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.

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A048619 a(n) = LCM(binomial(n,0), ..., binomial(n,n)) / binomial(n,floor(n/2)).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 3, 4, 2, 10, 5, 30, 15, 7, 7, 56, 28, 252, 126, 60, 30, 330, 165, 396, 198, 286, 143, 2002, 1001, 15015, 15015, 7280, 3640, 1768, 884, 15912, 7956, 3876, 1938, 38760, 19380, 406980, 203490, 99484, 49742, 1144066, 572033, 1961256, 980628
Offset: 0

Views

Author

Labos Elemer

Keywords

Examples

			If n=10 then A002944(10)=2520, A001405(10)=252, the quotient a(10)=10.

Links

Crossrefs

Cf. A001405, A002944.

Programs

  • Magma
    [Lcm([1..n+1]) div (Floor((n+3)/2)*Binomial(n+1,Floor((n+3)/2))): n in [0..50]]; // Vincenzo Librandi, Jul 10 2019
  • Mathematica
    Table[Apply[LCM, Binomial[n, Range[0, n]]]/Binomial[n, Floor[n/2]], {n, 0, 48}] (* Michael De Vlieger, Jun 29 2017 *)
  • PARI
    {A048619(n) = lcm(vector(n+1, i, i)) / binomial(n+1, (n+1)\2) / ((n+2)\2);}

Formula

a(n) = A002944(n)/A001405(n).
a(n) = lcm(1..n+1)/(floor((n+3)/2)*binomial(n+1,floor((n+3)/2))). - Paul Barry, Jul 03 2006
a(n) = lcm(1,2,...,n+1) / (ceiling((n+1)/2)*binomial(n+1,floor((n+1)/2))) = A003418(n+1) / A100071(n+1). - Max Alekseyev, Oct 23 2015
a(n) = A263673(n+1) / A110654(n+1) = A180000(n+1) / A152271(n). - Max Alekseyev, Oct 23 2015
a(2*n-1) = A068553(n) = A068550(n)/n.

Extensions

Definition corrected and a(0)=1 prepended by Max Alekseyev, Oct 23 2015

A141692 Triangle read by rows: T(n,k) = n*(binomial(n - 1, k - 1) - binomial(n - 1, k)), 0 <= k <= n.

Original entry on oeis.org

0, -1, 1, -2, 0, 2, -3, -3, 3, 3, -4, -8, 0, 8, 4, -5, -15, -10, 10, 15, 5, -6, -24, -30, 0, 30, 24, 6, -7, -35, -63, -35, 35, 63, 35, 7, -8, -48, -112, -112, 0, 112, 112, 48, 8, -9, -63, -180, -252, -126, 126, 252, 180, 63, 9, -10, -80, -270, -480, -420, 0, 420, 480, 270, 80, 10
Offset: 0

Views

Author

Roger L. Bagula, Sep 09 2008

Keywords

Comments

The row sums are zero.
Row n consists of the coefficients in the expansion of n*(x - 1)*(x + 1)^(n - 1). - Franck Maminirina Ramaharo, Oct 02 2018

Examples

			Triangle begins:
    0;
   -1,   1;
   -2,   0,    2;
   -3,  -3,    3,    3;
   -4,  -8,    0,    8,    4;
   -5, -15,  -10,   10,   15,   5;
   -6, -24,  -30,    0,   30,  24,   6;
   -7, -35,  -63,  -35,   35,  63,  35,   7;
   -8, -48, -112, -112,    0, 112, 112,  48,   8;
   -9, -63, -180, -252, -126, 126, 252, 180,  63,  9;
  -10, -80, -270, -480, -420,   0, 420, 480, 270, 80, 10;
  ...

Links

Crossrefs

Cf. A007318, A112467, A128433, A128434.

