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 1-4 of 4 results.

A366107 a(n) = Sum_{i=0..floor(q(n)/3)} binomial(n-3*(i+1), q(n)-3*i) with q(n) = ceiling((n-3)/2).

Original entry on oeis.org

1, 1, 2, 3, 6, 11, 21, 39, 75, 141, 273, 519, 1009, 1933, 3770, 7263, 14202, 27479, 53846, 104543, 205216, 399543, 785460, 1532779, 3017106, 5899167, 11624580, 22766607, 44905518, 88073091, 173863965, 341425551, 674506059, 1326019653, 2621371005, 5158412943, 10203609597
Offset: 3

Views

Author

Stefano Spezia, Sep 29 2023

Keywords

Crossrefs

Programs

  • Mathematica
    q[n_]:=Ceiling[(n-3)/2]; a[n_]:=Sum[Binomial[n-3(i+1),q[n]-3i], {i,0,Floor[q[n]/3]}]; Array[a,37,3]
  • PARI
    a(n) = my(q=ceil((n-3)/2)); sum(i=0, q\3, binomial(n-3*(i+1), q-3*i)); \\ Michel Marcus, Sep 30 2023

Formula

From Remark 3.4 at p. 5 in Czédli: (Start)
A366108(n)/a(n) ~ 7/4.
A366109(n)/a(n) ~ 7/6. (End)
a(n) ~ c*2^(n+1)/sqrt(n), with c = 1/(7*sqrt(2*Pi)) = (2/7)* A218708.

A366108 a(n) = floor(binomial(n-1, floor((n-1)/2))/2).

Original entry on oeis.org

1, 1, 3, 5, 10, 17, 35, 63, 126, 231, 462, 858, 1716, 3217, 6435, 12155, 24310, 46189, 92378, 176358, 352716, 676039, 1352078, 2600150, 5200300, 10029150, 20058300, 38779380, 77558760, 150270097, 300540195, 583401555, 1166803110, 2268783825, 4537567650, 8836315950
Offset: 3

Views

Author

Stefano Spezia, Sep 29 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:=Floor[Binomial[n-1,Floor[(n-1)/2]]/2]; Array[a,36,3]
  • PARI
    a(n) = binomial(n-1, (n-1)\2)\2; \\ Michel Marcus, Sep 30 2023

Formula

a(n)/A366107(n) ~ 7/4 (see Remark 3.4 at p. 5 in Czédli).
a(n) ~ c*2^n/sqrt(n), with c = 1/(2*sqrt(2*Pi)) = A218708.

A366109 a(n) = floor(n!*(3*floor(n/2)!*ceiling(n/2)! + 3*floor((n+2)/2)!*ceiling((n-2)/2)! - 6*floor(n/2)!*ceiling((n-2)/2)!)^(-1)).

Original entry on oeis.org

1, 1, 2, 4, 7, 13, 26, 46, 92, 168, 333, 616, 1225, 2288, 4558, 8580, 17107, 32413, 64664, 123170, 245832, 470288, 938943, 1802770, 3600207, 6933733, 13849778, 26744400, 53429368, 103411680, 206621384, 400720260, 800747232, 1555737480, 3109074130, 6050090200, 12091800773
Offset: 3

Views

Author

Stefano Spezia, Sep 29 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:=Floor[n!(3Floor[n/2]!Ceiling[n/2]! + 3Floor[(n+2)/2]!Ceiling[(n-2)/2]! - 6Floor[n/2]!Ceiling[(n-2)/2]!)^(-1)]; Array[a,37,3]

Formula

a(n)/A366107(n) ~ 7/6 (see Remark 3.4 at p. 5 in Czédli).
a(n) ~ c*2^n/sqrt(n), with c = 1/(3*sqrt(2*Pi)) = (2/3)*A218708.

A387635 a(n) = Sum_{k=0..n-1} binomial(2*n, k)^2.

Original entry on oeis.org

0, 1, 17, 262, 3985, 60626, 925190, 14168988, 217721745, 3355615450, 51855874642, 803232328548, 12467572005382, 193873026294052, 3019674502600220, 47101568276955512, 735663252850019217, 11503661742608944170, 180077229781765344602, 2821666487800835457300
Offset: 0

Views

Author

David Radcliffe, Sep 04 2025

Keywords

Comments

a(n) is the number of subsets of {1,...,4n} of size 2n containing at least n+1 elements from {1,...,2n}.
Also the maximum size of a family of 2n-subsets of a 4n-set such that every pairwise intersection has at least two elements. This was conjectured by Erdős, Ko, and Rado, and proved by Ahlswede and Khachatrian.

Crossrefs

Programs

  • Maple
    seq(add(binomial(2*n, k)^2, k=0..(n-1)), n=0..20);
    # or
    gf := (1/2)*((sqrt(1 + sqrt(1 - 16*x)))/(sqrt(2 - 32*x)) - hypergeom([1/2, 1/2], [1], 16*x)):
    ser := series(gf, x, 20): seq(coeff(ser, x, n), n = 0..19);  # Peter Luschny, Sep 05 2025
  • Mathematica
    Table[(Binomial[4n, 2n] - Binomial[2n, n]^2)/2, {n, 0, 20}]
    (* or *)
    gf[x_] := (Sqrt[1 + Sqrt[1 - 16 x]])/(2 Sqrt[2 - 32 x] ) - EllipticK[16 x]/Pi;
    CoefficientList[Series[gf[x], {x, 0, 19}], x]  (* Peter Luschny, Sep 05 2025 *)
    (* or *)
    CoefficientList[Series[(Sqrt[1 + Sqrt[1 - 16*x]])/(2*Sqrt[2 - 32*x]) - 1/(2*ArithmeticGeometricMean[1, Sqrt[1 - 16*x]]), {x, 0, 19}], x] (* Vaclav Kotesovec, Sep 06 2025 *)

Formula

a(n) = (1/2)*(C(4n, 2n) - C(2n, n)^2) = A071799(n)/2.
From Peter Luschny, Sep 05 2025: (Start)
a(n) = A036910(n) - A002894(n).
a(n) = [x^n]((1/2)*((sqrt(1 + sqrt(1 - 16*x)))/(sqrt(2 - 32*x)) - hypergeom([1/2, 1/2], [1], 16*x))).
a(n) = [x^n]((sqrt(1 + sqrt(1 - 16*x)))/(2*sqrt(2 - 32*x)) - EllipticK((4*sqrt(x))^m)/Pi) where m = 1 if the Maple conventions and m = 2 if the Mathematica conventions are followed.
a(n) ~ 16^n/sqrt(8*Pi*n) = A218708*16^n/sqrt(n). (End)
a(n) = [x^n] sqrt(1+sqrt(1-16*x))/(2*sqrt(2-32*x)) - 1/(2*AGM(1,sqrt(1-16*x))). - Vaclav Kotesovec, Sep 06 2025
Showing 1-4 of 4 results.