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A020921
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Triangle read by rows: T(m,n) = number of solutions to 1 <= a(1) < a(2) < ... < a(m) <= n, where gcd( a(1), a(2), ..., a(m), n) = 1.
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8
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1, 1, 1, 0, 1, 1, 0, 2, 3, 1, 0, 2, 5, 4, 1, 0, 4, 10, 10, 5, 1, 0, 2, 11, 19, 15, 6, 1, 0, 6, 21, 35, 35, 21, 7, 1, 0, 4, 22, 52, 69, 56, 28, 8, 1, 0, 6, 33, 83, 126, 126, 84, 36, 9, 1, 0, 4, 34, 110, 205, 251, 210, 120, 45, 10, 1, 0, 10, 55, 165, 330, 462, 462, 330, 165, 55, 11
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OFFSET
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0,8
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LINKS
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EXAMPLE
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Triangle begins
1
1 1
0 1 1
0 2 3 1
0 2 5 4 1
0 4 10 10 5 1
0 2 11 19 15 6 1
0 6 21 35 35 21 7 1
0 4 22 52 69 56 28 8 1
0 6 33 83 126 126 84 36 9 1
0 4 34 110 205 251 210 120 45 10 1
The inverse of the triangle is
1
-1 1
1 -1 1
-1 1 -3 1
1 -1 7 -4 1
-1 1 -15 10 -5 1
1 -1 31 -19 15 -6 1
-1 1 -63 28 -35 21 -7 1
1 -1 127 -28 71 -56 28 -8 1
-1 1 -255 1 -135 126 -84 36 -9 1
1 -1 511 80 255 -251 210 -120 45 -10 1
with row sums 1,0,1,-2,4,-9,22,-55,135,-319,721,...(cf. A038200).
(End)
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MAPLE
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A020921 := proc(n, k) option remember ; local divs ; if n <= 0 then 1 ; elif k > n then 0 ; else divs := numtheory[divisors](n) ; add(numtheory[mobius](op(i, divs))*binomial(n/op(i, divs), k), i=1..nops(divs)) ; fi ; end: nmax := 10 ; for row from 0 to nmax do for col from 0 to row do printf("%d, ", A020921(row, col)) ; od ; od ; # R. J. Mathar, Feb 12 2007
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MATHEMATICA
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nmax = 11; t[n_, k_] := Total[ MoebiusMu[#]*Binomial[n/#, k] & /@ Divisors[n]]; t[0, 0] = 1; Flatten[ Table[t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Oct 20 2011, after PARI *)
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PROG
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(PARI) {T(n, k) = if( n<=0, k==0 && n==0, sumdiv(n, d, moebius(d) * binomial(n/d, k)))}
(Sage) # uses[DivisorTriangle from A327029]
DivisorTriangle(moebius, binomial, 13) # Peter Luschny, Aug 24 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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