Displaying 1-10 of 18 results found.
Odd primes generated recursively: a(1) = 3, a(n) = Min {p is prime; p divides Q+2}, where Q is the product of previous terms in the sequence.
+10
20
3, 5, 17, 257, 65537, 641, 7, 318811, 19, 1747, 12791, 73, 90679, 67, 59, 113, 13, 41, 47, 151, 131, 1301297155768795368671, 20921, 1514878040967313829436066877903, 5514151389810781513, 283, 1063, 3027041, 29, 24040758847310589568111822987, 154351, 89
COMMENTS
The first five terms comprise the known Fermat primes: A019434.
EXAMPLE
a(7) = 7 is the smallest prime divisor of 3 * 5 * 17 * 257 * 65537 * 641 + 2 = 2753074036097 = 7 * 11 * 37 * 966329953.
MATHEMATICA
a={3}; q=1;
For[n=2, n<=20, n++,
q=q*Last[a];
AppendTo[a, Min[FactorInteger[q+2][[All, 1]]]];
];
Primes of the form 8*k + 3 generated recursively. Initial prime is 3. General term is a(n) = Min_{p is prime; p divides 2 + Q^2; p == 3 (mod 8)}, where Q is the product of previous terms in the sequence.
+10
19
3, 11, 1091, 1296216011, 2177870960662059587828905091, 76870667, 19, 257680660619, 73677606898727076965233531, 23842300525435506904690028531941969449780447746432390747, 35164737203
COMMENTS
2+Q^2 always has a prime divisor congruent to 3 modulo 8.
REFERENCES
D. M. Burton, Elementary Number Theory, McGraw-Hill, Sixth Edition (2007), p. 191.
EXAMPLE
a(3) = 1091 is the smallest prime divisor congruent to 3 mod 8 of 2+Q^2 = 1091, where Q = 3 * 11.
MATHEMATICA
a = {3}; q = 1;
For[n = 2, n ≤ 5, n++,
q = q*Last[a];
AppendTo[a, Min[Select[FactorInteger[2 + q^2][[All, 1]], Mod[#,
8] \[Equal] 3 &]]];
];
PROG
(PARI) lista(nn) = my(f, q=3); print1(q); for(n=2, nn, f=factor(2+q^2)[, 1]~; for(i=1, #f, if(f[i]%8==3, print1(", ", f[i]); q*=f[i]; break))); \\ Jinyuan Wang, Aug 05 2022
Primes of the form 22k+1 generated recursively. Initial prime is 23. General term is a(n) = Min {p is prime; p divides (R^11 - 1)/(R - 1); p == 1 (mod 11)}, where Q is the product of previous terms in the sequence and R = 11*Q.
+10
18
23, 4847239, 2971, 3936923, 9461, 1453, 331, 81373909, 89, 920771904664817214817542307, 353, 401743, 17088192002665532981, 11617
COMMENTS
All prime divisors of (R^11 - 1)/(R - 1) different from 11 are congruent to 1 modulo 22.
REFERENCES
M. Ram Murty, Problems in Analytic Number Theory, Springer-Verlag, NY, (2001), pp. 208-209.
EXAMPLE
a(3) = 2971 is the smallest prime divisor congruent to 1 mod 22 of (R^11-1)/(R-1) =
7693953366218628230903493622259922359469805176129784863956847906415055607909988155588181877
= 2971 * 357405886421 * 914268562437006833738317047149 * 7925221522553970071463867283158786415606996703, where Q = 23 * 4847239, and R = 11*Q.
MATHEMATICA
a={23}; q=1;
For[n=2, n<=2, n++,
q=q*Last[a]; r=11*q;
AppendTo[a, Min[Select[FactorInteger[(r^11-1)/(r-1)][[All, 1]], Mod[#, 11]==1 &]]];
];
Primes of the form 12k+7 generated recursively. Initial prime is 7. General term is a(n)=Min {p is prime; p divides 3+4Q^2; Mod[p,12]=7}, where Q is the product of previous terms in the sequence.
+10
2
7, 199, 7761799, 487, 67, 103, 1482549740515442455520791, 31, 139, 787, 19, 39266047, 1955959, 50650885759, 367, 185767, 62168707
COMMENTS
All prime divisors of 3+4Q^2 are congruent to 1 modulo 6.
At least one prime divisor of 3+4Q^2 is congruent to 3 modulo 4 and hence to 7 modulo 12.
The first six terms are the same as those of A057204.
EXAMPLE
a(3) = 1482549740515442455520791 is the smallest prime divisor congruent to 7 mod 12 of 3+4Q^2 = 5281642303363312989311974746340327 = 3562539697 * 1482549740515442455520791, where Q = 7 * 199 * 7761799 * 487 * 67 * 103.
MATHEMATICA
a={7}; q=1;
For[n=2, n<=7, n++,
q=q*Last[a];
AppendTo[a, Min[Select[FactorInteger[4*q^2+3][[All, 1]], Mod[#, 12]==7 &]]];
];
Primes of the form 10*k + 9 generated recursively. Initial prime is 19. General term is a(n) = Min_{p is prime; p divides 100*Q^2 - 5; p == 9 (mod 10)}, where Q is the product of previous terms in the sequence.
+10
2
19, 7219, 462739, 509, 129229, 295380580489, 9653956849, 149, 110212292237172705230749846071050188009093377022084806290042881946231583507557298889, 157881589, 60397967745386189, 1429, 79
COMMENTS
100Q^2-5 always has a prime divisor congruent to 9 modulo 10.
EXAMPLE
a(3) = 462739 is the smallest prime divisor congruent to 9 mod 10 of 100Q^2-5 = 1881313992095 = 5 * 462739 * 813121, where Q = 19 * 7219.
