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.

A256497 Triangle read by rows, sums of 2 distinct nonzero cubes: T(n,k) = (n+1)^3+k^3, 1 <= k <= n.

Original entry on oeis.org

9, 28, 35, 65, 72, 91, 126, 133, 152, 189, 217, 224, 243, 280, 341, 344, 351, 370, 407, 468, 559, 513, 520, 539, 576, 637, 728, 855, 730, 737, 756, 793, 854, 945, 1072, 1241, 1001, 1008, 1027, 1064, 1125, 1216, 1343, 1512, 1729
Offset: 1

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Author

Bob Selcoe, Mar 31 2015

Keywords

Comments

When n=k: T(n,k) = (2n+1)(n^2+n+1). Therefore, T(n,k)/(2n+1) = A002061(n+1).
A002383 is the sequence of all primes of the form T(n,k)/(2n+1), n=k.
When starting at T(n,k) n=k, diagonal sums are n^2*(2n+1)^2. For example, starting at T(4,4) = 189: 189+243+351+513 = 4^2*9^2 = 1296.
Coefficients in T(n,k) are multiples of n+k+1; therefore, coefficients in all diagonals starting at T(n,1) are multiples of n+2.
Let T"(n,k) = T(n,k)/(n+k+1). Then reading T"(n,k) by rows:
i. Row sums are A162147(n). For example, T"(3,k) = [65/5, 72/6, 91/7] = [13,12,13]. 13+12+13 = 38; A162147(3) = 38.
ii. Extend the triangle in A215631 to a symmetric array by reflection about the main diagonal, and let that array be T"215631(n,k). Then the diagonal starting with T"215631(n,1) is row n in T"(n,k). For example, the diagonal starting at T"215631(4,1) = [21,19,19,21]; T"(4,k) = [126/6, 133/7, 152/8, 189/9] = [21,19,19,21].
iii. Coefficients in T"(n,k) are a permutation of A024612.

Examples

			Triangle starts:
n\k   1    2    3    4    5    6    7     8    9   10 ...
1:    9
2:    28  35
3:    65  72   91
4:   126  133  152  189
5:   217  224  243  280  341
6:   344  351  370  407  468  559
7:   513  520  539  576  637  728  855
8:   730  737  756  793  854  945  1072 1241
9:   1001 1008 1027 1064 1125 1216 1343 1512 1729
10:  1332 1339 1358 1395 1456 1547 1674 1843 2060 2331
...
The successive terms are (2^3+1^3), (3^3+1^3), (3^3+2^3), (4^3+1^3), (4^3+2^3), (4^3+3^3), ...
		

Crossrefs

Formula

T(n,k) = (n+1)^3+k^3.
T(n,k) = (2k+1)(k^2+k+1) + Sum_{j=k+1..n} A003215(j), n>=k+1. For example, T(8,4) = 9*21 + 91 + 127 + 169 + 217 = 793.