A338754 Duplicate each decimal digit of n, so 0 -> 00, ..., 9 -> 99.

0, 11, 22, 33, 44, 55, 66, 77, 88, 99, 1100, 1111, 1122, 1133, 1144, 1155, 1166, 1177, 1188, 1199, 2200, 2211, 2222, 2233, 2244, 2255, 2266, 2277, 2288, 2299, 3300, 3311, 3322, 3333, 3344, 3355, 3366, 3377, 3388, 3399, 4400, 4411, 4422, 4433, 4444, 4455, 4466

Offset: 0

Author

Kevin Ryde, Nov 06 2020

Keywords

• base
• nonn

Comments

This is equivalent to changing decimal digits 0,1,..,9 to base 100 digits 0,11,..,99, so the sequence is numbers which can be written in base 100 using only digits 0,11,..,99. Also, numbers whose decimal digit runs are all even lengths (including 0 as no digits at all).

This sequence first differs from A044836 (apart from term 0) at a(100) = 110000 whereas A044836(100) = 10011, because A044836 allows odd length digit runs provided there are more even than odd.

Examples

			For n=5517, digits duplicate to a(n) = 55551177.
		

Links

• Kevin Ryde, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A051022 (0 above each digit), A044836.

Other bases: A001196, A338086.

Programs

PARI

a(n) = fromdigits(digits(n),100)*11;

Python

def A338754(n): return int(''.join(d*2 for d in str(n))) # Chai Wah Wu, May 07 2022

Formula

a(n) = Sum_{i=0..k} 11*d[i]*100^i where the decimal expansion of n is n = Sum_{i=0..k} d[i]*10^i with digits 0 <= d[i] <= 9.

a(n) = A051022(n)*11 for n > 0. - Kritsada Moomuang, Oct 20 2019