Programs

  • Maple
    a:=proc(n,k) n*(binomial(n-1,k-1)-binomial(n-1,k)); end proc: seq(seq(a(n,k),k=0..n),n=0..10); # Muniru A Asiru, Oct 03 2018
  • Mathematica
    Table[Table[n*(Binomial[n - 1, k - 1] - Binomial[n - 1, k]),{k, 0, n}],{n, 0, 12}]//Flatten
  • Maxima
    T(n, k) := n*(binomial(n - 1, k - 1) - binomial(n - 1, k))$
tabl(nn) := for n:0 thru nn do print(makelist(T(n, k), k, 0, n))$ /* Franck Maminirina Ramaharo, Oct 02 2018 */

Formula

T(n,k) = n*(B(1/2;n-1,k-1) - B(1/2;n-1,k))*2^(n - 1), where B(t;n,k) = binomial(n,k)*t^k*(1 - t)^(n - k) denotes the k-th Benstein basis polynomial of degree n.
T(n,k) = n*A112467(n,k).
From Franck Maminirina Ramaharo, Oct 02 2018: (Start)
T(n,k) = -T(n,n-k)
T(n,0) = -n.
T(n,1) = -A067998(n)
E.g.f.: (x*y - y)/(x*y + y - 1)^2.
Sum_{k=0..n} abs(T(n,k)) = 2*A100071(n).
Sum_{k=0..n} T(n,k)^2 = 2*A037965(n).
Sum_{k=0..n} k*T(n,k) = A001787(n).
Sum_{k=0..n} k^2*T(n,k) = A014477(n-1). (End)

Extensions

Edited, new name and offset corrected by Franck Maminirina Ramaharo, Oct 02 2018

A189230 Complementary Catalan triangle read by rows.

Original entry on oeis.org

0, 1, 0, 0, 2, 0, 3, 0, 3, 0, 0, 8, 0, 4, 0, 10, 0, 15, 0, 5, 0, 0, 30, 0, 24, 0, 6, 0, 35, 0, 63, 0, 35, 0, 7, 0, 0, 112, 0, 112, 0, 48, 0, 8, 0, 126, 0, 252, 0, 180, 0, 63, 0, 9, 0, 0, 420, 0, 480, 0, 270, 0, 80, 0, 10, 0, 462, 0, 990, 0, 825, 0, 385, 0, 99, 0, 11, 0
Offset: 0

Views

Author

Peter Luschny, May 01 2011

Keywords

Comments

T(n,k) = A189231(n,k)*((n - k) mod 2). For comparison: the classical Catalan triangle is A053121(n,k) = A189231(n,k)*((n-k+1) mod 2).
T(n,0) = A138364(n). Row sums: A100071.

Examples

			[0]  0,
[1]  1,  0,
[2]  0,  2,  0,
[3]  3,  0,  3,  0,
[4]  0,  8,  0,  4,  0,
[5] 10,  0, 15,  0,  5, 0,
[6]  0, 30,  0, 24,  0, 6, 0,
[7] 35,  0, 63,  0, 35, 0, 7, 0,
   [0],[1],[2],[3],[4],[5],[6],[7]

Links

Crossrefs

Cf. A053121, A162246, A057977, A189231.

Programs

  • Maple
    A189230 := (n,k) -> A189231(n,k)*modp(n-k,2):
seq(print(seq(A189230(n,k),k=0..n)),n=0..11);
  • Mathematica
    t[n_, k_] /; (k>n || k<0) = 0; t[n_, n_] = 1; t[n_, k_] := t[n, k] = t[n-1, k-1] + Mod[n-k, 2] t[n-1, k] + t[n-1, k+1];
T[n_, k_] := t[n, k] Mod[n-k, 2];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] (* Jean-François Alcover, Jun 24 2019 *)

A194589 a(n) = A194588(n) - A005043(n); complementary Riordan numbers.

Original entry on oeis.org

0, 0, 1, 1, 5, 11, 34, 92, 265, 751, 2156, 6194, 17874, 51702, 149941, 435749, 1268761, 3700391, 10808548, 31613474, 92577784, 271407896, 796484503, 2339561795, 6877992334, 20236257626, 59581937299, 175546527727, 517538571125, 1526679067331, 4505996000730
Offset: 0

Views

Author

Peter Luschny, Aug 30 2011

Keywords

Comments

The inverse binomial transform of a(n) is A194590(n).

Links

Crossrefs

Cf. A005043, A189912, A194588, A100071, A005717.

