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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A383120 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n*k,k).

Original entry on oeis.org

1, 2, 11, 139, 2885, 82381, 2979565, 130203494, 6664589321, 390857822425, 25832193906761, 1899273577364197, 153741850998047053, 13585520026454056279, 1301210398133681268381, 134270617908678099820891, 14849785991790603714043921, 1752283118795349858851381297
Offset: 0

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Author

Ilya Gutkovskiy, Apr 17 2025

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k] Binomial[n k, k], {k, 0, n}], {n, 0, 17}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k) * binomial(n*k,k)); \\ Michel Marcus, Apr 17 2025

Formula

a(n) = [x^n] ((1 + x)^n + x)^n.
a(n) ~ exp(n + exp(-1) - 1/2) * n^n / sqrt(2*Pi*n). - Vaclav Kotesovec, Apr 17 2025

A126177 Triangle read by rows: T(n,k) is number of hex trees with n edges and k leaves (n >= 1, 1 <= k <= 1 + floor(n/2)).

Original entry on oeis.org

3, 9, 1, 27, 9, 81, 54, 2, 243, 270, 30, 729, 1215, 270, 5, 2187, 5103, 1890, 105, 6561, 20412, 11340, 1260, 14, 19683, 78732, 61236, 11340, 378, 59049, 295245, 306180, 85050, 5670, 42, 177147, 1082565, 1443420, 561330, 62370, 1386, 531441, 3897234
Offset: 1

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Author

Emeric Deutsch, Dec 19 2006

Keywords

Comments

A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a middle child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read paper).
Also number of hex trees with n edges and k-1 nodes of outdegree 2.
Row n has 1 + floor(n/2) terms.
Sum of terms in row n = A002212(n+1).
T(n,1) = 3^n (A000244).
T(n,2) = A027472(n+1).
Sum_{k=1..1+floor(n/2)} k*T(n,k) = A026375(n).

Examples

			Triangle starts:
    3;
    9,   1;
   27,   9;
   81,  54,   2;
  243, 270,  30;
		

Crossrefs

Programs

  • Maple
    T:=(n,k)->3^(n-2*k+2)*binomial(2*k-2,k-1)*binomial(n,2*k-2)/k: for n from 1 to 13 do seq(T(n,k),k=1..1+floor(n/2)) od; # yields sequence in triangular form

Formula

T(n,k) = 3^(n-2k+2)*binomial(2k-2,k-1)*binomial(n,2k-2)/k. Proof: There are Catalan(k-1) full binary trees with k leaves. Each of them has 2k-2 edges. Additional n-2k+2 edges can be inserted as paths at the existing 2k-1 vertices in 3^(n-2k+2)*binomial(n,2k-2) ways.
G.f.: G=G(t,z) satisfies z^2*G^2-(1-3z-2tz^2)G+tz(3+tz)=0.

A293491 a(n) = n! * [x^n] exp((n+2)*x)*BesselI(0,2*x).

Original entry on oeis.org

1, 3, 18, 155, 1734, 23877, 390804, 7417377, 160256070, 3885021569, 104465601756, 3086353547433, 99399100528924, 3466411543407555, 130151205663179112, 5235127829223881895, 224609180728848273990, 10239557195235638377449, 494317596005491398892620, 25192788307121307053168673
Offset: 0

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Author

Ilya Gutkovskiy, Oct 10 2017

Keywords

Comments

The n-th term of the n-th binomial transform of A000984.

