A335078 Irregular triangle read by rows: T(n, k) is the number of rotationally inequivalent Ferris Wheel distributions of Omega(n) colored balls of specification number n into precisely k boxes of the Ferris Wheel, with no box empty (0 < k <= Omega(n)).
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 4, 2, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 3, 2, 1
Offset: 2
Examples
The triangle T(n, k) begins n\k| 1 2 2 4 ---+-------------------- 2 | 1 3 | 1 4 | 1 1 5 | 1 6 | 1 1 7 | 1 8 | 1 1 1 9 | 1 1 10 | 1 1 11 | 1 12 | 1 2 1 13 | 1 14 | 1 1 15 | 1 1 16 | 1 2 1 1 ...
References
- Richard Beekman, An Introduction to Number-Theoretic Combinatorics, Lulu Press 2017.
Links
- Stefano Spezia, First 3500 rows of the triangle, flattened
- Richard Beekman, A General Solution of the Ferris Wheel Problem, ResearchGate, 2020.
Programs
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Mathematica
tau[n_,k_]:=If[n==1,1,Product[Binomial[Extract[Extract[FactorInteger[n],i],2]+k,k],{i,Length[FactorInteger[n]]}]]; (* A334997 *) Nd[n_, m_]:=Sum[(-1)^k*Binomial[m, k]*tau[n, m-k-1], {k,0,m-1}]; (* A334996 *) T[n_,k_]:=1/k*DivisorSum[k,EulerPhi[#]*Nd[n^(1/#),k/#]&,IntegerQ[n^(1/#)]&]; Table[T[n, m], {n,2,43}, {m,PrimeOmega[n]}]//Flatten
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PARI
TT(n, k) = if (k==0, 1, sumdiv(n, d, TT(d, k-1))); \\ A334996 U(n, m) = sum(k=0, m-1, (-1)^k*binomial(m, k)*TT(n, m-k-1)); T(n, k) = my(p); (1/k)*sumdiv(k, d, if (ispower(n, d, &p), eulerphi(d)*U(p, k/d))); row(n) = vector(bigomega(n), k, T(n,k)); \\ Michel Marcus, May 25 2020
Formula
T(n, k) = (1/k)*Sum_{d divides k and n^(1/d) is a positive integer} phi(d)*A334996(n^(1/d), k/d) (see Theorem 4 in Beekman's article).