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A324608
Number of 1's in binary expansion of A308092(n).
2
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
1,3
COMMENTS
Conjecture: sequence is weakly increasing.
LINKS
FORMULA
a(n) = A000120(A308092(n)).
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 *)
PROG
(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
CROSSREFS
Sequence in context: A102680 A025791 A358474 * A237115 A362915 A069637
KEYWORD
nonn,base
AUTHOR
Peter Kagey, Jun 10 2019
STATUS
approved