Problems & Puzzles: Puzzles

Puzzle 1167  n!+239 G. L. Honaker, Jr. sent the following challenging puzzle. f(n) = n!+239 is prime for the following 21 n values: 2, 4, 5, 7, 8, 9, 10, 11, 18, 22, 38, 45, 48, 76, 90, 96, 99, 132, 136, 147, 225, ? But he and others have tried without success to get the 22th n integer such that f(n) is prime. Q. Can you try get this 22th n integer in the sequence?

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Well, this "challenging" puzzle became a piece of cake.

From March 16 to March 22, 2024 all the following puzzlers -Alessandro Casini, Ivan Ianakev, Gennady Gusev, Giorgos Kalogeropoulos, Michael Branicky, Emmanuel Vantieghem, Adam Stinchcombe, Metin Sariyar, J. R. Howell, Oscar Volpatti  - responded more or less the same:

"There are not any prime n!+239 for 225 < n < 239, and for all n=>239, n!+239 are composites because 239 is a common factor, so there is not any 22th prime for this function".

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Three puzzlers -unaware of the above reasoning- made effective calculations not finding any 22th prime: Ryan Kellar (n<=3000), V. F. Izquierdo (n<=864000) and Paul Cleary (n<=1000000)

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