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.

A283054 Triangle read by rows: T(n,k) = T(n,k-1) + T(n-1,k), T(n,0)=1, T(n,n) = T(n,n-1) + 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 1, 4, 8, 9, 1, 5, 13, 22, 23, 1, 6, 19, 41, 64, 65, 1, 7, 26, 67, 131, 196, 197, 1, 8, 34, 101, 232, 428, 625, 626, 1, 9, 43, 144, 376, 804, 1429, 2055, 2056, 1, 10, 53, 197, 573, 1377, 2806, 4861, 6917, 6918, 1, 11, 64, 261, 834, 2211, 5017, 9878, 16795, 23713, 23714, 1, 12, 76, 337, 1171, 3382, 8399, 18277, 35072, 58785, 82499, 82500
Offset: 0

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Author

Ely Golden, Feb 27 2017

Keywords

Comments

The left diagonals form polynomial sequences. This is due to the observation that diagonal 0 D_0(x) = 1, and D_n(x) = D_n(x-1)+D_(n-1)(x+1), with D_n(-1) = 1 which is a recurrence that can be solved.
These polynomials begin 1, x+2, (x(x+7)+8)/2, (x(x(x+15)+62)+54)/6, (x(x(x(x+26)+227)+730)+552)/24, etc., the first 3 of which correspond to A000012(n), A000027(n+2), and A034856(n+2), respectively.
The rightmost diagonal appears to follow A014137(n). The second rightmost appears to follow A014138(n+1), the third appears to follow A001453(n+2), the fourth appears to follow A114277(n), and the fifth appears to follow A143955(n+3).
A closed-form formula for T(n,k) would be very desirable.

Examples

			First 7 rows:
  1;
  1,   2;
  1,   3,   4;
  1,   4,   8,   9;
  1,   5,  13,  22,  23;
  1,   6,  19,  41,  64,  65;
  1,   7,  26,  67, 131, 196, 197;
		

Programs

  • Mathematica
    T[0, 0] = 1; T[n_, k_] := T[n, k] = Which[k == 0, 1, k == n, T[n, n - 1] + 1, True, T[n, k - 1] + T[n - 1, k]]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 27 2017 *)
  • PARI
    T(n,k)=if(k==0,return(1));if(k==n,return(T(n,n-1)+1));T(n,k-1)+T(n-1,k)
    for(n=0,10,for(k=0,n,print1(T(n,k),", "))) \\ Derek Orr, Feb 28 2017
  • SageMath
    def sideTriangleAt(a,b):
        if(b==0): return 1
        elif(b==a): return sideTriangleAt(a,b-1)+1
        else: return sideTriangleAt(a,b-1)+sideTriangleAt(a-1,b)
    def sideTriangle(size):
        li=[]
        for c in range(size):
            for d in range(c+1):
                if(d==0): li.append([1])
                elif(d==c): li[c].append(li[c][d-1]+1)
                else: li[c].append(li[c][d-1]+li[c-1][d])
        return li
    trig=sideTriangle(125)
    for c in range(len(trig)):
        print(str(trig[c])[1:-1].replace(",",""))