Crossrefs

Programs

  • Mathematica
    Table[n! SeriesCoefficient[Exp[(n + 2) x] BesselI[0, 2 x], {x, 0, n}], {n, 0, 19}]
    Table[SeriesCoefficient[1/Sqrt[(1 - n x) (1 - (n + 4) x)], {x, 0, n}], {n, 0, 19}]
    Join[{1}, Table[Sum[Binomial[n, k] Binomial[2 k, k] n^(n - k), {k, 0, n}], {n, 1, 19}]]
    Table[(n + 2)^n HypergeometricPFQ[{1/2 - n/2, -n/2}, {1}, 4/(2 + n)^2], {n, 0, 19}]

Formula

a(n) = [x^n] 1/sqrt((1 - n*x)*(1 - (n + 4)*x)).
a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*k,k)*n^(n-k).
a(n) ~ exp(2) * BesselI(0,2) * n^n. - Vaclav Kotesovec, Oct 16 2017

A302180 Number of 3D walks of type aad.

Original entry on oeis.org

1, 1, 3, 7, 23, 71, 251, 883, 3305, 12505, 48895, 193755, 783355, 3205931, 13302329, 55764413, 236174933, 1008773269, 4343533967, 18834033443, 82201462251, 360883031291, 1592993944723, 7066748314147, 31493800133173, 140953938878821, 633354801073571, 2856369029213263
Offset: 0

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Author

N. J. A. Sloane, Apr 09 2018

Keywords

Comments

See Dershowitz (2017) for precise definition.
Number of 3D walks of length n in the first octant using steps (1, 1, 0), (1, -1, 0), (1, 0, 1), (1, 0, -1) and (1, 0, 0) that start at the origin and end at (n, 0, 0). The analogous problem in 2D is given by the Motzkin numbers A001006. - Farzan Byramji, Mar 06 2021
Inverse binomial transform of A145867 (Number of 3D walks of type aae). - Mélika Tebni, Nov 05 2024

Crossrefs

Programs

  • Maple
    M := n-> add(binomial(n, 2*k)*binomial(2*k, k)/(k+1), k = 0 .. iquo(n,2)): # Motzkin numbers
    A302180 := n-> add((-1)^(n-k)*binomial(n, k)*add(binomial(k, j)*M(j)*M(k-j), j=0..k), k=0..n):  seq(A302180(n), n = 0 .. 26); # Mélika Tebni, Nov 05 2024

Extensions

a(14)-a(26) from Farzan Byramji, Mar 06 2021

A302182 Number of 3D walks of type abc.

Original entry on oeis.org

1, 1, 5, 12, 62, 200, 1065, 3990, 21714, 89082, 492366, 2147376, 12004740, 54718092, 308559537, 1454116950, 8255788970, 39935276810, 227976044010, 1126178350440, 6457854821340, 32456552441040, 186814834574550, 952569927106980, 5500292590186380, 28391993275117500
Offset: 0

Views

Author

N. J. A. Sloane, Apr 09 2018

Keywords

Comments

See Dershowitz (2017) for precise definition.

Crossrefs

Programs

  • Python
    from math import comb as binomial
    def row(n: int) -> list[int]:
        return sum(binomial(n, k)*binomial(k, k//2)//(k//2+1)*((k+1) %2)*binomial(n-k, (n-k)//2)**2 for k in range(n+1))
    for n in range(26): print(row(n)) # Mélika Tebni, Nov 27 2024

Formula

From Mélika Tebni, Nov 27 2024: (Start)
a(n) = Sum_{k=0..n} binomial(n, k)*A126120(k)*A018224(n-k).
a(2*n+1) = A135394(n) / (2*n+2).
a(2*n) = A302181(n). (End)

Extensions

a(13)-a(25) from Mélika Tebni, Nov 27 2024

A302184 Number of 3D walks of type abe.

Original entry on oeis.org

1, 2, 7, 26, 108, 472, 2159, 10194, 49396, 244328, 1229308, 6273896, 32410096, 169181664, 891181607, 4731912082, 25302648644, 136150941064, 736747902236, 4007011320808, 21893702201648, 120125750018656, 661630546993116, 3656966382542984, 20278320788680912, 112782556853239712
Offset: 0

Views

Author

N. J. A. Sloane, Apr 09 2018

Keywords

Comments

See Dershowitz (2017) for precise definition.

