A330733 - OEIS

%I #48 Jul 18 2021 19:26:53

%S 1,0,1,0,0,1,0,1,0,2,0,0,0,0,1,0,1,1,1,0,3,0,0,0,0,0,0,1,0,2,0,3,0,2,

%T 0,4,0,0,1,0,0,1,0,0,2,0,1,0,1,1,1,0,1,0,3,0,0,0,0,0,0,0,0,0,0,1,0,3,

%U 2,4,0,6,0,4,2,3,0,8,0,0,0,0,0,0,0,0,0,0,0,0,1

%N Triangle read by rows in which row n is the "complete rhythm" of n (see Comments for precise definition).

%C Define the "natural rhythm" of any positive integer n to be the sequence consisting of n-1 zeros followed by 1; e.g., the natural rhythm of 5 is [0, 0, 0, 0, 1].

%C Define the "complete rhythm" of any positive integer n to be the term-by-term sum of the natural rhythm of n and the complete rhythm of f for every proper divisor f of n, extended through n/f cycles so as to give n terms. (Thus the complete rhythm of any noncomposite number is simply its natural rhythm.)

%C E.g., n=4 has a unique proper factor f=2 (whose complete rhythm is simply its natural rhythm, since 2 is prime).

%C Thus, for 4, we must add the following two components:

%C   [0, 0, 0, 1]  (the natural rhythm of 4)

%C + [0, 1, 0, 1]  (the rhythm of 2, repeated to give 4 terms)

%C ==============

%C   [0, 1, 0, 2]  (the complete rhythm of 4).

%C Right diagonal is A002033 (conjectured).

%C Any prime column stripped of zeros also yields A002033 (conjectured).

%C From _Michael De Vlieger_, Dec 29 2019: (Start)

%C Positions of 0 in each row n > 1 are in the reduced residue system of n (A038566). Therefore the number of zeros in each row n > 1 is given by the Euler totient function (A000010). This arises because a nonzero addend is introduced for multiples of divisors of n; the numbers k < n such that gcd(k,n) = 1 remain 0.

%C Conversely, nonzero positions in each row n > 1 are in the cototient of n (A121998), their number given by row n of A051953. (End)

%e Here are the rhythms of the first thirteen positive integers:

%e    1 | 1

%e    2 | 0,  1

%e    3 | 0,  0,  1

%e    4 | 0,  1,  0,  2

%e    5 | 0,  0,  0,  0,  1

%e    6 | 0,  1,  1,  1,  0,  3

%e    7 | 0,  0,  0,  0,  0,  0,  1

%e    8 | 0,  2,  0,  3,  0,  2,  0,  4

%e    9 | 0,  0,  1,  0,  0,  1,  0,  0,  2

%e   10 | 0,  1,  0,  1,  1,  1,  0,  1,  0,  3

%e   11 | 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  1

%e   12 | 0,  3,  2,  4,  0,  6,  0,  4,  2,  3,  0,  8

%e   13 | 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  1

%e .

%e The complete rhythm of 12 is composed as follows:

%e 12 has a "natural rhythm" of

%e   12 | 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  1

%e 12 has proper divisors 2, 3, 4 and 6, whose complete rhythms are

%e    2 | 0,  1

%e    3 | 0,  0,  1

%e    4 | 0,  1,  0,  2

%e    6 | 0,  1,  1,  1,  0,  3

%e When the padded (i.e., repeated) rhythms of the proper factors are added to the natural rhythm of 12, we have

%e    2 | 0,  1,  0,  1,  0,  1,  0,  1,  0,  1,  0,  1

%e    3 | 0,  0,  1,  0,  0,  1,  0,  0,  1,  0,  0,  1

%e    4 | 0,  1,  0,  2,  0,  1,  0,  2,  0,  1,  0,  2

%e    6 | 0,  1,  1,  1,  0,  3,  0,  1,  1,  1,  0,  3

%e   12 | 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  1

%e   ===+==============================================

%e   12 | 0,  3,  2,  4,  0,  6,  0,  4,  2,  3,  0,  8

%t Nest[Function[{a, n, d}, Append[#1, Total@ Map[PadRight[a[[#]], n, a[[#]] ] &, d] + Append[ConstantArray[0, n - 1], 1]]] @@ {#1, #2, Most@ Rest@ Divisors[#2]} & @@ {#, Length@ # + 1} &, {{1}}, 12] // Flatten (* _Michael De Vlieger_, Dec 29 2019 *)

%o (Python)

%o def memoize(f):

%o     memo = {}

%o     def helper(x):

%o         if x not in memo:

%o             memo[x] = f(x)

%o         return memo[x]

%o     return helper

%o @memoize

%o def unique_factors_of(n):

%o     factors = []

%o     for candidate in range(2, n//2 + 1):

%o         if n % candidate == 0:

%o             factors.append(candidate)

%o     return factors

%o @memoize

%o def is_prime(n):

%o     if n <= 1:

%o         return False

%o     if n <= 3:

%o         return True

%o     if n % 2 == 0 or n % 3 == 0:

%o         return False

%o     i = 5

%o     while i * i <= n:

%o         if n % i == 0 or n % (i + 2) == 0:

%o             return False

%o         i = i + 6

%o     return True

%o @memoize

%o def rhythm(n):

%o     if n == 0:

%o         return [0]

%o     natural_rhythm_of_n = [0]*(n-1)

%o     natural_rhythm_of_n += [1]

%o     if is_prime(n):

%o         return natural_rhythm_of_n

%o     else:

%o         component_rhythms = [natural_rhythm_of_n]

%o         for divisor in unique_factors_of(n):

%o             component_rhythm = n//divisor * rhythm(divisor)

%o             component_rhythms.append(component_rhythm)

%o         return [sum(i) for i in zip(*component_rhythms)]

%o for i in range(0, 201):

%o     formatted_string = f'{str(i).rjust(3)}|'

%o     for note in rhythm(i):

%o         formatted_string += f'{str(note).rjust(4)}'

%o     print(formatted_string)

%Y Cf. A002033 (number of perfect partitions of n), A000040 (prime numbers), A000010, A038566, A051953, A121998.

%K nonn,tabl

%O 1,10

%A _Andrew Hood_, Dec 28 2019

%E Name clarified by _Omar E. Pol_ and _Jon E. Schoenfield_, Dec 31 2019