A305493 A binary encoding of the decimal representation of a number: for any number n >= 0, consider its decimal representation and replace each 9 with "111111111" and each other digit d with a "0" followed by d "1"s and interpret the result as a binary string.

0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 2, 5, 11, 23, 47, 95, 191, 383, 767, 1023, 6, 13, 27, 55, 111, 223, 447, 895, 1791, 2047, 14, 29, 59, 119, 239, 479, 959, 1919, 3839, 4095, 30, 61, 123, 247, 495, 991, 1983, 3967, 7935, 8191, 62, 125, 251, 503, 1007, 2015

Offset: 0

Author

Rémy Sigrist, Jun 02 2018

Keywords

• nonn
• base
• easy

Comments

This sequence is a permutation of the nonnegative numbers.

The inverse sequence, say b, satisfies b(n) = A001202(n+1) for n = 0..1022, but b(1023) = 19 whereas A001202(1024) = 10.

The first known fixed points are: 0, 1, 65010; they all belong to A037308.

This encoding can be applied to any base b > 1 (when b = 2, we obtain the identity function) as well as to the factorial base and to the primorial base.

Examples

			For n = 1972:
- the digit 1 is replaced by "01",
- the digit 9 is replaced by "111111111",
- the digit 7 is replaced by "01111111",
- the digit 2 is replaced by "011",
- hence we obtain the binary string "0111111111101111111011",
- and a(1972) = 2096123.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..10000
• Rémy Sigrist, Colored logarithmic scatterplot of the first 1000000 terms (where the color is function of the digital sum minus the number of nonleading digits different from 9)
• Index entries for sequences related to decimal expansion of n
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A000120, A001202, A002275, A002450, A007953, A037308, A051885.

Programs

Mathematica

tb=Table[n->PadRight[{0},n+1,1],{n,9}]/.PadRight[{0},10,1]-> PadRight[ {},9,1]; Table[FromDigits[IntegerDigits[n]/.tb//Flatten,2],{n,0,60}] (* Harvey P. Dale, Jul 31 2018 *)

PARI

a(n, base=10) = my (b=[], d=digits(n, base)); for (i=1, #d, if (d[i]!=base-1, b=concat(b, 0)); b=concat(b, vector(d[i], k, 1))); fromdigits(b, 2)
/* inverse */ b(n, base=10) = my (v=0, p=1); while (n, my (d = min(valuation(n+1, 2), base-1)); v += p * d; n \= 2^min(base-1, 1+d); p *= base); v

Formula

A000120(a(n)) = A007953(n).

a(A051885(k)) = 2^k - 1 for any k >= 0.

a(A002275(k)) = A002450(k) for any k >= 0.

a(10 * n) = 2 * a(n).