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A103528
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a(n) = Sum_{k = 1..n-1 such that n == k (mod 2^k)} 2^(k-1).
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4
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0, 0, 1, 0, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 17, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 17, 34, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8
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OFFSET
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1,6
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COMMENTS
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Is there a simpler closed form?
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LINKS
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David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
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FORMULA
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G.f.: Sum_{k>=1} 2^(k-1) x^(k+2^k)/(1 - x^(2^k)). - Robert Israel, Jan 21 2017
Conjecture: a(n) = (b(n) - b(n-1) - 1)/2 for n > 1 where b(n) = Sum_{k=0..A000523(n)} c(n-k, k) and c(n, m) = n - (n mod 2^m) with a(1) = 0. - Mikhail Kurkov, Jun 01 2022
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MAPLE
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f:=proc(n) local t1, k; t1:=0; for k from 1 to n-1 do if n mod 2^k = k then t1:=t1+2^(k-1); fi; od: t1; end;
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MATHEMATICA
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b[n_] := b[n] = If[n == 1, 1, BitXor[b[n-1], b[n-1] + n]];
a[n_] := (b[n] + n)/2 - 2^(n-1);
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PROG
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(PARI) a(n) = sum(k = 1, n-1, if ((n % 2^k) == k, 2^(k-1))); \\ Michel Marcus, May 06 2020
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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