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A161822
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a(n) = the smallest positive integer such that both n and a(n), when represented in binary, contain the same types of runs of 0's, the runs being in any order.
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5
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1, 2, 1, 4, 2, 2, 1, 8, 4, 10, 2, 4, 2, 2, 1, 16, 8, 18, 4, 18, 10, 10, 2, 8, 4, 10, 2, 4, 2, 2, 1, 32, 16, 34, 8, 36, 18, 18, 4, 34, 18, 42, 10, 18, 10, 10, 2, 16, 8, 18, 4, 18, 10, 10, 2, 8, 4, 10, 2, 4, 2, 2, 1, 64, 32, 66, 16, 68, 34, 34, 8, 68, 36, 74, 18, 36, 18, 18, 4, 66, 34, 74, 18
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OFFSET
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1,2
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COMMENTS
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Clarification of definition: Think of binary n and a(n) each as a string of 0's and 1's. Consider the "runs" of 0's in binary n and a(n), where each run is made up completely of 0's, and is bounded on both sides either by 1's or by the edge of the string. Now consider the lengths of each bounded run of 0's (the number of 0's in each run). Then a(n) is the smallest positive integer whose set of run-lengths is a permutation of the set of run-lengths for n. (See example.)
1 <= a(n) <= n. a(n) = n iff n = 2^m for some m. a(n) = 1 iff n = 2^m - 1 for some m. - Hagen von Eitzen, Jun 22 2009
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LINKS
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EXAMPLE
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84 in binary is 1010100. There are three runs of 0's, two runs of one 0 each and one run of two 0's. So we are looking for the smallest positive integer with two runs of one 0 each and one run of two 0's (and no other runs of 0's). For example, 16 in binary is 10000, which contains the runs, except that it is required that each run be bounded by 1's or the edge of the binary string. The next number that fits the requirements completely is 74 = 1001010 in binary. So a(84) = 74.
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MATHEMATICA
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f[n_] := Sort@ Map[Length, Select[Split@ IntegerDigits[n, 2], First@ # == 0 &]]; Table[Block[{k = 1}, While[f@ k != f@ n, k++]; k], {n, 83}] (* Michael De Vlieger, Aug 31 2017 *)
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PROG
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(PARI) a(n)=local(B=binary(n), L=[], ct=0); forstep(i=#B, 1, -1, if(B[i], if(ct, L=concat(L, ct); ct=0), ct++)); L=vecsort(L); forstep(i=#L, 1, -1, ct=(ct*2+1)*2^L[i]); if(ct>0, ct, 1) \\ Hagen von Eitzen, Jun 22 2009
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CROSSREFS
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KEYWORD
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base,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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