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A352459
First of four consecutive primes p,q,r,s such that p+q+r+s is divisible by A001414(p+q).
3
7, 53, 151, 263, 269, 311, 317, 457, 587, 761, 883, 919, 1049, 1303, 1423, 1579, 1637, 1759, 1993, 2063, 2111, 2789, 2851, 2903, 3137, 3533, 3593, 3797, 3889, 3967, 4153, 4219, 4463, 4507, 4663, 5179, 5843, 6353, 6737, 6829, 7937, 8563, 8999, 9059, 9241, 9437, 9439, 9629, 10039, 10831, 10847
OFFSET
1,1
COMMENTS
If x is a prime and 2*x-y, 2*x+y, 2*x+z and 2*x+32-z are consecutive primes where 0 < y < z < 16, then 2*x-y is a term. Thus Dickson's conjecture implies there are infinitely many terms.
LINKS
EXAMPLE
a(3) = 151 is a term because 151,157,163,167 are consecutive primes, and A001414(151+157) = A001414(2^2*7*11) = 2+2+7+11 = 22 divides 151+157+163+167 = 638.
MAPLE
q:= 2: r:= 3: s:= 5:
R:= NULL: count:= 0:
while count < 50 do
p:= q; q:= r; r:= s; s:= nextprime(s);
a:= add(t[1]*t[2], t = ifactors(p+q)[2]);
if (p+q+r+s) mod a = 0 then count:= count+1; R:= R, p fi
od:
R;
CROSSREFS
KEYWORD
nonn
AUTHOR
J. M. Bergot and Robert Israel, Mar 17 2022
STATUS
approved