A323233 -id:A323233 - cp's OEIS frontend

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

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, 41, 64, 65, 1, 0, 1, 7, 26, 67, 131, 196, 197, 1, 0, 1, 8, 34, 101, 232, 428, 625, 626, 1, 0, 1, 9, 43, 144, 376, 804, 1429, 2055, 2056, 1

Offset: 0

Peter Luschny, Jan 24 2019

Equals A096465 when the leading column (k = 0) is removed. - Georg Fischer, Jul 26 2023

			The square array starts:
   [n\k]  0  1   2   3    4     5     6      7       8       9
    ---------------------------------------------------------------
    [0]   0, 1,  1,  1,   1,    1,    1,     1,      1,      1, ... A057427
    [1]   0, 1,  2,  4,   9,   23,   65,   197,    626,   2056, ... A014137
    [2]   0, 1,  3,  8,  22,   64,  196,   625,   2055,   6917, ... A014138
    [3]   0, 1,  4, 13,  41,  131,  428,  1429,   4861,  16795, ... A001453
    [4]   0, 1,  5, 19,  67,  232,  804,  2806,   9878,  35072, ... A114277
    [5]   0, 1,  6, 26, 101,  376, 1377,  5017,  18277,  66727, ... A143955
    [6]   0, 1,  7, 34, 144,  573, 2211,  8399,  31655, 118865, ...
    [7]   0, 1,  8, 43, 197,  834, 3382, 13378,  52138, 201364, ...
    [8]   0, 1,  9, 53, 261, 1171, 4979, 20483,  82499, 327656, ...
    [9]   0, 1, 10, 64, 337, 1597, 7105, 30361, 126292, 515659, ...
.
Triangle given by ascending antidiagonals:
    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,  41,  64,  65,   1;
    0, 1, 7, 26,  67, 131, 196, 197,   1;
    0, 1, 8, 34, 101, 232, 428, 625, 626, 1;
.
The difference table of a column successively gives the preceding columns, here starting with column 6.
col(6) = 1, 65, 196, 428, 804, 1377, 2211, 3382, 4979, 7105, ...
col(5) =    64, 131, 232, 376,  573,  834, 1171, 1597, 2126, ...
col(4) =         67, 101, 144,  197,  261,  337,  426,  529, ...
col(3) =              34,  43,   53,   64,   76,   89,  103, ...
col(2) =                    9,   10,   11,   12,   13,   14, ...
col(1) =                          1,    1,    1,    1,    1, ...
col(0) =                                0,    0,    0,    0, ...
.
Example for the sum formula: C(0) = 1, C(1) = 1, C(2) = 2 and C(3) = 5.
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},
{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},
{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.

The coefficients of the polynomials generating the columns are in A323233.
Sums of antidiagonals and row 1 are A014137. Main diagonal is A242798.
Rows: A057427 (n=0), A014137 (n=1), A014138 (n=2), A001453 (n=3), A114277 (n=4), A143955 (n=5).
Columns: A000027 (k=2), A034856 (k=3), A323221 (k=4), A323220 (k=5).
Similar array based on central binomials is A323222.
Cf. A096465.

Row := proc(n, len) local C, ogf, ser; C := (1-sqrt(1-4*x))/(2*x);
ogf := C^n*x/(1-x); ser := series(ogf, x, (n+1)*len+1);
seq(coeff(ser, x, j), j=0..len) end:
for n from 0 to 9 do Row(n, 9) od;
# Alternatively by recurrence:
B := proc(n, k) option remember; if n <= 0 or k < 0 then 0
elif n = k then 1 else B(n-1, k) + B(n, k-1) fi end:
A := (n, k) -> B(n + k, k): seq(lprint(seq(A(n, k), k=0..9)), n=0..9);

(* Illustrating the sum formula, not efficient. *) T[0, K_] := Boole[K != 0];
T[N_, K_] := Module[{}, r[n_, k_] := FrobeniusSolve[ConstantArray[1, n], k];
X[n_] := Flatten[Table[r[N, j], {j, 0, n - 1}], 1];
Sum[Product[CatalanNumber[m[[i]]], {i, 1, N}], {m , X[K]}]];
Trow[n_] := Table[T[n, k], {k, 0, 9}]; Table[Trow[n], {n, 0, 9}]

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).

