A089849 -id:A089849 - cp's OEIS frontend

A063886 Number of n-step walks on a line starting from the origin but not returning to it.

1, 2, 2, 4, 6, 12, 20, 40, 70, 140, 252, 504, 924, 1848, 3432, 6864, 12870, 25740, 48620, 97240, 184756, 369512, 705432, 1410864, 2704156, 5408312, 10400600, 20801200, 40116600, 80233200, 155117520, 310235040, 601080390, 1202160780, 2333606220, 4667212440

Offset: 0

Henry Bottomley, Aug 28 2001

A Chebyshev transform of A007877(n+1). The g.f. is transformed to (1+x)/((1-x)(1+x^2)) under the mapping G(x)->(1/(1+x^2))G(1/(1+x^2)). - Paul Barry, Oct 12 2004
a(n-1) = 2*C(n-2, floor((n-2)/2)) is also the number of bit strings of length n in which the number of 00 substrings is equal to the number of 11 substrings. For example, when n = 4 we have 4 such bit strings: 0011, 0101, 1010, and 1100. - Angel Plaza, Apr 23 2009
Hankel transform is A120617. - Paul Barry, Aug 10 2009
The Hankel transform of a(n) is (-2)^C(n+1,2). The Hankel transform of (-1)^C(n+1,2)*a(n) is (-1)^C(n+1,2)*A164584(n). - Paul Barry, Aug 17 2009
For n > 1, a(n) is also the number of n-step walks starting from the origin and returning to it exactly once. - Geoffrey Critzer, Jan 24 2010
-a(n) is the Z-sequence for the Riordan array A130777. (See the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices). - Wolfdieter Lang, Jul 12 2011
Number of subsets of {1,...,n} in which the even elements appear as often at even positions as at odd positions. - Gus Wiseman, Mar 17 2018

			a(4) = 6 because there are six length four walks that do not return to the origin: {-1, -2, -3, -4}, {-1, -2, -3, -2}, {-1, -2, -1, -2}, {1, 2, 1, 2}, {1, 2, 3, 2}, {1, 2, 3, 4}. There are also six such walks that return exactly one time: {-1, -2, -1, 0}, {-1, 0, -1, -2}, {-1, 0, 1, 2}, {1, 0, -1, -2}, {1, 0, 1, 2}, {1, 2, 1, 0}. - _Geoffrey Critzer_, Jan 24 2010
The a(5) = 12 subsets in which the even elements appear as often at even positions as at odd positions: {}, {1}, {3}, {5}, {1,3}, {1,5}, {2,4}, {3,5}, {1,2,4}, {1,3,5}, {2,4,5}, {1,2,4,5}. - _Gus Wiseman_, Mar 17 2018

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
Emeric Deutsch, Problem 11424, The American Mathematical Monthly, Vol. 116, No. 3 (March 2009), p. 277.
D. Perrin, A conjecture on rational sequences, pp. 267-274 of R. M. Capocelli, ed., Sequences, Springer-Verlag, NY 1990.

Apart from initial terms, same as A182027.
Cf. A000108, A000246, A000712, A000984, A001405, A001700, A002420, A006232, A007877, A026010, A028329, A045931, A047073, A089849, A097613, A106180, A120617, A130777, A130780, A138364, A164584, A171966, A239241, A300787, A300788, A300789.
Cf. A307768 (complementary event).

[1] cat [2*Binomial(n-1, Floor((n-1)/2)): n in [1..40]]; // G. C. Greubel, Jun 07 2023

seq(seq(binomial(2*j,j)*i, i=1..2),j=0..16); # Zerinvary Lajos, Apr 28 2007
# second Maple program:
a:= proc(n) option remember; `if`(n<2, n+1,
       4*a(n-2) +2*(a(n-1) -4*a(n-2))/n)
    end:
seq(a(n), n=0..40);  # Alois P. Heinz, Feb 10 2014
# third program:
A063886 := series(BesselI(0, 2*x)*(1 + x*2 + x*Pi*StruveL(1, 2*x)) - Pi*x*BesselI(1, 2*x)*StruveL(0, 2*x), x = 0, 34): seq(n!*coeff(A063886, x, n), n = 0 .. 33); # Mélika Tebni, Jun 17 2024

