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%I #21 Dec 09 2023 23:44:16
%S 0,0,0,0,0,0,0,10,10,10,10,31,31,31,31,31,31,52,52,73,73,73,73,73,73,
%T 73,83,104,104,140,140,140,140,140,140,140,140,140,140,140,140,176,
%U 176,197,218,218,218,218,218,239,239,260,260,260,260,260,260,260,260
%N For each triprime (A014612) less than or equal to n, sum the positive integers less than or equal to the number of divisors of the triprime.
%C The largest difference between any pair of consecutive numbers in the sequence = 36, The second largest difference = 21, the third largest = 10, and the fourth (and last) possible difference is 0.
%H G. C. Greubel, <a href="/A211872/b211872.txt">Table of n, a(n) for n = 1..5000</a>
%F a(n) = Sum_{i=1..n} [Omega(i) = 3] * Sum_{j = 1..d(i)} j.
%F a(n) = Sum_{i=1..n} [Omega(i) = 3] * (omega(i) + 1) * (d(i) + 1).
%F a(n) = Sum_{i=1..n} [Omega(i) = 3] * (2*omega(i)^2 + 5*omega(i) + 3), where [ ] is the Iverson bracket.
%e a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = a(7) = 0. a(8) = 10 since 8 has 4 divisors, and the sum of all the numbers up to 4 is 1 + 2 + 3 + 4 = 10. The next triprime is 12, so a(8) = a(9) = a(10) = a(11) = 10. Since there are two triprimes less than or equal to 12, we sum the numbers from 1 to d(8) and 1 to d(12), then take the sum total. Thus, a(12) = 10 + 21 = 31.
%t sm = 0; Table[If[Total[Transpose[FactorInteger[n]][[2]]] == 3, d = DivisorSigma[0, n]; sm = sm + d (d + 1)/2]; sm, {n, 100}] (* _T. D. Noe_, Feb 14 2013 *)
%t Table[Sum[KroneckerDelta[PrimeOmega[i], 3]*Sum[j, {j, DivisorSigma[0, i]}], {i, n}], {n, 50}] (* _Wesley Ivan Hurt_, Oct 07 2014 *)
%Y Cf. A000005, A001221, A001222, A014612, A209323.
%K nonn
%O 1,8
%A _Wesley Ivan Hurt_, Feb 12 2013
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