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A186972
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Irregular triangle T(n,k), n>=1, 1<=k<=A186971(n), read by rows: T(n,k) is the number of k-element subsets of {1, 2, ..., n} containing n and having pairwise coprime elements.
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18
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1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 4, 5, 2, 1, 2, 1, 1, 6, 11, 8, 2, 1, 4, 6, 4, 1, 1, 6, 12, 10, 3, 1, 4, 5, 2, 1, 10, 31, 42, 26, 6, 1, 4, 6, 4, 1, 1, 12, 45, 79, 72, 33, 6, 1, 6, 14, 16, 9, 2, 1, 8, 21, 25, 14, 3, 1, 8, 24, 36, 29, 12, 2, 1, 16, 79, 183, 228, 157, 56, 8, 1, 6, 15, 20, 15, 6, 1
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OFFSET
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1,5
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COMMENTS
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T(n,k) = 0 for k>A186971(n). The triangle contains all positive values of T.
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LINKS
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EXAMPLE
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T(5,3) = 5 because there are 5 3-element subsets of {1,2,3,4,5} containing 5 and having pairwise coprime elements: {1,2,5}, {1,3,5}, {1,4,5}, {2,3,5}, {3,4,5}.
Irregular Triangle T(n,k) begins:
1;
1, 1;
1, 2, 1;
1, 2, 1;
1, 4, 5, 2;
1, 2, 1;
1, 6, 11, 8, 2;
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MAPLE
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with(numtheory):
s:= proc(m, r) option remember; mul(`if`(i<r, i, 1), i=factorset(m)) end:
a:= proc(n) option remember; `if`(n<4, n, pi(n)-nops(factorset(n))+2) end:
b:= proc(t, n, k) option remember; local c, d, h;
if k=0 or k>n then 0
elif k=1 then 1
elif k=2 and t=n then `if`(n<2, 0, phi(n))
else c:= 0;
d:= 2-irem(t, 2);
for h from 1 to n-1 by d do
if igcd(t, h)=1 then c:= c +b(s(t*h, h), h, k-1) fi
od; c
fi
end:
T:= proc(n, k) option remember; b(s(n, n), n, k) end:
seq(seq(T(n, k), k=1..a(n)), n=1..20);
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MATHEMATICA
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s[m_, r_] := s[m, r] = Product[If[i < r, i, 1], {i, FactorInteger[m][[All, 1]]}]; a[n_] := a[n] = If[n < 4, n, PrimePi[n] - Length[FactorInteger[n]]+2]; b[t_, n_, k_] := b[t, n, k] = Module[{c, d, h}, Which[k == 0 || k > n, 0, k == 1, 1, k == 2 && t == n, If[n < 2, 0, EulerPhi[n]], True, c = 0; d = 2-Mod[t, 2]; For[h = 1, h <= n-1, h = h+d, If[GCD[t, h] == 1, c = c+b[s[t*h, h], h, k-1]]]; c]]; t[n_, k_] := t[n, k] = b[s[n, n], n, k]; Table[Table[t[n, k], {k, 1, a[n]}], {n, 1, 20}] // Flatten (* Jean-François Alcover, Dec 17 2013, translated from Maple *)
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CROSSREFS
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Rightmost elements of rows give A186994.
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KEYWORD
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AUTHOR
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STATUS
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approved
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