

A005846


Primes of the form n^2 + n + 41.
(Formerly M5273)


118



41, 43, 47, 53, 61, 71, 83, 97, 113, 131, 151, 173, 197, 223, 251, 281, 313, 347, 383, 421, 461, 503, 547, 593, 641, 691, 743, 797, 853, 911, 971, 1033, 1097, 1163, 1231, 1301, 1373, 1447, 1523, 1601, 1847, 1933, 2111, 2203, 2297, 2393, 2591, 2693, 2797
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OFFSET

1,1


COMMENTS

The link to E. Wegrzynowski contains the following incorrect statement: "It is possible to find a polynomial of the form n^2 + n + B that gives prime numbers for n = 0, ..., A, A being any number." It is known that the maximum is A = 39 for B = 41.  Luis Rodriguez (luiroto(AT)yahoo.com), Jun 22 2008
Contrary to the last comment, Mollin's Theorem 2.1 shows that any A is possible if the Prime ktuples Conjecture is assumed.  T. D. Noe, Aug 31 2009
a(n) can be generated by a recurrence based on the gcd in the type of Eric Rowland and Aldrich Stevens. See the recurrence in PARI under PROG.  Mike Winkler, Oct 02 2013
These primes are not prime in O_(Q(sqrt(163)). Given p = n^2 + n + 41, we have ((2n + 1)/2  sqrt(163)/2)((2n + 1)/2 + sqrt(163)/2) = p, e.g., 1601 = 39^2 + 39 + 41 = (79/2  sqrt(163)/2)(79/2 + sqrt(163)/2).  Alonso del Arte, Nov 03 2017
The polynomial P(n) := n^2 + n + 41 takes distinct prime values for the 40 consecutive integers n = 0 to 39. It follows that the polynomial P(n40) takes prime values for the 80 consecutive integers n = 0 to 79, consisting of the 40 primes above each taken twice. We note two consequences of this fact.
1) The polynomial P(2*n40) = 4*n^2  158*n + 1601 also takes prime values for the 40 consecutive integers n = 0 to 39.
2) The polynomial P(3*n40) = 9*n^2  237*n + 1601 takes prime values for the 27 consecutive integers n = 0 to 26 ( = floor(79/3)). In addition, calculation shows that P(3*n40) also takes prime values for n from 13 to 1. Equivalently put, the polynomial P(3*n79) = 9*n^2  471*n + 6203 takes prime values for the 40 consecutive integers n = 0 to 39. This result is due to Higgins. Cf. A007635 and A048059. (End)


REFERENCES

O. Higgins, Another long string of primes, J. Rec. Math., 14 (1981/2) 185.
Paulo Ribenboim, The Book of Prime Number Records. SpringerVerlag, NY, 2nd ed., 1989, p. 137.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS



FORMULA



EXAMPLE

a(39) = 1601 = 39^2 + 39 + 41 is in the sequence because it is prime.
1681 = 40^2 + 40 + 41 is not in the sequence because 1681 = 41*41.


MAPLE

for y from 0 to 10 do
U := y^2+y+41;
if isprime(U) = true then print(U) end if ;
end do:


MATHEMATICA

Select[Table[n^2 + n + 41, {n, 0, 59}], PrimeQ] (* Alonso del Arte, Dec 08 2011 *)


PROG

(Haskell)
a005846 n = a005846_list !! (n1)
a005846_list = filter ((== 1) . a010051) a202018_list
(PARI) {k=2; n=1; for(x=1, 100000, f=x^2+x+41; g=x^2+3*x+43; a=gcd(f, gk); if(a>1, k=k+2); if(a==x+2k/2, print(n" "a); n++))} \\ Mike Winkler, Oct 02 2013
(GAP) Filtered(List([0..100], n>n^2+n+41), IsPrime); # Muniru A Asiru, Apr 22 2018
(Magma) [a: n in [0..55]  IsPrime(a) where a is n^2+n+ 41]; // Vincenzo Librandi, Apr 24 2018


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



