A133823 Triangle whose rows are sequences of increasing and decreasing cubes:1; 1,8,1; 1,8,27,8,1; ... .
1, 1, 8, 1, 1, 8, 27, 8, 1, 1, 8, 27, 64, 27, 8, 1, 1, 8, 27, 64, 125, 64, 27, 8, 1, 1, 8, 27, 64, 125, 216, 125, 64, 27, 8, 1, 1, 8, 27, 64, 125, 216, 343, 216, 125, 64, 27, 8, 1, 1, 8, 27, 64, 125, 216, 343, 512, 343, 216, 125, 64, 27, 8, 1, 1, 8, 27, 64, 125, 216, 343, 512, 729
Offset: 0
Examples
Triangle starts 1; 1, 8, 1; 1, 8, 27, 8, 1; 1, 8, 27, 64, 27, 8, 1; From _Boris Putievskiy_, Jan 13 2013: (Start) The start of the sequence as table: 1...1...1...1...1...1... 1...8...8...8...8...8... 1...8..27..27..27..27... 1...8..27..64..64..64... 1...8..27..64.125.125... 1...8..27..64.125.216... . . . The start of the sequence as triangle array read by rows: 1; 1,8,1; 1,8,27,8,1; 1,8,27,64,27,8,1; 1,8,27,64,125,64,27,8,1; 1,8,27,64,125,216,125,64,27,8,1; . . . Row number k contains 2*k-1 numbers 1,8,...,(k-1)^3,k^3,(k-1)^3,...,8,1. (End)
Links
- Harvey P. Dale, Table of n, a(n) for n = 0..9999
- Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Programs
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Mathematica
Table[Join[Range[n]^3,Range[n-1,1,-1]^3],{n,10}]//Flatten (* Harvey P. Dale, May 29 2019 *)
Formula
O.g.f.: (1+qx)(1+4qx+q^2x^2)/((1-x)(1-qx)^3(1-q^2x)) = 1 + x(1 + 8q + q^2) + x^2(1 + 8q + 27q^2 + 8q^3 + q^4) + ... .
From Boris Putievskiy, Jan 13 2013: (Start)
a(n) = (A004737(n))^3.
a(n) = (floor(sqrt(n-1)) - |n- floor(sqrt(n-1))^2- floor(sqrt(n-1))-1| +1)^3. (End)
Comments