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Old 2020-05-10, 15:29   #1
enzocreti
 
Mar 2018

52610 Posts
Default 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 A111466 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

Last fiddled with by enzocreti on 2020-09-29 at 09:00
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