A261036 Table read by rows: number of complete partitions of n with largest part = k.
1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 3, 2, 2, 1, 3, 4, 2, 1, 4, 5, 4, 2, 1, 4, 6, 5, 4, 1, 5, 8, 8, 5, 4, 1, 5, 10, 10, 8, 5, 1, 6, 11, 14, 10, 8, 5, 1, 6, 14, 16, 16, 10, 8, 1, 7, 16, 22, 20, 16, 10, 8, 1, 7, 18, 26, 27, 20, 16, 10, 1, 8, 21, 32, 34, 31
Offset: 1
Examples
T(8,2) = #{1+1+1+1+1+1+2, 1+1+1+1+2+2, 1+1+2+2+2} = 3; T(8,3) = #{1+1+1+1+1+3, 1+1+1+2+3, 1+1+3+3, 1+2+2+3} = 4; T(8,4) = #{1+1+1+1+4, 1+1+2+4} = 2; T(9,2) = #{+11+1+1+1+1+1+2, 1+1+1+1+1+2+2, 1+1+1+2+2+2, 1+2+2+2+2} = 4; T(9,3) = #{1+1+1+1+1+1+3, 1+1+1+1+2+3, 1+1+1+3+3, 1+1+2+2+3, 3,3,2,1} = 5; T(9,4) = #{1+1+1+1+1+4, 1+1+1+2+4, 1+1+3+4, 1+2+2+4} = 4; T(9,5) = #{1+1+1+1+5, 1+1+2+2+5} = 2. . ----------------------------------------------- . n | T(n,k), k = 1 .. [(n+1)/2] | A126796(n) . ----+------------------------------+----------- . 1 | 1 | 1 . 2 | 1 | 1 . 3 | 1 1 | 2 . 4 | 1 1 | 2 . 5 | 1 2 1 | 4 . 6 | 1 2 2 | 5 . 7 | 1 3 2 2 | 8 . 8 | 1 3 4 2 | 10 . 9 | 1 4 5 4 2 | 16 . 10 | 1 4 6 5 4 | 20 . 11 | 1 5 8 8 5 4 | 31 . 12 | 1 5 10 10 8 5 | 39 . 13 | 1 6 11 14 10 8 5 | 55 . 14 | 1 6 14 16 16 10 8 | 71 . 15 | 1 7 16 22 20 16 10 8 | 100 . 16 | 1 7 18 26 27 20 16 10 | 125 . 17 | 1 8 21 32 34 31 20 16 10 | 173 . 18 | 1 8 24 37 42 39 31 20 16 | 218 . 19 | 1 9 26 46 53 50 39 31 20 16 | 291 . 20 | 1 9 30 52 66 63 55 39 31 20 | 366
Links
- Reinhard Zumkeller, Rows n = 1..200 of triangle, flattened
- SeungKyung Park, Complete Partitions, Fibonacci Quarterly, Vol. 36 (1998), pp. 354-360.
Programs
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Haskell
import Data.MemoCombinators (memo2, integral, Memo) a261036 n k = a261036_tabf !! (n-1) !! (k-1) a261036_row n = a261036_tabf !! (n-1) a261036_tabf = zipWith (map . flip dMemo) [1..] a122197_tabf where dMemo = memo2 integral integral d d 0 _ = 0 d _ 0 = 0 d 1 _ = 1 d k n | n <= 2 * k - 2 = 0 | n <= 3 * k - 2 = dMemo (k - 1) (n - 1) | otherwise = dMemo (k - 1) (n - 1) + dMemo k (n - k)
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Mathematica
d[k_, n_] := d[k, n] = Which[n == 0 || k == 0, 0, k == 1, 1, n >= 3 k - 1, d[k - 1, n - 1] + d[k, n - k], 2 k - 1 <= n <= 3 k - 2, d[k - 1, n - 1], True, 0]; Table[d[k, n], {n, 17}, {k, Floor[(n + 1)/2]}] // Flatten (* Michael De Vlieger, Jul 13 2017 *)
Formula
According to the Park link, Theorem 3.7, p. 357f:
Let D_k(n) be the number of complete partitions of a positive integer n with largest part exactly k.
D_0(n) = 0 for all n, D_k(0) = 0 for all k, D_1(n)=1 for n>0, and for k>1:
D_k(n) = D_(k-1)(n-1) + D_k(n-k) if n >= 3*k-1, D_(k-1)(n-1) if 2*k-1 <= n <= 3*k-2, 0 if 1 <= n <= 2*k-2.
In the following, T(n,k) = D_k(n).
Comments