MATHEMATICA
a={19}; q=1;
For[n=2, n<=6, n++,
q=q*Last[a];
AppendTo[a, Min[Select[FactorInteger[100*q^2-5][[All, 1]], Mod[#, 10]==9&]]];
];
Primes of the form 12k+1 generated recursively. Initial prime is 13. General term is a(n)=Min {p is prime; p divides Q^4-Q^2+1}, where Q is the product of previous terms in the sequence.
+10
2
13, 28393, 128758492789, 73, 193, 37, 457, 8363172060732903211423577787181
COMMENTS
All prime divisors of Q^4 - Q^2 + 1 are congruent to 1 modulo 12.
REFERENCES
K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, NY, Second Edition (1990), p. 63.
EXAMPLE
a(3) = 128758492789 is the smallest prime divisor of Q^4 - Q^2 + 1 = 18561733755472408508281 = 128758492789 * 144159296629, where Q = 13 * 28393.
MATHEMATICA
a = {13}; q = 1;
For[n = 2, n ≤ 8, n++,
q = q*Last[a];
AppendTo[a, Min[Select[FactorInteger[q^4 - q^2 + 1][[All, 1]],
Mod[#, 12] == 1 &]]];
];
Primes of the form 8k+1 generated recursively. Initial prime is 17. General term is a(n)=Min {p is prime; p divides (2Q)^4 + 1}, where Q is the product of previous terms in the sequence.
+10
2
17, 1336337, 4261668267710686591310687815697, 41, 4390937134822286389262585915435960722186022220433, 241, 1553, 243537789182873, 97, 27673, 4289, 457, 137201, 73, 337, 569891669978849, 617, 1697, 65089, 1609, 761
COMMENTS
All prime divisors of (2Q)^4 + 1 are congruent to 1 modulo 8.
REFERENCES
G. A. Jones and J. M. Jones, Elementary Number Theory, Springer-Verlag, NY, (1998), p. 271.
EXAMPLE
a(3) = 4261668267710686591310687815697 is the smallest prime divisor of (2Q)^4 + 1 = 4261668267710686591310687815697, where Q = 17 * 1336337.
Primes of the form 12*k + 5 generated recursively. Initial prime is 5. General term is a(n) = Min_{p is prime; p divides 1 + 4*Q^2; p == 5 (mod 12)}, where Q is the product of previous terms in the sequence.
+10
1
5, 101, 1020101, 53, 29, 2507707213238852620996901, 449, 433361, 401, 925177698346131180901394980203075088053316845914981, 44876921, 17, 173
COMMENTS
All prime divisors of 1+4Q^2 are congruent to 1 modulo 4.
At least one prime divisor of 1+4Q^2 is congruent to 2 modulo 3 and hence to 5 modulo 12.
The first seven terms are the same as those of A057207. The next term is known but is too large to include.
EXAMPLE
a(8) = 433361 is the smallest prime divisor congruent to 5 mod 12 of 1+4Q^2 = 3179238942812523869898723304484664524974766291591037769022962819805514576256901 = 13 * 433361 * 42408853 * 2272998442375593325550634821 * 5854291291251561948836681114631909089, where Q = 5 * 101 * 1020101 * 53 * 29 * 2507707213238852620996901 * 449.
MATHEMATICA
a={5}; q=1;
For[n=2, n<=5, n++,
q=q*Last[a];
AppendTo[a, Min[Select[FactorInteger[4*q^2+1][[All, 1]], Mod[#, 12]==5 &]]];
];
Primes of the form 12k+5 generated recursively. Initial prime is 5. General term is a(n) = Min {p is prime; p divides 4+Q^2; p == 5 (mod 12)}, where Q is the product of previous terms in the sequence.
+10
1
5, 29, 17, 6076229, 1289, 78067083126343039013, 521, 8606045503613, 15837917, 1873731749, 809, 137, 2237, 17729
COMMENTS
Since Q is odd, all prime divisors of 4+Q^2 are congruent to 1 modulo 4.
At least one prime divisor of 4+Q^2 is congruent to 2 modulo 3 and hence to 5 modulo 12.
The first two terms are the same as those of A057208.
EXAMPLE
a(3) = 17 is the smallest prime divisor congruent to 5 mod 12 of 4+Q^2 = 21029 = 17 * 1237, where Q = 5 * 29.
MATHEMATICA
a={5}; q=1;
For[n=2, n<=5, n++,
q=q*Last[a];
AppendTo[a, Min[Select[FactorInteger[q^2+4][[All, 1]], Mod[#, 12]==5 &]]];
];
Primes of the form 10k+1 generated recursively. Initial prime is 11. General term is a(n)=Min {p is prime; p divides (R^5 - 1)/(R - 1); Mod[p,5]=1}, where Q is the product of previous terms in the sequence and R = 5Q.
+10
1
11, 211, 1031, 22741, 41, 15487770335331184216023237599647357572461782407557681, 311, 61, 55172461, 3541, 1381, 2851, 19841, 151, 9033671, 456802301, 1720715817015281, 19001, 71
COMMENTS
All prime divisors of (R^5 - 1)/(R - 1) different from 5 are congruent to 1 modulo 10.
REFERENCES
M. Ram Murty, Problems in Analytic Number Theory, Springer-Verlag, NY, (2001), pp. 208-209.
EXAMPLE
a(3) = 1031 is the smallest prime divisor congruent to 1 mod 10 of (R^5 - 1)/(R - 1) = 18139194759758381 = 1031 * 17593787351851, where Q = 11 * 211 and R = 5Q.
MATHEMATICA
a={11}; q=1;
For[n=2, n<=6, n++,
q=q*Last[a]; r=5*q;
AppendTo[a, Min[Select[FactorInteger[(r^5-1)/(r-1)][[All, 1]], Mod[#, 10]==1&]]];
];
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