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enzocreti 2020-05-10 15:29

Ec prime exponents
 
Ec primes are primes formed by concatenation in base 10 of two consecutive Mersenne numbers, 157 for example.


The form is

(2^k-1)*10^d + 2^(k-1)-1=P(k) where d is the number of decimal digits of 2^(k-1)-1

Let r(1) denote the least exponent such that P(k) is prime
r(2) the next exponent such that P(k) is prime...
and so on...

For example r(1)=2, r(2)=3,r(3)=4, r(4)=7...r(30)=92020...r(i)...


I noticed that if i divides r(i), then r(i) +1 is prime.

There are only 5 cases known so it could be coincidence

r(1)=2, 2 is divisible by 1 and 3 is prime
r(6)=12, 12 is divisible by 6 and 13 is prime
r(9)=36, 36 is divisible by 9 and 37 is prime
r(26)=56238, 56238 is divisible by 26 and 56239 is prime
r(27)=69660, 69660 is divisible by 27 and 69661 is prime

1,6,9,27 satisfy the recurrence

a(1)=1, a(2)=6,
a(n)=a(n-1)+3*a(n-2)

So I wonder if 54 divides r(54)...i think this is impossible to test




apart from 3


13, 37, 56239, 69661 are primes of the form 6k+1




I notice also that 36 is congruent to 6^2 mod 323
56238 is congruent to 6^2 mod 323
69660 is congruent to 6^3-1 mod 323




Now I consider the sum


1/2+6/12+9/36+26/56238+27/69660+...=const???
c'est a dire the sum of i/r(i) when i divides r(i) ...


I conjecture that this sum converges to a constant 1.02508...very close to 5/4....




Curiously 1,0,2,5,0,8 are the first six terms of Oeis sequence [URL="https://oeis.org/A111466"][COLOR=#000080]A111466[/COLOR][/URL] involving Fibonacci numbers




2,12,36,69660 (not multiple of 13) are also divisible by the sum of their digits


2 is divisible by 2
12 is divisible by (1+2)
36 is divisible by (3+6)
69660 is divisible by (6+9+6+6=27)




the multiple of 12 (12, 36, 69660) are divisible by the sum of their digits and by i. They are also divisible by a perfect square>1.


56238 is multiple of 13 and it is not divisible by the sum of their digits but by i=26. 56238 is not divisible by a square>1 infact 56238 is squarefree

So I could conjecture that if i divides r(i) and r(i) is a multiple of 13, then r(i) is square free, but also this conjecture is hard to test


56238 is divisible by the sum of his digits in base 6


Now I consider the sum

1/3+6/13+9/37+26/56239+27/69661+...
I don't know if this sum is infinite or not

The sum is taken over the primes (3,13,37,56239,69661) a(i) +1 when i divides a(i). The numerator (1,6,9,26,27) is the position in the vector of the exponents leading to a prime.
Using Wolphram I found that
1/3+6/13+9/37+26/56239+27/69661 is very close to (923/8768)*pi^2 or (9/5)*gamma where gamma is euler mascheroni constant




now I consider sigma(x) that is the sum of the divisors of x, for x>2


sigma(12)=28 is a multiple of 7
sigma(36)=91 is a multiple of 7
sigma(56238)=139776 is a multiple of 7
sigma(69660)=223608=21*22^3 is a multiple of 7




consider the Pell equation (3x)^2-10*y^2=-1


x=56239 and y=53353 is a solution
56239=56238+1 is prime


53353 is prime


x-y=2886


2886 is divisible by 3, 13 and 37


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