login
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.
16
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
1,8
COMMENTS
Permuting the symbols will not change the structure.
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
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).
EXAMPLE
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.)
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 *)
PROG
(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(); );
CROSSREFS
Columns 2-6 are A056481, A056482, A056483, A056484, A056485.
Partial row sums include A056476, A056477, A056478, A056479, A056480.
Row sums are A284841.
Cf. A284823.
Sequence in context: A357354 A166408 A327077 * A307752 A101548 A117430
KEYWORD
nonn,tabf
AUTHOR
Andrew Howroyd, Apr 03 2017
STATUS
approved