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T(n,k) = A093995(n,k) + A075362(n,k) + A133819(n,k) = 2*A070216(n,k) - A215630(n,k), 1 <= k <= n.
The diagonals might derive from 1st subdiagonal T(k+1,k) = A003215(k), the centered hexagonal numbers 3k^2 + 3*k + 1 (A003215). For example, 1st subdiagonal T(k+1,k) = A003215(k). Second subdiagonal T(k+2,k)= A003215(k)/2 + A003215((k+1)/2. Third subdiagonal T(k+3,k) = A003215(k)/3 + A003215((k+1)/3 + A003215(k+2)/3 and so on. - Avi Friedlich, May 26 2015
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Second differences show expansion of tri-digit zeros interlaced with an arithmetic progression of the natural numbers (A258133). By differentiating coordinates, a(6n-5) = octagonal pyramidal numbers (A002414), a(3*n+1) = second hexagonal numbers (A014105), a(3n) - a(3*n-1) = oblong numbers (A002378), a(3n) + a(3*n-1) = stella octagonal numbers (A007588), and a(3*n+1) - a(3*n-1) = the squares (A000290). - Avi Friedlich, Jun 04 2015
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From: _Second differences show expansion of tri-digit zeros interlaced with an arithmetic progression of the natural numbers (A258133). By differentiating coordinates, a(6n-5) = octagonal pyramidal numbers (A002414), a(3*n+1) = second hexagonal numbers (A014105), a(3n) - a(3*n-1) = oblong numbers (A002378), a(3n) + a(3*n-1) = stella octagonal numbers (A007588), and a(3*n+1) - a(3*n-1) = the squares (A000290). - _Avi Friedlich_, May 11 Jun 04 2015: (Start)
Interlaced formulae may be differentiated to further characterize the sequence. For example: a(6n-5) = octagonal pyramidal numbers: (A002414).
a(3*n+1) = second hexagonal numbers (A014105). a(3n) - a(3*n-1) = oblong numbers (A002378). a(3n) + a(3*n-1) = stella octagonal numbers (A007588). a(3*n+1) - a(3*n-1) = the squares (A000290). First differences reveal triangular numbers a,b,c,d in the order of a, b, b, b, b, c, b, c, c, c, c, d, c, and second differences show expansion of tri-digit zeros interlaced with an arithmetic progression of the natural numbers. (1,0,0,0, 1, -1,1, 0, 0, 0, 2, -2, 2, 0, 0, 0, 3, -3, 3, 0, 0, 0, 4, -4, 4, 0, 0, 0, 5, -5, 5, 0, 0, 0, 6, -6, 6, 0, 0, 0, 7... (A258133). (End)
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In the limit n -> infinity, a(n+1)/a(n) -> 10+6*sqrt(2). - _Avi Friedlich_, Jun 02 2015
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