A210394 a(n) = least integer m>1 such that m divides none of S_i!+S_j! with 0

2, 3, 7, 11, 19, 31, 43, 67, 79, 101, 131, 163, 199, 241, 283, 331, 383, 443, 503, 571, 641, 719, 797, 877, 967, 1061, 1163, 1277, 1373, 1481, 1597, 1721, 1871, 1999, 2129, 2281, 2437, 2593, 2749, 2927, 3089, 3271, 3457, 3643, 3833, 4057, 4229, 4441, 4673, 4889

Offset: 1

Author

Zhi-Wei Sun, Mar 20 2012

Keywords

• nonn

Comments

When n>1, we have S_n!+S_{n-1}!=0 (mod m) for all m=1,...,S_{n-1} and hence a(n)>S_{n-1}. Zhi-Wei Sun conjectured that a(n) is always a prime not exceeding S_n.

Examples

			We have a(3)=7, since  m=7 divides none of 2!+(2+3)!,2!+(2+3+5)!,(2+3)!+(2+3+5)! but this fails for m=2,3,4,5,6.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..225
• Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.

Crossrefs

Cf. A000040, A210393, A210186, A210144, A208494, A208643, A207982.

Programs

Mathematica

s[n_]:=s[n]=Sum[Prime[k],{k,1,n}]
f[n_]:=s[n]!
R[n_,m_]:=Product[If[Mod[f[k]+f[j],m]==0,0,1],{k,2,n},{j,1,k-1}]
Do[Do[If[R[n,m]==1,Print[n," ",m];Goto[aa]],{m,Max[2,s[n-1]],s[n]}];
   Print[n];Label[aa];Continue,{n,1,225}]