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A385988 a(1) = 1, and for any n > 1, a(n) is the largest k < n such that a(1) + ... + a(k) is a power of 2.

%I A385988 #16 Aug 07 2025 16:52:21
%S A385988 1,1,2,3,3,3,3,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,23,23,23,23,23,23,23,
%T A385988 23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,
%U A385988 23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23
%N A385988 a(1) = 1, and for any n > 1, a(n) is the largest k < n such that a(1) + ... + a(k) is a power of 2.
%C A385988 In other words: a(1) = 1, and for any n > 0, if a(1) + ... + a(n) is a power of 2 then a(n+1) = n, otherwise a(n+1) = a(n).
%C A385988 This sequence is unbounded: if a(1) + ... + a(n) = 2^i, let n = m * 2^j with m odd, note that n <= 2^i, so j <= i, let z be the multiplicative order of m modulo 2, let t = 2^(i-j) * (2^z - 1) / m, a(1) + ... + a(n) + t * n = 2^(z+i), so a(1) + ... + a(n) + u * n is a power of 2 for some u in the range n+1..t, and a(u+1) > a(n+1) = n.
%H A385988 Paolo Xausa, <a href="/A385988/b385988.txt">Table of n, a(n) for n = 1..10000</a>
%e A385988 Sequence begins:
%e A385988   n   a(n)  a(1)+...+a(n)  Power of 2?
%e A385988   --  ----  -------------  -----------
%e A385988    1     1              1  Yes
%e A385988    2     1              2  Yes
%e A385988    3     2              4  Yes
%e A385988    4     3              7  No
%e A385988    5     3             10  No
%e A385988    6     3             13  No
%e A385988    7     3             16  Yes
%e A385988    8     7             23  No
%e A385988    9     7             30  No
%e A385988   10     7             37  No
%e A385988   11     7             44  No
%e A385988   12     7             51  No
%t A385988 Module[{s = -1, a = 1}, Table[If[DigitSum[s += a, 2] == 1, a = n - 1]; a, {n, 100}]] (* _Paolo Xausa_, Aug 07 2025 *)
%o A385988 (PARI) { v = 1; t = 0; for (n = 1, 71, print1(v", "); if (hammingweight(t += v)==1, v = n;);); }
%Y A385988 See A385986 for a similar sequence.
%Y A385988 Cf. A386905 (distinct terms).
%K A385988 nonn
%O A385988 1,3
%A A385988 _Rémy Sigrist_, Jul 14 2025