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.

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, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23

Offset: 1

Author

Rémy Sigrist, Jul 14 2025

Keywords

• nonn

Comments

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).

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.

Examples

			Sequence begins:
  n   a(n)  a(1)+...+a(n)  Power of 2?
  --  ----  -------------  -----------
   1     1              1  Yes
   2     1              2  Yes
   3     2              4  Yes
   4     3              7  No
   5     3             10  No
   6     3             13  No
   7     3             16  Yes
   8     7             23  No
   9     7             30  No
  10     7             37  No
  11     7             44  No
  12     7             51  No
		

Links

• Paolo Xausa, Table of n, a(n) for n = 1..10000

Crossrefs

See A385986 for a similar sequence.

Cf. A386905 (distinct terms).

Programs

Mathematica

Module[{s = -1, a = 1}, Table[If[DigitSum[s += a, 2] == 1, a = n - 1]; a, {n, 100}]] (* Paolo Xausa, Aug 07 2025 *)

PARI

{ v = 1; t = 0; for (n = 1, 71, print1(v", "); if (hammingweight(t += v)==1, v = n;);); }