|
%I #12 Jan 17 2018 11:45:45
%S 6,6216,7626,9180,16836,19900,22366,29646,76636,89676,93096,114960,
%T 116886,118828,322806,365940,397386,422740,437580,471906,499500,
%U 574056,595686,626640,690900,743590,984906,1041846,1148370,1209790,1260078,1357128,1450956
%N The first of three consecutive hexagonal numbers the sum of which is equal to the sum of three consecutive primes.
%H Robert Israel, <a href="/A298272/b298272.txt">Table of n, a(n) for n = 1..10000</a> (first 100 terms from Colin Barker)
%e 6 is in the sequence because 6+15+28 (consecutive hexagonal numbers) = 49 = 13+17+19 (consecutive primes).
%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]:= mi*(2*mi-1);
%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, u*(2*u-1)))); Vec(L)
%Y Cf. A000040, A000384, A054643, A298073, A298168, A298169, A298222, A298223, A298250, A298251, A298273.
%K nonn
%O 1,1
%A _Colin Barker_, Jan 16 2018
|
