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 A110540 Invertible triangle: T(n,k) = number of k-ary Lyndon words of length n-k+1 with trace 1 modulo k. 1
 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, 30, 12, 3, 1, 0, 16, 104, 170, 125, 54, 16, 4, 1, 0, 28, 270, 585, 516, 259, 84, 21, 4, 1, 0, 51, 729, 2048, 2232, 1296, 480, 128, 27, 5, 1, 0, 93, 1960, 7280, 9750, 6665, 2792, 819 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,12 COMMENTS An invertible number triangle related to Lyndon words of trace 1. LINKS Andrew Howroyd, Table of n, a(n) for n = 1..1275 FORMULA T(n, k) = Sum_{d | n-k+1, gcd(d, k)=1} mu(d)*k^((n-k+1)/d))/(k*(n-k+1)). EXAMPLE Rows begin   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, 30, 12, 3, 1; MATHEMATICA 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 *) PROG (PARI) 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 CROSSREFS Columns include A000048, A046211, A054660, A054662, A054666. Sequence in context: A108619 A091327 A327758 * A339071 A083475 A211994 Adjacent sequences:  A110537 A110538 A110539 * A110541 A110542 A110543 KEYWORD easy,nonn,tabl AUTHOR Paul Barry, Jul 25 2005 EXTENSIONS Name clarified by Andrew Howroyd, Mar 26 2017 STATUS approved

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Last modified July 30 07:58 EDT 2021. Contains 346348 sequences. (Running on oeis4.)