A330673 - OEIS

%I #72 Jan 17 2021 02:21:01

%S 41,43,47,53,61,71,83,97,113,131,151,163,167,173,179,197,199,223,227,

%T 251,263,281,307,313,347,359,367,373,379,383,397,409,419,421,439,457,

%U 461,487,499,503,523,547,563,577,593,607,641,647,653,661,673,677,691,701,709,733

%N The possible v-factors for any A202018(k) (while A202018(k) = v * w, v and w are integers, w >= v >= 41, v = w iff w = 41, all such v-factors form the set V).

%C This is different from A257362: a(n) = A257362(n+1) for n=0..109, but a(110) = 1468 != 1471 = A257362(111). - _Alois P. Heinz_, Mar 02 2020

%C A kind of prime number sieve for the numbers of form x^2+x+41 (for so-called Euler primes, or A005846).

%C A set of all composite Euler numbers of form x^2+x+41 could be written as a 4-dimensional matrix m(i,j,t,u); a set of all terms of a(n) could be written as a 3-dimensional matrix v(i,j,t), since, for any integer u > -1, and for any w-factor that has the same values for i, j, t, we have the same v-factor (u = -1 iff w = 41); see formulas below.

%C Theorem. Let m be a term of A202018. Then m is composite iff m == 0 (mod v), where v is a term of a(n), v <= sqrt(m) (v = sqrt(m) iff m = 1681); otherwise, m is prime. Moreover, while m == 0 (mod p) (p is prime, p <= sqrt(m), p = sqrt(m) iff m = 1681), p is a term of a(n).

%C While i = 1, any v(i,t,j) is a term of both A202018 and a(n) (trivial).

%C Any w is a term of V and of a(n) which is the superset of V.

%H Sergey Pavlov, <a href="/A330673/b330673.txt">Table of n, a(n) for n = 0..312</a>

%H Sergey Pavlov, <a href="/A330673/a330673.pdf">The Euler primes sieve (for the primes of form x^2+x+41; draft paper)</a>

%F Let  j = {-1;0;-2;1;-3;2;...;-(n+1);n}, m(-1) = 41, m(0) = 41, etc. (while j is negative, m(j) = A202018(-(j+1)); while j is nonnegative, m(j) = A202018(j)). Any term of a(n) could be written at least once as v(i,t+1,j) = m(j) * i^2 + b + ja, where i, t, and j are integers (j could be negative), i > 2; a = (i^2 - 2i) - 2i(t - 1), b = a - ((i^2 - 4)/4 - ((t - 1)^2 + 2(t - 1))), 0 < t < (i/2), while i is even; a = (i^2 - i) - 2i(t - 1), b = a - ((i^2 - 1)/4 - ((t - 1)^2 + (t - 1))), 0 < t < ((i + 1)/2), while i is odd (Note: v(i,1,j) = v(i,i/2,j), while i is even; v(i,1,j) = v(i,(i + 1)/2,j), while i is odd); at i = 2, v(2,1,j) = 4 * m(j) + 3 + 4j (at i = 2, we use only j < 0); at i = 1, v(1,1,j) = m(j) (at i = 1, we use only j >= 0; trivial).

%e Let  i = 3, t = 1, j = -1. Then v(i,t,j) = m(j) * i^2 + b + ja = 41 * 3^2 + 4 - 6 = 41 * 9 - 2 = 367, and 367 is a term of a(n).

%e We could find all terms of a(n) v < 10^n and then all Euler primes p < 10^(2n) (for n > 1, number of all numbers m such that are terms of A202018 (and any m < 10^(2n)) is 10^n; trivial).

%e Let 2n = 10; it's easy to establish that, while i > 49, any v(i,t,j)^2 > 10^10; thus, we can use 0 < i < 50 to find all numbers v < 10^5. While m is a term of A202018, m < 10^10, m is composite iff there is at least one v such that m == 0 (mod v); otherwise, m is prime. We could easily remove all "false" numbers v that cannot be divisors of any m. Let p' be a regular prime (p' is a term of A000040, but not of a(n)) such that any 3p' < UB(i); in our case, any 3p' < 50. Thus, we could try any v with p' = {2,3,5,7,11,13}; if v == 0 (mod p'), it is "false"; otherwise, there is at least one m < 10^10 such that m == 0 (mod v).

%Y Cf. A005846, A145292, A202018, A257362.

%K nonn

%O 0,1

%A _Sergey Pavlov_, Dec 23 2019