1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 13, 7, 1, 1, 9, 25, 25, 9, 1, 1, 11, 41, 63, 41, 11, 1, 1, 13, 61, 129, 129, 61, 13, 1, 1, 15, 85, 231, 321, 231, 85, 15, 1, 1, 17, 113, 377, 681, 681, 377, 113, 17, 1, 1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1, 1, 21, 181, 833, 2241, 3653, 3653
Offset: 0
The square array D(i,j) (i >= 0, j >= 0) begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ... = A000012
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, ... = A005408
1, 5, 13, 25, 41, 61, 85, 113, 145, 181, ... = A001844
1, 7, 25, 63, 129, 231, 377, 575, 833, 1159, ... = A001845
1, 9, 41, 129, 321, 681, 1289, 2241, 3649, 5641, ... = A001846
...
For D(2,5) = 61, which is seen above in the row labeled A001844, we calculate the sum (9 + 11 + 41) of the 3 nearest terms above and/or to the left. - _Peter Munn_, Jan 01 2023
D(2,5) = 61 can also be obtained from the row labeled A005408 using a recurrence mentioned in the formula section: D(2,5) = D(1,5) + 2*Sum_{k=0..4} D(1,k), so D(2,5) = 11 + 2*(1+3+5+7+9) = 11 + 2*25. - _Shel Kaphan_, Jan 01 2023
As a triangular array (on its side) this begins:
0, 0, 0, 0, 1, 0, 11, 0, ...
0, 0, 0, 1, 0, 9, 0, 61, ...
0, 0, 1, 0, 7, 0, 41, 0, ...
0, 1, 0, 5, 0, 25, 0, 129, ...
1, 0, 3, 0, 13, 0, 63, 0, ...
0, 1, 0, 5, 0, 25, 0, 129, ...
0, 0, 1, 0, 7, 0, 41, 0, ...
0, 0, 0, 1, 0, 9, 0, 61, ...
0, 0, 0, 0, 1, 0, 11, 0, ...
[Edited by _Shel Kaphan_, Jan 01 2023]
From _Roger L. Bagula_, Dec 09 2008: (Start)
As a triangle T(n,k) (with rows n >= 0 and columns k = 0..n), this begins:
1;
1, 1;
1, 3, 1;
1, 5, 5, 1;
1, 7, 13, 7, 1;
1, 9, 25, 25, 9, 1;
1, 11, 41, 63, 41, 11, 1;
1, 13, 61, 129, 129, 61, 13, 1;
1, 15, 85, 231, 321, 231, 85, 15, 1;
1, 17, 113, 377, 681, 681, 377, 113, 17, 1;
1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1;
... (End)
Triangle T(n,k) recurrence: 63 = T(6,3) = 25 + 13 + 25 = T(5,2) + T(4,2) + T(5,3).
Triangle T(n,k) recurrence with A-sequence A112478: 63 = T(6,3) = 1*25 + 2*25 - 2*9 + 6*1 (T entries from row n = 5 only). [Here the formula T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j) is used with n = 6 and k = 3; i.e., T(6,3) = Sum_{j=0..3} A111478(j) * T(5, 2+j). - _Petros Hadjicostas_, Aug 05 2020]
From _Philippe Deléham_, Mar 29 2012: (Start)
Subtriangle of the triangle given by (1, 0, 1, -1, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
1;
1, 0;
1, 1, 0;
1, 3, 1, 0;
1, 5, 5, 1, 0;
1, 7, 13, 7, 1, 0;
1, 9, 25, 25, 9, 1, 0;
1, 11, 41, 63, 41, 11, 1, 0;
...
Subtriangle of the triangle given by (0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
1;
0, 1;
0, 1, 1;
0, 1, 3, 1;
0, 1, 5, 5, 1;
0, 1, 7, 13, 7, 1;
0, 1, 9, 25, 25, 9, 1;
0, 1, 11, 41, 63, 41, 11, 1;
... (End)
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