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 1-3 of 3 results.

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.

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.
Showing 1-3 of 3 results.

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