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Number of primes in the n-th row of the triangle in A117530.
+20
5
1, 2, 3, 3, 5, 3, 7, 3, 5, 6, 6, 6, 13, 3, 11, 8, 12, 8, 13, 10, 8, 7, 12, 10, 9, 21, 6, 22, 11, 7, 13, 12, 21, 13, 14, 16, 18, 7, 20, 17, 21, 20, 24, 14, 18, 20, 16, 16, 35, 10, 18, 29, 18, 30, 30, 26, 21, 18, 21, 29, 16, 22, 32, 40, 10, 27, 24, 25, 45, 18, 39, 40, 43, 11, 11
COMMENTS
1 <= a(n) <= n; conjecture: a(n) < n for n>13.
Primes of the form n^2 + n + 41.
(Formerly M5273)
+10
120
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
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 k-tuples 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(n-40) 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*n-40) = 4*n^2 - 158*n + 1601 also takes prime values for the 40 consecutive integers n = 0 to 39.
2) The polynomial P(3*n-40) = 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*n-40) also takes prime values for n from -13 to -1. Equivalently put, the polynomial P(3*n-79) = 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
R. K. Guy, Unsolved Problems Number Theory, Section A1.
O. Higgins, Another long string of primes, J. Rec. Math., 14 (1981/2) 185.
Paulo Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 137.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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 !! (n-1)
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, g-k); if(a>1, k=k+2); if(a==x+2-k/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
Primes of form n^2 + n + 17.
(Formerly M5069)
+10
56
17, 19, 23, 29, 37, 47, 59, 73, 89, 107, 127, 149, 173, 199, 227, 257, 359, 397, 479, 523, 569, 617, 719, 773, 829, 887, 947, 1009, 1277, 1423, 1499, 1657, 1823, 1997, 2087, 2179, 2273, 2467, 2879, 3209, 3323, 3557, 3677, 3923, 4049, 4177, 4987, 5273
COMMENTS
Note that the gaps between terms increases by 2*k from k = 1 to 15: 19 - 17 = 2, 23 - 19 = 4, 29 - 23 = 6 and so on until 257 - 227 = 30 then fails at 289 - 257 = 32 since 289 = 17^2. - J. M. Bergot, Mar 18 2017 The polynomial P(n):= n^2 + n + 17 takes distinct prime values for the 16 consecutive integers n = 0 to 15. It follows that the polynomial P(n - 16) takes prime values for the 32 consecutive integers n = 0 to 31, consisting of the 16 primes above each taken twice. We note two consequences of this fact.
1) The polynomial P(2*n - 16) = 4*n^2 - 62*n + 257 also takes prime values for the 16 consecutive integers n = 0 to 15.
2)The polynomial P(3*n - 16) = 9*n^2 - 93*n + 257 takes prime values for the 11 consecutive integers n = 0 to 10 ( = floor(31/3)). In addition, calculation shows that P(3*n-16) also takes prime values for n from -5 to -1. Equivalently put, the polynomial P(3*n-31) = 9*n^2 - 183*n + 947 takes prime values for the 16 consecutive integers n = 0 to 15. Cf. A005846 and A048059. (End) The primes in this sequence are not primes in the ring of integers of Q(sqrt(-67)). If p = n^2 + n + 17, then ((2n + 1)/2 - sqrt(-67)/2)((2n + 1)/2 + sqrt(-67)/2) = p. For example, 3^2 + 3 + 17 = 29 and (7/2 - sqrt(-67)/2)(7/2 + sqrt(-67)/2) = 29 also. - Alonso del Arte, Nov 27 2019
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 96.
MATHEMATICA
Select[Table[n^2 + n + 17, {n, 0, 99}], PrimeQ] (* Alonso del Arte, Nov 27 2019 *)
PROG
(Magma) [a: n in [0..250]|IsPrime(a) where a is n^2+n+17] // Vincenzo Librandi, Dec 23 2010 (Python)
from sympy import isprime
it = (n**2 + n + 17 for n in range(250))
4, 5, 7, 9, 13, 15, 19, 21, 25, 31, 33, 39, 43, 45, 49, 55, 61, 63, 69, 73, 75, 81, 85, 91, 99, 103, 105, 109, 111, 115, 129, 133, 139, 141, 151, 153, 159, 165, 169, 175, 181, 183, 193, 195, 199, 201, 213, 225, 229, 231, 235, 241, 243, 253, 259
PROG
(Haskell)
(Sage) [nth_prime(n) +2 for n in (1..70)] # G. C. Greubel, May 20 2019 (GAP) Filtered([1..300], k-> IsPrime(k) ) +2 # G. C. Greubel, May 20 2019
AUTHOR
Simon Colton (simonco(AT)cs.york.ac.uk), Jan 24 2000
Euler's "Lucky" numbers: n such that m^2-m+n is prime for m=0..n-1.
+10
26
COMMENTS
Same as n such that 4n-1 is a Heegner number 1,2,3,7,11,19,43,67,163 (see A003173 and Conway and Guy's book).
REFERENCES
J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 225.
J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 41, p. 16, Ellipses, Paris 2008.
I. N. Herstein and I. Kaplansky, Matters Mathematical, Chelsea, NY, 2nd. ed., 1978, see p. 38.
F. Le Lionnais, Les Nombres Remarquables. Paris: Hermann, pp. 88 and 144, 1983.
MATHEMATICA
A003173 = Union[Select[-NumberFieldDiscriminant[Sqrt[-#]] & /@ Range[200], NumberFieldClassNumber[Sqrt[-#]] == 1 &] /. {4 -> 1, 8 -> 2}]; a[n_] := ( A003173[[n + 4]] + 1)/4; Table[a[n], {n, 0, 5}] (* Jean-François Alcover, Jul 16 2012, after M. F. Hasler *)
Select[Range[50], AllTrue[Table[m^2-m+#, {m, 0, #-1}], PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, May 12 2017 *)
Least x>0 such that x^2+x+n is not prime.
+10
1
2, 4, 1, 2, 1, 4, 1, 1, 1, 2, 1, 10, 1, 1, 1, 2, 1, 16, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 4, 1, 1, 1, 2, 1, 40, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 6, 1, 1, 1, 2, 1, 1, 1, 1
COMMENTS
By definition, a(n)>0 for all n, and a(n)>1 if n+2 is prime.
EXAMPLE
a(0)=2 since 1^2+1+0=2 is prime, but 2^2+2+0=6 is composite.
a(1)=4 since 1^2+1+1=2, 2^2+2+1=7 and 3^2+3+1=13 are prime, but 4^2+4+1=21 is composite.
MATHEMATICA
lx[n_]:=Module[{x=1}, While[PrimeQ[x^2+x+n], x++]; x]; Array[lx, 90, 0] (* Harvey P. Dale, Aug 14 2013 *)
PROG
(PARI) a(n)=for( x=1, n+3, isprime(x^2+x+n) | return(x))
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