A330091 Inverse permutation to A329303.

0, 1, 2, 3, 4, 5, 6, 7, 8, 11, 10, 9, 12, 13, 14, 15, 16, 23, 22, 19, 18, 21, 20, 17, 24, 27, 26, 25, 28, 29, 30, 31, 32, 47, 46, 39, 38, 45, 44, 35, 34, 41, 42, 37, 36, 43, 40, 33, 48, 55, 54, 51, 50, 53, 52, 49, 56, 59, 58, 57, 60, 61, 62, 63, 64, 95, 94, 79

Offset: 0

Author

Rémy Sigrist, Dec 01 2019

Keywords

• nonn
• base

Comments

If the run lengths in binary expansion of n are (r(1), ..., r(w)), then the run lengths in binary expansion of a(n) are (r(1), r(w), r(2), r(w-1), ...); this corresponds to a "milk shuffle".

Examples

			A329303(43) = 45, hence a(45) = 43.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..8192
• Index entries for sequences related to binary expansion of n
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A329303.

Programs

PARI

torl(n) = { my (rr=[]); while (n, my (r=valuation(n+(n%2),2)); rr = concat(r, rr); n\=2^r); rr }
unshuffle(v) = { my (w=vector(#v), o=0, e=#v+1); for (k=1, #v, w[k]=v[if (k%2, o++, e--)]); w }
fromrl(rr) = { my (v=0); for (k=1, #rr, v = (v+(k%2))*2^rr[k]-(k%2)); v }
a(n) = fromrl(unshuffle(torl(n)))