A047848 Array A read by diagonals; n-th difference of (A(k,n), A(k,n-1),..., A(k,0)) is (k+2)^(n-1), for n=1,2,3,...; k=0,1,2,...

1, 2, 1, 5, 2, 1, 14, 6, 2, 1, 41, 22, 7, 2, 1, 122, 86, 32, 8, 2, 1, 365, 342, 157, 44, 9, 2, 1, 1094, 1366, 782, 260, 58, 10, 2, 1, 3281, 5462, 3907, 1556, 401, 74, 11, 2, 1, 9842, 21846, 19532, 9332, 2802, 586, 92, 12, 2, 1, 29525, 87382, 97657, 55988, 19609, 4682, 821, 112, 13, 2, 1

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• tabl
• easy

Examples

			Array, A(n, k), begins as:
  1, 2,  5,  14,   41, ... = A007051.
  1, 2,  6,  22,   86, ... = A047849.
  1, 2,  7,  32,  157, ... = A047850.
  1, 2,  8,  44,  260, ... = A047851.
  1, 2,  9,  58,  401, ... = A047852.
  1, 2, 10,  74,  586, ... = A047853.
  1, 2, 11,  92,  821, ... = A047854.
  1, 2, 12, 112, 1112, ... = A047855.
  1, 2, 13, 134, 1465, ... = A047856.
  1, 2, 14, 158, 1886, ... = A196791.
  1, 2, 15, 184, 2381, ... = A196792.
Downward antidiagonals, T(n, k), begins as:
      1;
      2,     1;
      5,     2,     1;
     14,     6,     2,     1;
     41,    22,     7,     2,     1;
    122,    86,    32,     8,     2,    1;
    365,   342,   157,    44,     9,    2,   1;
   1094,  1366,   782,   260,    58,   10,   2,   1;
   3281,  5462,  3907,  1556,   401,   74,  11,   2,  1;
   9842, 21846, 19532,  9332,  2802,  586,  92,  12,  2, 1;
  29525, 87382, 97657, 55988, 19609, 4682, 821, 112, 13, 2, 1;
		

Links

• G. C. Greubel, Antidiagonals n = 0..50, flattened

Crossrefs

Cf. A047857 (row sums), A196793 (main diagonal).

Programs

Magma

A:= func< n,k | ((n+3)^k +n+1)/(n+2) >; // array A047848
[A(k,n-k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 11 2025

Mathematica

A[n_, k_]:= ((n+3)^k +n+1)/(n+2);
A047848[n_, k_]:= A[k,n-k];
Table[A047848[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 11 2025 *)

Python

def A(n,k): return (pow(n+3,k) +n+1)//(n+2) # array A047848
print(flatten([[A(k,n-k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 11 2025

Formula

A(n, k) = ((n+3)^k + n + 1)/(n+2). - Ralf Stephan, Feb 14 2004

From G. C. Greubel, Jan 11 2025: (Start)

T(n, k) = ((k+3)^(n-k) + k + 1)/(k+2) (antidiagonal triangle).

T(n, n) = A196793(n).

Sum_{k=0..n} T(n, k) = A047857(n). (End)