A299869 The sum of the first n terms of the sequence is the concatenation of the first n digits of the sequence, and a(1) = 6.
6, 60, 594, 5945, 59454, 594535, 5945351, 59453514, 594535135, 5945351351, 59453513510, 594535135104, 5945351351035, 59453513510351, 594535135103509, 5945351351035091, 59453513510350914, 594535135103509135, 5945351351035091351, 59453513510350913508, 594535135103509135082, 5945351351035091350820
Offset: 1
Examples
6 + 60 = 66 which is the concatenation of 6 and 6. 6 + 60 + 594 = 660 which is the concatenation of 6, 6 and 0. 6 + 60 + 594 + 5945 = 6605 which is the concatenation of 6, 6, 0 and 5. From n = 3 on, a(n) can be computed directly as c(n) - c(n-1), cf. formula: a(3) = 660 - 66 = 594, a(4) = 6605 - 660 = 5945, etc. - _M. F. Hasler_, Feb 22 2018
Links
- Jean-Marc Falcoz, Table of n, a(n) for n = 1..300
Crossrefs
Programs
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PARI
a(n,show=1,a=6,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) ~ 0.66*10^n, a(n) ~ 0.59*10^n. See A300000 for the proof. - M. F. Hasler, Feb 22 2018
Comments