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%I #11 Jan 17 2018 11:45:54
%S 13,6427,7873,9439,17203,20287,22783,30133,77417,90523,93949,115903,
%T 117841,119797,324403,367649,399163,424573,439441,473839,501493,
%U 576193,597859,628861,693223,746023,987697,1044733,1151399,1212889,1263247,1360417,1454351
%N The first of three consecutive primes the sum of which is equal to the sum of three consecutive hexagonal numbers.
%H Robert Israel, <a href="/A298273/b298273.txt">Table of n, a(n) for n = 1..10000</a> (first 100 terms from Colin Barker)
%e 13 is in the sequence because 13+17+19 (consecutive primes) = 49 = 6+15+28 (consecutive hexagonal numbers).
%p N:= 100: # to get a(1)..a(100)
%p count:= 0:
%p mmax:= floor((sqrt(24*N-87)-9)/12):
%p for i from 1 while count < N do
%p mi:= 2*i;
%p m:= 6*mi^2+9*mi+7;
%p r:= ceil((m-8)/3);
%p p1:= prevprime(r+1);
%p p2:= nextprime(p1);
%p p3:= nextprime(p2);
%p while p1+p2+p3 > m do
%p p3:= p2; p2:= p1; p1:= prevprime(p1);
%p od:
%p if p1+p2+p3 = m then
%p count:= count+1;
%p A[count]:= p1;
%p fi
%p od:
%p seq(A[i],i=1..count); # _Robert Israel_, Jan 16 2018
%o (PARI) L=List(); forprime(p=2, 2000000, q=nextprime(p+1); r=nextprime(q+1); t=p+q+r; if(issquare(24*t-87, &sq) && (sq-9)%12==0, u=(sq-9)\12; listput(L, p))); Vec(L)
%Y Cf. A000040, A000384, A054643, A298073, A298168, A298169, A298222, A298223, A298250, A298251, A298272.
%K nonn
%O 1,1
%A _Colin Barker_, Jan 16 2018
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