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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A371835 Triangle read by rows: T(n,k) is the number of points (x,y,z) satisfying |x|+|y|+|z|<=n and max(|x|,|y|,|z|)<=k; 0<=k<=n.

Original entry on oeis.org

1, 1, 7, 1, 19, 25, 1, 27, 57, 63, 1, 27, 93, 123, 129, 1, 27, 117, 195, 225, 231, 1, 27, 125, 263, 341, 371, 377, 1, 27, 125, 311, 461, 539, 569, 575, 1, 27, 125, 335, 569, 719, 797, 827, 833, 1, 27, 125, 343, 649, 895, 1045, 1123, 1153, 1159
Offset: 0

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Author

Peter Kagey, Apr 07 2024

Keywords

Comments

For all pairs of positive integers (a,b), T(a*m,b*m) satisfies a cubic polynomial in m.

Examples

			Table begins:
  n\k| 0  1   2   3   4    5    6    7    8    9   10
  ---+-----------------------------------------------
   0 | 1
   1 | 1  7
   2 | 1 19  25
   3 | 1 27  57  63
   4 | 1 27  93 123 129
   5 | 1 27 117 195 225  231
   6 | 1 27 125 263 341  371  377
   7 | 1 27 125 311 461  539  569  575
   8 | 1 27 125 335 569  719  797  827  833
   9 | 1 27 125 343 649  895 1045 1123 1153 1159
  10 | 1 27 125 343 697 1051 1297 1447 1525 1555 1561
		

Formula

T(n,k) = 8*n^3 + 12*n^2 + 6*n + 1 = A016755(k) if k <= n/3.
T(m,m) = (4*n^3 + 6*n^2 + 8*n + 3)/3 = A001845(m).
T(2m,m) = (20*n^3 + 24*n^2 + 10*n + 3)/3 = A371532(m).
T(3m,2m) = 32*n^3 + 18*n^2 + 6*n + 1 = A371515(m).
T(4m,3m) = (244*n^3 + 96*n^2 + 26*n + 3)/3.
T(5m,2m) = (188*m^3 + 132*m^2 + 28*m + 3)/3.
T(5m,3m) = (404*m^3 + 150*m^2 + 28*m + 3)/3.
T(5m,4m) = (488*m^3 + 150*m^2 + 34*m + 3)/3.
Conjectures:
T(n,k) = (-84*k^3 + 108*k^2*n - 72*k^2 - 36*k*n^2 + 72*k*n - 6*k + 4*n^3 - 12*n^2 + 8*n + 3)/3 for (n-2)/3 <= k <= n/2.
T(n,k) = (12*k^3 - 36*k^2*n + 36*k*n^2 + 6*k - 8*n^3 + 6*n^2 + 2*n + 3)/3 for (n-1)/2 <= k <= n.
The two conjectures are true. See links. - Sela Fried, Jul 05 2024
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