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A120858
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Dispersion of the Beatty sequence ([r*n]: n >= 1), where r = 3 + 8^(1/2): square array D(n,m) (n, m >= 1), read by ascending antidiagonals.
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5
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1, 2, 5, 3, 11, 29, 4, 17, 64, 169, 6, 23, 99, 373, 985, 7, 34, 134, 577, 2174, 5741, 8, 40, 198, 781, 3363, 12671, 33461, 9, 46, 233, 1154, 4552, 19601, 73852, 195025, 10, 52, 268, 1358, 6726, 26531, 114243, 430441, 1136689, 12, 58, 303, 1562
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OFFSET
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1,2
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COMMENTS
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Every positive integer occurs exactly once in array D and every pair of rows are mutually interspersed. That is, beginning at the first term of any row in D having greater initial term than that of another row, all the following terms individually separate the individual terms of the other row.
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LINKS
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FORMULA
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(1) Column 1 is the sequence ([s*n]: n >= 1) where 1/r + 1/s = 1. The numbers in all the other columns, arranged in increasing order, form the sequence ([r*n]: n >= 1).
(2) Every row satisfies these recurrences: x(n+1) = [r*x(n)] and x(n+2) = 6*x(n+1) - x(n). (Here [a] is the floor of number a.)
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EXAMPLE
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Northwest corner:
1, 5, 29, 169, 985, ...
2, 11, 64, 373, 2174, ...
3, 17, 99, 577, 3363, ...
4, 23, 134, 781, 4552, ...
6, 34, 198, 1154, 6726, ...
...
In row 1, we have 5 = [r], 29 = [5*r], 169 = [29*r], etc., where r = 3 + 8^(1/2); each new row starts with the least "new" number n, followed by [n*r], [[n*r]*r], [[[n*r]*r]*r], and so on.
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PROG
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(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)); for(m=2, nn, for(n=1, nn, D[n, m]=floor(r*D[n, m-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
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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