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.

A133826 Triangle whose rows are sequences of increasing and decreasing tetrahedral numbers: 1; 1,4,1; 1,4,10,4,1; ... .

Original entry on oeis.org

1, 1, 4, 1, 1, 4, 10, 4, 1, 1, 4, 10, 20, 10, 4, 1, 1, 4, 10, 20, 35, 20, 10, 4, 1, 1, 4, 10, 20, 35, 56, 35, 20, 10, 4, 1, 1, 4, 10, 20, 35, 56, 84, 56, 35, 20, 10, 4, 1, 1, 4, 10, 20, 35, 56, 84, 120, 84, 56, 35, 20, 10, 4, 1, 1, 4, 10, 20, 35, 56, 84, 120, 165, 120, 84, 56, 35, 20, 10, 4, 1
Offset: 0

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Author

Peter Bala, Sep 25 2007

Keywords

Comments

Reading the triangle by rows produces the sequence 1,1,4,1,1,4,10,4,1,..., analogous to A004737.
T(n,k) = min(n*(n+1)*(n+2)/6, k*(k+1)*(k+2)/6) n, k > 0. The order of the list T(n,k) is by sides of squares from T(1,n) to T(n,n), then from T(n,n) to T(n,1). - Boris Putievskiy, Jan 13 2013

Examples

			Triangle T(n,k) starts:
  1;
  1,  4,  1;
  1,  4, 10,  4,  1;
  1,  4, 10, 20, 10,  4,  1;
  1,  4, 10, 20, 35, 20, 10,  4,  1;
  1,  4, 10, 20, 35, 56, 35, 20, 10,  4,  1;
  1,  4, 10, 20, 35, 56, 84, 56, 35, 20, 10,  4,  1;
  ...
		

Crossrefs

Cf. A000292, A002415 (row sums), A004737, A124258, A133825.

Programs

  • Maple
    T:= n-> (f-> (f(i)$i=1..n, f(n-i)$i=1..n-1))(t-> t*(t+1)*(t+2)/6):
    seq(T(n), n=1..10);  # Alois P. Heinz, Feb 17 2022
  • Mathematica
    Module[{nn=10,tet},tet=Table[(n(n+1)(n+2))/6,{n,nn}];Table[Join[Take[ tet,k], Reverse[ Take[tet,k-1]]],{k,nn}]]//Flatten (* Harvey P. Dale, Oct 22 2017 *)
    Table[Series[(1-h^(2*N+4))^2/(1-h^2)^4-((2+N)^2 *h^(2N+2))/(1-h^2)^2, {h, 0, 4*N}], {N,0,5}] // Normal (* Sergii Voloshyn, Sep 09 2022 *)

Formula

O.g.f.: (1+q*x)/((1-x)*(1-q*x)^3*(1-q^2x)) = 1 + x*(1 + 4*q + q^2) + x^2*(1 + 4*q + 10*q^2 + 4*q^3 + q^4) + ... .
From Boris Putievskiy, Jan 13 2013: (Start)
a(n) = A004737(n)*(A004737(n)+1)*(A004737(n)+2)/2.
a(n) = z*(z+1)*(z+2)/6, where z = floor(sqrt(n-1)) - |n - floor(sqrt(n-1))^2 - floor(sqrt(n-1)) - 1| + 1. (End)