Crossrefs

Programs

  • Maple
    a := n -> 2*add(binomial(n, k)*binomial(k, k/2)*binomial(2*(n-k), n-k)/(k+2), k = 0..n, 2): seq(a(n), n = 0..25);  # Peter Luschny, Nov 30 2024
  • Python
    from math import comb as binomial
    def a(n: int):
        return sum(binomial(n, k)*binomial(k, k//2)//(k//2+1)*((k+1) %2)*binomial(2*(n-k), n-k) for k in range(n+1))
    print([a(n) for n in range(26)]) # Mélika Tebni, Nov 30 2024

Formula

a(n) = Sum_{k=0..n} binomial(n, k)*A126120(k)*A000984(n-k). - Mélika Tebni, Nov 30 2024

Extensions

a(12)-a(25) from Mélika Tebni, Nov 30 2024

A349001 The number of Lyndon words of size n from an alphabet of 5 letters and 1st and 2nd letter of the alphabet with equal frequency in the words.

Original entry on oeis.org

1, 3, 4, 14, 46, 174, 656, 2640, 10790, 45340, 193600, 839820, 3686424, 16353924, 73187456, 330052646, 1498335650, 6841899606, 31404443032, 144814450188, 670552118244, 3116578216310, 14534401932712, 67992210407514, 318969964124256, 1500268062754830
Offset: 0

Views

Author

R. J. Mathar, Nov 05 2021

Keywords

Comments

Counts a subset of the Lyndon words in A001692. Here there is no requirement of how often the 3rd to 5th letter of the alphabet are in the admitted word, only on the frequency of the 1st and 2nd letter of the alphabet.
Let T(n,k,M) be the number of words of length n drawn from an alphabet of size M where the first k letters of the alphabet appear with the same frequency f in each word. Then T(n,k,M) = Sum_{f=0..floor(n/k)} (M-k)^(n-f*k) * Product_{i=0..k-1} binomial(n-i*f,f) and T(n,2,5) = A026375(n), T(n,3,6) = A294035(n), T(n,2,6) = A081671(n). Removing the words with cycles by the inclusion-exclusion principle by a Mobius Transform gives words of length n of that type without cycles and division through n the Lyndon words of that type. - R. J. Mathar, Nov 07 2021

Examples

			Examples for the alphabet {0,1,2,3,4}:
a(0)=1 counts (), the empty word.
a(3)=14 counts (021) (031) (041) (012) (013) (223) (233) (243) (014) (224) (234) (334) (244) (344), words of length 3 where the letters 0 and the 1 occur both either not or once.
a(4)=46 counts (0011) (0221) (0321) (0421) (0231) (0331) (0431) (0241) (0341) (0441) (0212) (0312) (0412) (0122) (0132) (0142) (0213) (0313) (0413) (0123) (2223) (0133) (2233) (2333) (2433) (0143) (2243) (2343) (2443) (0214) (0314) (0414) (0124) (2224) (2324) (0134) (2234) (2334) (3334) (2434) (0144) (2244) (2344) (3344) (2444) (3444).
		

Crossrefs

Cf. A022553 (alphabet of 2 letters), A290277 (of 3 letters), A060165 (of 4 letters), A026375.

Programs

  • PARI
    a(n) = if(n>0, sumdiv(n, d, moebius(n/d)*sum(k=0, d, binomial(d,k)*binomial(2*k,k)))/n, n==0) \\ Andrew Howroyd, Jan 14 2023

Formula

n*a(n) = Sum_{d|n} mu(d)*A026375(n/d) where mu = A008683.

Extensions

Terms a(16) and beyond from Andrew Howroyd, Jan 14 2023

A370285 Coefficient of x^n in the expansion of ( (1+x)^2 + x^3 )^n.

