A320435 Regular triangle read by rows where T(n,k) is the number of relatively prime k-subsets of {1,...,n}, 1 <= k <= n.
1, 1, 1, 1, 3, 1, 1, 5, 4, 1, 1, 9, 10, 5, 1, 1, 11, 19, 15, 6, 1, 1, 17, 34, 35, 21, 7, 1, 1, 21, 52, 69, 56, 28, 8, 1, 1, 27, 79, 125, 126, 84, 36, 9, 1, 1, 31, 109, 205, 251, 210, 120, 45, 10, 1, 1, 41, 154, 325, 461, 462, 330, 165, 55, 11, 1, 1, 45, 196
Offset: 1
Examples
Triangle begins: 1 1 1 1 3 1 1 5 4 1 1 9 10 5 1 1 11 19 15 6 1 1 17 34 35 21 7 1 1 21 52 69 56 28 8 1 1 27 79 125 126 84 36 9 1 1 31 109 205 251 210 120 45 10 1 1 41 154 325 461 462 330 165 55 11 1 1 45 196 479 786 923 792 495 220 66 12 1 1 57 262 699 1281 1715 1716 1287 715 286 78 13 1 The T(6,2) = 11 sets are: {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,5}, {3,4}, {3,5}, {4,5}, {5,6}. Missing from this list are: {2,4}, {2,6}, {3,6}, {4,6}.
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)
Crossrefs
Programs
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Mathematica
Table[Length[Select[Subsets[Range[n],{k}],GCD@@#==1&]],{n,10},{k,n}]
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PARI
T(n,k) = sum(d=1, n\k, moebius(d)*binomial(n\d, k)) \\ Andrew Howroyd, Jan 19 2023
Formula
T(n,k) = Sum_{d=1..floor(n/k)} mu(d)*binomial(floor(n/d), k). - Andrew Howroyd, Jan 19 2023
Comments