A162853 Take the binary representation of n. Reduce by one digit every run (completely of either 0's or 1's) of an even number of digits. Increase by one digit every run of an odd number of digits in the binary representation of n (where this added digit has the same value that makes up the rest of the run's digits). a(n) = the decimal equivalent of the result.
0, 3, 12, 1, 6, 51, 4, 15, 48, 27, 204, 25, 2, 19, 60, 7, 24, 195, 108, 13, 102, 819, 100, 207, 16, 11, 76, 9, 30, 243, 28, 63, 192, 99, 780, 97, 54, 435, 52, 111, 816, 411, 3276, 409, 50, 403, 828, 103, 8, 67, 44, 5, 38, 307, 36, 79, 240, 123, 972, 121, 14, 115, 252, 31, 96
Offset: 0
Examples
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 reduce the two runs of two digits each to one digit; and we add a digit (a 1) to the first run of one 1, and a digit (a 0) to the last run of three 0's, to get 11010000. So a(152) is the decimal equivalent of this, which is 208.
Links
Programs
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Maple
rerun := proc(L) if nops(L) mod 2 = 0 then subsop(1=NULL,L) ; else [op(L),op(1,L)] ; fi; end: Lton := proc(L) local i; add( op(i,L)*2^(i-1),i=1..nops(L)) ; end: A162853 := 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(A162853(n),n=1..100) ; # R. J. Mathar, Aug 01 2009
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Mathematica
Table[FromDigits[Flatten[If[OddQ[Length[#]],Join[{First[#]},#],Drop[#,1]]& /@Split[ IntegerDigits[ n,2]]], 2],{n,70}] (* Harvey P. Dale, Jun 20 2011 *)
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PARI
a(n) = if (n==0, 0, my (b=n%2, r=valuation(n+b,2), rr=if (r%2, r+1, r-1)); (a(n\2^r)+b)*2^rr-b) \\ Rémy Sigrist, Oct 09 2018
Formula
Extensions
Extended beyond a(13) by R. J. Mathar, Aug 01 2009
a(0) added by Rémy Sigrist, Oct 09 2018
Comments