Displaying 1-10 of 23 results found.
Number of primes of the form P(k) = k^2 + k + 41 for k <= 10^n, where P(k) is Euler's prime-generating polynomial A202018.
+20
4
2, 11, 87, 582, 4149, 31985, 261081, 2208197, 19132653, 168806741, 1510676803
EXAMPLE
a(0) = 2 because 41 and 43 are the 2 primes generated for k <= 1 = 10^0.
a(1) = 11 because 41, 43, 47, 53, 61, 71, 83, 97, 113, 131, 151 are the 11 primes generated for k <= 10^1, ( A202018(10) = 151). a(3) = 87 because 87 terms of A202018(0..100) are prime. The 14 composites occur for k = A007634(1..14): 40, 41, 44, 49, 56, ...
PROG
(PARI) n=0; m=1; for(k=0, 10^7, my(j=k^2+k+41); if(isprime(j), n++); if(k==m, m*=10; print1(n, ", ")))
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).
+20
1
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
COMMENTS
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.
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).
EXAMPLE
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).
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
Hardy-Littlewood constant for x^2+x+41.
+10
41
3, 3, 1, 9, 7, 7, 3, 1, 7, 7, 4, 7, 1, 4, 2, 1, 6, 6, 5, 3, 2, 3, 5, 5, 6, 8, 5, 7, 6, 4, 9, 8, 8, 7, 9, 6, 6, 4, 6, 8, 5, 5, 4, 5, 8, 5, 6, 5, 2, 9, 8, 5, 8, 4, 9, 1, 5, 3, 9, 4, 0, 7, 2, 7, 9, 5, 0, 2, 6, 3, 3, 1, 0, 4, 2, 6, 1, 1, 8, 1, 4, 9, 7, 3, 7, 5, 5
REFERENCES
Henri Cohen, Number Theory, Vol II: Analytic and Modern Tools, Springer (Graduate Texts in Mathematics 240), 2007.
EXAMPLE
3.31977317747142166532355685764988796646855...
PROG
(PARI) \\ See Belabas, Cohen link. Run as HardyLittlewood2(x^2+x+41)/2 after setting the required precision.
Numbers n such that n^2 + n + 41 is prime.
+10
23
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 42, 43, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 77, 78
COMMENTS
Among first 100000 terms, the only run of 13 subsequent values >39 is 219..231. - Zak Seidov, Jan 28 2009 Number of terms less than 10^n: 1, 10, 86, 581, 4149, 31985, 261081, 2208197, 19132652, ... . - Robert G. Wilson v, Apr 20 2015
REFERENCES
P. Hoffman, Archimedes' Revenge, pp. 39-40,Penguin Books 1988.
FORMULA
a(n) = (sqrt(4* A005846(n)-163)-1)/2.
EXAMPLE
39 is in the sequence because 39^2+39+41=1601 which is prime but 40 is not because 40^2+40+41=1681=41*41.
MAPLE
select(t -> isprime(t^2+t+41), [$0..100]); # Robert Israel, Apr 20 2015
PROG
(Haskell)
a056561 n = a056561_list !! (n-1)
a056561_list = filter ((== 1) . a010051' . a202018) [0..]
Composite numbers generated by the Euler polynomial x^2 + x + 41.
+10
17
1681, 1763, 2021, 2491, 3233, 4331, 5893, 6683, 6847, 7181, 7697, 8051, 8413, 9353, 10547, 10961, 12031, 13847, 14803, 15047, 15293, 16043, 16297, 17071, 18673, 19223, 19781, 20633, 21797, 24221, 25481, 26123, 26447, 26773, 27101, 29111
COMMENTS
The Euler polynomial x^2 + x + 41 gives primes for consecutive x from 0 to 39.
For numbers x for which x^2 + x + 41 is not prime see A007634. Let P(x)=x^2 + x + 41. In view of identity P(x+P(x))=P(x)*P(x+1), all values of P(x+P(x)) are in the sequence. - Vladimir Shevelev, Jul 16 2012
MATHEMATICA
a = {}; Do[If[PrimeQ[x^2 + x + 41], null, AppendTo[a, x^2 + x + 41]], {x, 0, 500}]; a
Select[Table[x^2+x+41, {x, 200}], CompositeQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 21 2018 *)
PROG
(Haskell)
a145292 n = a145292_list !! (n-1)
a145292_list = filter ((== 0) . a010051) a202018_list
a(n) = n^2 - 79*n + 1601.