Programs

  • Maple
    # First method, describes the derivation:
A056040 := n -> n!/iquo(n,2)!^2:
A057977 := n -> A056040(n)/(iquo(n,2)+1);
A001006 := n -> add(binomial(n,k)*A057977(k)*irem(k+1,2),k=0..n):
A005043 := n -> `if`(n=0,1,A001006(n-1)-A005043(n-1)):
A189912 := n -> add(binomial(n,k)*A057977(k),k=0..n):
A194588 := n -> `if`(n=0,1,A189912(n-1)-A194588(n-1)):
A194589 := n -> A194588(n)-A005043(n):
# Second method, more efficient:
A100071 := n -> A056040(n)*(n/2)^(n-1 mod 2):
A194589 := proc(n) local k;
(n mod 2)+(1/2)*add((-1)^k*binomial(n,k)*A100071(k+1),k=1..n) end:
# Alternatively:
a := n -> `if`(n<3,iquo(n,2),hypergeom([1-n/2,-n,3/2-n/2],[1,2-n],4)): seq(simplify(a(n)), n=0..30); # Peter Luschny, Mar 07 2017
  • Mathematica
    sf[n_] := With[{f = Floor[n/2]}, Pochhammer[f+1, n-f]/f!]; a[n_] := Mod[n, 2] + (1/2)*Sum[(-1)^k*Binomial[n, k]*2^-Mod[k, 2]*(k+1)^Mod[k, 2]*sf[k+1], {k, 1, n}]; Table[a[n], {n, 0, 10}] (* Jean-François Alcover, Jul 30 2013, from 2nd method *)
Table[If[n < 3, Quotient[n, 2], HypergeometricPFQ[{1 - n/2, -n, 3/2 - n/2}, {1, 2-n}, 4]], {n,0,30}] (* Peter Luschny, Mar 07 2017 *)
  • Maxima
    a(n):=sum(binomial(n+2,k)*binomial(n-k,k),k,0,(n)/2); /* Vladimir Kruchinin, Sep 28 2015 */
  • PARI
    a(n) = sum(k=0, n/2, binomial(n+2,k)*binomial(n-k,k));
vector(30, n, a(n-3)) \\ Altug Alkan, Sep 28 2015

Formula

a(n) = sum_{k=0..n} C(n,k)*A194590(k).
a(n) = (n mod 2)+(1/2)*sum_{k=1..n} (-1)^k*C(n,k)*(k+1)$*((k+1)/2)^(k mod 2). Here n$ denotes the swinging factorial A056040(n).
a(n) = PSUMSIGN([0,0,1,2,6,16,45,..] = PSUMSIGN([0,0,A005717]) where PSUMSIGN is from Sloane's "Transformations of integer sequences". - Peter Luschny, Jan 17 2012
A(x) = B'(x)*(1/x^2-1/(B(x)*x)), where B(x)/x is g.f. of A005043. - Vladimir Kruchinin, Sep 28 2015
a(n) = Sum_{k=0..n/2} C(n+2,k)*C(n-k,k). - Vladimir Kruchinin, Sep 28 2015
a(n) = hypergeom([1-n/2,-n,3/2-n/2],[1,2-n],4) for n>=3. - Peter Luschny, Mar 07 2017
a(n) ~ 3^(n + 1/2) / (8*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 17 2024

A304933 a(0) = 0, a(1) = 1 and a(n) = 2*a(n-1)/(n-1) + 16*a(n-2) for n > 1.

Original entry on oeis.org

0, 1, 2, 18, 44, 310, 828, 5236, 14744, 87462, 255340, 1450460, 4349160, 23932220, 73268440, 393382440, 1224746032, 6447212294, 20354432076, 105417000268, 336767439560, 1720348748244, 5552121770888, 28030318314712, 91271367318096, 456091040311900
Offset: 0

Views

Author

Seiichi Manyama, May 21 2018

Keywords

Comments

Let a(0) = 0, a(1) = 1 and a(n) = 2*m*a(n-1)/(n-1) + k^2*a(n-2) for n > 1.
Then G.f. is x/(2*m) * d/dx ((1 + k*x)/(1 - k*x))^(m/k).

Links

Crossrefs

a(n) = 2*a(n-1)/(n-1) + b^2*a(n-2): A001477 (b=1), A100071 (b=2), this sequence (b=4), A304934 (b=8).
Cf. A303537.

Formula

a(n) = n*A303537(n)/2.
G.f.: x/(1-4*x)^2 * ((1-4*x)/(1+4*x))^(3/4).

A304934 a(0) = 0, a(1) = 1 and a(n) = 2*a(n-1)/(n-1) + 64*a(n-2) for n > 1.

