A120859 Dispersion of the sequence ([r*n] + 1: n >= 1), where r = 3 + 8^(1/2): square array D(n,m) (n, m >= 1), read by ascending antidiagonals.

1, 2, 6, 3, 12, 35, 4, 18, 70, 204, 5, 24, 105, 408, 1189, 7, 30, 140, 612, 2378, 6930, 8, 41, 175, 816, 3567, 13860, 40391, 9, 47, 239, 1020, 4756, 20790, 80782, 235416, 10, 53, 274, 1393, 5945, 27720, 121173, 470832, 1372105, 11, 59, 309, 1597, 8119

Offset: 1

Author

Clark Kimberling, Jul 09 2006

Keywords

• nonn
• tabl

Comments

Every positive integer occurs exactly once in array D and every pair of rows of D are mutually interspersed. That is, beginning at the first term of any row of array D having greater initial term than that of another row, all the following terms individually separate the individual terms of the other row.

Examples

			Northwest corner:
  1,  6,  35,  204, 1189, ...
  2, 12,  70,  408, 2378, ...
  3, 18, 105,  612, 3567, ...
  4, 24, 140,  816, 4756, ...
  5, 30, 175, 1020, 5945, ...
  ... [Corrected by _Petros Hadjicostas_, Jul 07 2020]
In row 1, we have 6 = [r] + 1, 35 = [6*r], 204 = [35*r] + 1, etc., where r = 3 + 8^(1/2); each new row starts with the least "new" number n, followed by [n*r] + 1, [[n*r + 1]*r + 1], [[[n*r + 1]*r + 1]*r] + 1, and so on.
		

Links

• Clark Kimberling, The equation (j+k+1)^2 - 4*k = Q*n^2 and related dispersions, Journal of Integer Sequences, 10 (2007), Article #07.2.7.
• N. J. A. Sloane, Classic Sequences.
• Eric Weisstein's World of Mathematics, Beatty sequence.
• Wikipedia, Beatty sequence.

Crossrefs

Cf. A120858, A120860, A120861, A120862, A120863.

Programs

PARI

tabls(nn)={default("realprecision", 1000); my(D=matrix(nn, nn));  r = 3 +  8^(1/2); s=r/(r-1); for(n=1, nn, D[n, 1]=floor(s*(n-1))+1); for(m=2, nn, for(n=1, nn, D[n, m]=floor(r*D[n, m-1])+1)); D}
/* To print the array flattened */
flat(nn)={D=tabls(nn); for(n=1, nn, for(m=1, n, print1(D[n+1-m, m], ", ")))}
/* To print the square array */
square(nn)={D=tabls(nn); for(n=1, nn, for(m=1, nn, print1(D[n, m], ", ")); print())} \\ Petros Hadjicostas, Jul 07 2020

Formula

(1) Column 1 is the sequence ([s*(n-1)] + 1: n >= 1), where 1/r + 1/s = 1. The numbers in all the other columns, arranged in increasing order, form the sequence ([r*n] + 1: n >= 1).

(2) Every row satisfies these recurrences: x(n+1) = [r*x(n)] + 1 and x(n+2) = 6*x(n+1) - x(n).

Extensions

Name edited by Petros Hadjicostas, Jul 07 2020