%I #32 May 02 2020 15:39:49
%S 0,0,2,2,2,1,2,2,2,2,3,2,3,2,3,4,3,3,3,4,5,2,4,3,5,5,2,6,3,5,5,5,5,6,
%T 4,3,7,5,6,6,5,6,7,4,5,6,6,7,6,7,9,6,6,7,8,5,6,7,9,7,7,8,7,11,7,8,8,7,
%U 6,11,5,12,7,7,7,11,11,7,12,10,9,10,7,9,9,8,10,12,10,7,10,9,12,9,11,10,13,14,10,7
%N Number of numbers of the form 2*p or 3*p between consecutive triangular numbers T(n - 1) < {2,3}*p <= T(n) with p prime.
%C 1) It is conjectured that for k >= 1 each left-sided half-open interval (T(2*k - 1), T(2*k + 1)] and (T(2*k), T(2*(k + 1))] contains at least one composite c_2 = 2*p_i and c_3 = 3*p_j each, p_i, p_j prime, i != j.
%C 2) It is conjectured that for k >= 3 each left-sided half-open interval (T(k - 1), T(k)] contains at least one composite c_2 = 2*p_i or c_3 = 3*p_j, p_i, p_j prime, i != j.
%C 3) It is conjectured that for k >= 2 each left-sided half-open interval (T(2*k - 1), T(2*k)] contains at least one composite c_3 = 3*p_j, p_j prime.
%C 4) It is conjectured that for k >= 1 each left-sided half-open interval (T(2*k), T(2*k + 1)] contains at least one composite c_2 = 2*p_i, p_i prime.
%H Harvey P. Dale, <a href="/A319506/b319506.txt">Table of n, a(n) for n = 1..1000</a>
%e a(3) = 2 since (T(3 - 1),T(3)] = {4 = 2*2,5,6 = 2*3 = 3*2}, 2,3 prime.
%t Table[Count[
%t Select[Range[(n - 1) n/2 + 1, n (n + 1)/2],
%t PrimeQ[#/2] || PrimeQ[#/3] &], _Integer], {n, 1, 100}]
%t p23[{a_,b_}]:=Module[{r=Range[a+1,b]},Count[Union[Join[r/2,r/3]], _?PrimeQ]]; p23/@Partition[Accumulate[Range[0,100]],2,1] (* _Harvey P. Dale_, May 02 2020 *)
%o (PARI) isok1(n, k) = ((n%k) == 0) && isprime(n/k);
%o isok2(n) = isok1(n,2) || isok1(n,3);
%o t(n) = n*(n+1)/2;
%o a(n) = sum(i=t(n-1)+1, t(n), isok2(i)); \\ _Michel Marcus_, Oct 12 2018
%Y Cf. A000040 (primes), A000217 (triangular numbers).
%K nonn
%O 1,3
%A _Ralf Steiner_, Sep 21 2018