Original entry on oeis.org

0, 1, 2, 66, 172, 4310, 12732, 280084, 894872, 18149094, 61304940, 1173803004, 4136934888, 75812881404, 276427353048, 4891514031720, 18343552465968, 315349842088326, 1211087339244108, 20316955153568876, 79648216569893320, 1308249951485397396
Offset: 0

Views

Author

Seiichi Manyama, May 21 2018

Keywords

Comments

Let a(0) = 0, a(1) = 1 and a(n) = 2*m*a(n-1)/(n-1) + k^2*a(n-2) for n > 1.
Then G.f. is x/(2*m) * d/dx ((1 + k*x)/(1 - k*x))^(m/k).

Links

Crossrefs

a(n) = 2*a(n-1)/(n-1) + b^2*a(n-2): A001477 (b=1), A100071 (b=2), A304933 (b=4), this sequence (b=8).
Cf. A303538.

Formula

a(n) = n*A303538(n)/2.
G.f.: x/(1-8*x)^2 * ((1-8*x)/(1+8*x))^(7/8).

A305032 a(0) = 0, a(1) = 1 and a(n) = 6*a(n-1)/(n-1) + 4*a(n-2) for n > 1.

Original entry on oeis.org

0, 1, 6, 22, 68, 190, 500, 1260, 3080, 7350, 17220, 39732, 90552, 204204, 456456, 1012440, 2230800, 4886310, 10647780, 23094500, 49884120, 107343236, 230205976, 492156392, 1049212528, 2230928700, 4732273000, 10015777800, 21154820400, 44596287000, 93846099600
Offset: 0

Views

Author

Seiichi Manyama, May 24 2018

Keywords

Comments

Let a(0) = 0, a(1) = 1 and a(n) = 2*m*a(n-1)/(n-1) + k^2*a(n-2), for n > 1, then the g.f. is x/(2*m) * d/dx ((1 + k*x)/(1 - k*x))^(m/k).

Links

Crossrefs

Cf. A100071, A304944, A305031.

Programs

  • Magma
    [n le 2 select n-1 else 2*(3*Self(n-1) + 2*(n-2)*Self(n-2))/(n-2): n in [1..40]]; // G. C. Greubel, Jun 07 2023
  • Mathematica
    CoefficientList[Series[x*Sqrt[1-4*x^2]/(1-2*x)^3, {x,0,40}], x] (* G. C. Greubel, Jun 07 2023 *)
  • SageMath
    @CachedFunction
def a(n): # b = A305032
    if n<2: return n
    else: return 2*(3*a(n-1) + 2*(n-1)*a(n-2))//(n-1)
[a(n) for n in range(41)] # G. C. Greubel, Jun 07 2023

Formula

a(n) = n*A305031(n)/6.
G.f.: x*sqrt(1-4*x^2)/(1-2*x)^3.

A331552 Expansion of (1 + 2*x)/(1 + 4*x^2)^(3/2).

Original entry on oeis.org

1, 2, -6, -12, 30, 60, -140, -280, 630, 1260, -2772, -5544, 12012, 24024, -51480, -102960, 218790, 437580, -923780, -1847560, 3879876, 7759752, -16224936, -32449872, 67603900, 135207800, -280816200, -561632400, 1163381400, 2326762800, -4808643120, -9617286240, 19835652870
Offset: 0

Views

Author

Seiichi Manyama, Jan 20 2020

Keywords

Links

Crossrefs

Column 1 of A331511.
Cf. A100071.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 33); Coefficients(R!( (1 + 2*x)/(1 + 4*x^2)^(3/2))); // Marius A. Burtea, Jan 20 2020
  • Magma
    [&+[(-1)^k*(k+1)*Binomial(n+1, k+1)^2:k in [0..n]]:n in [0..33]]; // Marius A. Burtea, Jan 20 2020
  • Mathematica
    a[n_] := Sum[(-1)^k * (k + 1) * Binomial[n + 1, k + 1]^2, {k, 0, n}]; Array[a, 33, 0] (* Amiram Eldar, Jan 20 2020 *)
  • PARI
    N=66; x='x+O('x^N); Vec((1+2*x)/(1+4*x^2)^(3/2))
  • PARI
    {a(n) = sum(k=0, n, (-2)^(n-k)*(n+k+1)*binomial(n, k)*binomial(n+k, k))}
  • PARI
    {a(n) = sum(k=0, n, (-1)^k*(k+1)*binomial(n+1, k+1)^2)}

Formula

|a(n)| = A100071(n+1).
a(n) = Sum_{k=0..n} (-2)^(n-k) * (n+k+1) * binomial(n,k) * binomial(n+k,k).
a(n) = Sum_{k=0..n} (-1)^k * (k+1) * binomial(n+1,k+1)^2.
n * (2*n-1) * a(n) = 2 * a(n-1) - 4 * n * (2*n+1) * a(n-2) for n>1.
E.g.f.: (1 + 2*x)*BesselJ(0,2*x) - 2*x*BesselJ(1,2*x). - Ilya Gutkovskiy, Mar 04 2021

A107233 An inverse Chebyshev transform of n^3.

