OFFSET
1,3
COMMENTS
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.
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.
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.
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.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
EXAMPLE
a(3) = 2 since (T(3 - 1),T(3)] = {4 = 2*2,5,6 = 2*3 = 3*2}, 2,3 prime.
MATHEMATICA
Table[Count[
Select[Range[(n - 1) n/2 + 1, n (n + 1)/2],
PrimeQ[#/2] || PrimeQ[#/3] &], _Integer], {n, 1, 100}]
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 *)
PROG
(PARI) isok1(n, k) = ((n%k) == 0) && isprime(n/k);
isok2(n) = isok1(n, 2) || isok1(n, 3);
t(n) = n*(n+1)/2;
a(n) = sum(i=t(n-1)+1, t(n), isok2(i)); \\ Michel Marcus, Oct 12 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Ralf Steiner, Sep 21 2018
STATUS
approved