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A321332
Duration of Morse code representation of n.
2
19, 17, 15, 13, 11, 9, 11, 13, 15, 17, 39, 37, 35, 33, 31, 29, 31, 33, 35, 37, 37, 35, 33, 31, 29, 27, 29, 31, 33, 35, 35, 33, 31, 29, 27, 25, 27, 29, 31, 33, 33, 31, 29, 27, 25, 23, 25, 27, 29, 31, 31, 29, 27, 25, 23, 21, 23, 25, 27, 29, 33, 31, 29, 27, 25, 23, 25, 27, 29, 31, 35, 33, 31, 29, 27, 25, 27
OFFSET
0,1
COMMENTS
In the Morse Code (ITU) the time unit is the duration of a dot. A dash has duration of 3 dots. The space (s) between dots (d) and dashes (D) within a Morse code of a letter (here digit of a number) has duration of 1 dot. The separation (S) between two letter codes has duration of 3 dots. (The duration between two words (numbers) is 7 dots.)
Only odd numbers >= 9 appear.
There are duration twins for pairs (n-1, n) with n ending with digits 10, 20, 30, 40 or 50, except for n = 10. The digits 1 and 9, 2 and 8, and 3 and 7 are pairs with identical duration (of 17, 15, and 13, respectively).
FORMULA
a(n) = S(n) + s(n) + dD(n), where S(n) = 3*(nrdigits(n) - 1), with nrdigits(n) the number of digits of n, s(n) = 4*nrdigits(n), and dD(n) = Sum_{j=1.. nrdigits(n)} 1*nrd(d_j) + 3*nrD(dj) = 1*A280913(n) + 3*A280916(n), with nrd(dj) the number of dots of the code of the j-th digits of n, and nrD(dj) the number of dashes of the code of the j-th digits of n.
EXAMPLE
n = 10: dsDsDsDsDSDsDsDsDsD, with a(10) = 3*(2-1) + 4*2 + ((1*1 + 3*4) + (1*0 + 3*5)) = 3 + 8 + (1*1 + 3*9) = 39.
MATHEMATICA
nd[n_] := 15 - 2 * If[n<5, n, 10-n]; a[n_] := Module[{d = IntegerDigits[n]}, 7 * Length[d] - 3 + Total[nd/@ d]]; Array[a, 100, 0] (* Amiram Eldar, Dec 04 2018 *)
CROSSREFS
KEYWORD
nonn,word,easy
AUTHOR
Wolfdieter Lang, Dec 03 2018
STATUS
approved