A307693 - OEIS

%I #17 Oct 31 2019 21:41:32

%S 1,3,1,2,3,1,6,2,2,1,7,6,4,3,1,5,7,5,2,3,1,4,5,3,6,2,2,1,12,4,8,7,5,4,

%T 2,1,13,12,9,5,6,5,4,3,1,15,13,10,4,4,3,3,2,3,1,14,15,7,12,10,8,7,6,2,

%U 3,1,10,14,6,13,11,9,9,7,6,2,2,1,11,10,16

%N Rectangular quotient array, R, of A003188 read by descending antidiagonals; see Comments.

%C Suppose that P = (p(m)) is a permutation of the positive integers, such as A038722.  For each n >= 1, let q(n,k) be the k-th index m such that n divides p(m), and let r(n) = p(q(n,k))/n.  Let R be the array having (r(n)) as row n.  We call R the quotient array of P.  Every row of R is a permutation of the positive integers.

%C In the present case that P = A003188, every row occurs infinitely many times.  Specifically, if p is a prime (A000040), then for every multiple m*p of p, the rows numbered m*p are identical.  See A327314 for the array that results by deleting duplicate rows from R.

%e A003188 = (1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8, 24, 25, 27, 26, 30, 31, 29, 28, 20, ...)

%e Row 1 of R is just A003188. To get row 2 of R, skip the odds in A003188 and divide the evens by 2; row 2 equals row 1. Generally, to get row n, divide A003188 by n and then delete the non-integers.

%e ________________

%e Northwest corner of R:

%e   1   3   2   6   7   5    4   12   13   15

%e   1   3   2   6   7   5    4   12   13   15

%e   1   2   4   5   3   8    9   10    7    6

%e   1   3   2   6   7   5    4   12   13   15

%e   1   3   2   5   6   4   10   11   12    8

%e   1   2   4   5   3   8    9   10    7    6

%t s = Table[BitXor[n, Floor[n/2]], {n, 300}]  (* A003188 *)

%t g[n_] := Flatten[Position[Mod[s, n], 0]];

%t u[n_] := s[[g[n]]]/n;

%t TableForm[Table[Take[u[n], 10], {n, 1, 20}]]  (* A307693 array *)

%t v[n_, k_] := u[n][[k]]

%t Table[v[n - k + 1, k], {n, 14}, {k, n, 1, -1}] // Flatten (* A307693 sequence *)

%Y Cf. A000040, A003188, A327314.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Oct 26 2019