cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

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

Views

Author

Alexander Adamchuk, Oct 03 2006

Keywords

Comments

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).

Crossrefs

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

Formula

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

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

Views

Author

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

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

Cf. A065507, A065854, A084738, A084742, A096059, A126659, A128164, A246005.
Cf. A066180, A085398, A103795, A123487, A123627.

Programs

  • PARI
    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

Extensions

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

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

Views

Author

Jonathan Sondow, Feb 04 2014

Keywords

Comments

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).

Examples

			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.

Links

Crossrefs

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

Programs

  • Mathematica
    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

Formula

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

Views

Author

Jonathan Sondow, Feb 04 2014

Keywords

Comments

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, ... .

Examples

			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.

Links

Crossrefs

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

Programs

  • Mathematica
    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

Formula

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

Views

Author

Jonathan Sondow, Feb 05 2014

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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

Programs

  • Mathematica
    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

Formula

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

Views

Author

Jonathan Sondow, Feb 06 2014

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

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

Programs

  • Mathematica
    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

Formula

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

A252503 Smallest prime p such that Phi_n(p) is also prime, where Phi is the cyclotomic polynomial, or 0 if no such p exists.

Original entry on oeis.org

3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 7, 2, 11, 3, 2, 113, 2, 43, 2, 2, 5, 151, 2, 2, 2, 2, 2, 179, 3, 61, 2, 23, 2, 53, 2, 89, 137, 11, 2, 5, 5, 2, 7, 73, 11, 307, 7, 7, 2, 5, 7, 19, 3, 2, 2, 3, 0, 2, 53, 491, 197, 2, 3, 3, 3, 11, 19, 59, 7, 2, 2, 271, 2, 191, 61, 41, 7, 2, 2, 59, 5, 2, 2
Offset: 1

Views

Author

Eric Chen, Dec 18 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n != 2^k (k>5).
Clearly, if n is a power of 2, and Phi_n(2) is not prime, then a(n) = 0.
Records: 3, 5, 7, 11, 113, 151, 179, 307, 491, 839, 1427, 2411, 5987, 6389, 8933, 11813, 18587, 31721, 40763, 46349, ..., . - Robert G. Wilson v, May 21 2017

Links

Crossrefs

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

Programs

  • Mathematica
    Do[n=1; p=Prime[n]; cp=Cyclotomic[k, p]; While[!PrimeQ[cp], n=n+1; p=Prime[n]; cp=Cyclotomic[k, p]]; Print[p], {k, 1, 300}]
  • PARI
    a(n)=if(n>>valuation(n,2)==1 && n>32, if(ispseudoprime(2^(n/2)+1), 2, 0), my(P=polcyclo(n)); forprime(p=2,, if(ispseudoprime(subst(P,'x,p)), return(p)))) \\ Charles R Greathouse IV, Dec 18 2014
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