A103795 -id:A103795 - cp's OEIS frontend

A085398 Let Cn(x) be the n-th cyclotomic polynomial; a(n) is the least k>1 such that Cn(k) is prime.

3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 6, 2, 4, 3, 2, 10, 2, 22, 2, 2, 4, 6, 2, 2, 2, 2, 2, 14, 3, 61, 2, 10, 2, 14, 2, 15, 25, 11, 2, 5, 5, 2, 6, 30, 11, 24, 7, 7, 2, 5, 7, 19, 3, 2, 2, 3, 30, 2, 9, 46, 85, 2, 3, 3, 3, 11, 16, 59, 7, 2, 2, 22, 2, 21, 61, 41, 7, 2, 2, 8, 5, 2, 2

Offset: 1

Don Reble, Jun 28 2003

nonn

Conjecture: a(n) is defined for all n. - Eric Chen, Nov 14 2014

Existence of a(n) is implied by Bunyakovsky's conjecture. - Robert Israel, Nov 13 2014

			a(11) = 5 because C11(k) is composite for k = 2, 3, 4 and prime for k = 5.
a(37) = 61 because C37(k) is composite for k = 2, 3, 4, ..., 60 and prime for k = 61.

Jinyuan Wang, Table of n, a(n) for n = 1..5000 (terms 1..1500 from Eric Chen)
Wikipedia, Bunyakowsky conjecture

Cf. A117544, A066180, A085399, A103795, A056993, A153438, A246119, A246120, A246121, A206418, A205506, A181980.

Cf. A008864, A006093, A002384, A005574, A049409, A055494, A100330, A000068, A153439, A246392, A162862, A246397, A217070, A006314, A217071, A164989, A217072, A217073, A153440, A217074, A217075, A006313, A097475.

f:= proc(n) local k;
for k from 2 do if isprime(numtheory:-cyclotomic(n,k)) then return k fi od
end proc:
seq(f(n), n = 1 .. 100); # Robert Israel, Nov 13 2014

Table[k = 2; While[!PrimeQ[Cyclotomic[n, k]], k++]; k, {n, 300}] (* Eric Chen, Nov 14 2014 *)

a(n) = k=2; while(!isprime(polcyclo(n, k)), k++); k; \\ Michel Marcus, Nov 13 2014

a(A072226(n)) = 2. - Eric Chen, Nov 14 2014
a(n) = A117544(n) except when n is a prime power, since if n is a prime power, then A117544(n) = 1. - Eric Chen, Nov 14 2014
a(prime(n)) = A066180(n), a(2*prime(n)) = A103795(n), a(2^n) = A056993(n-1), a(3^n) = A153438(n-1), a(2*3^n) = A246120(n-1), a(3*2^n) = A246119(n-1), a(6^n) = A246121(n-1), a(5^n) = A206418(n-1), a(6*A003586(n)) = A205506(n), a(10*A003592(n)) = A181980(n).

A084742 Least k such that (n^k+1)/(n+1) is prime, or 0 if no such prime exists.

Original entry on oeis.org

3, 3, 3, 5, 3, 3, 0, 3, 5, 5, 5, 3, 7, 3, 3, 7, 3, 17, 5, 3, 3, 11, 7, 3, 11, 0, 3, 7, 139, 109, 0, 5, 3, 11, 31, 5, 5, 3, 53, 17, 3, 5, 7, 103, 7, 5, 5, 7, 1153, 3, 7, 21943, 7, 3, 37, 53, 3, 17, 3, 7, 11, 3, 0, 19, 7, 3, 757, 11, 3, 5, 3, 7, 13, 5, 3, 37, 3, 3, 5, 3, 293, 19, 7, 167, 7, 7, 709, 13, 3, 3, 37, 89, 71, 43, 37

