A097070 -id:A097070 - cp's OEIS frontend

A088218 Total number of leaves in all rooted ordered trees with n edges.

1, 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052

Offset: 0

Michael Somos, Sep 24 2003

Essentially the same as A001700, which has more information.
Note that the unique rooted tree with no edges has no leaves, so a(0)=1 is by convention. - Michael Somos, Jul 30 2011
Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic, Sep 15 2004
Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,126; 10,35,126,462]) = 4. - Philippe Deléham, Apr 13 2007
a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] if xA045992(n). - Geoffrey Critzer, Apr 02 2009
Hankel transform of the aeration of this sequence is A000027 doubled: 1,1,2,2,3,3,... - Paul Barry, Sep 26 2009
The Fi1 and Fi2 triangle sums of A039599 are given by the terms of this sequence. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Alternating row sums of Riordan triangle A094527. See the Philippe Deléham formula. - Wolfdieter Lang, Nov 22 2012
(-2)*a(n) is the Z-sequence for the Riordan triangle A110162. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 22 2012
From Gus Wiseman, Jun 27 2021: (Start)
Also the number of integer compositions of 2n with alternating (or reverse-alternating) sum 0 (ranked by A344619). This is equivalent to Ran Pan's comment at A001700. For example, the a(0) = 1 through a(3) = 10 compositions are:
  ()  (11)  (22)    (33)
            (121)   (132)
            (1111)  (231)
                    (1122)
                    (1221)
                    (2112)
                    (2211)
                    (11121)
                    (12111)
                    (111111)
For n > 0, a(n) is also the number of integer compositions of 2n with alternating sum 2.
(End)
Number of terms in the expansion of (x_1+x_2+...+x_n)^n. - César Eliud Lozada, Jan 08 2022

			G.f. = 1 + x + 3*x^2 + 10*x^3 + 35*x^4 + 126*x^5 + 462*x^6 + 1716*x^7 + ...
The five rooted ordered trees with 3 edges have 10 leaves.
..x........................
..o..x.x..x......x.........
..o...o...o.x..x.o..x.x.x..
..r...r....r....r.....r....

L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Anwar Al Ghabra, K. Gopala Krishna, Patrick Labelle, and Vasilisa Shramchenko, Enumeration of multi-rooted plane trees, arXiv:2301.09765 [math.CO], 2023.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
B. Dasarathy and C. Yang, A transformation on ordered trees, Comput. J. 23 (1980) 161-164. - _Nachum Dershowitz_, Sep 01 2016
Toufik Mansour and Mark Shattuck, A statistic on n-color compositions and related sequences, Proc. Indian Acad. Sci. (Math. Sci.) Vol. 124, No. 2, May 2014, pp. 127-140.
Anthony Mansuy, Preordered forests, packed words and contraction algebras, J. Algebra 411 (2014) 259-311.
Mircea Merca, A Note on Cosine Power Sums, J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
A. Vogt, Resummation of small-x double logarithms in QCD: semi-inclusive electron-positron annihilation, arXiv preprint arXiv:1108.2993 [hep-ph], 2011.

Same as A001700 modulo initial term and offset.
First differences are A024718.
Main diagonal of A071919 and of A305161.
A signed version is A110556.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A003242 counts anti-run compositions.
A025047 counts wiggly compositions (ascend: A025048, descend: A025049).
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A106356 counts compositions by number of maximal anti-runs.
A124754 gives the alternating sum of standard compositions.
A345197 counts compositions by sum, length, and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218 (this sequence), ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218 (this sequence), ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000027, A000070, A000097, A000108, A001622, A006232, A008965, A039599, A045992, A058696, A094527, A097070, A110162, A110555, A180662, A238279, A239830, A325534, A325535, A333213, A344607, A344611, A344617.

[Binomial(2*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014

seq(binomial(2*n-1, n),n=0..24); # Peter Luschny, Sep 22 2014

a[ n_] := SeriesCoefficient[(1 - x)^-n, {x, 0, n}];
c = (1 - (1 - 4 x)^(1/2))/(2 x);CoefficientList[Series[1/(1-(c-1)),{x,0,20}],x] (* Geoffrey Critzer, Dec 02 2010 *)
Table[Binomial[2 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ (1 + BesselI[0, 2 x]) / 2, {x, 0, m}]]]; (* Michael Somos, Nov 22 2014 *)

{a(n) = sum( i=0, n, binomial(n+i-2,i))};

{a(n) = if( n<0, 0, polcoeff( (1 + 1 / sqrt(1 - 4*x + x * O(x^n))) / 2, n))};

{a(n) = if( n<0, 0, polcoeff( 1 / (1 - x + x * O(x^n))^n, n))};

{a(n) = if( n<0, 0, binomial( 2*n - 1, n))};

{a(n) = if( n<1, n==0, polcoeff( subst((1 - x) / (1 - 2*x), x, serreverse( x - x^2 + x * O(x^n))), n))};

def A088218(n):
    return rising_factorial(n,n)/falling_factorial(n,n)
[A088218(n) for n in (0..24)]  # Peter Luschny, Nov 21 2012

