A327029 T(n, k) = Sum_{d|n} phi(d) * A008284(n/d, k) for n >= 1, T(0, 0) = 1. Triangle read by rows for 0 <= k <= n.
1, 0, 1, 0, 2, 1, 0, 3, 1, 1, 0, 4, 3, 1, 1, 0, 5, 2, 2, 1, 1, 0, 6, 6, 4, 2, 1, 1, 0, 7, 3, 4, 3, 2, 1, 1, 0, 8, 8, 6, 6, 3, 2, 1, 1, 0, 9, 6, 9, 6, 5, 3, 2, 1, 1, 0, 10, 11, 10, 10, 8, 5, 3, 2, 1, 1, 0, 11, 5, 10, 11, 10, 7, 5, 3, 2, 1, 1, 0, 12, 17, 19, 19, 14, 12, 7, 5, 3, 2, 1, 1
Offset: 0
Examples
Triangle starts: [0] [1] [1] [0, 1] [2] [0, 2, 1] [3] [0, 3, 1, 1] [4] [0, 4, 3, 1, 1] [5] [0, 5, 2, 2, 1, 1] [6] [0, 6, 6, 4, 2, 1, 1] [7] [0, 7, 3, 4, 3, 2, 1, 1] [8] [0, 8, 8, 6, 6, 3, 2, 1, 1] [9] [0, 9, 6, 9, 6, 5, 3, 2, 1, 1]
Links
- Alois P. Heinz, Rows n = 0..200, flattened
Crossrefs
Programs
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SageMath
def DivisorTriangle(f, T, Len, w = None): D = [[1]] for n in (1..Len-1): r = lambda k: [f(d)*T(n//d,k) for d in divisors(n)] L = [sum(r(k)) for k in (0..n)] if w != None: L = [*map(lambda v: v * w(n), L)] D.append(L) return D DivisorTriangle(euler_phi, A008284, 10)
Formula
From Richard L. Ollerton, May 07 2021: (Start)
For n >= 1, T(n,k) = Sum_{i=1..n} A008284(gcd(n,i),k).
For n >= 1, T(n,k) = Sum_{i=1..n} A008284(n/gcd(n,i),k)*phi(gcd(n,i))/phi(n/gcd(n,i)). (End)
Comments