Table[Length[Select[Map[Accumulate, Strings[{-1, 1}, n]], Count[ #, 0] == 0 &]], {n, 0, 20}] (* Geoffrey Critzer, Jan 24 2010 *)
CoefficientList[Series[Sqrt[(1+2x)/(1-2x)],{x,0,40}],x] (* Harvey P. Dale, Apr 28 2016 *)

PARI
```
a(n)=(n==0)+2*binomial(n-1,(n-1)\2)
```

a(n) = 2^n*prod(k=0,n-1,(k/n+1/n)^((-1)^k)); \\ Michel Marcus, Dec 03 2013

from math import ceil
from sympy import binomial
def a(n):
    if n==0: return 1
    return 2*binomial(n-1,(n-1)//2)
print([a(n) for n in range(18)])
# David Nacin, Feb 29 2012

[2*binomial(n-1, (n-1)//2) + int(n==0) for n in range(41)] # G. C. Greubel, Jun 07 2023

G.f.: sqrt((1+2*x)/(1-2*x)).
a(n+1) = 2*C(n, floor(n/2)) = 2*A001405(n); a(2n) = C(2n, n) = A000984(n) = 4*a(2n-2)-|A002420(n)| = 4*a(2n-2)-2*A000108(n-1) = 2*A001700(n-1); a(2n+1) = 2*a(2n) = A028329(n).
2*a(n) = A047073(n+1).
a(n) = Sum_{k=0..n} abs(A106180(n,k)). - Philippe Deléham, Oct 06 2006
a(n) = Sum_{k=0..n} (k+1)binomial(n, (n-k)/2) ( 1-cos((k+1)*Pi/2) (1+(-1)^(n-k))/(n+k+2) ). - Paul Barry, Oct 12 2004
G.f.: 1/(1-2*x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+ ... (continued fraction). - Paul Barry, Aug 10 2009
G.f.: 1 + 2*x/(G(0)-x+x^2) where G(k)= 1 - 2*x^2 - x^4/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Aug 10 2012
D-finite with recurrence: n*a(n) = 2*a(n-1) + 4*(n-2)*a(n-2). - R. J. Mathar, Dec 03 2012
From Sergei N. Gladkovskii, Jul 26 2013: (Start)
G.f.: 1/G(0), where G(k) = 1 - 2*x/(1 + 2*x/(1 + 1/G(k+1) )); (continued fraction).
G.f.: G(0), where G(k) = 1 + 2*x/(1 - 2*x/(1 + 1/G(k+1) )); (continued fraction).
G.f.: W(0)/2*(1+2*x), where W(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)/(x*(2*k+1))/W(k+1) )), abs(x) < 1/2; (continued fraction). (End)
a(n) = 2^n*Product_{k=0..n-1} (k/n + 1/n)^((-1)^k). - Peter Luschny, Dec 02 2013
G.f.: G(0), where G(k) = 1 + 2*x*(4*k+1)/((2*k+1)*(1+2*x) - (2*k+1)*(4*k+3)*x*(1+2*x)/((4*k+3)*x + (k+1)*(1+2*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 19 2014
From Peter Bala, Mar 29 2024: (Start)
a(n) = 2^n * Sum_{k = 0..n} (-1)^(n+k)*binomial(1/2, k)*binomial(- 1/2, n-k) = 2^n * A000246(n)/n!.
a(n) = (1/2^n) * binomial(2*n, n) * hypergeom([-1/2, -n], [1/2 - n], -1). (End)
E.g.f.: BesselI(0, 2*x)*(1 + x*(2 + Pi)*StruveL(1, 2*x)) - Pi*x*BesselI(1, 2*x)*StruveL(0, 2*x). - Stefano Spezia, May 11 2024
a(n) = A089849(n) + A138364(n). - Mélika Tebni, Jun 17 2024
From Amiram Eldar, Aug 15 2025: (Start)
Sum_{n>=0} 1/a(n) = Pi/(3*sqrt(3)) + 2.
Sum_{n>=0} (-1)^n/a(n) = 2/3 + Pi/(9*sqrt(3)). (End)

A062196 Triangle read by rows, T(n, k) = binomial(n, k)*binomial(n + 2, k).