G.f.: (1 + 1 / sqrt(1 - 4*x)) / 2.
a(n) = binomial(2*n - 1, n).
a(n) = (n+1)*A000108(n)/2, n>=1. - B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Feb 05 2002 (in A060150)
a(n) = (0^n + C(2n, n))/2. - Paul Barry, May 21 2004
a(n) is the coefficient of x^n in 1 / (1 - x)^n and also the sum of the first n coefficients of 1 / (1 - x)^n. Given B(x) with the property that the coefficient of x^n in B(x)^n equals the sum of the first n coefficients of B(x)^n, then B(x) = B(0) / (1 - x).
G.f.: 1 / (2 - C(x)) = (1 - x*C(x))/sqrt(1-4*x) where C(x) is g.f. for Catalan numbers A000108. Second equation added by Wolfdieter Lang, Nov 22 2012.
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..n} binomial(2*n, k)*cos((n-k)*Pi);
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1+(-1)^(n-k))*cos(k*Pi/2)/2 (with interpolated zeros);
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*cos((n-2*k)*Pi/2) (with interpolated zeros); (End)
a(n) = A110556(n)*(-1)^n, central terms in triangle A110555. - Reinhard Zumkeller, Jul 27 2005
a(n) = Sum_{0<=k<=n} A094527(n,k)*(-1)^k. - Philippe Deléham, Mar 14 2007
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-x/(1-2x/(1-(1/2)x/(1-(3/2)x/(1-(2/3)x/(1-(4/3)x/(1-(3/4)x/(1-(5/4)x/(1-... (continued fraction);
E.g.f.: (of aerated sequence) (1 + Bessel_I(0, 2*x))/2. (End)
a(n + 1) = A001700(n). a(n) = A024718(n) - A024718(n - 1).
E.g.f.: E(x) = 1+x/(G(0)-2*x) ; G(k) = (k+1)^2+2*x*(2*k+1)-2*x*(2*k+3)*((k+1)^2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 21 2011
a(n) = Sum_{k=0..n}(-1)^k*binomial(2*n,n+k). - Mircea Merca, Jan 28 2012
a(n) = rf(n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
D-finite with recurrence: n*a(n) +2*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = hypergeom([1-n,-n],[1],1). - Peter Luschny, Sep 22 2014
G.f.: 1 + x/W(0), where W(k) = 4*k+1 - (4*k+3)*x/(1 - (4*k+1)*x/(4*k+3 - (4*k+5)*x/(1 - (4*k+3)*x/W(k+1)  ))) ; (continued fraction). - Sergei N. Gladkovskii, Nov 13 2014
a(n) = A000984(n) + A001791(n). - Gus Wiseman, Jun 28 2021
E.g.f.: (1 + exp(2*x) * BesselI(0,2*x)) / 2. - Ilya Gutkovskiy, Nov 03 2021
From Amiram Eldar, Mar 12 2023: (Start)
Sum_{n>=0} 1/a(n) = 5/3 + 4*Pi/(9*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/5 - 8*log(phi)/(5*sqrt(5)), where phi is the golden ratio (A001622). (End)
a(n) ~ 2^(2*n-1)/sqrt(n*Pi). - Stefano Spezia, Apr 17 2024

A014107 a(n) = n(2n-3).

Original entry on oeis.org

0, -1, 2, 9, 20, 35, 54, 77, 104, 135, 170, 209, 252, 299, 350, 405, 464, 527, 594, 665, 740, 819, 902, 989, 1080, 1175, 1274, 1377, 1484, 1595, 1710, 1829, 1952, 2079, 2210, 2345, 2484, 2627, 2774, 2925, 3080, 3239, 3402, 3569, 3740, 3915, 4094, 4277

Offset: 0

N. J. A. Sloane

Positive terms give a bisection of A000096. - Omar E. Pol, Dec 16 2016

G. C. Greubel, Table of n, a(n) for n = 0..5000
Emily Barnard and Nathan Reading, Coxeter-biCatalan combinatorics, arXiv:1605.03524 [math.CO], 2016. See p. 51.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A000108, A014105, A033537, A097070, A100035, A100036, A100037, A100038, A100039, A100345, A002378, A000384.

