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.

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A087037 Smallest integer x > 0 such that x^x + n is prime, or 0 if no such x exists.

Original entry on oeis.org

1, 1, 2, 1, 444, 1, 2
Offset: 1

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Author

Hugo Pfoertner, Jul 31 2003

Keywords

Comments

The sequence with the unknown terms a(8), a(11), a(17), a(24), a(41), a(53), a(59), a(65) indicated by ? (each of which exceeds 6000) begins: 1, 1, 2, 1, 444, 1, 2, ?, 2, 1, ?, 1, 2, 3, 2, 1, ?, 1, 2, 3, 4, 1, 6, ?, 2, 3, 2, 1, 30, 1, 6, 3, 2, 3, 6, 1, 2, 5, 2, 1, ?, 1, 2, 3, 58, 1, 6, 7, 2, 3017, 4, 1, ?, 35, 2, 3, 2, 1, ?, 1, 4, 3, 2, 19, ?, 1, 2, 27, 2, 1, 6, 1, 8, 3, 2, ..., where the value a(50)=3017 corresponds to a probable prime. [extended by Jon E. Schoenfield, Mar 17 2018, Mar 19 2018]
It is conjectured that such x always exists. - Dean Hickerson
From Farideh Firoozbakht and M. F. Hasler, Nov 27 2009: (Start)
We can show that for all n=(6k-1)^3, k > 0, there is no such x, which disproves the conjecture:
Since n=(6k-1)^3 is odd, x must be even, else x^x+n is even and composite.
If x == +-1 (mod 3), then x^x + n == (+-1)^2 + (-1)^3 == 0 (mod 3), i.e., divisible by 3 and therefore composite.
Finally, if x == 0 (mod 3), then x^x + n = (x^(x/3) + 6k-1)*(x^(2x/3) - x^(x/3)*(6k-1) + (6k-1)^2) is again composite. (End)
a(8) >= 36869. - Max Alekseyev, Sep 16 2013

Examples

			a(7)=2 because 2^2 + 7 = 11 is prime.
		

Crossrefs

Cf. A000312 (n^n), A087038 (x^x+n is prime, x>1).
Cf. A166853 (x^x-n is prime). - Farideh Firoozbakht and M. F. Hasler, Nov 27 2009

Programs

  • PARI
    a(n) = {my(x = 1); while (!isprime(x^x+n), x++); x;} \\ Michel Marcus, Mar 20 2018

Extensions

Name edited by Altug Alkan, Apr 01 2018

A166853 a(n) is the smallest number m such that m^m-n is prime, or zero if there is no such m.

Original entry on oeis.org

2, 2, 8, 3, 4, 5, 6, 3, 0, 3, 78, 13, 6, 3, 4, 3, 4, 17, 12, 3, 118, 3, 4, 3, 3
Offset: 1

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Author

Keywords

Comments

The sequence with the unknown terms a(n) indicated by -n:
(0's occur for n=9, 49, 81, 121....)
2,2,8,3,4,5,6,3,0,3,78,13,6,3,4,3,4,17,12,3,118,3,4,3,3,
-26,4,-28,4,487,90,9,4,-34,24,5,6,271,28,969,-41,5,-43,7,4,5,32,37,0,621,
20,15,34,7,6,9,4,5,4,7,-61,7,4,5,4,-66,6,63,134,27,10,35,102,31,4,
5,4,569,-79,13,0,15,4,5,-85,7,110,5,4,131,1122,7,4,11,8,7,6,9,4,-100,
22,5,-103,-104,4,5,4,11,12,39,-111,...
If they exist, the first two unknown terms, a(26) and a(28), they are greater than 10000. All other unknown terms a(n), for n<112 are greater than 4000.
If it exists, a(26) > 25000. - Robert Price, Apr 26 2019

Examples

			We have a(1)=2 since 1^1-1 is not prime, but 2^2-1 is prime.
a(9)=0 since 2^2-9 is not prime, and if m is an even number greater than 2 then m^m-9=(m^(m/2)-3)*(m^(m/2)+3) is composite. So there is no number m such that m^m-9 is prime. The same applies to any odd square > 25.
We have a(25)=3 since 3^3-25=2 is prime. But 25 is the only known square of the form m^m-2, so a(n)=0 for other odd squares > 25, e.g., n = 49,81,121,....
a(115)=2736 is the largest known term. 2736^2736-115 is a probable prime.
		

Crossrefs

Formula

a(n)=0 if n=3^2 or n=(2k+1)^2 > 25, or n = (6k+1)^3 = A016923(k) with k>0.
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