Original entry on oeis.org

1, 2, 6, 23, 94, 392, 1659, 7107, 30734, 133880, 586576, 2582142, 11411371, 50597900, 224986467, 1002867878, 4479814606, 20049099908, 89878609344, 403521966942, 1814102538624, 8165526187128, 36794746597494, 165968135843522, 749314496125451, 3385881647958442
Offset: 0

Views

Author

Seiichi Manyama, Feb 14 2024

Keywords

Crossrefs

Similar to A082759.

Programs

  • Maple
    a := n -> binomial(2*n, n) * hypergeom([(1-n)/3, (2-n)/3, -n/3], [1/2-n, n+1], 27/4):
    seq(simplify(a(n)), n = 0..25);  # Peter Luschny, Jan 04 2025
  • PARI
    a(n) = sum(k=0, n\3, binomial(n, k)*binomial(2*n-2*k, n-3*k));

Formula

a(n) = Sum_{k=0..floor(n/3)} binomial(n,k) * binomial(2*n-2*k,n-3*k).
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x / ((1+x)^2 + x^3) ). See A369212.

A110221 Triangle read by rows: T(n,k) (0<=k<=floor(n/2)) is the number of Delannoy paths of length n, having k ED's.

Original entry on oeis.org

1, 3, 11, 2, 45, 18, 195, 120, 6, 873, 720, 90, 3989, 4110, 870, 20, 18483, 22806, 6930, 420, 86515, 124264, 49560, 5320, 70, 408105, 668520, 331128, 52920, 1890, 1936881, 3562830, 2111760, 456120, 29610, 252, 9238023, 18850590, 13020480, 3575880
Offset: 0

Views

Author

Emeric Deutsch, Jul 20 2005

Keywords

Comments

A Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1).
Row n has 1+floor(n/2) terms. Row sums are the central Delannoy numbers (A001850). Column 0 yields A026375. Sum(k*T(n,k),k=0..floor(n/2))=2*A002695(n).

Examples

			T(2,1)=2 because we have NED and EDN.
Triangle begins:
1;
3;
11,2;
45,18;
195,120,6;
		

Crossrefs

Programs

  • Maple
    R:=(1-z-sqrt(1-6*z+5*z^2-4*z^2*t))/2/z/(1-z+t*z): G:=1/(1-z-2*t*z^2*R-2*z*R+2*z^2*R): Gser:=simplify(series(G,z=0,15)): P[0]:=1: for n from 1 to 12 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 12 do seq(coeff(t*P[n],t^k),k=1..1+floor(n/2)) od; # yields sequence in triangular form

Formula

G.f.: 1/(1-z-2tz^2*R-2zR+2z^2*R), where R=[1-z-sqrt(1-6z+5z^2-4tz^2)]/[2z(1-z+tz)].

A168216 Riordan array (1/(1-x),xc(x)/(1-xc(x))) where c(x)is the g.f. of A000108.It factorizes as A007318*A106566.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 8, 5, 1, 1, 23, 19, 7, 1, 1, 74, 69, 34, 9, 1, 1, 262, 256, 147, 53, 11, 1, 1, 993, 986, 615, 265, 76, 13, 1, 1, 3943, 3935, 2571, 1235, 431, 103, 15, 1, 1, 16178, 16169, 10862, 5591, 2216, 653, 134, 17, 1
Offset: 0

Views

Author

Philippe Deléham, Nov 20 2009

Keywords

Comments

Inverse is Riordan array (1/(1+x-x^2),x(1-x)/(1+x-x^2)) = [ -1,-1,1,0,0,0,0,0,...]DELTA[1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. Unsigned version of A091698.

Examples

			Triangle begins : 1 ; 1,1 ; 1,3,1 ; 1,8,5,1 ; 1,23,19,7,1 ; ...
		

Crossrefs

Formula

Sum_{k, 0<=k<=n}T(n,k)*x^k = A000012(n), A007317(n+1), A026375(n) for x = 0, 1, 2 respectively.
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