|
|
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.
|
|
5
|
|
|
|
OFFSET
|
1,1
|
|
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.
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
The smallest square > 17# = 510510 is 715^2 = 17# + 715, so 17 is a member.
|
|
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 *)
|
|
PROG
|
(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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,more
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|