cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A120861 Fixed-k dispersion for Q = 8: Square array D(g,h) (g, h >= 1), read by ascending antidiagonals.

Original entry on oeis.org

1, 2, 7, 3, 12, 41, 4, 19, 70, 239, 5, 24, 111, 408, 1393, 6, 31, 140, 647, 2378, 8119, 8, 36, 181, 816, 3771, 13860, 47321, 9, 48, 210, 1055, 4756, 21979, 80782, 275807, 10, 53, 280, 1224, 6149, 27720, 128103, 470832, 1607521, 11, 60, 309, 1632, 7134
Offset: 1

Views

Author

Clark Kimberling, Jul 09 2006

Keywords

Comments

For each positive integer n, there exists a unique pair (j,k) of positive integers such that (j + k + 1)^2 - 4*k = 8*n^2; in fact, j(n) = A087056(n) and k(n) = A087059(n).
Suppose g >= 1 and let k = k(g). The numbers in row g of array D are among those n for which (j + k + 1)^2 - 4*k = 8*n^2 for some j; that is, k stays fixed and j and n vary - hence the name "fixed-k dispersion". (The fixed-j dispersion for Q = 8 is A120860.)
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 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,  7,  41,  239, 1393,  8119,  47321, ...
  2, 12,  70,  408, 2378, 13860,  80782, ...
  3, 19, 111,  647, 3771, 21979, 128103, ...
  4, 24, 140,  816, 4756, 27720, 161564, ...
  5, 31, 181, 1055, 6149, 35839, 208885, ...
  6, 36, 210, 1224, 7134, 41580, 242346, ...
... [Edited by _Petros Hadjicostas_, Jul 07 2020]
		

Crossrefs

Cf. A087056, A087059, A120858, A120859, A120860, A120862, A120863, A336109 (first column), A002315 (first row), A001542 (2nd row), A253811 (3rd row).

Programs

  • PARI
    f(n) = 3*n + 2*sqrtint(2*n^2) + 2;
    unused(listus) = {my(v=vecsort(Vec(listus))); for (i=1, vecmax(v), if (!vecsearch(v, i), return (i)););};
    D(nb) = {my(m = matrix(nb, nb), t); my(listus = List); for (g=1, nb, if (g==1, t = 1, t = unused(listus)); m[g,1]=t; listput(listus, t); t = f(t); m[g,2]=t; listput(listus, t); for (h=3, nb, t = 6*m[g,h-1] - m[g,h-2]; m[g, h] = t; listput(listus, t););); m;}; \\ Michel Marcus, Jul 08 2020

Formula

Define f(n) = 3*n + 2*floor(n*sqrt(2)) + 2. Let D(g,h) be the term in row g and column h of the array to be defined:
D(1,1) = 1; D(1,2) = f(1); and D(1,h) = 6*D(1,h-1) - D(1,h-2) for h >= 3.
For arbitrary g >= 1, once row g is defined, define D(g+1,1) = least positive integer not in rows 1,2,...,g; D(g+1,2) = f(D(g+1,1)); and D(g+1,h) = 6*D(g+1,h-1) - D(g+1,h-2) for h >= 3. All rows after row 1 are thus inductively defined. [Corrected by Petros Hadjicostas, Jul 07 2020]

Extensions

Name edited by Petros Hadjicostas, Jul 07 2020

A120860 Fixed-j dispersion for Q = 8: Square array D(g,h) (g, h >= 1), read by ascending antidiagonals.

Original entry on oeis.org

1, 2, 5, 3, 10, 29, 4, 17, 58, 169, 6, 22, 99, 338, 985, 7, 34, 128, 577, 1970, 5741, 8, 39, 198, 746, 3363, 11482, 33461, 9, 46, 227, 1154, 4348, 19601, 66922, 195025, 11, 51, 268, 1323, 6726, 25342, 114243, 390050, 1136689, 12, 63, 297, 1562, 7711
Offset: 1

Views

Author

Clark Kimberling, Jul 09 2006

Keywords

Comments

For each positive integer n, there exists a unique pair (j,k) of positive integers such that (j + k + 1)^2 - 4*k = 8*n^2; in fact, j(n) = A087056(n) and k(n) = A087059(n). Suppose g >= 1 and let j = j(g).
The numbers in row g of array D are among those n for which (j + k + 1)^2 - 4*k = 8*n^2 for some k; that is, j stays fixed and k and n vary - hence the name "fixed-j dispersion". (The fixed-k dispersion for Q = 8 is A120861.)
Every positive integer occurs exactly once in D and every pair of rows are mutually interspersed. That is, beginning at the first term of any row having greater initial term than that of another row, all the following terms individually separate the individual terms of the other row. Possibly, D is the dispersion of A098021.
From Petros Hadjicostas, Jul 08 2020: (Start)
It appears that the first column of the dispersion array D is A187968. That is, the first column of D consists of those positive integers m such that A187967(m) = 1; i.e., those m for which floor(sqrt(2)*m + 2*sqrt(2)) - floor(sqrt(2)*m) - floor(2*sqrt(2)) = 1.
Indeed, Clark Kimberling, the author of this sequence and of sequence A187968, conjectured that A187968 is the complement of A098021, and that A098021 is the complement of the first column of array D here. (End)

