A329320 a(n) = Sum_{k=0..floor(log_2(n))} 1 - A035263(1 + floor(n/2^k)).

0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 2, 3, 1, 2, 2, 2, 2, 3, 2, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 3

Offset: 0

Author

Mikhail Kurkov, Nov 10 2019 [verification needed]

Keywords

• nonn
• easy

Comments

Sequence which arise from attempts to simplify computing of A329319.

For all positive integers k, the subsequence a(2^k) to a(3*2^(k-1)-1) is identical to the subsequence a(3*2^(k-1)) to a(2^(k+1)-1). Also subsequences a(2^k) to a(3*2^(k-1)-1) and a(0) to a(2^(k-1)-1) always differ by 1.

Links

• Mikhail Kurkov, Table of n, a(n) for n = 0..8191 [verification needed]

Crossrefs

Cf. A035263, A329319.

Programs

PARI

a(n) = if (n==0, 0, a(floor(n/2)) + valuation(n+1, 2) %  2); \\ Michel Marcus, Nov 13 2019

PARI

a(n)=my(s,t); while(n, n>>=valuation(n,2); t=valuation(n+1,2); s+=(t+1)\2; n>>=t); s \\ Charles R Greathouse IV, Oct 14 2021

Formula

a(n) = a(floor(n/2)) + 1 - A035263(n+1) for n>0 with a(0)=0.

a(2^m+k) = a(k mod 2^(m-1)) + 1 for 0<=k<2^m, m>0 with a(0)=0, a(1)=1.