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%I #18 Dec 12 2020 17:33:01
%S 199,35951,46351,69221,88427,230291,490481,707573,829883,1088419,
%T 1129693,1258109,1736101,1918157,1976243,2456939,2741159,2753351,
%U 2822881,3249419,4603351,5121713,5528623,6186407,6664429,6945559,6964949,7094839,7120963,7147121
%N The first of three consecutive primes the sum of which is equal to the sum of three consecutive pentagonal numbers.
%H Robert Israel, <a href="/A298251/b298251.txt">Table of n, a(n) for n = 1..2352</a>
%e 199 is in the sequence because 199+211+223 (consecutive primes) = 633 = 176+210+247 (consecutive pentagonal numbers).
%p N:= 10^8: # to get all terms where the sums <= N
%p Res:= NULL:
%p mmax:= floor((sqrt(8*N-23)-5)/6):
%p M3:= map(t->9/2*t^2+15/2*t+6, [seq(seq(4*i+j,j=2..3),i=0..mmax/4)]):
%p for m in M3 do
%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 Res:= Res, p1
%p fi
%p od:
%p Res; # _Robert Israel_, Jan 16 2018
%t Module[{nn=50000,pn},pn=Total/@Partition[PolygonalNumber[5,Range[ Ceiling[ (1+Sqrt[1+24 Prime[nn]])/6]]],3,1];Select[Partition[ Prime[ Range[ nn]],3,1],MemberQ[pn,Total[#]]&]][[All,1]] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Dec 12 2020 *)
%o (PARI) L=List(); forprime(p=2, 8000000, q=nextprime(p+1); r=nextprime(q+1); t=p+q+r; if(issquare(72*t-207, &sq) && (sq-15)%18==0, u=(sq-15)\18; listput(L, p))); Vec(L)
%Y Cf. A000040, A000326, A054643, A298073, A298168, A298169, A298222, A298223, A298250.
%K nonn
%O 1,1
%A _Colin Barker_, Jan 15 2018
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