A215658 Primes p such that the smallest positive integer k for which p# + k is square satisfies p# + k = k^2, where p# = 2*3*5*7*11*...*p is a primorial.

2, 3, 5, 7, 17

Offset: 1

Author

Jonathan Sondow, Sep 02 2012

Keywords

• nonn
• more

Comments

The corresponding values of k are 2, 3, 6, 15, 715 = A215659.

The equation p# + k = k^2 has an integer solution k if and only if 1 + 4*p# is a square.

Conjecture: Not the same sequence as A192579, which is finite.

When p is in this sequence, p# = k(k-1) is in A161620, the intersection of A002110 and A002378. - Jeppe Stig Nielsen, Mar 27 2018

Examples

			The smallest square > 17# = 510510 is 715^2 = 17# + 715, so 17 is a member.
		

Links

• C. Aebi and G. Cairns, Partitions of primes, Parabola 45, Issue 1 (2009); see p. 5.

Crossrefs

Cf. A002110, A060797, A145781, A216144, A215659, A161620.

Programs

Mathematica

t = {}; pm = 1; Do[pm = pm*p; s = Floor[Sqrt[pm]]; If[pm == s*(s+1), AppendTo[t, p]], {p, Prime[Range[100]]}]; t (* T. D. Noe, Sep 05 2012 *)

PARI

for (n=1, 10, if (ceil(sqrt(prod(i=1, n, prime(i))))^2 - prod(i=1, n, prime(i)) - ceil(sqrt(prod(i=1, n, prime(i)))) == 0, print(prime(n)));); \\ Michel Marcus, Sep 05 2012

Python

from sympy import primorial, integer_nthroot, prime
A215658_list = [prime(i) for i in range(1,10**2) if integer_nthroot(4*primorial(i)+1,2)[1]] # Chai Wah Wu, Apr 01 2021

Formula

A145781(n) = A216144(n) if and only if prime(n) is a member.

a(n)# = A215659(n)*(A215659(n)-1).