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 A073122 Minimal reversing binary representation of n. Converting sum of powers of 2 in binary representation of a(n) to alternating sum gives n. See A072339. 6
 1, 2, 5, 4, 13, 10, 9, 8, 25, 26, 29, 20, 21, 18, 17, 16, 49, 50, 53, 52, 61, 58, 57, 40, 41, 42, 45, 36, 37, 34, 33, 32, 97, 98, 101, 100, 109, 106, 105, 104, 121, 122, 125, 116, 117, 114, 113, 80, 81, 82, 85, 84, 93, 90, 89, 72, 73, 74, 77, 68, 69, 66, 65, 64, 193 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS The minimal representation is unique. The number of powers of 2 can be either even or odd. Compare with A065621, in which the number of powers of 2 is odd. The Mathematica program computes the representation for numbers 1 to 2^m. a(0) = 0. No term has odd part congruent to 3 modulo 4. - Charlie Neder, Oct 28 2018 REFERENCES D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1981, Vol. 2 (Second Edition), p. 196, (exercise 4.1. Nr. 27) LINKS Clark Kimberling, Table of n, a(n) for n = 1..1000 FORMULA a(2n) = a(n). Express n as a sum of terms 2^x - 2^y, x > y, such that each term defines a run of 1's in n's binary expansion. Then a(n) is the sum of all 2^x + 2^y, with the exception that a term 2^(x+1) - 2^x at the end of a representation becomes 2^x. - Charlie Neder, Oct 28 2018 EXAMPLE a(11) = 29 because 29 = 16 + 8 + 4 + 1 and 16 - 8 + 4 - 1 = 11. a(100) = 164 because 100 in binary is 1100100. The two runs of ones correspond to 2^7 - 2^5 and 2^3 - 2^2, but since 2^3 - 2^2 is the last term of the representation, it can be replaced with 2^2. Therefore, a(100) = 2^7 + 2^5 + 2^2. - Charlie Neder, Oct 28 2018 MATHEMATICA Needs["DiscreteMath`Combinatorica`"]; sumit[s_List] := Module[{i, ss=0}, Do[If[OddQ[i], ss+=s[[ -i]], ss-=s[[ -i]]], {i, Length[s]}]; ss]; m=7; powers=Table[2^i, {i, 0, m}]; lst=Table[2m, {2^m}]; lst2=Table[0, {2^m}]; Do[t=NthSubset[i, powers]; len=Length[t]; st=sumit[t]; If[len

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Last modified February 20 23:24 EST 2019. Contains 320362 sequences. (Running on oeis4.)