A330673 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).

41, 43, 47, 53, 61, 71, 83, 97, 113, 131, 151, 163, 167, 173, 179, 197, 199, 223, 227, 251, 263, 281, 307, 313, 347, 359, 367, 373, 379, 383, 397, 409, 419, 421, 439, 457, 461, 487, 499, 503, 523, 547, 563, 577, 593, 607, 641, 647, 653, 661, 673, 677, 691, 701, 709, 733

Offset: 0

Author

Sergey Pavlov, Dec 23 2019

Keywords

• nonn

Comments

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

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

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.

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).

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

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

Examples

			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).
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).
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).
		

Links

• Sergey Pavlov, Table of n, a(n) for n = 0..312
• Sergey Pavlov, The Euler primes sieve (for the primes of form x^2+x+41; draft paper)

Crossrefs

Cf. A005846, A145292, A202018, A257362.

Formula

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).