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Integers m such that ceiling(sqrt(m!)) is prime.
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%I #27 Jul 29 2022 20:43:13

%S 2,3,4,5,7,21,2132,3084,9301

%N Integers m such that ceiling(sqrt(m!)) is prime.

%C This sequence includes the known solutions of Brocard's problem as of 2016 (see A146968).

%e 3 is in the sequence because 3! = 6, sqrt(6) = 2.449489742783178..., the ceiling of which is 3, which is prime.

%e 4 is in the sequence because 4! = 24, sqrt(24) = 4.898979485566356..., the ceiling of which is 5, which is prime.

%t Select[Range[3200]], PrimeQ[Ceiling[Sqrt[#!]]] &]

%o (Python)

%o from math import isqrt, factorial

%o from itertools import count, islice

%o from sympy import isprime

%o def A273932_gen(): # generator of terms

%o return filter(lambda n:isprime(1+isqrt(factorial(n)-1)),count(1))

%o A273932_list = list(islice(A273932_gen(),7)) # _Chai Wah Wu_, Jul 29 2022

%Y Cf. A055228 (ceiling(sqrt(n!))), A146968.

%K nonn,more

%O 1,1

%A _Salvador Cerdá_, Jun 04 2016

%E a(9) from _Giovanni Resta_, Jun 20 2016