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A323224 A(n, k) = [x^k] C^n*x/(1 - x) where C = 2/(1 + sqrt(1 - 4*x)), square array read by ascending antidiagonals with n >= 0 and k >= 0.

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%I A323224 #34 Jul 26 2023 01:14:29
%S A323224 0,0,1,0,1,1,0,1,2,1,0,1,3,4,1,0,1,4,8,9,1,0,1,5,13,22,23,1,0,1,6,19,
%T A323224 41,64,65,1,0,1,7,26,67,131,196,197,1,0,1,8,34,101,232,428,625,626,1,
%U A323224 0,1,9,43,144,376,804,1429,2055,2056,1
%N A323224 A(n, k) = [x^k] C^n*x/(1 - x) where C = 2/(1 + sqrt(1 - 4*x)), square array read by ascending antidiagonals with n >= 0 and k >= 0.
%C A323224 Equals A096465 when the leading column (k = 0) is removed. - _Georg Fischer_, Jul 26 2023
%F A323224 For n>0 and k>0 let X(n, k) denote the set of all tuples of length n with elements from {0, ..., k-1} with sum < k. Let C(m) denote the m-th Catalan number. Then: A(n, k) = Sum_{(j1,...,jn) in X(n, k)} C(j1)*C(j2)*...*C(jn).
%F A323224 A(n, k) = T(n + k, k) with T(n, k) = T(n-1, k) + T(n, k-1) with T(n, k) = 0 if n <= 0 or k < 0 and T(n, n) = 1.
%e A323224 The square array starts:
%e A323224    [n\k]  0  1   2   3    4     5     6      7       8       9
%e A323224     ---------------------------------------------------------------
%e A323224     [0]   0, 1,  1,  1,   1,    1,    1,     1,      1,      1, ... A057427
%e A323224     [1]   0, 1,  2,  4,   9,   23,   65,   197,    626,   2056, ... A014137
%e A323224     [2]   0, 1,  3,  8,  22,   64,  196,   625,   2055,   6917, ... A014138
%e A323224     [3]   0, 1,  4, 13,  41,  131,  428,  1429,   4861,  16795, ... A001453
%e A323224     [4]   0, 1,  5, 19,  67,  232,  804,  2806,   9878,  35072, ... A114277
%e A323224     [5]   0, 1,  6, 26, 101,  376, 1377,  5017,  18277,  66727, ... A143955
%e A323224     [6]   0, 1,  7, 34, 144,  573, 2211,  8399,  31655, 118865, ...
%e A323224     [7]   0, 1,  8, 43, 197,  834, 3382, 13378,  52138, 201364, ...
%e A323224     [8]   0, 1,  9, 53, 261, 1171, 4979, 20483,  82499, 327656, ...
%e A323224     [9]   0, 1, 10, 64, 337, 1597, 7105, 30361, 126292, 515659, ...
%e A323224 .
%e A323224 Triangle given by ascending antidiagonals:
%e A323224     0;
%e A323224     0, 1;
%e A323224     0, 1, 1;
%e A323224     0, 1, 2,  1;
%e A323224     0, 1, 3,  4,   1;
%e A323224     0, 1, 4,  8,   9,   1;
%e A323224     0, 1, 5, 13,  22,  23,   1;
%e A323224     0, 1, 6, 19,  41,  64,  65,   1;
%e A323224     0, 1, 7, 26,  67, 131, 196, 197,   1;
%e A323224     0, 1, 8, 34, 101, 232, 428, 625, 626, 1;
%e A323224 .
%e A323224 The difference table of a column successively gives the preceding columns, here starting with column 6.
%e A323224 col(6) = 1, 65, 196, 428, 804, 1377, 2211, 3382, 4979, 7105, ...
%e A323224 col(5) =    64, 131, 232, 376,  573,  834, 1171, 1597, 2126, ...
%e A323224 col(4) =         67, 101, 144,  197,  261,  337,  426,  529, ...
%e A323224 col(3) =              34,  43,   53,   64,   76,   89,  103, ...
%e A323224 col(2) =                    9,   10,   11,   12,   13,   14, ...
%e A323224 col(1) =                          1,    1,    1,    1,    1, ...
%e A323224 col(0) =                                0,    0,    0,    0, ...
%e A323224 .
%e A323224 Example for the sum formula: C(0) = 1, C(1) = 1, C(2) = 2 and C(3) = 5.
%e A323224 X(3, 4) = {{0,0,0}, {0,0,1}, {0,1,0}, {1,0,0}, {0,0,2}, {0,1,1}, {0,2,0}, {1,0,1},
%e A323224 {1,1,0}, {2,0,0}, {0,0,3}, {0,1,2}, {0,2,1}, {0,3,0}, {1,0,2}, {1,1,1}, {1,2,0},
%e A323224 {2,0,1}, {2,1,0}, {3,0,0}}. T(3,4) = 1+1+1+1+2+1+2+1+1+2+5+2+2+5+2+1+2+2+2+5 = 41.
%p A323224 Row := proc(n, len) local C, ogf, ser; C := (1-sqrt(1-4*x))/(2*x);
%p A323224 ogf := C^n*x/(1-x); ser := series(ogf, x, (n+1)*len+1);
%p A323224 seq(coeff(ser, x, j), j=0..len) end:
%p A323224 for n from 0 to 9 do Row(n, 9) od;
%p A323224 # Alternatively by recurrence:
%p A323224 B := proc(n, k) option remember; if n <= 0 or k < 0 then 0
%p A323224 elif n = k then 1 else B(n-1, k) + B(n, k-1) fi end:
%p A323224 A := (n, k) -> B(n + k, k): seq(lprint(seq(A(n, k), k=0..9)), n=0..9);
%t A323224 (* Illustrating the sum formula, not efficient. *) T[0, K_] := Boole[K != 0];
%t A323224 T[N_, K_] := Module[{}, r[n_, k_] := FrobeniusSolve[ConstantArray[1, n], k];
%t A323224 X[n_] := Flatten[Table[r[N, j], {j, 0, n - 1}], 1];
%t A323224 Sum[Product[CatalanNumber[m[[i]]], {i, 1, N}], {m , X[K]}]];
%t A323224 Trow[n_] := Table[T[n, k], {k, 0, 9}]; Table[Trow[n], {n, 0, 9}]
%Y A323224 The coefficients of the polynomials generating the columns are in A323233.
%Y A323224 Sums of antidiagonals and row 1 are A014137. Main diagonal is A242798.
%Y A323224 Rows: A057427 (n=0), A014137 (n=1), A014138 (n=2), A001453 (n=3), A114277 (n=4), A143955 (n=5).
%Y A323224 Columns: A000027 (k=2), A034856 (k=3), A323221 (k=4), A323220 (k=5).
%Y A323224 Similar array based on central binomials is A323222.
%Y A323224 Cf. A096465.
%K A323224 nonn,tabl
%O A323224 0,9
%A A323224 _Peter Luschny_, Jan 24 2019