Offset: 2

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jun 15 2003

nonn

When (n^k+1)/(n+1) is prime, k must be prime. As mentioned by Dubner and Granlund, when n is a perfect power (the power is greater than 2), then (n^k+1)/(n+1) will usually be composite for all k, which is the case for n = 8, 27, 32 and 64. a(n) are only probable primes for n = {53, 124, 150, 182, 205, 222, 296}.
a(n) = 0 if n = {8, 27, 32, 64, 125, 243, ...}. - Eric Chen, Nov 18 2014
More terms: a(124) = 16427, a(150) = 6883, a(182) = 1487, a(205) = 5449, a(222) = 1657, a(296) = 1303. For n up to 300, a(n) is currently unknown only for n = {97, 103, 113, 175, 186, 187, 188, 220, 284}. All other terms up to a(300) are less than 1000. - Eric Chen, Nov 18 2014
a(97) > 31000. - Eric Chen, Nov 18 2014
a(311) = 2707, a(313) = 4451. - Eric Chen, Nov 20 2014
a(n)=3 if and only if n^2-n+1 is a prime; that is, n belongs to A055494. - Thomas Ordowski, Sep 19 2015
From Altug Alkan, Sep 29 2015: (Start)
a(n)=5 if and only if Phi(10, n) is prime and Phi(6, n) is composite. n belongs to A246392.
a(n)=7 if and only if Phi(14, n) is prime, and Phi(10, n) and Phi(6, n) are both composite. n belongs to A250174.
a(n)=11 if and only if Phi(22, n) is prime, and Phi(14, n), Phi(10, n) and Phi(6, n) are all composite. n belongs to A250178.
Where Phi(k, n) is the k-th cyclotomic polynomial. (End)
a(97) > 800000 (or a(97) = 0). - Wang Runsen, May 10 2023

			a(5) = 5 as (5^5 + 1)/(5 + 1) = 1 - 5 + 5^2 - 5^3 + 5^4 = 521 is a prime.
a(7) = 3 as (7^3 + 1)/(7 + 1) = 1 - 7 + 7^2 = 43 is a prime.

Eric Chen, Table of known a(n) up to a(300)
H. Dubner and T. Granlund, Primes of the Form (b^n+1)/(b+1), J. Integer Sequences, 3 (2000), #P00.2.7.
Eric Weisstein's World of Mathematics, Repunit
Robert G. Wilson v, Letter to N. J. A. Sloane, circa 1991.

Cf. A084741, A065507, A084740, A128164, A126659, A103795.

a(n) = {l=List([8, 27, 32, 64, 125, 243, 324, 343]); for(q=1, #l, if(n==l[q], return(0))); k=2; while(k, s=(n^prime(k)+1)/(n+1); if(ispseudoprime(s), return(prime(k))); k++)}
n=2; while(n<361, print1(a(n), ", "); n++) \\ Eric Chen, Nov 25 2014

More terms from T. D. Noe, Jan 22 2004

A084740 Least k such that (n^k-1)/(n-1) is prime, or 0 if no such prime exists.

Original entry on oeis.org

2, 3, 2, 3, 2, 5, 3, 0, 2, 17, 2, 5, 3, 3, 2, 3, 2, 19, 3, 3, 2, 5, 3, 0, 7, 3, 2, 5, 2, 7, 0, 3, 13, 313, 2, 13, 3, 349, 2, 3, 2, 5, 5, 19, 2, 127, 19, 0, 3, 4229, 2, 11, 3, 17, 7, 3, 2, 3, 2, 7, 3, 5, 0, 19, 2, 19, 5, 3, 2, 3, 2, 5, 5, 3, 41, 3, 2, 5, 3, 0, 2, 5, 17, 5, 11, 7, 2, 3, 3, 4421, 439, 7, 5, 7, 2, 17, 13, 3, 2, 3, 2, 19, 97, 3, 2, 17, 2, 17, 3, 3, 2, 23, 29, 7, 59

Offset: 2

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jun 15 2003

nonn

When (n^k-1)/(n-1) is prime, k must be prime. As mentioned by Dubner, when n is a perfect power, then (n^k-1)/(n-1) will usually be composite for all k, which is the case for n = 9, 25, 32, 49, 64, 81, 121, 125, 144, 169, 216, 225, 243, 289, 324, 343, ... - T. D. Noe, Jan 30 2004
a(152) > prime(1100) or 0. - Derek Orr, Nov 29 2014
a(n)=2 if and only if n=p-1, where p is an odd prime; that is, n belongs to A006093, except 2. - Thomas Ordowski, Sep 19 2015
Probably a(152) = 270217 since (152^270217-1)/(152-1) has been shown to be probably prime. - Michael Stocker, Jan 24 2019

			a(7) = 5 as (7^5 - 1 )/(7 - 1) = 2801 = 1 + 7 + 7^2 + 7^3 + 7^4 is a prime but no smaller partial sum yields a prime.

Eric Chen, Table of known a(n) up to a(360) [with a(316) corrected by Jon E. Schoenfield]
H. Dubner, Generalized repunit primes, Math. Comp., 61 (1993), 927-930.
Andy Steward, Titanic Prime Generalized Repunits
Eric Weisstein's World of Mathematics, Repunit
Robert G. Wilson v, Letter to N. J. A. Sloane, circa 1991.
Robert G. Wilson v, Table of known a(n) from n = 2 to 1000.

