cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A319777 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.

Original entry on oeis.org

1, 15, 100, 436, 1459, 4069, 9929, 21871, 44426, 84494, 152171, 261749, 432906, 692102, 1074198, 1624314, 2399943, 3473337, 4934182, 6892578, 9482341, 12864643, 17232007, 22812673, 29875352, 38734384, 49755317, 63360923, 80037668, 100342652, 124911036
Offset: 0

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Author

Peter Kagey, Sep 26 2018

Keywords

Comments

A019298(n) is the analogous sequence if the three sets must each have exactly n elements.

Examples

			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:
|A|         = |{1, 2}|    = 2 = |{1, 8}|    = |A'|
|B|         = |{1, 2, 3}| = 3 = |{1, 4, 8}| = |B'|
|C|         = |{1, 4}|    = 2 = |{5, 8}|    = |C'|
|A & B|     = |{1, 2}|    = 2 = |{1, 8}|    = |A' & B'|
|A & C|     = |{1}|       = 1 = |{8}|       = |A' & C'|
|B & C|     = |{1}|       = 1 = |{8}|       = |B' & C'|
|A & B & C| = |{1}|       = 1 = |{8}|       = |A' & B' & C'|
		

Crossrefs

Cf. A000330(n-1) is analogous, but with pairs instead of triples.

Programs

  • 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
    
  • Maple
    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
  • 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
    
  • 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

Formula

a(n) = Sum_{k=0..n} A244865(k). [corrected by Michel Marcus, Dec 27 2018]
From Colin Barker, Dec 27 2018: (Start)
G.f.: (1 + 8*x + 15*x^2 + 8*x^3 + x^4) / ((1 - x)^8*(1 + x)).
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.
(End)
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