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A327029
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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.
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9
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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
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refs;
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history;
text;
internal format)
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OFFSET
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0,5
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COMMENTS
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LINKS
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FORMULA
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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)
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EXAMPLE
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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]
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PROG
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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)
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CROSSREFS
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Cf. A000041 (where reversed rows converge to).
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KEYWORD
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AUTHOR
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STATUS
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approved
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