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%I #11 May 19 2020 19:12:55
%S 0,0,1,4,1,17,1,20,8,20,1,103,1,20,27,54,1,109,1,112,27,20,1,315,8,20,
%T 27,112,1,315,1,112,27,20,27,481,1,20,27,324,1,321,1,112,125,20,1,695,
%U 8,112,27,112,1,321,27,324,27,20,1,1285,1,20
%N Number of triples (d1,d2,d3) where each element is a divisor of n and d1 + d2 + d3 <= n.
%C It appears that a(n) depends on both parity of n and its prime signature. For instance a(odd prime)=1, a(even semiprime)=20, a(odd semiprime)=27, a(odd prime cube)=27, a(odd prime fourth power)=64. Maybe it is possible to find a formula for a(n). Similar sequences with pairs, quadruples, ... instead of triples can be envisioned. - _Michel Marcus_, Aug 21 2013
%C There's more to the story above. It seems that a(A233819(n)) gives the largest possible value per prime signature. Some prime signatures may have more than two possible values for a(n). - _David A. Corneth_, May 19 2020
%H Antti Karttunen, <a href="/A093035/b093035.txt">Table of n, a(n) for n = 1..16384</a>
%H Antti Karttunen, <a href="/A093035/a093035.txt">Data supplement: n, a(n) computed for n = 1..65537</a>
%e a(9) = 8 because the divisors of 9 are {1,3,9} making the valid triples (1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3).
%o (PARI) a(n) = {nb = 0; d = divisors(n); for (i = 1, #d, for (j = 1, #d, for (k = 1, #d, if (d[i]+d[j]+d[k] <= n, nb++);););); nb;} \\ _Michel Marcus_, Aug 21 2013
%Y Cf. A233819.
%K easy,nonn
%O 1,4
%A Jonathan A. Cohen (cohenj02(AT)tartarus.uwa.edu.au), May 08 2004
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