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 1-2 of 2 results.

A380661 Rectangular array neg(i,j,1,2) read by descending antidiagonals: pos() and neg() denote the positive part and negative part of a determinant; see Comments.

Original entry on oeis.org

6, 20, 30, 56, 72, 90, 132, 156, 182, 210, 272, 306, 342, 380, 420, 506, 552, 600, 650, 702, 756, 870, 930, 992, 1056, 1122, 1190, 1260, 1406, 1482, 1560, 1640, 1722, 1806, 1892, 1980, 2162, 2256, 2352, 2450, 2550, 2652, 2756, 2862, 2970, 3192, 3306, 3422
Offset: 1

Views

Author

Clark Kimberling, Feb 04 2025

Keywords

Comments

Suppose that (m(i,j)) is a rectangular array of infinitely many rows and infinitely many columns. For integers s>=1 and n>=1, let M(i,j,s,n) be the nXn matrix (m(i+h*s,j+k*s)), where h=0..n-1, k=0..n-1.
Let D(i,j,s,n) and P(i,j,s,n) denote the determinant and permanent of M(i,j,s,n), respectively. Define arrays pos(i,j,s,n) and neg(i,j,s,n) by pos(i,j,s,n) = (P(i,j,s,n)+D(i,j,s,n))/2 and neg(i,j,s,n) = (P(i,j,s,n)-D(i,j,s,n))/2, so that P(i,j,s,n) = pos(i,j,s,n)+neg(i,j,s,n) and D(i,j,s,n) = pos(i,j,s,n)-neg(i,j,s,n).
A definition of determinant of an nXn matrix (a(i,j)) is the sum of the products (-1)^p(u) a(1,j(1))*a(2,j(2))*...*a(n,j(n)) over the n! permutations u = (j(1),j(2),...,j(n)) of (1,2,...,n), where p(u) is the parity of u; i.e., p(u) = 0 or 1 according as u is an even or odd permutation; see Lang, pp. 452-3, especially Proposition 4.8.
We have:
pos(i,j,s,n) is the sum of the n!/2 products for which p(u) = 0, and
neg(i,j,s,n) is the sum of the n!/2 products for which p(u) = 1.
Here, the foundational array (m(i,j)) is the natural number array (see A000027, A185787, A144112). The row sequences of pos(i,j,s,n) and neg(i,j,s,n) are linearly recurrent with signature (5, -10, 10, -5, 1).

Examples

			Corner of neg(i,j,1,2):
     6     20    56     132    272    506    870   1406   2162   3192
    30     72    156    306    552    930   1482   2256   3306   4692
    90    182    342    600    992   1560   2352   3422   4830   6642
   210    380    650   1056   1640   2450   3540   4970   6806   9120
   420    702   1122   1722   2550   3660   5112   6972   9312  12210
   756   1190   1806   2652   3782   5256   7140   9506  12432  16002
  1260   1892   2756   3906   5402   7310   9702  12656  16256  20592
  1980   2862   4032   5550   7482   9900  12882  16512  20880  26082
  2970   4160   5700   7656  10100  13110  16770  21170  26406  32580
  4290   5852   7832  10302  13340  17030  21462  26732  32942  40200
  6006   8010  10506  13572  17292  21756  27060  33306  40602  49062
  8190  10712  13806  17556  22052  27390  33672  41006  49506  59292
M(1,1,1,2) is the matrix with (row 1) = (1,2), (row 2) =(3,5), so that
pos(1,1,1,2) = 1*5 = 5; neg(1,1,1,2) = 2*3 = 6; D(1,1,1,2) = -1; P(1,1,1,2) = 11.
		

References

  • S. Lang, Algebra, 2nd ed., Addison-Wesley, 1984, 452-453.

Crossrefs

Programs

  • Mathematica
    s = 1; n = 2; z = 12;
    r[n_, k_] := n + (n + k - 2)*(n + k - 1)/2 (* Array A000027 *)
    Grid[Table[r[n, k], {n, 1, z}, {k, 1, z}]]
    t[i_, j_] := Table[r[i, j + k*s], {k, 0, n - 1}];
    d[i_, j_] := Det[Table[t[i + k*s, j], {k, 0, n - 1}]];  (* D(i,j,s,n) *)
    p[i_, j_] := Permanent[Table[t[i + k*s, j], {k, 0, n - 1}]];  (* P(i,j,s,n) *)
    pos[i_, j_] := (p[i, j] + d[i, j])/2;
    neg[i_, j_] := (p[i, j] - d[i, j])/2;
    Grid[Table[pos[i, j], {i, 1, z}, {j, 1, z}]]  (* A380660 array *)
    Grid[Table[neg[i, j], {i, 1, z}, {j, 1, z}]]  (* A380661 array *)
    FindLinearRecurrence[Table[pos[1, k], {k, 1, 20}]] (* row recurrence, all rows *)
    FindLinearRecurrence[Table[neg[7, k], {k, 1, 20}]] (* row recurrence, all rows *)
    Table[pos[k, m - k], {m, 2, z}, {k, 1, m - 1}] // Flatten (* A380660 sequence *)
    Table[neg[k, m - k], {m, 2, z}, {k, 1, m - 1}] // Flatten (* A380661 sequence *)