Examples

			Northwest corner:
  1,  5,  29,  169,  985,  5741,  33461,  195025, ...
  2, 10,  58,  338, 1970, 11482,  66922,  390050, ...
  3, 17,  99,  577, 3363, 19601, 114243,  665857, ...
  4, 22, 128,  746, 4348, 25342, 147704,  860882, ...
  6, 34, 198, 1154, 6726, 39202, 228486, 1331714, ...
  7, 39, 227, 1323, 7711, 44943, 261947, 1526739, ...
  ...
		

Crossrefs

Programs

  • PARI
    f(n) = 3*n + 2*sqrtint(2*n^2); \\ A098021
    unused(listus) = {my(v=vecsort(Vec(listus))); for (i=1, vecmax(v), if (!vecsearch(v, i), return (i)););};
    D(nb) = {my(m = matrix(nb, nb), t); my(listus = List); for (g=1, nb, if (g==1, t = 1, t = unused(listus)); m[g,1]=t; listput(listus, t); t = f(t); m[g,2]=t; listput(listus, t); for (h=3, nb, t = 6*m[g,h-1] - m[g,h-2]; m[g, h] = t; listput(listus, t););); m;}; \\ Michel Marcus, Jul 08 2020

Formula

Define f(n) = 3*n + 2*floor(n*2^(1/2)). Let D(g,h) be the term in row g and column h of the array to be defined:
D(1,1) = 1; D(1,2) = f(1); D(1,h) = 6*D(1,h-1) - D(1,h-2) for h >= 3.
For arbitrary g >= 1, once row g is defined, define D(g+1,1) = least positive integer not in rows 1, 2, ..., g; D(g+1,2) = f(D(g+1,1)); and D(g+1,h) = 6*D(g+1,h-1) - D(g+1,h-2) for h >= 3. All rows after row 1 are thus inductively defined. [Corrected by Petros Hadjicostas, Jul 07 2020]

Extensions

Name edited by Petros Hadjicostas, Jul 07 2020

A120862 Fixed-j dispersion for Q = 13: array D(g,h) (g, h >= 1), read by ascending antidiagonals.

Original entry on oeis.org

1, 2, 10, 3, 20, 109, 4, 30, 218, 1189, 5, 43, 327, 2378, 12970, 6, 53, 469, 3567, 25940, 141481, 7, 63, 578, 5116, 38910, 282962, 1543321, 8, 76, 687, 6305, 55807, 424443, 3086642, 16835050, 9, 86, 829, 7494, 68777, 608761, 4629963, 33670100, 183642229
Offset: 1

Views

Author

Clark Kimberling, Jul 09 2006

Keywords

Comments

For each positive integer n, there exists a unique pair (j,k) of positive integers such that (j+k+1)^2 - 4*k = 13*n^2; in fact, j(n) = A120869(n) and k(n) = A120870(n).
Suppose g >= 1 and let j = j(g). The numbers in row g of array D are among those n for which (j+k+1)^2 - 4*k = 13*n^2 for some k; that is, j stays fixed and k and n vary - hence the name "fixed-j dispersion". (The fixed-k dispersion for Q=13 is A120863.)
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 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, 10, 109, 1189, ...
  2, 20, 218, 2378, ...
  3, 30, 327, 3567, ...
  4, 43, 469, 5116, ...
  5, 53, 578, 6305, ...
  6, 63, 687, 7494, ...
  ...
		