Original entry on oeis.org

1, 1, 3, 1, 8, 6, 1, 15, 30, 10, 1, 24, 90, 80, 15, 1, 35, 210, 350, 175, 21, 1, 48, 420, 1120, 1050, 336, 28, 1, 63, 756, 2940, 4410, 2646, 588, 36, 1, 80, 1260, 6720, 14700, 14112, 5880, 960, 45, 1, 99, 1980, 13860, 41580, 58212, 38808, 11880, 1485, 55

Offset: 0

Wolfdieter Lang, Jun 19 2001

Also the coefficient triangle of certain polynomials N(2; m,x) := Sum_{k=0..m} T(m,k)*x^k. The e.g.f. of the m-th (unsigned) column sequence without leading zeros of the generalized (a=2) Laguerre triangle L(2; n+m,m) = A062139(n+m,m), n >= 0, is N(2; m,x)/(1-x)^(3+2*m), with the row polynomials N(2; m,x).

			Triangle starts:
  n\k 0...1.....2......3..... 4.....;
  [0] 1;
  [1] 1,  3;
  [2] 1,  8,    6;
  [3] 1, 15,   30,    10;
  [4] 1, 24,   90,    80,    15;
  [5] 1, 35,  210,   350,   175,    21;
  [6] 1, 48,  420,  1120,  1050,   336,    28;
  [7] 1, 63,  756,  2940,  4410,  2646,   588,    36;
  [8] 1, 80, 1260,  6720, 14700, 14112,  5880,   960,   45;
  [9] 1, 99, 1980, 13860, 41580, 58212, 38808, 11880, 1485, 55.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Family of polynomials (see A062145): A008459 (c=1), A132813 (c=2), this sequence (c=3), A062145 (c=4), A062264 (c=5), A062190 (c=6).
Sums include: A001791 (row), (-1)^n*A089849(n+1) (alternating sign row).
Diagonals: A000217 (k=n), A002417 (k=n-1), A001297 (k=n-2), A105946 (k=n-3), A105947 (k=n-4), A105948 (k=n-5), A107319 (k=n-6).
Columns: A005563 (k=1), A033487 (k=2), A027790 (k=3), A107395 (k=4), A107396 (k=5), A107397 (k=6), A107398 (k=7), A107399 (k=8).
Cf. A000108, A089849.

A062196:= func;
[A062196(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 21 2025

T := (n, k) -> binomial(n, k)*binomial(n + 2, k);
seq(seq(T(n, k), k=0..n), n=0..9); # Peter Luschny, Sep 30 2021

A062196[n_, k_]:= Binomial[n, k]*Binomial[n+2, k];
Table[A062196[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 21 2025 *)

def A062196(n,k): return binomial(n,k)*binomial(n+2,k)
print(flatten([[A062196(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Feb 21 2025

T(m, k) = [x^k] N(2; m, x), where N(2; m, x) = ((1-x)^(3+2*m))*(d^m/dx^m)(x^m/(m!*(1-x)^(m+3))).
N(2; m, x) = Sum_{j=0..m} ((binomial(m, j)*(2*m+2-j)!/((m+2)!*(m-j)!)*(x^(m-j)))*(1-x)^j).
T(n,m) = binomial(n, m)*(binomial(n+1, m) + binomial(n+1, m-1)). - Vladimir Kruchinin, Apr 06 2018
From G. C. Greubel, Feb 21 2025: (Start)
T(2*n, n) = (n+1)^2*A000108(n)*A000108(n+1).
T(2*n-1, n) = (4*n^2 - 1)*A000108(n-1)*A000108(n), n >= 1.
T(2*n+1, n) = (1/2)*binomial(n+2,2)*A000108(n+1)*A000108(n+2). (End)

New name by Peter Luschny, Sep 30 2021

A089880 Number of cycles in range [A014137(n-1)..A014138(n-1)] of permutation A069772.

