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A284826
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Irregular triangle T(n,k) for 1 <= k <= (n+1)/2: T(n,k) = number of primitive (aperiodic) palindromic structures of length n using exactly k different symbols.
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16
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1, 0, 0, 1, 0, 1, 0, 3, 1, 0, 2, 1, 0, 7, 6, 1, 0, 6, 6, 1, 0, 14, 25, 10, 1, 0, 12, 24, 10, 1, 0, 31, 90, 65, 15, 1, 0, 27, 89, 65, 15, 1, 0, 63, 301, 350, 140, 21, 1, 0, 56, 295, 349, 140, 21, 1, 0, 123, 965, 1701, 1050, 266, 28, 1
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OFFSET
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1,8
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COMMENTS
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Permuting the symbols will not change the structure.
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REFERENCES
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M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]
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LINKS
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FORMULA
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T(n, k) = (Sum_{j=0..k} (-1)^j * binomial(k, j) * A284823(n, k-j)) / k!.
T(n, k) = Sum_{d | n} mu(n/d) * stirling2(ceiling(d/2), k).
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EXAMPLE
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Triangle starts:
1
0
0 1
0 1
0 3 1
0 2 1
0 7 6 1
0 6 6 1
0 14 25 10 1
0 12 24 10 1
0 31 90 65 15 1
0 27 89 65 15 1
0 63 301 350 140 21 1
0 56 295 349 140 21 1
0 123 965 1701 1050 266 28 1
0 120 960 1700 1050 266 28 1
0 255 3025 7770 6951 2646 462 36 1
0 238 2999 7760 6950 2646 462 36 1
0 511 9330 34105 42525 22827 5880 750 45 1
0 495 9305 34095 42524 22827 5880 750 45 1
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For n=5, structures with 2 symbols are aabaa, ababa and abbba, so T(5,2) = 3.
For n=6, structures with 2 symbols are aabbaa and abbbba, so T(6,2) = 2.
(In this case, the structure abaaba is excluded because it is not primitive.)
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MATHEMATICA
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T[n_, k_] := DivisorSum[n, MoebiusMu[n/#]*StirlingS2[Ceiling[#/2], k]&];
Table[T[n, k], {n, 1, 15}, {k, 1, Floor[(n+1)/2]}] // Flatten (* Jean-François Alcover, Jun 12 2017, from 2nd formula *)
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PROG
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(PARI)
b(n, k) = sumdiv(n, d, moebius(n/d) * k^(ceil(d/2)));
a(n, k) = sum(j=0, k, b(n, k-j)*binomial(k, j)*(-1)^j)/k!;
for(n=1, 20, for(k=1, ceil(n/2), print1( a(n, k), ", "); ); print(); );
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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