Crossrefs

Programs

  • PARI
    f(n) = floor((11 + 3*sqrt(13))/2*n) - floor(3*frac((1 + sqrt(13))*n/2));
    unused(listus) = {my(v=vecsort(Vec(listus))); for (i=1, vecmax(v), if (!vecsearch(v, i), return (i)); ); };
    D(nb) = {my(m = matrix(nb, nb), t); my(listus = List); for (g=1, nb, if (g==1, t = 1, t = unused(listus)); m[g, 1]=t; listput(listus, t); t = f(t); m[g, 2]=t; listput(listus, t); for (h=3, nb, t = 11*m[g, h-1] - m[g, h-2]; m[g, h] = t; listput(listus, t); ); ); m; };
    lista(nb) = {my(m=D(nb)); for (n=1, nb, for (j=1, n, print1(m[n-j+1, j], ", ");););} \\ Michel Marcus, Jul 09 2020

Formula

Define f(n) = floor(r*n) - floor(3*F(n)), where r = (11 + 3*sqrt(13))/2 and F(n) is the fractional part of (1 + sqrt(13))*n/2. Let D(g,h) be the term in row g and column h of the array to be defined:
D(1,1) = 1; D(1,2) = f(1); and D(1,h) = 11*D(1,h-1) - D(1,h-2) for h >= 3.
For arbitrary g >= 1, once row g is defined, define D(g+1,1) = least positive integer not in rows 1,2,...,g; D(g+1,2) = f(D(g+1,1)); and D(g+1,h) = 11*D(g+1,h-1) - D(g+1,h-2) for h >= 3. All rows after row 1 are thus inductively defined. [Corrected using Conjecture 1 in Kimberling (2007) by Petros Hadjicostas, Jul 07 2020]

Extensions

Name edited by Petros Hadjicostas, Jul 07 2020
More terms from Michel Marcus, Jul 09 2020

A120863 Fixed-k dispersion for Q = 13: array D(g,h) (g, h >= 1), read by ascending antidiagonals.

Original entry on oeis.org

1, 2, 13, 3, 23, 142, 4, 33, 251, 1549, 5, 46, 360, 2738, 16897, 6, 56, 502, 3927, 29867, 184318, 7, 66, 611, 5476, 42837, 325799, 2010601, 8, 79, 720, 6665, 59734, 467280, 3553922, 21932293, 9, 89, 862, 7854, 72704, 651598, 5097243, 38767343, 239244622
Offset: 1

Views

Author

Clark Kimberling, Jul 09 2006

Keywords

Comments

For each positive integer n, there exists a unique pair (j,k) of positive integers such that (j+k+1)^2 - 4*k = 13*n^2; in fact, j(n) = A120869(n) and k(n) = A120870(n).
Suppose g >= 1 and let k = k(g). The numbers in row g of array D are among those n for which (j+k+1)^2 - 4*k = 13*n^2 for some j; that is, k stays fixed and j and n vary - hence the name "fixed-k dispersion". (The fixed-j dispersion for Q = 13 is A120862.)
Every positive integer occurs exactly once in the array D and every pair of rows are mutually interspersed. That is, beginning at the first term of any row of 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, 13, 142, 1549, ...
  2, 23, 251, 2738, ...
  3, 33, 360, 3927, ...
  4, 46, 502, 5476, ...
  5, 56, 611, 6665, ...
  6, 66, 720, 7854, ...
  ...
		

Crossrefs

Programs

  • PARI
    f(n) = floor((11 + 3*sqrt(13))/2*n) - floor(3*frac((1 + sqrt(13))*n/2)) + 3;
    unused(listus) = {my(v=vecsort(Vec(listus))); for (i=1, vecmax(v), if (!vecsearch(v, i), return (i)); ); };
    D(nb) = {my(m = matrix(nb, nb), t); my(listus = List); for (g=1, nb, if (g==1, t = 1, t = unused(listus)); m[g, 1]=t; listput(listus, t); t = f(t); m[g, 2]=t; listput(listus, t); for (h=3, nb, t = 11*m[g, h-1] - m[g, h-2]; m[g, h] = t; listput(listus, t); ); ); m; };
    lista(nb) = {my(m=D(nb)); for (n=1, nb, for (j=1, n, print1(m[n-j+1, j], ", ");););} \\ Michel Marcus, Jul 09 2020

Formula

Define f(n) = floor(r*n) - floor(3*F(n)) + 3, where r = (11 + 3*sqrt(13))/2 and F(n) is the fractional part of (1 + sqrt(13))*n/2. Let D(g,h) be the term in row g and column h of the array to be defined:
D(1,1) = 1; D(1,2) = f(1); and D(1,h) = 11*D(1,h-1) - D(1,h-2) for h >= 3.
For arbitrary g >= 1, once row g is defined, define D(g+1,1) = least positive integer not in rows 1,2,...,g; D(g+1,2) = f(D(g+1,1)); and D(g+1,h) = 11*D(g+1,h-1) - D(g+1,h-2) for h >= 3. All rows after row 1 are thus inductively defined. [Corrected using Conjecture 2 in Kimberling (2007) by Petros Hadjicostas, Jul 07 2020]

Extensions

Name edited by Petros Hadjicostas, Jul 07 2020
More terms from Michel Marcus, Jul 09 2020

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.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jul 09 2006

Keywords

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.
		

Crossrefs

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
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