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A173601
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Greatest inverse of A071542, i.e., a(n) = maximal i such that A071542(i) = n.
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10
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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
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OFFSET
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0,3
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COMMENTS
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What is s = lim sup a(n)/(n log_2(n))? A counting argument suggests s >= 1/2, and in any case s <= 1.
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LINKS
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FORMULA
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a(n)/log_2(a(n)) < n < a(n) for n > 1.
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PROG
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(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+)))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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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
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STATUS
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approved
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