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A210394 a(n) = least integer m>1 such that m divides none of S_i!+S_j! with 0<i<j<=n where S_k is the sum of the first k primes. 6

%I

%S 2,3,7,11,19,31,43,67,79,101,131,163,199,241,283,331,383,443,503,571,

%T 641,719,797,877,967,1061,1163,1277,1373,1481,1597,1721,1871,1999,

%U 2129,2281,2437,2593,2749,2927,3089,3271,3457,3643,3833,4057,4229,4441,4673,4889

%N a(n) = least integer m>1 such that m divides none of S_i!+S_j! with 0<i<j<=n where S_k is the sum of the first k primes.

%C 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.

%H Zhi-Wei Sun, <a href="/A210394/b210394.txt">Table of n, a(n) for n = 1..225</a>

%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2013.02.003">On functions taking only prime values</a>, J. Number Theory 133(2013), no.8, 2794-2812.

%e 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.

%t s[n_]:=s[n]=Sum[Prime[k],{k,1,n}]

%t f[n_]:=s[n]!

%t R[n_,m_]:=Product[If[Mod[f[k]+f[j],m]==0,0,1],{k,2,n},{j,1,k-1}]

%t Do[Do[If[R[n,m]==1,Print[n," ",m];Goto[aa]],{m,Max[2,s[n-1]],s[n]}];

%t Print[n];Label[aa];Continue,{n,1,225}]

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

%K nonn

%O 1,1

%A _Zhi-Wei Sun_, Mar 20 2012

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Last modified May 29 04:33 EDT 2020. Contains 334697 sequences. (Running on oeis4.)