 mersenneforum.org Ec prime exponents
 Register FAQ Search Today's Posts Mark Forums Read 2020-05-10, 15:29 #1 enzocreti   Mar 2018 10248 Posts 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 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 Last fiddled with by enzocreti on 2020-11-25 at 10:35  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post biwema Math 10 2021-11-28 04:05 Bobby Jacobs Wagstaff PRP Search 2 2019-03-03 19:37 science_man_88 Miscellaneous Math 3 2010-10-13 14:32 HuzursuZ PrimeNet 3 2005-03-09 06:16 GP2 Data 22 2004-02-03 22:59

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