A133825 Triangle whose rows are sequences of increasing and decreasing triangular numbers: 1; 1,3,1; 1,3,6,3,1; ... .

1, 1, 3, 1, 1, 3, 6, 3, 1, 1, 3, 6, 10, 6, 3, 1, 1, 3, 6, 10, 15, 10, 6, 3, 1, 1, 3, 6, 10, 15, 21, 15, 10, 6, 3, 1, 1, 3, 6, 10, 15, 21, 28, 21, 15, 10, 6, 3, 1, 1, 3, 6, 10, 15, 21, 28, 36, 28, 21, 15, 10, 6, 3, 1, 1, 3, 6, 10, 15, 21, 28, 36, 45, 36, 28, 21, 15, 10, 6, 3, 1, 1, 3, 6, 10

Offset: 0

Author

Peter Bala, Sep 25 2007

Keywords

• easy
• nonn
• tabf

Comments

Reading the triangle by rows produces the sequence 1,1,3,1,1,3,6,3,1,..., analogous to A004737.

T(n,k) = min(n*(n+1)/2,k*(k+1)/2), 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 starts
  1;
  1, 3, 1;
  1, 3, 6, 3, 1;
  1, 3, 6, 10, 6, 3, 1;
From _Boris Putievskiy_, Jan 13 2013: (Start)
The start of the sequence as table:
  1...1...1...1...1...1...
  1...3...3...3...3...3...
  1...3...6...6...6...6...
  1...3...6..10..10..10...
  1...3...6..10..15..15...
  1...3...6..10..15..21...
  1...3...6..10..15..21...
  . . .
The start of the sequence as triangle array read by rows:
  1,
  1, 3, 1,
  1, 3, 6, 3, 1,
  1, 3, 6, 10, 6, 3, 1,
  1, 3, 6, 10, 15, 10, 6, 3, 1,
  1, 3, 6, 10, 15, 21, 15, 10, 6, 3, 1,
  1, 3, 6, 10, 15, 21, 28, 21, 15, 10, 6, 3, 1,
  . . .
Row number k contains 2*k-1 numbers 1,3,...,k*(k-1)/2,k*(k+1)/2,k*(k-1)/2,...,3,1. (End)
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.

Crossrefs

Cf. A000330 (row sums), A004737, A124258, A133826, A106255.

Programs

Mathematica

Module[{nn=10,ac},ac=Accumulate[Range[nn]];Table[Join[Take[ ac,n],Reverse[ Take[ac,n-1]]],{n,nn}]]//Flatten (* Harvey P. Dale, Apr 18 2019 *)

Formula

O.g.f.: (1+qx)/((1-x)(1-qx)^2(1-q^2x)) = 1 + x(1 + 3q + q^2) + x^2(1 + 3q + 6q^2 + 3q^3 + q^4) + ... .

From Boris Putievskiy, Jan 13 2013: (Start)

a(n) = A004737(n)*(A004737(n)+1)/2.

a(n) = z*(z+1)/2, where z = floor(sqrt(n-1)) - |n- floor(sqrt(n-1))^2- floor(sqrt(n-1))-1| +1. (End)