A245714 -id:A245714 - cp's OEIS frontend

A245723 a(n) = position of the first occurrence of n in A245714.

1, 3, 7, 19, 109, 509, 241, 317, 181, 1471, 2503, 2491, 7151, 11779, 3361, 2927, 1733, 5881, 15893, 16943, 11639, 31897, 25939, 12011, 17123, 6283, 10369, 63949, 8471, 125261, 64579, 117541, 21859, 58879, 44711, 216829, 64081, 67159, 73273, 181931, 139709

Offset: 1

Vaclav Kotesovec, Jul 30 2014

nonn

Least m > 0 such that m+n! is the smallest prime of form m+k!. - Jens Kruse Andersen, Jul 30 2014

			a(2) = 3 since 3+2! is the smallest prime of the form 3+k!, and 3 is the least such number. While 1+2! is also prime, there is a smaller prime 1+1! in that case so a(2) is not 1. - _Jens Kruse Andersen_, Jul 30 2014

Vaclav Kotesovec and Jens Kruse Andersen, Table of n, a(n) for n = 1..200 (first 92 terms from Vaclav Kotesovec)

Cf. A245714.

nmax=2000; Table[nn=1; k=0; While[k!=n && nnnn,k=0]; nn++]; If[nn==nmax,0,nn-1],{n,1,10}]

a(n)=for(k=1,n,if(ispseudoprime(n+k!),return(k)))
b(n)=for(k=1,10^6,if(a(k)==n,return(k)))
n=1;while(n<150,print1(b(n),", ");n++) \\ Derek Orr, Jul 30 2014

A245716 Least number k > 0 such that n + k! and n - k! are both prime, or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 1, 2, 1, 0, 0, 2, 0, 3, 1, 3, 0, 2, 0, 3, 1, 0, 0, 2, 0, 3, 0, 3, 0, 0, 0, 4, 1, 0, 0, 0, 0, 3, 0, 3, 0, 2, 0, 0, 1, 4, 0, 2, 0, 3, 0, 0, 0, 0, 0, 3, 0, 4, 0, 0, 0, 0, 1, 0, 0, 0, 0, 3, 0, 3, 0, 2, 0, 0, 1, 3, 0, 0, 0, 3, 0, 0, 0, 2, 0, 4, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 2, 0, 0, 1, 3, 0, 2, 0, 3, 1, 0, 0, 2, 0, 4, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Derek Orr, Jul 30 2014

nonn

For a(n) > 0, a(n)! < n for all n. Thus a(n) = 0 is definite.

			13 + 1! and 13 - 1! are not both prime.
13 + 2! and 13 - 2! are not both prime.
13 + 3! and 13 - 3! are both prime (19 and 7). Thus a(13) = 3.

Jens Kruse Andersen, Table of n, a(n) for n = 1..10000

Cf. A245714, A245715.

a(n)=for(k=1,n,if(ispseudoprime(n-k!)&&ispseudoprime(n+k!),return(k)))
vector(150,n,a(n))

A241423 Largest number k > 0 such that n + k! is prime, or 0 if no such k exists.

Original entry on oeis.org

1, 2, 1, 4, 1, 6, 0, 2, 1, 10, 1, 6, 0, 2, 1, 11, 1, 14, 0, 2, 1, 16, 0, 3, 0, 2, 1, 20, 1, 22, 0, 0, 0, 4, 1, 33, 0, 2, 1, 25, 1, 38, 0, 2, 1, 44, 0, 6, 0, 2, 1, 52, 0, 4, 0, 2, 1, 27, 1, 50, 0, 0, 0, 4, 1, 64, 0, 2, 1, 55, 1, 67, 0, 0, 0, 6, 1, 73, 0, 2, 1, 68, 0, 4, 0, 2, 1, 52, 0, 6

Offset: 2

Derek Orr, Aug 08 2014

nonn

If k >= n, then n + k! is divisible by n and is not prime.
a(n) < A020639(n), because if prime p divides n then p divides n + k! for k >= p. - Robert Israel, Aug 10 2014
There is no term for n = 1 since factorial primes 1 + k! can probably be arbitrarily large (A002981 shows k values). - Jens Kruse Andersen, Aug 13 2014

Jens Kruse Andersen, Table of n, a(n) for n = 2..1000

Cf. A245714, A125162.

a:= proc(n)
local k;
for k from min(numtheory:-factorset(n)) to 1 by -1 do
  if isprime(n+k!)  then return(k) fi
od:
0
end proc:
seq(a(n),n=2..100); # Robert Israel, Aug 10 2014

a[n_] := Module[{k}, For[k = FactorInteger[n][[1, 1]], k >= 1, k--, If[PrimeQ[n + k!], Return[k]]]; 0];
a /@ Range[2, 100] (* Jean-François Alcover, Jul 27 2020, after Maple *)

a(n)=forstep(k=n,1,-1,if(ispseudoprime(n+k!),return(k)))
n=2;while(n<150,print1(a(n),", ");n++)

cp's OEIS Frontend

A245723 a(n) = position of the first occurrence of n in A245714.

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A245716 Least number k > 0 such that n + k! and n - k! are both prime, or 0 if no such k exists.

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A241423 Largest number k > 0 such that n + k! is prime, or 0 if no such k exists.

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Author

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Programs

Maple

Mathematica

PARI