A306759 - OEIS
COMMENTS
The name "constant of Brazilian primes" is used in the article "Les nombres brésiliens" in link, théorème 4, page 36. Brazilian primes are in A085104.
Let S(k) be the sum of reciprocals of Brazilian primes < k. These values below come from different calculations by Jon, Michel, Daniel and Davis.
q S(10^q)
== ========================
1 0.1428571428571428571... (= 1/7)
2 0.2889927283868234859...
3 0.3229022355626914481...
4 0.3295236806353669357...
5 0.3312171311946179843...
6 0.3316038696349217289...
7 0.3317139158654747333...
8 0.3317434191078170412...
9 0.3317513267394988538...
10 0.3317535651668937256...
11 0.3317542057931842329...
12 0.3317543906772274268...
13 0.3317544444033188051...
14 0.3317544601136967527...
15 0.3317544647354485208...
16 0.3317544661014868080...
17 0.3317544665073451951...
18 0.3317544666282877863...
19 0.3317544666644601817...
20 0.3317544666753095766...
According to the Goormaghtigh conjecture, there are only two Brazilian primes which are twice Brazilian: 31 = (111)_5 = (11111)_2 and 8191 = (111)_90 = (1111111111111)_2. The reciprocals of these two numbers are counted only once in the sum.
REFERENCES
Daniel Lignon, Dictionnaire de (presque) tous les nombres entiers, Ellipses, Paris, 2012, page 175.
LINKS
Bernard Schott, Les nombres brésiliens, Quadrature, no. 76, avril-juin 2010, pages 30-38; included here with permission from the editors of Quadrature.
FORMULA
Equals Sum_{n>=1} 1/A085104(n).
EXAMPLE
1/7 + 1/13 + 1/31 + 1/43 + 1/73 + 1/127 + 1/157 + ... = 0.33175...
PROG
(PARI) brazil(N, L=List())=forprime(K=3, #binary(N+1)-1, for(n=2, sqrtnint(N-1, K-1), if(isprime((n^K-1)/(n-1)), listput(L, (n^K-1)/(n-1))))); Set(L);
brazilcons(lim, nbd) = r=brazil(10^lim); x=sum(M=1, #r, 1./r[M]); for(n=1, nbd, print1(floor(x*10^n)%10, ", ")); \\ Davis Smith, Mar 10 2019
(PARI) cons(lim)=my(v=List(), t, k); for(n=2, sqrt(lim), t=1+n; k=1; while((t+=n^k++)<=lim, if(isprime(t), listput(v, t)))); v = vecsort(Vec(v), , 8); sum(k=1, #v, 1./v[k]); \\ Michel Marcus, Mar 11 2019