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A352451
2^k appears in the binary expansion of a(n) iff 2^k appears in the binary expansion of n and k+1 divides n.
5
0, 1, 2, 1, 0, 1, 6, 1, 8, 1, 2, 1, 12, 1, 2, 5, 0, 1, 2, 1, 16, 5, 2, 1, 8, 17, 2, 1, 8, 1, 22, 1, 0, 1, 2, 1, 36, 1, 2, 5, 8, 1, 34, 1, 8, 5, 2, 1, 32, 1, 18, 1, 0, 1, 38, 17, 8, 1, 2, 1, 60, 1, 2, 5, 0, 1, 2, 1, 0, 5, 66, 1, 8, 1, 2, 1, 8, 65, 6, 1, 16, 1
OFFSET
0,3
COMMENTS
The idea is to keep the 1's in the binary expansion of a number whose positions are related in some way to that number.
FORMULA
a(n) <= n.
EXAMPLE
For n = 42:
- 42 = 2^5 + 2^3 + 2^1,
- 5+1 divides 42,
- 3+1 does not divide 42,
- 1+1 divides 42,
- so a(42) = 2^5 + 2^1 = 34.
PROG
(PARI) a(n) = { my (v=0, m=n, k); while (m, m-=2^k=valuation(m, 2); if (n%(k+1)==0, v+=2^k)); v }
CROSSREFS
See A352449, A352450, A352452, A352458 for similar sequences.
Sequence in context: A302978 A108723 A291584 * A349618 A321615 A011126
KEYWORD
nonn,base
AUTHOR
Rémy Sigrist, Mar 16 2022
STATUS
approved