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%I #30 Aug 08 2022 11:58:05
%S 1,1,1,1,2,1,1,3,2,1,1,4,4,2,1,1,5,6,4,2,1,1,6,9,7,4,2,1,1,7,12,11,7,
%T 4,2,1,1,8,16,16,12,7,4,2,1,1,9,20,23,18,12,7,4,2,1,1,10,25,31,27,19,
%U 12,7,4,2,1,1,11,30,41,38,29,19,12,7,4,2,1,1,12,36,53,53,42,30,19,12,7,4,2,1
%N Triangle, read by rows, such that the partial sums of the n-th row form the n-th diagonal, for n>=0, where each row begins with 1.
%C Row sums form A000070, which is the partial sums of the partition numbers (A000041). Rows read backwards converge to the row sums (A000070).
%C From _Alford Arnold_, Feb 07 2010: (Start)
%C The table can also be generated by summing sequences embedded within Table A008284
%C For example,
%C 1 1 1 1 ... yields 1 2 3 4 ...
%C 1 1 2 2 3 3 ... yields 1 2 4 6 9 12 ...
%C 1 1 2 3 4 5 7 ... yields 1 2 4 7 11 16 ...
%C (End)
%C T(n,k) is also count of all 'replacable' cells in the (Ferrers plots of) the partitions on n in exactly k parts. [_Wouter Meeussen_, Sep 16 2010]
%C From _Wolfdieter Lang_, Dec 03 2012: (Start)
%C The triangle entry T(n,k) is obtained from triangle A072233 by summing the entries of column k up to n (see the partial sum type o.g.f. given by Vladeta Jovovic in the formula section).
%C Therefore, the o.g.f. for the sequence in column k is x^k/((1-x)* product(1-x^j,j=1..k)).
%C The triangle with entry a(n,m) = T(n-1,m-1), n >= 1, m = 1, ..., n, is obtained from the partition array A103921 when in row n all entries belonging to part number m are summed (a conjecture). (End)
%H V. V. Kruchinin, <a href="https://doi.org/10.1134/S0001434609090260">The number of partitions of a natural number n into parts each of which is not less than m</a>, Math. Notes 86 (4) (2009) 505-509
%H R. J. Mathar, <a href="/A293482/a293482.pdf">Size of the set of residues of integer powers of fixed exponent</a>, (2017), Table 11.
%F T(n, k) = sum_{j=0..k} T(n-k, j), with T(n, 0) = 1 for all n>=0. A000070(n) = sum_{k=0..n} T(n, k).
%F O.g.f.: (1/(1-y))*(1/Product(1-x*y^k, k=1..infinity)). - _Vladeta Jovovic_, Jan 29 2005
%e The fourth row (n=3) is {1,3,2,1} and the fourth diagonal is the partial sums of the fourth row: {1,4,6,7,7,7,7,7,...}.
%e The triangle T(n,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 ...
%e 0 1
%e 1 1 1
%e 2 1 2 1
%e 3 1 3 2 1
%e 4 1 4 4 2 1
%e 5 1 5 6 4 2 1
%e 6 1 6 9 7 4 2 1
%e 7 1 7 12 11 7 4 2 1
%e 8 1 8 16 16 12 7 4 2 1
%e 9 1 9 20 23 18 12 7 4 2 1
%e 10 1 10 25 31 27 19 12 7 4 2 1
%e 11 1 11 30 41 38 29 19 12 7 4 2 1
%e 12 1 12 36 53 53 42 30 19 12 7 4 2 1
%e ... Reformatted by _Wolfdieter Lang_, Dec 03 2012
%e T(5,3)=4 because the partitions of 5 in exactly 3 parts are 221 and 311, and they give rise to partitions of 4 in four ways: 221->22 and 211, 311->211 and 31, since both their Ferrers plots have 2 'mobile cells' each. [_Wouter Meeussen_, Sep 16 2010]
%e T(5,3) = a(6,4) = 4 because the partitions of 6 with 4 parts are 1113 and 1122, with the number of distinct parts 2 and 2, respectively, summing to 4 (see the array A103921). An example for the conjecture given as comment above. - _Wolfdieter Lang_, Dec 03 2012
%p T(n,k)=if(n<k|k<0,0,if(n==k|k==0,1,sum(j=0,min(k,n-k),T(n-k,j))))
%t (*Needs["DiscreteMath`Combinatorica`"]; partitionexact[n_, m_] := TransposePartition /@ (Prepend[ #1, m] & ) /@ Partitions[n - m, m] *); mobile[p_?PartitionQ]:=1+Count[Drop[p,-1]-Rest[p],_?Positive]; Table[Tr[mobile/@partitionexact[n,k]],{n,12},{k,n}] (* _Wouter Meeussen_, Sep 16 2010 *)
%Y Antidiagonal sums form the partition numbers (A000041).
%Y Cf. A000070.
%Y Cf. A008284. [_Alford Arnold_, Feb 07 2010]
%Y Columns: A087811, A000601, A002621, A002622, A288341 - A288345.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Mar 12 2004
%E Several corrections by _Wolfdieter Lang_, Dec 03 2012
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