Original entry on oeis.org

0, 1, 8, 30, 96, 270, 720, 1820, 4480, 10710, 25200, 58212, 133056, 300300, 672672, 1492920, 3294720, 7220070, 15752880, 34179860, 73902400, 159074916, 341429088, 730122120, 1557593856, 3312591100, 7030805600, 14883258600, 31451414400
Offset: 0

Views

Author

Paul Barry, May 13 2005

Keywords

Comments

Image of n^3 under the mapping of g(x)->(1/sqrt(1-4x^2))g(xc(x^2)) where c(x) is the g.f. of A000108.

Crossrefs

Cf. A000108, A100071, A001787.

Programs

  • Mathematica
    CoefficientList[Series[x*(1 + 4*x) / ((1 - 2*x)^(5/2) * Sqrt[1 + 2*x]), {x, 0, 30}], x] (* Vaclav Kotesovec, Nov 04 2017 *)

Formula

G.f.: 4x(sqrt(1-4x^2)-1)^2(4x+1)/(sqrt(1-4x^2)(sqrt(1-4x^2)+2x-1)^4); a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*(n-2k)^3.
D-finite with recurrence (n-1)*a(n)+4*(n-4)*a(n-1) -4*(n+4)*a(n-2) +16*(2-n)*a(n-3)=0. - R. J. Mathar, Nov 09 2012
From Vaclav Kotesovec, Nov 04 2017: (Start)
G.f.: x*(1 + 4*x) / ((1 - 2*x)^(5/2) * sqrt(1 + 2*x)).
a(n) ~ 2^(n + 1/2) * n^(3/2) / sqrt(Pi). (End)

A191791 Triangle read by rows: T(n,k) is the number of length n left factors of Dyck paths having k UDUD's, where U=(1,1) and D=(1,-1).

Original entry on oeis.org

1, 1, 2, 3, 5, 1, 8, 2, 15, 4, 1, 25, 8, 2, 46, 19, 4, 1, 79, 36, 9, 2, 147, 76, 24, 4, 1, 256, 146, 48, 10, 2, 477, 304, 109, 29, 4, 1, 841, 578, 224, 60, 11, 2, 1570, 1180, 499, 144, 34, 4, 1, 2791, 2244, 1002, 312, 72, 12, 2, 5217, 4525, 2172, 731, 181, 39, 4, 1, 9336, 8588, 4347, 1530, 410, 84, 13, 2
Offset: 0

Views

Author

Emeric Deutsch, Jun 18 2011

Keywords

Comments

Row n>=2 contains floor(n/2) entries.
Sum of entries in row n is binomial(n, floor(n/2)) =A001405(n).
T(n,0)=A191792(n).
Sum(k*T(n,k), k>=0)=A100071(n-3).

Examples

			T(7,2)=2 because we have (UD[UD)UD]U and U(UD[UD)UD], where U=(1,1) and D=(1,-1) (the UDUD's are shown between parentheses).
Triangle starts:
1;
1;
2;
3;
5,1;
8,2;
15,4,1;
25,8,2;

Crossrefs

Cf. A001405, A191792, A100071.

Programs

  • Maple
    eq := z^2*(1+z^2-t*z^2)*C^2-(1+z^2+z^4-t*z^2-t*z^4)*C+1+z^2-t*z^2: C := RootOf(eq, C): G := C/(1-z*C): Gser := simplify(series(G, z = 0, 20)): for n from 0 to 17 do P[n] := sort(coeff(Gser, z, n)) end do: 1; 1; for n from 2 to 17 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)-1) end do; # yields sequence in triangular form

Formula

G.f.: G(t,z) = C/(1-z*C), where C=C(t,z) is given by z^2*(1+z^2-t*z^2)*C^2 - (1+z^2+z^4- t*z^2-t*z^4)*C + 1 + z^2 - t*z^2 = 0.
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