+10
8
1601, 1523, 1447, 1373, 1301, 1231, 1163, 1097, 1033, 971, 911, 853, 797, 743, 691, 641, 593, 547, 503, 461, 421, 383, 347, 313, 281, 251, 223, 197, 173, 151, 131, 113, 97, 83, 71, 61, 53, 47, 43, 41, 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, 1681
COMMENTS
a(n) is prime for 0 <= n <= 79. a(80) = 1681 = 41^2.
More than the usual number of terms are shown in order to display the initial 80 primes.
First 80 prime entries are palindromically distributed because a(n) = P(x) = x^2 + x + 41, with x=n-40 and we observe that P(x) generates primes ( A005846) for x = 0 through 39, along with the fact that P(-x) = P(x-1). - Lekraj Beedassy, Apr 24 2006
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 6.
C. Stanley Ogilvy and John T. Anderson, Excursions in Number Theory, Dover Publications, NY, 1966, p. 37, 147.
FORMULA
G.f.: (1601 - 3280*x + 1681*x^2)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
(End)
MATHEMATICA
Table[n^2-79*n+1601, {n, 100}] (* or *) LinearRecurrence[{3, -3, 1}, {1523, 1447, 1373}, 100] (* Harvey P. Dale, Jan 14 2017 *)
Primes p such that p^2+p+41 is also prime.
+10
4
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 97, 101, 103, 107, 113, 131, 137, 139, 149, 151, 157, 167, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 241, 257, 263, 269, 277, 281, 293, 307, 311, 313, 317, 337, 353
COMMENTS
n^2 + n + 41 is Euler’s prime generating polynomial.
The first 12 terms in the sequence are the first 12 consecutive primes.
EXAMPLE
13 is in the sequence because 13 is prime and 13^2+13+41 = 223 is also prime.
113 is in the sequence because 113 is prime and 113^2+113+41 = 12923 is also prime.
MAPLE
with(numtheory):KD := proc() local a, b; a:=ithprime(n); b:=a^2+a+41; if isprime(b) then RETURN (a); fi; end: seq(KD(), n=1..500);
MATHEMATICA
Select[Prime[Range[200]], PrimeQ[#^2+#+41]&]
PROG
(PARI) s=[]; forprime(p=2, 1000, if(isprime(p^2+p+41), s=concat(s, p))); s \\ Colin Barker, Feb 20 2014 (Magma) [p: p in PrimesUpTo(400)| IsPrime(p^2+p+41)]; // Vincenzo Librandi, Feb 22 2014
Odd primes modulo which -163 is a square.
+10
4
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
COMMENTS
Contains A005846. The first members that are not in A005846 are 163 and 167. Primes that divide some member of A202018. Primes congruent to x^2 mod 163 for some x, 0 <= x <= 162.
Primes of the form x^2 + xy + 41y^2. Also, primes of the form x^2 - xy + 41y^2 with x and y nonnegative. - Jianing Song, Feb 19 2021
MAPLE
select(p -> isprime(p) and (p=163 or numtheory:-legendre(-163, p)=1), [seq(2*i+1, i=1..1000)]);
MATHEMATICA
Reap[For[p=3, p<1000, p = NextPrime[p], If[p==163 || KroneckerSymbol[-163, p] == 1, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Apr 29 2019 *)
Least p for which min { x >= 0 | p + (2n+1)*x + x^2 is composite } reaches the (local) maximum given in A273770.
+10
4
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, 73303, 73361, 73421, 73483, 3443897, 3071069, 3071137, 15949847, 76553693, 365462323, 365462399, 2204597, 9721, 1842719, 246407633, 246407719, 246407807, 246407897, 246407989
COMMENTS
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.)
PROG
(PARI) A273756(n, p=2*n+1, L=10^(5+n\10), m=0, Q)={forprime(q=1, L, for(x=1, oo, ispseudoprime(q+p*x+x^2)&& next; x>m&& [Q=q, m=x]; break)); Q}
EXTENSIONS
a(27) corrected and more terms from Don Reble, Feb 15 2018
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