A391958 - OEIS

%I #50 Jan 13 2026 13:40:39

%S 0,0,3,1,13,7,12,4,41,27,32,16,46,26,33,11,113,83,88,56,102,66,73,35,

%T 138,94,101,55,119,69,78,26,289,227,232,168,246,178,185,115,282,206,

%U 213,135,231,149,158,74,370,278,285,191,303,205,214,114,347,241,250

%N First column of A391957.

%C Powers of 2 in the factorization of A391956.

%C The first 82 terms were obtained from a brute force calculation of A391956. From these results a regular pattern is observed to determine more terms.

%C See also A391957 for a Python program. _A.H.M. Smeets_, Jan 13 2026

%H A.H.M. Smeets, <a href="/A391958/b391958.txt">Table of n, a(n) for n = 0..10000</a>

%F Conjectured: this sequence can be obtained by the following two rules:

%F (Eq.1) a(2^n) = 2*a(2^(n-1)) + 2^(n+1) - 1, a(1) = 0 (see also A392083) and

%F (Eq.2) a(n*2^m + 2^(m-1)) = a(n*2^m) + a(2^(m-1)) - 2^m*Sum_{i = 1..n} A001511(i).

%F The second rule (Eq.2) is a generalization of formula like:

%F a(2*n+1) = a(2*n) - 2*Sum_{i = 1..n} A001511(i),

%F a(4*n+2) = a(4*n) + 3 - 4*Sum_{i = 1..n} A001511(i) and

%F a(8*n+4) = a(8*n) + 13 - 8*Sum_{i = 1..n} A001511(i).

%o (Python)

%o def SumA001511(n):

%o     s, i, Sum = [1], 2, 0

%o     while len(s) < n: s, i = s+[i]+s, i+1

%o     while n > 0: Sum, n = Sum+s[n-1], n-1

%o     return Sum

%o def A391958(n):

%o     d, k = 2, 0

%o     while n%d == 0: d, k = 2*d, k+1

%o     if n == d//2: return A2power(k)

%o     else:  # Eq.2 in formula section

%o         return A391958(n-d//2) + A392083(k)- 2*2**k*SumA001511((n-d//2)//(2*2**k))

%Y Cf. A001511, A391956, A391957, A392083.

%K nonn

%O 0,3

%A _A.H.M. Smeets_, Dec 23 2025

%E Terms a(21), a(22) and a(23) inserted by _A.H.M. Smeets_, Jan 08 2026