A256497 - OEIS

A256497

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

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

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

1,1

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.

LINKS

Table of n, a(n) for n=1..45.

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.

EXAMPLE

Triangle starts:

n\k 1 2 3 4 5 6 7 8 9 10 ...

1: 9

2: 28 35

3: 65 72 91