1,1

Clarification: The consecutive "runs" (mentioned in the definition) alternate between those completely of 1's and those completely of 0's.

This sequence is not a permutation of the positive integers.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

152 in binary is: 10011000 There is a run of one 1, followed by a run of two 0's, followed by a run of two 1's, followed by a run of three 0's. We halve the two runs of two digits each to one digit each; and we double the number of digits (with a 1) in the first run of one 1, and double the number of digits (with 0's) in the last run of three 0's, to get 1101000000. a(152) is the decimal equivalent of this, which is 832.

rerun := proc(L) if nops(L) mod 2 = 0 then [op(1..nops(L)/2, L)] ; else [op(L), op(L)] ; fi; end: Lton := proc(L) local i; add( op(i, L)*2^(i-1), i=1..nops(L)) ; end: A162854 := proc(n) local strt, en, L, dgs, i; strt := 1; en := -1; L := [] ; dgs := convert(n, base, 2) ; for i from 2 to nops(dgs) do if op(i, dgs) <> op(i-1, dgs) then en := i-1 ; L := [op(L), op(rerun([op(strt..en, dgs)])) ] ; strt := i; fi; od: en := nops(dgs) ; L := [op(L), op(rerun([op(strt..en, dgs)])) ] ; Lton(L) ; end: seq(A162854(n), n=1..100) ; # R. J. Mathar, Aug 01 2009

Table[FromDigits[Flatten[If[EvenQ[Length[#]], Take[#, Length[#]/2], Join[ #, #]]&/@ Split[IntegerDigits[n, 2]]], 2], {n, 60}] (* Harvey P. Dale, May 30 2018 *)

Sequence in context: A354568 A072117 A162853 * A342787 A110121 A321710

Adjacent sequences: A162851 A162852 A162853 * A162855 A162856 A162857

base,nonn

Leroy Quet, Jul 14 2009

Extended beyond a(16) by R. J. Mathar, Aug 01 2009

approved