A331364 If the set of nonzero digits of n in some base of the form 2^2^k (with k >= 0) has exactly two elements, let b be the least such base and u and v the corresponding two nonzero digits; the base b representation of a(n) is obtained by replacing the u's by v's and vice versa in the base b representation of n; otherwise a(n) = n.

0, 1, 2, 3, 4, 5, 9, 13, 8, 6, 10, 14, 12, 7, 11, 15, 16, 17, 33, 49, 65, 81, 41, 61, 36, 38, 37, 177, 52, 55, 225, 53, 32, 18, 34, 50, 24, 26, 25, 114, 130, 22, 162, 62, 56, 210, 59, 58, 48, 19, 35, 51, 28, 31, 99, 29, 44, 147, 47, 46, 195, 23, 43, 243, 64

Offset

0,3

Comments

This sequence is a self-inverse permutation of the nonnegative integers. See A332520 for the corresponding fixed points.

For any m > 1, we can devise a similar sequence by considering bases of the form m^2^k (with k >= 0).

Links

Rémy Sigrist, Table of n, a(n) for n = 0..65536

Rémy Sigrist, Scatterplot of the first 2^2^4 terms

Rémy Sigrist, Colored scatterplot of the first 2^2^4 terms (where the color denotes the base b if any)

Index entries for sequences that are permutations of the natural numbers

Formula

a(n) < 2^2^k iff n < 2^2^k for any n, k >= 0.

a(2^k) = 2^k for any k >= 0.

a(2^2^k-1) = 2^2^k-1 for any k >= 0.

Example

For n = 73:
- the base 2^2^0 representation of 73 is "1001001" which has only one kind of nonzero digits,
- the base 2^2^1 representation of 73 is "1021" which has exactly two kinds of nonzero digits, "1" and "2",
- so the base 2^2^1 representation of a(73) is "2012",
- and a(73) = 134.

Prog

(PARI) a(n) = { for (x=0, oo, my (b=2^2^x, d=if (n, digits(n, b), [0])); if (#d==1, return (n), my (uv=select(sign, Set(d))); if (#uv==2, return (
fromdigits(apply (t -> if (t==0, 0, t==uv[1], uv[2], uv[1]), d), b))))) }

Crossrefs

Cf. A001146, A332520 (fixed points).

Sequence in context: A239742 A101526 A264994 * A319720 A140298 A253181

Adjacent sequences: A331361 A331362 A331363 * A331365 A331366 A331367

Keyword

nonn,base

Author

Rémy Sigrist, Jun 24 2020

Status

approved