A299867 The sum of the first n terms of the sequence is the concatenation of the first n digits of the sequence, and a(1) = 4.
4, 40, 396, 3963, 39636, 396357, 3963567, 39635676, 396356757, 3963567567, 39635675670, 396356756706, 3963567567057, 39635675670567, 396356756705673, 3963567567056727, 39635675670567276, 396356756705672757, 3963567567056727567, 39635675670567275672, 396356756705672756722
Offset: 1
Examples
4 + 40 = 44 which is the concatenation of 4 and 4. 4 + 40 + 396 = 440 which is the concatenation of 4, 4, and 0. 4 + 40 + 396 + 3963 = 4403 which is the concatenation of 4, 4, 0 and 3. From n = 3 on, a(n) can be computed directly as c(n) - c(n-1), cf. formula: a(3) = 440 - 4 = 396, a(4) = 4403 - 440 = 3963, 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=4,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) ~ 0.44*10^n, a(n) ~ 0.396*10^n. See A300000 for the proof. - M. F. Hasler, Feb 22 2018
Comments