cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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.

Original entry on oeis.org

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

Views

Author

Peter Luschny, Jan 24 2019

Keywords

Comments

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

Examples

			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.

Crossrefs

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.

Programs

  • Maple
    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);
  • Mathematica
    (* 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}]

Formula

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.

A277178 a(n) = Sum_{k=0..n} k*binomial(2*k,k)/2.

Original entry on oeis.org

0, 1, 7, 37, 177, 807, 3579, 15591, 67071, 285861, 1209641, 5089517, 21314453, 88918353, 369734553, 1533115953, 6341759073, 26177411943, 107853629643, 443633635743, 1822098923943, 7473806605563, 30618895206483, 125303348573883, 512274592771083, 2092407173242983, 8539348101568335
Offset: 0

Views

Author

Vladimir Reshetnikov, Oct 02 2016

Keywords

Links

Crossrefs

Row 3 of A323222.
Cf. A000984, A002457.

Programs

  • Maple
    a:=n->sqrt(-1/27)-((n+1)/2)*binomial(2*(n+1),n+1)*hypergeom([1,n+3/2],[n+1],4):
seq(simplify(a(n)), n=0..26); # Peter Luschny, Oct 03 2016
  • Mathematica
    Table[Binomial[2 n, n] (2 n + 1 - Hypergeometric2F1[1, -n, 1/2 - n, 1/4])/3, {n, 0, 30}]
  • PARI
    {a(n) = sum(k=0, n, k*binomial(2*k, k))/2} \\ Seiichi Manyama, Jan 29 2019

Formula

a(n) = binomial(2*n,n) * (2*n + 1 - hypergeom([1,-n], [1/2-n], 1/4))/3.
a(n+1) - a(n) = A002457(n) = (2*n+1)!/n!^2.
Recurrence: (5*n + 2) * a(n) = (4*n + 2) * a(n-1) + n * a(n+1).
a(n) ~ sqrt(n) * 2^(2*n+1) / (3*sqrt(Pi)). - Vaclav Kotesovec, Jan 29 2019
G.f.: x/(1-x) * (1-4*x)^(-3/2). - Seiichi Manyama, Jan 29 2019

A323223 a(n) = [x^n] x/((1 - x)*(1 - 4*x)^(5/2)).

Original entry on oeis.org

0, 1, 11, 81, 501, 2811, 14823, 74883, 366603, 1752273, 8218733, 37964449, 173172249, 781607349, 3496163949, 15517771749, 68412846069, 299828796219, 1307168814519, 5672308893819, 24511334499219, 105519144602439, 452695473616239, 1936085243038839, 8256615564926439
Offset: 0

Views

Author

Peter Luschny, Jan 26 2019

Keywords

Links

Crossrefs

Row 5 of A323222.
Cf. A002802.

Programs

  • Maple
    A323223List := proc(len) local ogf, ser; ogf := (1 - 4*x)^(-5/2)*x/(1 - x);
ser := series(ogf, x, (n+1)*len+1); seq(coeff(ser, x, j), j=0..len) end:
A323223List(24);
# Alternative:
a := proc(n) option remember; `if`(n<2,n,((5*n+1)*a(n-1)-(4*n+2)*a(n-2))/(n-1)) end: seq(a(n), n=0..24);

Formula

a(n) = ((5*n + 1)*a(n-1) - (4*n + 2)*a(n-2))/(n - 1) for n >= 2.
a(n) = -(-4)^n*binomial(-5/2, n)*hypergeom([1, n+5/2], [n+1], 4) - i*sqrt(3)/27.
a(n) ~ 2^(2*n+2) * n^(3/2) / (9*sqrt(Pi)). - Vaclav Kotesovec, Jan 29 2019
a(n+1) - a(n) = A002802(n). - Seiichi Manyama, Jan 29 2019

A323218 a(n) = (4*n^3 + 30*n^2 + 50*n)/3 + 1.

Original entry on oeis.org

1, 29, 85, 177, 313, 501, 749, 1065, 1457, 1933, 2501, 3169, 3945, 4837, 5853, 7001, 8289, 9725, 11317, 13073, 15001, 17109, 19405, 21897, 24593, 27501, 30629, 33985, 37577, 41413, 45501, 49849, 54465, 59357, 64533, 70001, 75769, 81845, 88237, 94953, 102001
Offset: 0

Views

Author

Peter Luschny, Jan 26 2019

Keywords

Links

Crossrefs

Column 4 of A323222.

Programs

  • Maple
    a := n -> (4*n^3 + 30*n^2 + 50*n)/3 + 1: seq(a(n), n = 0..40);
  • Mathematica
    Table[(4n^3+30n^2+50n)/3+1,{n,0,40}] (* Harvey P. Dale, May 24 2019 *)

A323219 a(n) = [x^n] (1 - 4*x)^(-n/2)*x/(1 - x).

Original entry on oeis.org

0, 1, 5, 37, 313, 2811, 26093, 247311, 2377905, 23104441, 226289605, 2230309533, 22093913449, 219786279909, 2194096906461, 21969023675097, 220538907003489, 2218881134793411, 22368588800763701, 225891901214751423, 2284746661102951833, 23140953249273852519
Offset: 0

Views

Author

Peter Luschny, Jan 26 2019

Keywords

Crossrefs

Central diagonal of A323222.
Cf. A242798.

Programs

  • Maple
    ogf := n -> (1 - 4*x)^(-n/2)*x/(1 - x):
ser := n -> series(ogf(n), x, 46):
seq(coeff(ser(n), x, n), n=0..21);

Formula

a(n) = 1/(-3)^(n/2) - 4^n * Pochhammer(n/2,n)/n! * hypergeom([1,3*n/2],[n+1],4). - Robert Israel, Jan 28 2019
From Vaclav Kotesovec, Jan 29 2019: (Start)
Recurrence: 3*(n-2)*(n-1)*(65*n - 213)*a(n) = (20995*n^3 - 152844*n^2 + 347783*n - 238614)*a(n-2) + 12*(3*n - 10)*(3*n - 8)*(65*n - 83)*a(n-4).
a(n) ~ 2^(n - 1/2) * 3^((3*n - 1)/2) / (5*sqrt(Pi*n)). (End)
G.f.: -(24*x*cos(arcsin(216*x^2-1)/3))/(sqrt(3-324*x^2)*(2*sin(arcsin(216*x^2-1)/3)-11)). - Vladimir Kruchinin, Oct 27 2021
Showing 1-5 of 5 results.

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