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All visible sequence terms give exactly 3 prime factors. The smallest composite of the form p(n)=n^2+n+41 with 4 prime factors occurs for p(1721)=2963603=43*41^3. Smallest n with 4 distinct prime factors: p(2911)=8476873=83*53*47*41, smallest n with 5 prime factors: p(14144)=200066921=47^4*41, smallest n with 5 distinct prime factors: p(38913)=1514260523=173*71*61*47*43.

n^2 + n + 41 is Euler’s prime generating polynomial.

The first 12 terms in the sequence are the first 12 consecutive primes.

This is a subsequence of A273756 which considers all odd numbers (2n+1) instead of only prime(n) as coefficients of the linear term.

All terms are necessarily prime, since this is necessary and sufficient to get a prime for x = 0.

The respective minima (= number of consecutive primes for x = 0, 1, 2, ...) are given in A273597.

It has been pointed out by Don Reble that the prime k-tuple conjecture predicts infinitely long sequences of primes of the given form, therefore we consider the "local" maxima, for q below some appropriate (large) limit: see sequences A273756 & A273770 for further details. - M. F. Hasler, Feb 17 2020

All terms are prime, since this is necessary and sufficient to get a prime for x = 0.

The values given in A273770 are the number of consecutive primes obtained for x = 0, 1, 2, ....

Sequence A273595 is the subsequence of terms for which 2n+1 is prime.

For even coefficients of the linear term, the answer would always be q=2, the only choice that yields a prime for x=0 and also for x=1 if (coefficient of the linear term)+3 is prime.

The initial term a(n=0) = 41 corresponds to Euler's famous prime-generating polynomial 41+x+x^2. Some subsequent terms are equal to the primes this polynomial takes for x=1,2,3,.... This stems from the fact that adding 2 to the coefficient of the linear term is equivalent to shifting the x-variable by 1. Since here we require x >= 0, we find a reduced subset of the previous sequence of primes, missing the first one, starting with q equal to the second one. (It is known that there is no better prime-generating polynomial of this form than Euler's, see the MathWorld page and A014556. "Better" means a larger p producing p-1 primes in a row. However, the prime k-tuple conjecture suggests that there should be arbitrarily long runs of primes of this form (for much larger p), i.e., longer than 41, but certainly much less than the respective p. Therefore we speak of local maxima.)

The values for p are given in A273756 which is the main entry, see there for further information and (cross)references.

From the initial values, the sequence seems strictly decreasing, with a(n) = 40-n, however, this property does not persist beyond a(27) = 13.

The upper limit on p ensures that we have a well-defined sequence: The prime k-tuple conjecture predicts existence of arbitrarily long sequences of primes of the given form, and thus unbounded minimal value of x. However, the corresponding prime tuples are expected to appear for much larger values of p. The given limit should be understood as "below the first/next such prime tuple", and in general the values a(n) should not change if that limit would be increased by some orders of magnitude. There might be counterexamples, which would be interesting. The given limit was chosen for lack of a more natural expression, and is relatively small. It could be replaced by a more appropriate function of n if a proposal is available, which should not affect the values given so far. - M. F. Hasler, Jan 22 2018, edited Feb 17 2020

Motivated by the fact that n+(17-n)^2 = 1+16^2, 2+15^2, ..., 16+1^2, 17+0^2, 18+1^2, 19+2^2, ..., 32+15^2 are all prime. This has an explanation through Heegener numbers, similar to Euler's prime-generating polynomial, cf. A002837 and related crossrefs.

See A273595 for further information and (cross)references.

From the initial values, the sequence seems strictly decreasing, with a(n+1) - a(n) = (prime(n+1) - prime(n))/2; however, this property does not persist beyond n = 16.

This is the subsequence of A273770 with indices n corresponding to odd primes 2n+1, see formula. - M. F. Hasler, Feb 17 2020

n^2+n+41: Euler’s prime generating polynomial.

First 6 terms in the sequence are first 6 consecutive squares.

Motivated by the fact that the first 32 terms of this sequence are primes. This has an explanation through Heegener numbers, similar to Euler's prime-generating polynomial (cf. A002837 and related crossrefs).

See also A007635 for the primes in this sequence, A260678 for indices k for which a(k) is composite.

Sequence provides all numbers m for which 4*m - 67 is a square. - Bruno Berselli, Nov 16 2015