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Triangle T read by rows: T(n,k) is the number of partitions of n in which k is the least integer that is not a part and there are more parts > k than there are < k (n >= k > 0).
4

%I #15 Aug 13 2023 02:24:39

%S 0,1,0,1,0,0,2,0,0,0,2,0,0,0,0,4,0,0,0,0,0,4,1,0,0,0,0,0,7,1,0,0,0,0,

%T 0,0,8,2,0,0,0,0,0,0,0,12,3,0,0,0,0,0,0,0,0,14,5,0,0,0,0,0,0,0,0,0,21,

%U 6,0,0,0,0,0,0,0,0,0,0,24,10,0,0,0,0,0,0,0,0,0,0,0

%N Triangle T read by rows: T(n,k) is the number of partitions of n in which k is the least integer that is not a part and there are more parts > k than there are < k (n >= k > 0).

%H George E. Andrews and Mircea Merca, <a href="https://doi.org/10.1016/j.jcta.2012.05.001">The truncated pentagonal number theorem</a>, Journal of Combinatorial Theory, Series A, Volume 119, Issue 8, 2012, Pages 1639-1643.

%H K. Banerjee and M. G. Dastidar, <a href="https://doi.org/10.35011/risc.22-20">Inequalities for the partition function arising from truncated theta series</a>, RISC Report Series No. 22-20, 2023. See Theorem 1.1 at p. 2.

%F T(n,k) = (-1)^(k-1)*Sum_{j=0..k-1} (-1)^j*(p(n - j*(3*j + 1)/2) - p(n - j*(3*j + 5)/2 - 1)), where p(n) = A000041(n) is the number of partitions of n (see Theorem 1.1 in Andrews and Merca).

%F 1st column: T(n,1) = A002865(n) for n > 0.

%e T(9,2) = 3 from 6 + 3 = 5 + 3 + 1 = 4 + 4 + 1 = 3 + 3 + 3.

%e The triangle T(n, k) begins

%e n\k| 1 2 3 4 5 6 7 8 9

%e ---+------------------------------------

%e 1 | 0

%e 2 | 1 0

%e 3 | 1 0 0

%e 4 | 2 0 0 0

%e 5 | 2 0 0 0 0

%e 6 | 4 0 0 0 0 0

%e 7 | 4 1 0 0 0 0 0

%e 8 | 7 1 0 0 0 0 0 0

%e 9 | 8 2 0 0 0 0 0 0 0

%e ...

%t T[n_,k_]:=(-1)^(k-1)*Sum[(-1)^j*(PartitionsP[n-j*(3*j+1)/2]-PartitionsP[n-j*(3*j+5)/2-1]),{j,0,k-1}]; Flatten[Table[T[n,k],{n,1,15},{k,1,n}]]

%o (PARI)

%o T(n,k) = (-1)^(k-1)*sum(j=0, k-1, (-1)^j*(numbpart(n-j*(3*j+1)/2)-numbpart(n-j*(3*j+5)/2-1)));

%o tabl(nn) = for(n=1, nn, for(k=1, n, print1(T(n,k), ", ")); print);

%o tabl(10) \\ yields sequence in triangular form

%Y Cf. A000041, A002865, A325434 (row sums).

%K nonn,tabl

%O 1,7

%A _Stefano Spezia_, Apr 27 2019