Cf. A065507, A065854, A084738, A084742, A096059, A126659, A128164, A246005.

Cf. A066180, A085398, A103795, A123487, A123627.

a(n) = {l=List([9, 25, 32, 49, 64, 81, 121, 125, 144, 169, 216, 225, 243, 289, 324, 343]); for(q=1, #l, if(n==l[q], return(0))); k=1; while(k, s=(n^prime(k)-1)/(n-1); if(ispseudoprime(s), return(prime(k))); k++)}
n=2; while(n<361, print1(a(n), ", "); n++) \\ Derek Orr, Jul 13 2014

More terms from T. D. Noe, Jan 23 2004

A123627 Smallest prime q such that (q^p+1)/(q+1) is prime, where p = prime(n); or 0 if no such prime q exists.

Original entry on oeis.org

0, 2, 2, 2, 2, 2, 2, 2, 2, 7, 2, 19, 61, 2, 7, 839, 89, 2, 5, 409, 571, 2, 809, 227, 317, 2, 5, 79, 23, 4073, 2, 281, 89, 739, 1427, 727, 19, 19, 2, 281, 11, 2143, 2, 1013, 4259, 2, 661, 1879, 401, 5, 4099, 1579, 137, 43, 487, 307, 547, 1709, 43, 3, 463, 2161, 353, 443, 2

Offset: 1

Alexander Adamchuk, Oct 04 2006, Aug 05 2008

nonn

a(1) = 0 because such a prime does not exist, Mod[n^2+1,n+1] = 2 for n>1.
Corresponding primes (q^p+1)/(q+1), where prime q = a(n) and p = Prime[n], are listed in A123628[n] = {1,3,11,43,683,2731,43691,174763,2796203,402488219476647465854701,715827883,...}.
a(n) coincides with A103795[n] when A103795[n] is prime.
a(n) = 2 for n = PrimePi[A000978[k]] = {2,3,4,5,6,7,8,9,11,14,18,22,26,31,39,43,46,65,69,126,267,380,495,762,1285,1304,1364,1479,1697,4469,8135,9193,11065,11902,12923,13103,23396,23642,31850,...}.
Corresponding primes of the form (2^p + 1)/3 are the Wagstaff primes that are listed in A000979[n] = {3,11,43,683,2731,43691,174763,2796203,715827883,...}.

Robert Israel, Table of n, a(n) for n = 1..240

Cf. A123628, A103795, A123487, A123488, A000978, A000979.

f:= proc(n) local p,q;
  p:= ithprime(n);
  q:= 1;
  do
   q:= nextprime(q);
   if isprime((q^p+1)/(q+1)) then return q fi
  od
end proc:
f(1):= 0:
map(f, [$1..70]); # Robert Israel, Jul 31 2019

a(1) = 0, for n>1 Do[p=Prime[k]; n=1; q=Prime[n]; cp=(q^p+1)/(q+1); While[ !PrimeQ[cp], n=n+1; q=Prime[n]; cp=(q^p+1)/(q+1)]; Print[q], {k, 2, 61}]
Do[p=Prime[k]; n=1; q=Prime[n]; cp=(q^p+1)/(q+1); While[ !PrimeQ[cp], n=n+1; q=Prime[n]; cp=(q^p+1)/(q+1)]; Print[{k,q}], {k, 1, 134}]
spq[n_]:=Module[{p=Prime[n],q=2},While[!PrimeQ[(q^p+1)/(q+1)],q=NextPrime[ q]]; q]; Join[{0},Array[spq,70,2]] (* Harvey P. Dale, Mar 23 2019 *)

A123628(n) = (a(n)^prime(n) + 1) / (a(n) + 1).

A123628 Smallest prime of the form (q^p+1)/(q+1), where p = prime(n) and q is also prime (q = A123627(n)); or 1 if such a prime does not exist.

