A065159 Binary string self-substitutions: a(n) is obtained by substituting the binary expansion of n for each 1-bit in the binary expansion of n.

0, 1, 4, 15, 16, 85, 108, 511, 64, 585, 660, 5819, 816, 7085, 7644, 65535, 256, 4369, 4644, 78451, 5200, 87381, 91564, 1531639, 6336, 105625, 109876, 1825659, 118384, 1961821, 2029500, 33554431, 1024, 33825, 34884, 1149155, 37008, 1217189, 1250124, 41056743

Offset: 0

Author

Marc LeBrun, Oct 18 2001

Keywords

• base
• easy
• nonn

Examples

			a(5): 5 = 101 -> (101)0(101) = 1010101 = 85.
		

Links

• Paul Tek, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A007088, A065157, A065158, A065160.

Programs

Mathematica

bss[n_]:=Module[{idn2=IntegerDigits[n,2]},FromDigits[Flatten[idn2/.{1-> idn2}],2]]; Array[bss,40,0] (* Harvey P. Dale, Aug 15 2017 *)

Python

def a(n): b = bin(n)[2:]; return int(b.replace("1", b), 2)
print([a(n) for n in range(40)]) # Michael S. Branicky, Aug 05 2022

Formula

a(0) = 0. a(2^n) = 4^n. a(4n+2) = (4n+2)*(1+a(4n+1)/(4n+1)).

a(n) = A065157(n,n) = A065158(n,n)*n = A065160(n)*n.

a(n) =z(n, n) with z(u, v) = if u=0 then 0 else if u mod 2 = 0 then z(u/2, v)*2 else z([u/2], v)*A062383(v)+v. - Reinhard Zumkeller, Feb 15 2004

Extensions

Name clarified by Michael S. Branicky, Aug 05 2022