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.

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A083362 Square table, read by antidiagonals, of least distinct positive integers such that the sum of any two consecutive terms in any row is a square number.

Original entry on oeis.org

1, 3, 2, 6, 7, 4, 10, 9, 5, 8, 15, 16, 11, 17, 12, 21, 20, 14, 19, 13, 18, 28, 29, 22, 30, 23, 31, 24, 36, 35, 27, 34, 26, 33, 25, 32, 45, 46, 37, 47, 38, 48, 39, 49, 40, 55, 54, 44, 53, 43, 52, 42, 51, 41, 50, 66, 67, 56, 68, 57, 69, 58, 70, 59, 71, 60, 78, 77, 65, 76, 64, 75
Offset: 0

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Author

Paul D. Hanna, Apr 27 2003

Keywords

Comments

A permutation of the natural numbers.

Examples

			Table begins:
   1   3   6  10  15  21  28  36  45  55  66 ...
   2   7   9  16  20  29  35  46  54  67  77 ...
   4   5  11  14  22  27  37  44  56  65  79 ...
   8  17  19  30  34  47  53  68  76  93 103 ...
  12  13  23  26  38  43  57  64  80  89 107 ...
  18  31  33  48  52  69  75  94 102 123 133 ...
  24  25  39  42  58  63  81  88 108 117 139 ...
  32  49  51  70  74  95 101 124 132 157 167 ...
  40  41  59  62  82  87 109 116 140 149 175 ...
  50  71  73  96 100 125 131 158 166 195 205 ...
  60  61  83  86 110 115 141 148 176 185 215 ...
  72  97  99 126 130 159 165 196 204 237 247 ...
		

Crossrefs

Cf. A083363 (diagonal), A083364 (antidiagonal sums).
Cf. A000217 (1st row), A080476 (1st column).

Formula

T(0, k) = (k+1)*(k+2)/2 for k>=0, T(n, 0) = floor((n+1)^2/2) for n>0, T(n, k+1) = (2*floor((n+1)/2) + k+1)^2 - T(n, k) for n>0 and k>=0.

A083364 Antidiagonal sums of table A083362.

Original entry on oeis.org

1, 5, 17, 32, 71, 105, 187, 248, 389, 485, 701, 840, 1147, 1337, 1751, 2000, 2537, 2853, 3529, 3920, 4751, 5225, 6227, 6792, 7981, 8645, 10037, 10808, 12419, 13305, 15151, 16160, 18257, 19397, 21761, 23040, 25687, 27113, 30059, 31640, 34901, 36645
Offset: 0

Views

Author

Paul D. Hanna, Apr 27 2003

Keywords

Comments

A083362 is the square table of least distinct positive integers such that the sum of any two consecutive terms in any row form a square.

Crossrefs

Programs

  • Magma
    [(4*n^3+12*n^2+18*n+9+(2*n^2+2*n-1)*(-1)^n)/8 : n in [0..40]]; // Wesley Ivan Hurt, Sep 26 2014
  • Maple
    A083364:=n->(4*n^3+12*n^2+18*n+9+(2*n^2+2*n-1)*(-1)^n)/8: seq(A083364(n), n=0..40); # Wesley Ivan Hurt, Sep 26 2014
  • Mathematica
    Table[(4 n^3 + 12 n^2 + 18 n + 9 + (2 n^2 + 2 n - 1) (-1)^n)/8, {n,0,50}] (* Wesley Ivan Hurt, Sep 26 2014 *)
    LinearRecurrence[{1,3,-3,-3,3,1,-1},{1,5,17,32,71,105,187},50] (* Harvey P. Dale, Aug 16 2021 *)
  • PARI
    Vec((x^5+6*x^4+3*x^3+9*x^2+4*x+1)/((x-1)^4*(x+1)^3) + O(x^100)) \\ Colin Barker, Sep 26 2014
    

Formula

a(2n) = n(n+1)(4n+3)+(2n+1), a(2n+1) = ((n+1)^2)(4n+3)+(2n+2), for n>=0. - Paul D. Hanna, Apr 30 2003
a(n) = a(n-1)+3*a(n-2)-3*a(n-3)-3*a(n-4)+3*a(n-5)+a(n-6)-a(n-7). - Colin Barker, Sep 26 2014
G.f.: (x^5+6*x^4+3*x^3+9*x^2+4*x+1) / ((x-1)^4*(x+1)^3). - Colin Barker, Sep 26 2014
a(n) = (4*n^3+12*n^2+18*n+9+(2*n^2+2*n-1)*(-1)^n)/8. - Wesley Ivan Hurt, Sep 26 2014
Showing 1-2 of 2 results.