A308121 Irregular triangle read by rows: T(n,k) = A109395(n)*k-A076512(n)*A038566(n,k).
0, 1, 1, 2, 1, 1, 1, 2, 3, 4, 2, 1, 1, 2, 3, 4, 5, 6, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 2, 1, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 4, 5, 6, 1, 2, 3, 7, 14, 13, 4, 11, 2, 1, 8
Offset: 1
Examples
The sequence as an irregular triangle: n/k 1, 2, 3, 4, ... 1: 0 2: 1 3: 1, 2 4: 1, 1 5: 1, 2, 3, 4 6: 2, 1 7: 1, 2, 3, 4, 5, 6 8: 1, 1, 1, 1 9: 1, 2, 1, 2, 1, 2 10: 3, 4, 1, 2 11: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 12: 2, 1, 2, 1 13: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 14: 4, 5, 6, 1, 2, 3 15: 7, 14, 13, 4, 11, 2, 1, 8 ... Row sums: 0, 1, 3, 2, 10, 3, 21, 4, 9, 10, 55, 6, 78, 21, 60. T(14,5) = A109395(14)*5 - A076512(14)*A038566(14,5) = 7*5 - 3*11 = 2. T(210,2) = A109395(210)*2 - A076512(210)*A038566(210,2) = 35*2 - 8*11 = -18.
Links
- Jamie Morken, Table of n, a(n) for n = 1..13413 (Rows n = 1..210 of triangle, flattened)
Programs
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Mathematica
Flatten@ Table[With[{a = n/GCD[n, #], b = Numerator[#/n]}, MapIndexed[a First@ #2 - b #1 &, Flatten@ Position[GCD[Table[Mod[k, n], {k, n - 1}], n], 1] /. {} -> {1}]] &@ EulerPhi@ n, {n, 15}] (* Michael De Vlieger, Jun 06 2019 *)
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PARI
vtot(n) = select(x->(gcd(n, x)==1), vector(n, k, k)); row(n) = my(q = eulerphi(n)/n, v = vtot(n)); vector(#v, k, denominator(q)*k - numerator(q)*v[k]); \\ Michel Marcus, May 14 2019
Comments