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%I #34 May 03 2019 08:39:33
%S 1,0,1,0,1,1,0,1,1,1,0,2,3,2,1,0,3,6,5,2,1,0,5,16,16,8,3,1,0,9,39,51,
%T 30,12,3,1,0,16,104,170,125,54,16,4,1,0,28,270,585,516,259,84,21,4,1,
%U 0,51,729,2048,2232,1296,480,128,27,5,1,0,93,1960,7280,9750,6665,2792,819
%N Invertible triangle: T(n,k) = number of k-ary Lyndon words of length n-k+1 with trace 1 modulo k.
%C An invertible number triangle related to Lyndon words of trace 1.
%H Andrew Howroyd, <a href="/A110540/b110540.txt">Table of n, a(n) for n = 1..1275</a>
%H F. Ruskey, <a href="http://combos.org/TSlyndon">Number of q-ary Lyndon words with given trace mod q</a>
%H F. Ruskey, <a href="http://combos.org/Tpoly">Number of monic irreducible polynomials over GF(q) with given trace</a>
%H F. Ruskey, <a href="http://combos.org/TlyndonFk">Number of Lyndon words over GF(q) with given trace</a>
%F T(n, k) = Sum_{d | n-k+1, gcd(d, k)=1} mu(d)*k^((n-k+1)/d))/(k*(n-k+1)).
%e Rows begin
%e 1;
%e 0, 1;
%e 0, 1, 1;
%e 0, 1, 1, 1;
%e 0, 2, 3, 2, 1;
%e 0, 3, 6, 5, 2, 1;
%e 0, 5, 16, 16, 8, 3, 1;
%e 0, 9, 39, 51, 30, 12, 3, 1;
%t T[n_, k_]:=Sum[Boole[GCD[d, k] == 1] MoebiusMu[d] k^((n - k + 1)/d), {d, Divisors[n - k + 1]}] /(k(n - k + 1)); Flatten[Table[T[n, k], {n, 12}, {k, n}]] (* _Indranil Ghosh_, Mar 27 2017 *)
%o (PARI)
%o for(n=1, 11, for(k=1, n, print1( sum(d=1,n-k+1, if(Mod(n-k+1, d)==0 && gcd(d, k)==1, moebius(d)*k^((n-k+1)/d), 0)/(k*(n-k+1)) ),", ");); print();) \\ _Andrew Howroyd_, Mar 26 2017
%Y Columns include A000048, A046211, A054660, A054662, A054666.
%K easy,nonn,tabl
%O 1,12
%A _Paul Barry_, Jul 25 2005
%E Name clarified by _Andrew Howroyd_, Mar 26 2017
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