Original entry on oeis.org

1, 1, 2, 3, 10, 22, 76, 217, 750, 2438, 8524, 29414, 104468, 371516, 1338936, 4847637, 17685270, 64823110, 238843660, 883634026, 3282152588, 12233141908, 45741634536, 171529836218, 644953425740, 2430973304732, 9183681736376

Offset: 0

Antti Karttunen, Nov 29 2003

The number of orbits to which the corresponding automorphism(s) partitions the set of A000108(n) binary trees with n internal nodes.

A. Karttunen, C-program for computing the initial terms of this sequence

(define (A089880 n) (/ (+ (A000108 n) (A089849 n)) 2))

a(n) = (A000108(n)+A089849(n))/2

A345013 Triangle read by rows, related to clusters of type D.

Original entry on oeis.org

1, 4, 3, 15, 20, 6, 56, 105, 60, 10, 210, 504, 420, 140, 15, 792, 2310, 2520, 1260, 280, 21, 3003, 10296, 13860, 9240, 3150, 504, 28, 11440, 45045, 72072, 60060, 27720, 6930, 840, 36

Offset: 1

F. Chapoton, Sep 30 2021

Let C_{n+1} be the cyclic quiver with n+1 vertices. Empirically, the n-th row is related to the green-mutation partial order on clusters for this quiver, restricted to clusters that do not meet the initial seed.
Apparently, value of the associated polynomials at -2 is A089849, up to sign.
By evaluating the associated polynomials at x-1, one apparently gets A062196.
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)^2*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)*(x+n+2) / (n! * (n+2)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 31 2022
Chapoton's observation above is correct: the precise expansion is (x+1)^2*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)*(x+n+2) / (n! * (n+2)!) = Sum_{k = 0..n} (-1)^k*T(n+1,k)*binomial(x+2*n+2-k, 2*n+2-k), as can be verified using the WZ algorithm. For example, n = 2 gives (x+1)^2*(x+2)^2*(x+3)*(x+4)/(2!*4!) = 15*binomial(x+6,6) - 20*binomial(x+5,5) + 6*binomial(x+4,4). - Peter Bala, Jun 24 2023

			Triangle begins:
[1] 1
[2] 4,    3
[3] 15,   20,    6
[4] 56,   105,   60,    10
[5] 210,  504,   420,   140,  15
[6] 792,  2310,  2520,  1260, 280,  21
[7] 3003, 10296, 13860, 9240, 3150, 504, 28
...

Cf. A001791 (T(n,1)), A000217 (T(n,n)), A026002 (row sums), A000012 (alternating row sum), A051924 (number of clusters of type D_n).

Cf. A089849, A062196, A063007, A253283.

row(n) = vector(n, k, k--; (n-k)*binomial(n,k)*binomial(2*n-k, n-1)/n); \\ Michel Marcus, Sep 30 2021

def T_row(n):
    return [(n-k)*binomial(n,k)*binomial(2*n-k,n-1)//n for k in range(n)]
for n in range(1, 8): print(T_row(n))

T(n, k) = (n-k)*binomial(n,k)*binomial(2*n-k, n-1)/n, for n >= 1 and 0 <= k < n.
From Peter Bala, Jun 24 2023: (Start)
As conjectured above by Chapoton we have
Sum_{k = 0..n-1} T(n,k)*(x - 1)^k = Sum_{k = 0..n-1} A062196(n-1,k)*x^k and
Sum_{k = 0..n-1} T(n,k)*(-2)^k = (-1)^floor(n/2)*A089849(n) for n >= 1 (both easily verified using the WZ algorithm). (End)

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A063886 Number of n-step walks on a line starting from the origin but not returning to it.

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A062196 Triangle read by rows, T(n, k) = binomial(n, k)*binomial(n + 2, k).

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A089880 Number of cycles in range [A014137(n-1)..A014138(n-1)] of permutation A069772.

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A345013 Triangle read by rows, related to clusters of type D.

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