OFFSET
2,18
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.
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).
EXAMPLE
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
...
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
PROG
(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
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Stefano Spezia, May 23 2020
STATUS
approved