A103447 Triangle read by rows: T(n,k) = Moebius(binomial(n,k)) (0 <= k <= n).
1, 1, 1, 1, -1, 1, 1, -1, -1, 1, 1, 0, 1, 0, 1, 1, -1, 1, 1, -1, 1, 1, 1, 1, 0, 1, 1, 1, 1, -1, 1, 1, 1, 1, -1, 1, 1, 0, 0, 0, -1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, -1, 1, -1, 1, 1, 1, 1, -1, 1, -1, 1, 1, 0, -1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1, 1, -1, -1, -1, -1, 0, 0, 0, 0, -1, -1, -1, -1, 1
Offset: 0
Examples
T(3,2)=-1 because binomial(3,2)=3 and Moebius(3)=-1. Triangle begins: 1; 1, 1; 1, -1, 1; 1, -1, -1, 1; 1, 0, 1, 0, 1; 1, -1, 1, 1, -1, 1;
Links
- G. C. Greubel, Rows n = 0..100 of the triangle, flattened
- A. Granville and O. Ramaré, Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients, Mathematika 43 (1996), 73-107, [DOI].
Programs
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Magma
[MoebiusMu(Binomial(n, k)): k in [0..n], n in [0..15]]; // G. C. Greubel, Jun 16 2021
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Maple
with(numtheory):T:=proc(n,k) if k<=n then mobius(binomial(n,k)) else 0 fi end: for n from 0 to 15 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
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Mathematica
T[n_, k_]:= MoebiusMu[Binomial[n, k]]; Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 16 2021 *)
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PARI
T(n,k) = moebius(binomial(n,k)) for(n=0, 15, for(k=0, n, print1(T(n,k)", "))) \\ Charles R Greathouse IV, Nov 03 2014
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Sage
def T(n, k): return moebius(binomial(n, k)) flatten([[T(n, k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jun 16 2021
Comments