

A075758


Smallest k such that n!+k!1 is prime, or 0 if no such k exists.


4



2, 1, 2, 3, 5, 4, 7, 4, 4, 6, 2, 8, 14, 8, 8, 4, 17, 7, 26, 7, 6, 12, 6, 14, 14, 16, 2, 27, 6, 33, 6, 36, 26, 7, 33, 18, 2, 18, 6, 36, 2, 18, 14, 20, 25, 14, 25, 0, 22, 24, 34, 16, 46, 61, 18, 7, 25, 38, 47, 47, 54, 79, 157, 97, 28, 23, 7, 137, 24, 46, 36, 25, 2, 214, 94, 40, 2, 96
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OFFSET

1,1


COMMENTS

It is possible to prove a(48)=0: since 67 divides 48!1 and thus all 48!+k!1 for k>=67 are composite, it is sufficient to test all k<67 which was done by Robert G. Wilson v. Likewise 113 divides 111!1, 797 divides 128!1, 137 divides 135!1, 163 divides 161!1 etc. and therefore a(111)=a(128)=a(135)=a(161)=0.
From the previous comments follows that a(n)=0 if there is no k smaller than the smallest prime factor of n!1 such that n!+k!1 is prime.


LINKS



EXAMPLE

4!=24, 1!=1 but 24+11=24 is not prime. 24+2!1=25, not prime 24+3!1=29, prime, so a[4]=3


MATHEMATICA

a = {}; Do[k = 1; While[ ! PrimeQ[n! + k!  1], k++ ]; a = Append[a, k], {n, 1, 47}]


PROG

(PARI) for (a=1, 100, c=0; for (b=1, 200, if (isprime(a!+b!1), c=b; break)); if (c>0, print1(c, ", "), print1("0, ")))


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



