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 1-4 of 4 results.

A216148 Primes of the form 2*k^k + 1 = A216147(k).

Original entry on oeis.org

3, 17832200896513, 78692816150593075150849
Offset: 1

Views

Author

M. F. Hasler, Sep 02 2012

Keywords

Comments

The sequence should be extended through A110932, which lists the corresponding values of k: The next term, 2*251^251 + 1 = A216147(A110932(4)) ~ 4.16*10^602, is too large to include here.

Crossrefs

Cf. A110932.
A subsequence of A133663, with b=a and c=1.

Programs

  • Mathematica
    Select[Table[2n^n+1,{n,20}],PrimeQ] (* Harvey P. Dale, Mar 27 2016 *)
  • PARI
    for(n=1,999, ispseudoprime(p=n^n*2+1) & print1(p","))

Formula

a(2) = A216147(12) = A005109(95) = A070855(12) = A058383(89) = A133663(18).
a(3) = A216147(18) = A005109(183)= A070855(18) = A058383(177)= A133663(36).

A174711 Composites of the form 2*n^n + 1 = A216147(n).

Original entry on oeis.org

9, 55, 513, 6251, 93313, 1647087, 33554433, 774840979, 20000000001, 570623341223, 605750213184507, 22224013651116033, 875787780761718751, 36893488147419103233, 1654480523772673528355, 3956839311320627178247959
Offset: 1

Views

Author

Michel Lagneau, Mar 27 2010

Keywords

Comments

If p = n+2 is prime, then p divides 2*n^n + 1. Proof: Let p = n+2 prime. Then, according to Fermat's theorem, n^(p-1) == 1 (mod p). Because p-1 = n+1, n^(n+1) == 1 (mod p), and with n = p-2 == -2 (mod p), we obtain successively: n*n^n == 1 (mod p), -2*n^n == 1 (mod p), 2*n^n == -1 (mod p) => p divides 2*n^n + 1.

Examples

			a(2) = 9 = 3^2, a(3) = 55 = 5*11, a(4) = 513 = 3 ^ 3 * 19.
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.
  • J. M. De Koninck, A. Mercier, 1001 problemes en theorie classique des nombres, Ellipses 2004, p. 52.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 2.

Crossrefs

Complement of A216148 in A216147. - M. F. Hasler, Sep 02 2012

Programs

  • Maple
    with(numtheory):for n from 0 to 50 do: x:=2*n^n + 1 : if type(x,prime)=false then print (x):else fi:od:
  • Mathematica
    Select[Table[2n^n+1,{n,20}],CompositeQ] (* Harvey P. Dale, Jun 21 2015 *)

A110932 Numbers k such that 2*k^k + 1 is prime.

Original entry on oeis.org

0, 1, 12, 18, 251, 82992
Offset: 0

Views

Author

Ray G. Opao, Sep 25 2005

Keywords

Comments

As a "list of numbers such that ...", the sequence should have offset 1, but to preserve the validity of formulas referring to this sequence, the offset was set to 0 when the initial value a(0)=0 was added. - M. F. Hasler, Sep 02 2012

Crossrefs

Cf. A110931, A121270 (= primes in A014566), A088790, A160360, A160600.
The primes 2n^n+1, for k<4, n=a(k)<251, are listed at A216148(k) = A216147(a(k)). - M. F. Hasler, Sep 02 2012

Programs

  • Mathematica
    Join[{0}, Select[Range[1000], PrimeQ[2*#^# + 1] &]] (* Robert Price, Mar 27 2019 *)
  • PARI
    is_A110932(n)=ispseudoprime(n^n*2+1) \\ M. F. Hasler, Sep 02 2012

Extensions

a(5) from Serge Batalov, Apr 08 2018

A160600 Numbers k such that 3*(2k)^(2k)+1 is prime.

Original entry on oeis.org

1, 2, 3, 5, 143, 225
Offset: 1

Views

Author

M. F. Hasler, Jul 10 2009

Keywords

Comments

This corresponds to the numbers such that 3m^m+1 is prime, but these must all be even, m=2k, and therefore it is more natural to record the sequence of k=m/2.
Next term > 15000. - Matevz Markovic, Oct 09 2012

Examples

			a(1) = 1, because 2^2*3+1 = 13 is the smallest prime of this form.
a(2) = 2, because 4^4*3+1 = 769 is the next smallest prime of this form. a(3) = 3, because 6^6*3+1 = 139969 is again a prime.
		

Crossrefs

Cf. A160360 (3n^n+2 is prime), A121270 = primes among Sierpinski numbers A014566(n)=n^n+1; A216148 = A216147(A110932): primes 2n^n+1; A088790, A065798.

Programs

  • Maple
    q:= k-> isprime(3*(2*k)^(2*k)+1):
    select(q, [$1..225])[];  # Alois P. Heinz, Aug 04 2025
  • PARI
    for(i=1,9999,ispseudoprime(i^i*3+1)&print1(i/2,","))
Showing 1-4 of 4 results.