A221915 Array for a certain labeled Morse code, recorded in standard order.
1, 1, 2, 1, 6, 3, 1, 24, 12, 4, 2, 1, 120, 60, 20, 10, 6, 5, 3, 1, 720, 360, 120, 60, 36, 24, 30, 18, 12, 6, 4, 2, 1, 5040, 2520, 840, 420, 252, 168, 120, 210, 126, 84, 60, 42, 28, 20, 14, 10, 6, 7, 5, 3, 1, 40320, 20160, 6720, 3360, 2016, 1344, 960, 720, 1680, 1008, 672, 480, 360, 336, 224, 160, 120, 112, 80, 60, 48, 36, 24, 56, 40, 30, 24, 18, 12, 8, 6, 4, 2, 1
Offset: 0
Examples
The array a(n,k) begins: n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 ... 0: 1 1: 1 2: 2 1 3: 6 3 1 4: 24 12 4 2 1 5: 120 60 20 10 6 5 3 1 6: 720 360 120 60 36 24 30 18 12 6 4 2 1 ... Row n=7: 5040 2520, 840, 420, 252, 168, 120, 210, 126, 84, 60, 42, 28, 20, 14, 10, 6, 7, 5, 3, 1; Row n=8: 40320, 20160, 6720, 3360, 2016, 1344, 960, 720, 1680, 1008, 672, 480, 360, 336, 224, 160, 120, 112, 80, 60, 48, 36, 24, 56, 40, 30, 24, 18, 12, 8, 6, 4, 2, 1. a(5,0) = 5! from dot dot dot dot dot, the first code, with labels 1*2*3*4*5*x^5. a(5,3) = 10 because the (3+1)-th Morse code over {1,2,...,5} in standard order has m = 1 dash which starts at position 3: dot dot dash dot with label 1*2*5*x^1= (5!/(3*4))*x^1 = 10*x. This code belongs to the combination [3] of binomial(5-1,1) = binomial(4,1) which is the third one. Labeled Morse code row polynomial Q(3,x) = 6*x^0 + (3 + 1)*x^1 = 6 + 4*x.
References
- Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2n-d ed.; Addison-Wesley, 1994.
Crossrefs
Cf. A084950.
Formula
For k >= 1: a(n,k) = product of dot positions of the k-th Morse code (dashes and dots on {1, 2, ..., n}), ordered in a standard way with nondecreasing dash number m, 0 <= m < = floor(n/2), and lexicographic order based on the increasing start positions of the dashes of the codes with m dashes. In addition a(n,0) := n!.
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