A321163 Square array A(n,k), n >= 1, k >= 1, read by antidiagonals, where A(n,k) is the sum of distinct products Product_{j=1..k} b_j with 1 <= b_j<= n.
1, 1, 3, 1, 7, 6, 1, 15, 25, 10, 1, 31, 90, 61, 15, 1, 63, 301, 310, 136, 21, 1, 127, 966, 1441, 990, 244, 28, 1, 255, 3025, 6370, 6391, 2220, 440, 36, 1, 511, 9330, 27301, 38325, 17731, 5300, 680, 45, 1, 1023, 28501, 114670, 218926, 130851, 54831, 9660, 1022, 55
Offset: 1
Examples
In case of (n,k) = (3,2):
| 1 2 3
--+--------
1 | 1, 2, 3
2 | 2, 4, 6
3 | 3, 6, 9
Distinct products are 1,2,3,4,6,9. So A(3,2) = 1+2+3+4+6+9 = 25.
Square array begins:
1, 1, 1, 1, 1, 1, ...
3, 7, 15, 31, 63, 127, ...
6, 25, 90, 301, 966, 3025, ...
10, 61, 310, 1441, 6370, 27301, ...
15, 136, 990, 6391, 38325, 218926, ...
21, 244, 2220, 17731, 130851, 916714, ...
28, 440, 5300, 54831, 514668, 4519390, ...
36, 680, 9660, 116991, 1280916, 13092430, ...
45, 1022, 17130, 242091, 3070935, 36184072, ...
55, 1472, 28670, 467391, 6807045, 91765822, ...
Links
- Seiichi Manyama, Antidiagonals n = 1..25, flattened
Crossrefs
Programs
-
Mathematica
A[n_, k_] := Module[{b, bb}, bb = Array[b, k]; Table[Times @@ bb, Evaluate[Sequence @@ ({#, n}& /@ bb)]] // Flatten // Union // Total]; Table[A[n-k+1, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 25 2020 *)