A173601 Greatest inverse of A071542, i.e., a(n) = maximal i such that A071542(i) = n.

0, 1, 3, 5, 7, 9, 11, 15, 17, 19, 23, 27, 31, 33, 35, 39, 43, 47, 51, 55, 59, 63, 65, 67, 71, 75, 79, 83, 87, 91, 95, 99, 103, 107, 111, 115, 119, 125, 127, 129, 131, 135, 139, 143, 147, 151, 155, 159, 163, 167, 171, 175, 179, 183, 189, 191, 195, 199, 203, 207

Offset: 0

Author

Charles R Greathouse IV, Oct 30 2012

Keywords

• nonn

Comments

What is s = lim sup a(n)/(n log_2(n))? A counting argument suggests s >= 1/2, and in any case s <= 1.

Essentially also the partial sums of A086876. - Antti Karttunen, Nov 10 2012 (per personal mail from Carl R. White, Nov 02 2012)

Links

• Charles R Greathouse IV and Antti Karttunen, Table of n, a(n) for n = 0..10000 (terms 1-10000 from Greathouse, Karttunen recomputed with prepended a(0)=0)

Crossrefs

See A213708 for the least inverse. A086876 gives the first differences. Also, a(n)=A213708(n)+A086876(n)-1. Cf. A071542, A179016, A218604, A218608.

Programs

PARI

v=vectorsmall(10^3);v[1]=1;for(n=2,#v,v[n]=v[n-hammingweight(n)]+1); u=vector(solve(x=1,#v,x*log(x)/log(2)-#v)\1);for(i=1,#v,if(v[i]<=#u,u[v[i]]=i)); u

Scheme

;; With Antti Karttunen's intseq-library:
(define A173601 (PARTIALSUMS 1 0 (compose-funs A086876 1+)))

Formula

a(n)/log_2(a(n)) < n < a(n) for n > 1.

Extensions

Changed the starting offset by prepending a(0)=0 (with the indexing of the rest of terms thus not changed) - Antti Karttunen, Nov 10 2012