A299868 The sum of the first n terms of the sequence is the concatenation of the first n digits of the sequence, and a(1) = 5.

5, 50, 495, 4954, 49545, 495446, 4954459, 49544595, 495445946, 4954459459, 49544594590, 495445945905, 4954459459046, 49544594590459, 495445945904591, 4954459459045909, 49544594590459095, 495445945904590946, 4954459459045909459, 49544594590459094590, 495445945904590945902, 4954459459045909459018

Offset: 1

Author

Eric Angelini and Jean-Marc Falcoz, Feb 21 2018

Keywords

• nonn
• base

Comments

The sequence starts with a(1) = 5 and is always extended with the smallest integer not yet present in the sequence and not leading to a contradiction.

Examples

			5 + 50 = 55 which is the concatenation of 5 and 5.
5 + 50 + 495 = 550 which is the concatenation of 5, 5 and 0.
5 + 50 + 495 + 4954 = 5504 which is the concatenation of 5, 5, 0 and 4.
From n = 3 on, a(n) can be computed directly as c(n) - c(n-1), cf. formula: a(3) = 550 - 55 = 495, a(4) = 5504 - 550 = 4954, etc. - _M. F. Hasler_, Feb 22 2018
		

Links

• Jean-Marc Falcoz, Table of n, a(n) for n = 1..300

Crossrefs

A300000 is the lexicographically first sequence of this type, with a(1) = 1.

Cf. A299865, ..., A299872 for variants with a(1) = 2, ..., 9.

Programs

PARI

a(n,show=1,a=5,c=a,d=[a])={for(n=2,n,show&&print1(a",");a=-c+c=c*10+d[1];d=concat(d[^1],if(n>2,digits(a))));a} \\ M. F. Hasler, Feb 22 2018

Formula

a(n) = c(n) - c(n-1), where c(n) = concatenation of the first n digits, c(n) ~ 0.55*10^n, a(n) ~ 0.495*10^n. See A300000 for the proof. - M. F. Hasler, Feb 22 2018