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.

A323221 a(n) = n(n + 5)(n + 7)/6 + 1.

Original entry on oeis.org

1, 9, 22, 41, 67, 101, 144, 197, 261, 337, 426, 529, 647, 781, 932, 1101, 1289, 1497, 1726, 1977, 2251, 2549, 2872, 3221, 3597, 4001, 4434, 4897, 5391, 5917, 6476, 7069, 7697, 8361, 9062, 9801, 10579, 11397, 12256, 13157, 14101, 15089, 16122, 17201, 18327, 19501

Offset: 0

Peter Luschny, Jan 25 2019

a(n) is related to the total angular defect of certain polytopes. See Hilton and Pedersen, Cor. 1; compare A275874.

			For n = 2 the sum formula gives:
I(2) = {{0,0}, {0,1}, {1,0}, {0,2}, {1,1}, {2,0}, {0,3}, {1,2}, {2,1}, {3,0}};
a(2) = 1 + 1 + 1 + 2 + 1 + 2 + 5 + 2 + 2 + 5 = 22.

P. Hilton and J. Pedersen, Descartes, Euler, Poincaré, Pólya and Polyhedra, L'Enseign. Math., 27 (1981), 327-343.

Çf. A323224 (column 4), A323233 (row 4), A034856 (first difference), A275874.

a := n -> n*(35 + 12*n + n^2)/6 + 1:
seq(a(n), n = 0..45);

a[n_] := n (35 + 12 n + n^2)/6 + 1;
Table[a[n], {n, 0, 45}]

Let I(n) denote the set of all tuples of length n with elements from {0, 1, 2, 3} with sum <= 3 and C(m) denote the m-th Catalan number. Then for n > 0
a(n) = Sum_{(j1,...,jn) in I(n)} C(j1)*C(j2)*...*C(jn).
a(n) = [x^n] (3*x^3 - 8*x^2 + 5*x + 1)/(x - 1)^4.
a(n) = n! [x^n] exp(x)*(x^3 + 15*x^2 + 48*x + 6)/6.
a(n) = a(n - 1)*(n*(n + 5)*(n + 7) + 6)/(n*(n + 2)*(n + 7) - 18) for n > 0.
a(n) = A323224(n, 4).
a(n) = A275874(n+4) + 1.

A323220 a(n) = n(n + 5)(n + 7)*(n + 10)/24 + 1.

Original entry on oeis.org

1, 23, 64, 131, 232, 376, 573, 834, 1171, 1597, 2126, 2773, 3554, 4486, 5587, 6876, 8373, 10099, 12076, 14327, 16876, 19748, 22969, 26566, 30567, 35001, 39898, 45289, 51206, 57682, 64751, 72448, 80809, 89871, 99672, 110251, 121648, 133904, 147061, 161162, 176251

Offset: 0

Peter Luschny, Jan 25 2019

Cf. A323224 (column 5), A323233 (row 5), A323221 (first diff.), A034856 (second diff.).

a := n -> (n^4 + 22*n^3 + 155*n^2 + 350*n + 24)/24:
seq(a(n), n=0..40);

a(n) = [x^n] (8*x^4 - 31*x^3 + 41*x^2 - 18*x - 1)/(x - 1)^5.
a(n) = n! [x^n] exp(x)*(x^4 + 28*x^3 + 228*x^2 + 528*x + 24)/24.
a(n) = (1/3)*((2*n + 17)*a(n-3) - (3*n + 25)*a(n-2) + (n + 15)*a(n-1)) for n >= 3.
a(n) = A323224(n, 5).

cp's OEIS Frontend

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

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A323221 a(n) = n(n + 5)(n + 7)/6 + 1.

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A323220 a(n) = n(n + 5)(n + 7)*(n + 10)/24 + 1.

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cp's OEIS Frontend

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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Maple

Mathematica

Formula

A323221 a(n) = n*(n + 5)*(n + 7)/6 + 1.

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Maple

Mathematica

Formula

A323220 a(n) = n*(n + 5)*(n + 7)*(n + 10)/24 + 1.

Views

Author

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Crossrefs

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Maple

Formula

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

A323221 a(n) = n(n + 5)(n + 7)/6 + 1.

A323220 a(n) = n(n + 5)(n + 7)*(n + 10)/24 + 1.