Original entry on oeis.org

1, 3, 11, 43, 683, 2731, 43691, 174763, 2796203, 402488219476647465854701, 715827883, 10300379826060720504760427912621791994517454717, 254760179343040585394724919772965278539769280548173566545431025735121201

Offset: 1

Alexander Adamchuk, Oct 03 2006

nonn

a(1) = 1 because such a prime does not exist; (n^2+1) mod (n+1) = 2 for n > 1. a(n) = (A103795(n)^prime(n)+1)/(A103795(n)+1) when A103795(n) is prime. Corresponding smallest primes q such that (q^p+1)/(q+1) is prime, where p = prime(n), are listed in A123627(n) = {0, 2, 2, 2, 2, 2, 2, 2, 2, 7, 2, 19, 61, 2, 7, 839, 1459, 2, 5, 409, 571, 2, ...}. All Wagstaff primes or primes of form (2^p + 1)/3 belong to a(n). Wagstaff primes are listed in A000979(n) = {3, 11, 43, 683, 2731, 43691, 174763, 2796203, 715827883, ...}. Corresponding indices n such that a(n) = (2^prime(n) + 1)/3 are PrimePi(A000978(n)) = {2, 3, 4, 5, 6, 7, 8, 9, 11, 14, 18, 22, 26, 31, 39, 43, 46, 65, 69, 126, 267, 380, 495, 762, 1285, 1304, 1364, 1479, 1697, 4469, 8135, 9193, 11065, 11902, 12923, 13103, 23396, 23642, 31850, ...}. All primes with prime indices in the Jacobsthal sequence A001045(n) belong to a(n).

Cf. A123627, A103795, A123487, A123488, A000978, A000979, A001045, A049883, A107036.

a(n) = (A123627(n)^prime(n) + 1) / (A123627(n) + 1).

A103794 Smallest number b such that b^prime(n) - (b-1)^prime(n) is prime.

Original entry on oeis.org

2, 2, 2, 2, 6, 2, 2, 2, 6, 3, 2, 40, 7, 5, 13, 3, 3, 2, 7, 18, 47, 8, 6, 2, 26, 3, 42, 2, 13, 8, 2, 8, 328, 8, 9, 45, 27, 13, 76, 15, 52, 111, 5, 15, 50, 287, 16, 5, 40, 23, 110, 368, 23, 68, 28, 96, 81, 150, 3, 143, 4, 12, 403, 4, 45, 11, 83, 21, 96, 5, 109, 350, 128, 304, 38, 4, 163

Offset: 1

Lei Zhou, Feb 24 2005

nonn

Conjecture: sequence is defined for all positive indices.

For p=prime(n), Eisenstein's irreducibility criterion can be used to show that the polynomial (x+1)^p-x^p is irreducible, which is a necessary (but not sufficient) condition for a(n) to exist. - T. D. Noe, Dec 05 2005

			2^prime(1)-1^prime(1)=3 is prime, so a(1)=2;
2^prime(5)-1^prime(5)=2047 has a factor of 23;
...
6^prime(5)-5^prime(5)=313968931 is prime, so a(5)=6;

Cf. A103795, A066180, A058013, A222119.

f:= proc(n) local p,b;
  p:= ithprime(n);
  for b from 2 do
    if isprime(b^p - (b-1)^p) then return b fi
  od
end proc:
map(f, [$1..80]); # Robert Israel, Jun 04 2024

Do[p=Prime[k]; n=2; nm1=n-1; cp=n^p-nm1^p; While[ !PrimeQ[cp], n=n+1; nm1=n-1; cp=n^p-nm1^p]; Print[n], {k, 1, 200}]

a(n) = A222119(n) + 1. - Ray Chandler, Feb 26 2017

A237114 Smallest semiprime of the form k^prime(n)+1, or 0 if no such semiprime exists.

Original entry on oeis.org

10, 9, 33, 129, 2049, 8193, 131073, 524289, 8388609, 21214052113249267732127817825945098816023915043832462900000000000000000000000000001, 2147483649, 356811923176489970264571492362373784095686657, 1821119122882338858450163704901509732674059569636703920027007853793548503164173361298060584748698304513

Offset: 1

Jonathan Sondow, Feb 04 2014

nonn

For n > 1, smallest number k^p+1 with both (k^p+1)/(k+1) and k+1 prime, where p = prime(n); the corresponding primes (k^p+1)/(k+1) for n > 1 are A237116(n) = 3, 11, 43, 683, 2731, 43691, 174763, 2796203, ... and the corresponding primes k+1 are A237115(n) = 3, 3, 3, 3, 3, 3, 3, 3, 691, 3, 17, ... .
a(n) == its smaller prime factor A237115(n) (mod prime(n)). Proof: 10 == 2 (mod 2), so true for n=1. For n>1, true by Fermat's little theorem: k^p+1 == k+1 (mod p).
a(n) is in A006881 (squarefree semiprimes), except for a(2) = 9 = 3^2. Proof: True for n=1. For n>1, if k^p+1 = (k+1)^2, then k^(p-1) = k+2, so k*(k^(p-2)-1) = 2. Now k>1 implies k=2 and p=3, so that n=2.
It appears that a(n) mod p > 0 for all n > 2 (see A237117), where p = prime(n). If true, then the larger prime factor A237116(n) of a(n) is == 1 (mod p), since a(n) == its smaller prime factor (mod p).

