A064895 Binary concentration of n. Replace 2^e_k with 2^(e_k/g(n)) in binary expansion of n, where g(n) = GCD of exponents e_k = A064894(n).

0, 1, 2, 3, 2, 3, 6, 7, 2, 3, 10, 11, 12, 13, 14, 15, 2, 3, 18, 19, 6, 7, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 2, 3, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 2, 3, 66, 67, 10, 11, 70, 71, 6, 7, 74, 75

Offset: 0

Author

Marc LeBrun, Oct 11 2001

Keywords

• base
• easy
• nonn

Examples

			577 = 2^0 + 2^6 + 2^9, GCD(0,6,9) = 3, a(577) = 2^(0/3)+2^(6/3)+2^(9/3) = 13.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..8192

Crossrefs

Cf. A000079, A064894.

Programs

Mathematica

A064895[n_] := With[{e = Flatten[Position[Reverse[IntegerDigits[n, 2]], 1]] - 1}, Total[2^(e/Max[Apply[GCD, e], 1])]];
Array[A064895, 100, 0] (* Paolo Xausa, Feb 13 2024 *)

PARI

a(n) = { my (b=vector(hammingweight(n))); for (i=1, #b, n-=2^b[i]=valuation(n,2);); b /= max(1, gcd(b)); sum(i=1, #b, 2^b[i]); } \\ Rémy Sigrist, Oct 16 2022

Formula

If n = 2^(g(n)e0) + 2^(g(n)e1) +... then a(n) = 2^e0 + 2^e1 +...