537 A162538 A162539 KEYWORD nonn,base AUTHOR Leroy Quet, Jul 05 2009 EXTENSIONS More terms from Sean A. Irvine, Jan 26 2011 More terms from Giovanni Resta, Aug 11 2019 STATUS approved

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Last modified February 18 04:48 EST 2020. Contains 332011 sequences. (Running on oeis4.)