A380649 Rectangular array ((-1)*D(i,j,1,2)) read by descending antidiagonals: D(i,j,s,n) denotes the determinant of the matrix described in Comments.

Original entry on oeis.org

1, 4, 3, 8, 7, 6, 13, 12, 11, 10, 19, 18, 17, 16, 15, 26, 25, 24, 23, 22, 21, 34, 33, 32, 31, 30, 29, 28, 43, 42, 41, 40, 39, 38, 37, 36, 53, 52, 51, 50, 49, 48, 47, 46, 45, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66
Offset: 1

Views

Author

Clark Kimberling, Jan 31 2025

Keywords

Comments

Suppose that (m(i,j)) is a rectangular array of infinitely many rows and infinitely many columns. For integers s>=1 and n>=1, let M(i,j,s,n) be the nXn matrix (m(i+h*s,j+k*s)), where h=0..n-1, k=0..n-1.
Let D(i,j,s,n) and P(i,j,s,n) denote the determinant and permanent of M(i,j,s,n), respectively. For A380649, we take (m(i,j)) to be the natural number array (see A000027, A185787, and A144112), and ((-1)*D(i,j,1,2)) is as shown below in Example.
* D(i,j,1,1) = M(i,j,1,1) = m(i,j) has linearly recurrent row sequences, all with signature (3,-3,1).
* D(i,j,1,2) has linearly recurrent row sequences, all with signature (3,-3,1).
* (-1)*D(i,j,s,3) is the constant array in which every term is s^6, for all i,j,s.
* D(i,j,s,n) is the constant 0 array for all n>=4, for all i,j,s.
* P(i,j,s,n) depends only on n, and the rows all have the following linear recurrence signature:
b(2n+1,1), - b(2n+1,2), b(2n+1-3),..., -(2n+1,2n), 1, where b=binomial.
((-1)*D(i,j,1,2)) includes, exactly once, every positive integer not in A000096. The order array of ((-1)*D(i,j,1,2)) is the array in Example in A038722; see A333029 for the definition of order array.

Examples

			Corner of (-1)*D(i,j,1,2):
   1   4    8   13   19   26   34   43   53   64   76   89
   3   7   12   18   25   33   42   52   63   75   88  102
   6  11   17   24   32   41   51   62   74   87  101  116
  10  16   23   31   40   50   61   73   86  100  115  131
  15  22   30   39   49   60   72   85   99  114  130  147
  21  29   38   48   59   71   84   98  113  129  146  164
  28  37   47   58   70   83   97  112  128  145  163  182
  36  46   57   69   82   96  111  127  144  162  181  201
  45  56   68   81   95  110  126  143  161  180  200  221
  55  67   80   94  109  125  142  160  179  199  220  242
  66  79   93  108  124  141  159  178  198  219  241  264
  78  92  107  123  140  158  177  197  218  240  263  287
m(1,1) = 1, so M(1,1,1,2) is the matrix having (row 1) = (1,2) and (row 2) = (3,5), so (-1)*D(1,1,1,2) = -(1*5-2*3) = 1.
		

Crossrefs

Programs

  • Mathematica
    s = 1; n = 2; z = 12;
    r[n_, k_] := n + (n + k - 2)*(n + k - 1)/2;
    Grid[Table[r[n, k], {n, 1, z}, {k, 1, z}]]  (* Array A000027 *)
    FindLinearRecurrence[Table[r[1, k], {k, 1, 20}]]
    t[i_, j_] := Table[r[i, j + k*s], {k, 0, n - 1}];
    d[i_, j_] := -Det[Table[t[i + k*s, j], {k, 0, n - 1}]];  (* (-1)*D(i,j,s,n) *)
    Grid[Table[d[i, j], {i, 1, z}, {j, 1, z}]]  (* array *)
    FindLinearRecurrence[Table[d[12, k], {k, 1, 20}]]
    Table[d[k, m - k], {m, 2, z}, {k, 1, m - 1}] // Flatten (* sequence *)
Showing 1-2 of 2 results.