A284826 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.
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
Offset: 1
Examples
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 -------------------------------------------- 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.)
References
- 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]
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..930
Crossrefs
Programs
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Mathematica
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 *) -
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(););
Formula
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).
Comments