A324608 Number of 1's in binary expansion of A308092(n).

1, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 10, 11, 11, 11, 13, 13, 14, 14, 14, 16, 16, 16, 17, 17, 17, 19, 19, 19, 19, 20, 20, 20, 22, 22, 22, 22, 22, 23, 23, 23, 25, 25, 25, 25, 25, 25, 26, 26, 26, 28, 28, 28, 28, 28, 28, 28, 29, 29, 30, 31, 31, 31, 31, 31, 31, 31, 31

Offset: 1

Author

Peter Kagey, Jun 10 2019

Keywords

• nonn
• base

Comments

Conjecture: sequence is weakly increasing.

Links

• Robert Israel, Table of n, a(n) for n = 1..3000

Crossrefs

Cf. A000120, A308092.

Programs

Maple

S:= "110":
b("0"):= 0: b("1"):= 1:
A308092[1]:= 1: A308092[2]:= 2: t:= 3:
for n from 3 to 300 do
  tp:= add(b(S[i])*2^(n-i),i=1..n);
  A308092[n]:= tp - t;
  t:= tp;
  S:= cat(S,convert(A308092[n],binary));
od:
seq(convert(convert(A308092[n],base,2),`+`), n=1..300); # Robert Israel, Jun 12 2019

Mathematica

a[1]=1;a[2]=2;a[n_]:=a[n]=FromDigits[Flatten[IntegerDigits[#,2]&/@Table[a[k],{k,n-1}]][[;;n]],2]-Total@Table[a[m],{m,n-1}]
Count[#,1]&/@Table[IntegerDigits[a[l],2],{l,70}] (* Giorgos Kalogeropoulos, Mar 30 2021 *)

Python

def aupton(terms):
  alst, bstr = [1, 1], "110"
  for n in range(3, terms+1):
    an = int(bstr[:n], 2) - int(bstr[:n-1], 2)
    binan = bin(an)[2:]
    alst, bstr = alst + [binan.count('1')], bstr + binan
  return alst
print(aupton(68)) # Michael S. Branicky, Mar 30 2021

Formula

a(n) = A000120(A308092(n)).