A299872 The sum of the first n terms of the sequence is the concatenation of the first n digits of the sequence, and a(1) = 9.
9, 90, 891, 8918, 89181, 891802, 8918027, 89180271, 891802702, 8918027027, 89180270270, 891802702701, 8918027027002, 89180270270027, 891802702700263, 8918027027002637, 89180270270026371, 891802702700263702, 8918027027002637027, 89180270270026370262, 891802702700263702622, 8918027027002637026226
Offset: 1
Examples
9 + 90 = 99 which is the concatenation of 9 and 9. 9 + 90 + 891 = 990 which is the concatenation of 9, 9 and 0. 9 + 90 + 891 + 8918 = 9908 which is the concatenation of 9, 9, 0 and 8. From n = 3 on, a(n) can be computed directly as c(n) - c(n-1), cf. formula: a(3) = 990 - 99 = 891, a(4) = 9908 - 990 = 8918, 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=9,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) ~ 0.99*10^n, a(n) ~ 0.89*10^n. See A300000 for the proof. - M. F. Hasler, Feb 22 2018
Comments