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A239829 Triangular array: T(n,k) = number of partitions of 2n - 1 that have alternating sum 2k - 1.

%I A239829 #17 Oct 27 2023 20:46:29
%S A239829 1,2,1,4,2,1,7,5,2,1,12,10,5,2,1,19,19,10,5,2,1,30,33,20,10,5,2,1,45,
%T A239829 57,36,20,10,5,2,1,67,92,64,36,20,10,5,2,1,97,147,107,65,36,20,10,5,2,
%U A239829 1,139,227,177,110,65,36,20,10,5,2,1,195,345,282,184
%N A239829 Triangular array:  T(n,k) = number of partitions of 2n - 1 that have alternating sum 2k - 1.
%C A239829 Suppose that p, with parts x(1) >= x(2) >= ... >= x(k), is a partition of n.  Define AS(p), the alternating sum of p, by x(1) - x(2) + x(3) - ... + ((-1)^(k-1))*x(k); note that AS(p) has the same parity as n.  Column 1 is given by T(n,1) = (number of partitions of 2n-1 having AS(p) = 1) = A000070(n) for n >= 1.  Columns 2 and 3 are essentially A000098 and A103924, and the limiting column (after deleting initial 0's), A000712.  The sum of numbers in row n is A000041(2n-1).  The corresponding array for partitions into distinct parts is given by A152157 (defined as the number of partitions of 2n+1 into 2k+1 odd parts).
%H A239829 Alois P. Heinz, <a href="/A239829/b239829.txt">Rows n = 1..141, flattened</a> (first 20 rows from Clark Kimberling)
%e A239829 First nine rows:
%e A239829 1
%e A239829 2 ... 1
%e A239829 4 ... 2 ... 1
%e A239829 7 ... 5 ... 2 ... 1
%e A239829 12 .. 10 .. 5 ... 2 ... 1
%e A239829 19 .. 19 .. 10 .. 5 ... 2 ... 1
%e A239829 30 .. 33 .. 20 .. 10 .. 5 ... 2 ... 1
%e A239829 45 .. 57 .. 36 .. 20 .. 10 .. 5 ... 2 ... 1
%e A239829 67 .. 92 .. 64 .. 36 .. 20 .. 10 .. 5 ... 2 ... 1
%e A239829 The partitions of 5 are 5, 41, 32, 311, 221, 2111, 11111, with respective alternating sums 5, 3, 1, 3, 1, 1, 1, so that row 2 of the array is 4 .. 2 .. 1.
%p A239829 b:= proc(n, i, t) option remember; `if`(n=0, x^(1/2), `if`(i<1, 0,
%p A239829       expand(b(n, i-1, t)+`if`(i>n, 0, b(n-i, i, -t)*x^((t*i)/2)))))
%p A239829     end:
%p A239829 T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(2*n-1$2, 1)):
%p A239829 seq(T(n), n=1..14);  # _Alois P. Heinz_, Mar 30 2014
%t A239829 z = 15; s[w_] := s[w] = Total[Take[#, ;; ;; 2]] - Total[Take[Rest[#], ;; ;; 2]] &[w]; c[n_] := c[n] = Table[s[IntegerPartitions[n][[k]]], {k, 1, PartitionsP[n]}]; t[n_, k_] := Count[c[2 n - 1], 2 k - 1]; u = Table[t[n, k], {n, 1, z}, {k, 1, n}]
%t A239829 TableForm[u]  (* A239829, array *)
%t A239829 Flatten[u]    (* A239829, sequence *)
%t A239829 (* _Peter J. C. Moses_, Mar 21 2014 *)
%t A239829 b[n_, i_, t_] := b[n, i, t] = If[n == 0, x^(1/2), If[i<1, 0, Expand[b[n, i-1, t] + If[i>n, 0, b[n-i, i, -t]*x^((t*i)/2)]]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[2n-1, 2n-1, 1]]; Table[T[n], {n, 1, 14}] // Flatten (* _Jean-François Alcover_, Aug 27 2016, after _Alois P. Heinz_ *)
%Y A239829 Cf. A239830, A239832, A239833.
%K A239829 nonn,tabl,easy
%O A239829 1,2
%A A239829 _Clark Kimberling_, Mar 28 2014