			Prime(1)=2 and the smallest semiprime of the form k^2+1 is a(1) = 3^2+1 = 10 = 2*5.
Prime(2)=3 and the smallest semiprime of the form k^3+1 is a(2) = 2^3+1 = 9 = 3*3.

Eric Weisstein's World of Mathematics, Semiprime
Wikipedia, Semiprime

Cf. A001358, A006881, A103795, A123627, A123628, A237040, A237115, A237116, A237117.

L = {10}; Do[p = Prime[k]; n = 1; q = Prime[n] - 1; cp = (q^p + 1)/(q + 1); While[! PrimeQ[cp], n = n + 1; q = Prime[n] - 1; cp = (q^p + 1)/(q + 1)]; L = Append[L, q^p + 1], {k, 2, 12}]; L

a(n) = A237115(n)*A237116(n), for n > 0.
a(n) = (A237115(n)-1)^prime(n)+1, for n > 1.
a(n) == A237115(n) (mod prime(n)), for n > 0.
a(n) mod prime(n) = A237117(n), if a(n) > 0.

A237115 Lesser prime factor of the smallest semiprime of the form k^prime(n)+1, or 0 if no such semiprime exists.

Original entry on oeis.org

2, 3, 3, 3, 3, 3, 3, 3, 3, 691, 3, 17, 313, 3, 7, 11, 7, 3, 11, 47, 19, 3, 1499, 17, 71, 3, 97, 7, 13, 823, 3, 97, 1163, 31, 17, 199, 1907, 53, 3, 17, 1231, 1013, 3, 13, 53, 3, 67, 47, 23, 1013, 787, 127, 347, 17, 37, 97, 683, 631, 73, 4549, 173, 11, 17, 1039, 3, 17, 47, 6389, 3, 461, 23, 673, 37, 29, 331, 7451, 1433, 4561

Offset: 1

Jonathan Sondow, Feb 04 2014

nonn

For n > 1, smallest prime p such that ((p-1)^prime(n)+1)/p is prime; the corresponding primes ((p-1)^prime(n)+1)/p are A237116(n) = 3, 11, 43, 683, 2731, 43691, 174763, 2796203, ... and the corresponding semiprimes (p-1)^prime(n)+1 are A237114(n) = 9, 33, 129, 2049, 8193, 131073, 524289, 8388609, ... .

			Prime(1)=2 and the smallest semiprime of the form k^2+1 is 3^2+1 = 10 = 2*5, so a(1) = 2.
Prime(2)=3 and the smallest semiprime of the form k^3+1 is 2^3+1 = 9 = 3*3, so a(2) = 3.

Eric Weisstein's World of Mathematics, Semiprime
Wikipedia, Semiprime

Cf. A001358, A103795, A123627, A123628, A237040, A237114, A237116, A237117.

L = {2}; Do[p = Prime[k]; n = 1; q = Prime[n] - 1; cp = (q^p + 1)/(q + 1); While[! PrimeQ[cp], n = n + 1; q = Prime[n] - 1; cp = (q^p + 1)/(q + 1)]; L = Append[L, q + 1], {k, 2, 78}]; L

a(n) = A237114(n)/A237116(n), for n > 0.
(a(n)-1)^prime(n) = A237114(n)-1, for n > 1.
a(n) == A237114(n) (mod prime(n)) (for a proof, see A237114).
a(n) mod prime(n) = A237117(n), if a(n) > 0.

A237116 Larger prime factor of the smallest semiprime of the form k^prime(n)+1, or 0 if no such semiprime exists.

Original entry on oeis.org

5, 3, 11, 43, 683, 2731, 43691, 174763, 2796203, 30700509570548867919143006984001590182379037690061451374819102749638205499276411, 715827883, 20988936657440586486151264256610222593863921, 5818271958090539483866337715340286685859615238455923067178938830011337070812055467405944360219483401

Offset: 1

Jonathan Sondow, Feb 05 2014

nonn

For n > 1, smallest prime of the form ((p-1)^prime(n)+1)/p, where p is prime; the corresponding primes p are A237115(n) = 3, 3, 3, 3, 3, 3, 3, 3, 691, 3, 17, ... and the corresponding semiprimes (p-1)^prime(n)+1 are A237114(n) = 9, 33, 129, 2049, 8193, 131073, 524289, 8388609, ... .

