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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).

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 easily 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).

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Theorem. Let m is 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).

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editing

reviewed

approved

proposed

reviewed

editing

proposed

Theorem. Let m is 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).

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 non-negative, 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).

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