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%I #27 Jan 26 2019 11:21:19
%S 1,15,100,436,1459,4069,9929,21871,44426,84494,152171,261749,432906,
%T 692102,1074198,1624314,2399943,3473337,4934182,6892578,9482341,
%U 12864643,17232007,22812673,29875352,38734384,49755317,63360923,80037668,100342652,124911036
%N a(n) is the number of equivalence classes of triples of sets each with n or fewer elements where two triples are equivalent if the number of elements in all intersections is the same.
%C A019298(n) is the analogous sequence if the three sets must each have exactly n elements.
%H Muniru A Asiru, <a href="/A319777/b319777.txt">Table of n, a(n) for n = 0..5000</a>
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (7,-20,28,-14,-14,28,-20,7,-1).
%F a(n) = Sum_{k=0..n} A244865(k). [corrected by _Michel Marcus_, Dec 27 2018]
%F From _Colin Barker_, Dec 27 2018: (Start)
%F G.f.: (1 + 8*x + 15*x^2 + 8*x^3 + x^4) / ((1 - x)^8*(1 + x)).
%F a(n) = 7*a(n-1) - 20*a(n-2) + 28*a(n-3) - 14*a(n-4) - 14*a(n-5) + 28*a(n-6) - 20*a(n-7) + 7*a(n-8) - a(n-9) for n>8.
%F (End)
%e The triple (A, B, C) = ({1, 2}, {1, 2, 3}, {1, 4}) is equivalent to the triple (A', B', C') = ({1, 8}, {1, 4, 8}, {5, 8}) because all intersections of the sets in a triple are equal:
%e |A| = |{1, 2}| = 2 = |{1, 8}| = |A'|
%e |B| = |{1, 2, 3}| = 3 = |{1, 4, 8}| = |B'|
%e |C| = |{1, 4}| = 2 = |{5, 8}| = |C'|
%e |A & B| = |{1, 2}| = 2 = |{1, 8}| = |A' & B'|
%e |A & C| = |{1}| = 1 = |{8}| = |A' & C'|
%e |B & C| = |{1}| = 1 = |{8}| = |B' & C'|
%e |A & B & C| = |{1}| = 1 = |{8}| = |A' & B' & C'|
%p a:=n->add((15*(127+(-1)^k)+6432*k+8936*k^2+6480*k^3+2570*k^4+528*k^5+44*k^6)/1920,k=0..n): seq(a(n),n=0..30); # _Muniru A Asiru_, Sep 28 2018
%o (GAP) List([0..30],n->Sum([0..n],k->(15*(127+(-1)^k)+6432*k+8936*k^2+6480*k^3+2570*k^4+528*k^5+44*k^6)/1920)); # _Muniru A Asiru_, Sep 28 2018
%o (PARI) a(n) = sum(k=0, n, (15*(127+(-1)^k) + 6432*k + 8936*k^2 + 6480*k^3 + 2570*k^4 + 528*k^5 + 44*k^6) / 1920); \\ _Michel Marcus_, Dec 27 2018
%o (PARI) Vec((1 + 8*x + 15*x^2 + 8*x^3 + x^4) / ((1 - x)^8*(1 + x)) + O(x^40)) \\ _Colin Barker_, Dec 28 2018
%Y Cf. A019298, A244865.
%Y Cf. A000330(n-1) is analogous, but with pairs instead of triples.
%K nonn
%O 0,2
%A _Peter Kagey_, Sep 26 2018