It appears that a(n) == 1 (mod prime(n)), for all n <> 2. See 4th comment in A237114.

			Prime(1)=2 and the smallest semiprime of the form k^2+1 is 3^2+1 = 10 = 2*5, so a(1) = 5.
Prime(2)=3 and the smallest semiprime of the form k^3+1 is 2^3+1 = 9 = 3*3, so a(2) = 3.

Eric Weisstein's World of Mathematics, Semiprime
Wikipedia, Semiprime

Cf. A001358, A103795, A123627, A123628, A237040, A237114, A237115.

L = {5}; Do[p = Prime[k]; n = 1; q = Prime[n] - 1; cp = (q^p + 1)/(q + 1); While[! PrimeQ[cp], n = n + 1; q = Prime[n] - 1; cp = (q^p + 1)/(q + 1)]; L = Append[L, cp], {k, 2, 13}]; L

a(n) = A237114(n)/A237115(n), for n > 0.

a(n) = ((A237115(n)-1)^prime(n)+1)/A237115(n), for n > 1.

A237117 Remainder mod p of the smallest semiprime of the form k^p+1, where p = prime(n); or -1 if no such semiprime exists.

Original entry on oeis.org

0, 0, 3, 3, 3, 3, 3, 3, 3, 24, 3, 17, 26, 3, 7, 11, 7, 3, 11, 47, 19, 3, 5, 17, 71, 3, 97, 7, 13, 32, 3, 97, 67, 31, 17, 48, 23, 53, 3, 17, 157, 108, 3, 13, 53, 3, 67, 47, 23, 97, 88, 127, 106, 17, 37, 97, 145, 89, 73, 53, 173, 11, 17, 106, 3, 17, 47, 323, 3, 112, 23, 314, 37, 29, 331, 174, 266, 194, 226, 397, 29, 16, 176, 45, 44, 152, 373, 349, 101, 143, 53, 386, 133, 29, 345, 1

Offset: 1

Jonathan Sondow, Feb 06 2014

nonn

It appears that a(n) > 0 for all n > 2. See the comments in A237114.

			Prime(2)=3 and the smallest semiprime of the form k^3+1 is 2^3+1 = 9 = 3*3, so a(2) = 9 mod 3 = 0.
Prime(3)=5 and the smallest semiprime of the form k^5+1 is 2^5+1 = 33 = 3*11, so a(3) = 33 mod 5 = 3.

Eric Weisstein's World of Mathematics, Semiprime
Wikipedia, Semiprime

Cf. A001358, A006881, A103795, A123627, A123628, A237040, A237114, A237115, A237116.

L = {0}; Do[p = Prime[k]; n = 1; q = Prime[n] - 1; cp = (q^p + 1)/(q + 1); While[! PrimeQ[cp], n = n + 1; q = Prime[n] - 1; cp = (q^p + 1)/(q + 1)]; L = Append[L, Mod[q^p + 1, p]], {k, 2, 87}]; L

a(n) = A237114(n) mod prime(n) = A237115(n) mod prime(n), if A237114(n)>0.

cp's OEIS Frontend

A085398 Let Cn(x) be the n-th cyclotomic polynomial; a(n) is the least k>1 such that Cn(k) is prime.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A084742 Least k such that (n^k+1)/(n+1) is prime, or 0 if no such prime exists.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Extensions

A084740 Least k such that (n^k-1)/(n-1) is prime, or 0 if no such prime exists.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Extensions

A123627 Smallest prime q such that (q^p+1)/(q+1) is prime, where p = prime(n); or 0 if no such prime q exists.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A123628 Smallest prime of the form (q^p+1)/(q+1), where p = prime(n) and q is also prime (q = A123627(n)); or 1 if such a prime does not exist.

Views

Author

Keywords

Comments

Crossrefs

Formula

A103794 Smallest number b such that b^prime(n) - (b-1)^prime(n) is prime.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A237114 Smallest semiprime of the form k^prime(n)+1, or 0 if no such semiprime exists.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A237115 Lesser prime factor of the smallest semiprime of the form k^prime(n)+1, or 0 if no such semiprime exists.

Views

Author

Keywords