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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A201631 a(n) is the number of Fibonacci meanders of length m*n and central angle 360/m degrees where m = 2.

Original entry on oeis.org

1, 3, 6, 13, 30, 70, 167, 405, 992, 2450, 6090, 15214, 38165, 96069, 242530, 613811, 1556856, 3956316, 10070871, 25674210, 65541142, 167517654, 428635032, 1097874434, 2814611701, 7221917871, 18544968768, 47655572191, 122544150258, 315313433594, 811792614547
Offset: 1

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Author

Peter Luschny, Jan 15 2012

Keywords

Comments

Empirically the partial sums of A051291. - Sean A. Irvine, Jul 13 2022
The above conjecture was proved by Baril et al., which also give a formal definition of the Fibonacci meanders and describe a bijection with a certain class of peakless grand Motzkin paths of length n. - Peter Luschny, Mar 16 2023

Examples

			a(3) = 6 = card({100001, 100100, 110000, 111001, 111100, 111111}).
		

Crossrefs

Programs

  • Maple
    A201631 := n -> add(A202411(k),k=0..2*n-1): seq(A201631(i),i=1..9);
    # Alternative, using the g.f. of Baril et al.:
    S := (x^2 - x + 1 - R)/((x - 1)*(x^2 - x - 1 + R)*R):
    R := (((x - 3)*x + 1)*(x^2 + x + 1))^(1/2): ser := series(S, x, 33):
    seq(coeff(ser, x, n), n = 1..31); # Peter Luschny, Mar 16 2023
    # Using a recurrence:
    a := proc(n) option remember; if n < 5 then return [0, 1, 3, 6, 13][n + 1] fi;
    (n*(2*n - 1)*(2*n - 3)*(n - 5)*a(n - 5) - (n - 4)*(2*n - 1)^2*(3*n - 5)*a(n - 4) + (2*n - 5)*(n - 3)*(2*n^2 - 3*n + 2)*a(n - 3) - (2*n - 3)*(n - 2)*(2*n^2 - 3*n + 5)*a(n - 2) + (3*n - 4)*(2*n - 1)*(2*n - 5)*(n - 1)*a(n - 1))/(n*(2*n - 3)*(2*n - 5)*(n - 1)) end: seq(a(n), n = 1..31); # Peter Luschny, Mar 16 2023
  • Mathematica
    a[n_] := Sum[A202411[k], {k, 0, 2 n - 1}];
    Array[a, 31] (* Jean-François Alcover, Jun 29 2019 *)

Formula

a(n) = Sum_{k=0..2n-1} A202411(k).
a(n) = [x^n] (x^2 - x + 1 - R)/((x - 1)*(x^2 - x - 1 + R) * R), where R = (((x - 3)*x + 1)*(x^2 + x + 1))^(1/2). (This is Theorem 21 in Baril et al.) - Peter Luschny, Mar 16 2023

A110197 Number triangle of sums of squared binomial coefficients.

Original entry on oeis.org

1, 2, 1, 3, 5, 1, 4, 14, 10, 1, 5, 30, 46, 17, 1, 6, 55, 146, 117, 26, 1, 7, 91, 371, 517, 251, 37, 1, 8, 140, 812, 1742, 1476, 478, 50, 1, 9, 204, 1596, 4878, 6376, 3614, 834, 65, 1, 10, 285, 2892, 11934, 22252, 19490, 7890, 1361, 82, 1, 11, 385, 4917, 26334, 66352, 82994, 51990, 15761, 2107, 101, 1
Offset: 0

Views

Author

Paul Barry, Jul 15 2005

Keywords

Comments

Alternatively, number square T(n,k) = Sum_{i=0..n} binomial(i+k,k)^2 read by antidiagonals.

Examples

			Rows start:
  1;
  2,   1;
  3,   5,   1;
  4,  14,  10,   1;
  5,  30,  46,  17,   1;
  6,  55, 146, 117,  26,   1;
  ...
		

Crossrefs

Row sums are A006134.
Antidiagonal sums are A110198.
T(2n,n) gives A112029.

Programs

  • PARI
    T(n,k) = sum(i=0, n-k, binomial(i+k,k)^2);
    tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print();); \\ Michel Marcus, Dec 03 2016

Formula

T(n,k) = Sum_{i=0..n-k} binomial(i+k,k)^2.
G.f.: 1/((1-x)*sqrt(x^2*y^2-2*x^2*y-2*x*y+x^2-2*x+1)). - Vladimir Kruchinin, Mar 20 2025
Showing 1-2 of 2 results.