A086171 Continued fraction of sum(prime(n)/10^b(n)), where b(n) = 1 + the total number of digits of the first n-1 primes, A068670.
0, 4, 4, 6, 2, 5, 18, 1, 3, 4, 1, 2, 1, 2, 4, 7, 4, 1, 21, 2, 1, 3, 5, 2, 1, 1, 27, 1, 1, 5, 12, 1, 18, 1, 1, 1, 1, 1, 1, 9, 3, 1, 1, 1, 5, 1, 5, 2, 2, 1, 53, 1, 8, 1, 23, 6, 2, 2, 1, 1, 3, 1, 1, 25, 2, 2, 7, 1, 2, 3, 1, 4, 3, 12, 1, 2, 7, 1, 68, 1, 19, 1, 2, 2, 14, 4, 6, 2, 1, 2, 58, 2, 16, 1, 1, 1, 2, 2, 1
Offset: 1
Examples
r=0.235811317192329313741434753596167717... = 2/10^1 + 3/10^2 + 5/10^3 + 7/10^4 + 11/10^5 + 13/10^7 + ..., the exponents being increased by the length of the previous prime. - _M. F. Hasler_, Oct 17 2013
Programs
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Mathematica
(* number of powers of ten in the Primes as a sequence*) byte[n_Integer?Positive] := byte[n] =byte[n-1]+Floor[Log[Prime[n-1]]/Log[10]]+1 byte[0]=byte[1] = 1 b=Table[N[Prime[n]*10^(-byte[n]), Digits], {n, 1, Digits}] r=Apply[Plus, b]
Formula
r = sum_{n=1..infinity} prime(n)/10^b(n), where b(n+1)=b(n)+floor(log[10] prime(n))+1, b(1)=1. (Edited by M. F. Hasler, Oct 17 2013)
Extensions
Edited by M. F. Hasler, Oct 17 2013
Comments