A014107:=n->n*(2*n-3); seq(A014107(n), n=0..100); # Wesley Ivan Hurt, Nov 19 2013

Table[2n^2 - 3n, {n, 0, 49}] (* Alonso del Arte, Dec 15 2012 *)
LinearRecurrence[{3,-3,1},{0,-1,2},50] (* Harvey P. Dale, Sep 18 2018 *)

PARI
```
a(n)=n*(2*n-3)
```

a(n) = A100345(n, n - 3) for n > 2.
a(n) = A033537(n) - 8*n^2; A100035(a(n)) = 2 for n > 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = A014106(-n) for all n in Z. - Michael Somos, Nov 06 2005
From Michael Somos, Nov 06 2005: (Start)
G.f.: x*(-1 + 5*x)/(1 - x)^3.
E.g.f: x*(-1 + 2*x)*exp(x). (End)
a(n) = A097070(n)/A000108(n - 2), n >= 2. - Philippe Deléham, Apr 12 2007
a(n) = 2*a(n-1) - a(n-2) + 4, n > 1; a(0) = 0, a(1) = -1, a(2) = 2. - Zerinvary Lajos, Feb 18 2008
a(n) = a(n-1) + 4*n - 5 with a(0) = 0. - Vincenzo Librandi, Nov 20 2010
a(n) = (2*n-1)*(n-1) - 1. Also, with an initial offset of -1, a(n) = (2*n-1)*(n+1) = 2*n^2 + n - 1. - Alonso del Arte, Dec 15 2012
(a(n) + 1)^2 + (a(n) + 2)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2n - 1)^2 starting with a(1) = -1. - Jeffreylee R. Snow, Sep 17 2013
a(n) = A014105(n-1) - 1 for all n in Z. - Michael Somos, Nov 23 2021
From Amiram Eldar, Feb 20 2022: (Start)
Sum_{n>=1} 1/a(n) = -2*(1 - log(2))/3.
Sum_{n>=1} (-1)^n/a(n) = Pi/6 + log(2)/3 + 2/3. (End)
For n > 0, A002378(a(n)) = A000384(n-1)*A000384(n). - Charlie Marion, May 21 2023

A189911 Row sums of the extended Catalan triangle A189231.

Original entry on oeis.org

1, 2, 4, 9, 18, 40, 80, 175, 350, 756, 1512, 3234, 6468, 13728, 27456, 57915, 115830, 243100, 486200, 1016158, 2032316, 4232592, 8465184, 17577014, 35154028, 72804200, 145608400, 300874500, 601749000, 1240940160, 2481880320, 5109183315, 10218366630

Offset: 0

Peter Luschny, May 01 2011

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
Peter Luschny, The lost Catalan numbers.

Cf. A037965, A097070, A056040, A212303.

A189911 := proc(n) local a,b,d; if n = 0 then 1 else
a := GAMMA(n-floor(n/2)); b := GAMMA(floor(n/2+3/2));
d := GAMMA(floor(n/2+1))^2; GAMMA(n+1)*(a*b+d)/(a*b*d) fi end: seq(A189911(n),n=0..32);
A189911 := proc(n) h:=irem(n,2); g:=iquo(n,2); (g+h+1)*binomial(2*g+h,g+h) end; # Peter Luschny, Oct 24 2013

a[n_] := Module[{q, r}, {q, r} = QuotientRemainder[n, 2]; (q+r+1)*Pochhammer[q+1, q+r]/(q+r)!]; Table[a[n], {n, 0, 32}] (* Jean-François Alcover, Jan 09 2014 *)

def A189911():
    r, n = 1, 1
    while True:
        yield r
        h = n//2
        r *= 2 if is_even(n) else (h+2)*(2*h+1)/(h+1)^2
        n += 1
a = A189911(); [next(a) for i in range(16)]  # Peter Luschny, Oct 24 2013

Let a = Gamma(n-floor(n/2)), b = Gamma(floor(n/2+3/2)), d = Gamma( floor(n/2+1))^2, c = Gamma(n+1). Then a(n) = c*(a*b+d)/(a*b*d).
a(n) = A162246(n,n) + A162246(n,n+1) for n > 0.
From Peter Luschny, Oct 24 2013 : (Start)
E.g.f.: (x+1)*(BesselI(0, 2*x)+BesselI(1, 2*x)).
O.g.f.: I*(2*x^2-1)/(2*sqrt(2*x+1)*x*(2*x-1)^(3/2))-1/(2*x).
Recurrence: a(0) = 1; a(n) = a(n-1)*2 if n is even else ([n/2]+2)*(2*[n/2]+1)/([n/2]+1)^2. ([.] the floor brackets.)
a(n) = A056040(n) + A212303(n) = n$*(1+[(n+1)/2]^((-1)^n)), where n$ is the swinging factorial.
a(2*n) = (n+1)*C(2*n, n) (A037965);
a(2*n+1) = (n+2)*C(2*n+1, n+1) (A097070). (End)
Sum_{n>=0} 1/a(n) = 4*Pi/sqrt(3) - Pi^2/3 - 2. - Amiram Eldar, Aug 20 2022
D-finite with recurrence: (n-2)*(n+1)^2*a(n) - (2*(n-2)^2+2*n-12)*a(n-1) - 4*(n+2)*(n-1)^2*a(n-2) = 0. - Georg Fischer, Nov 25 2022

A371400 Triangle read by rows: T(n, k) = binomial(k + n, k)binomial(2n - k, n).

Original entry on oeis.org

1, 2, 2, 6, 9, 6, 20, 40, 40, 20, 70, 175, 225, 175, 70, 252, 756, 1176, 1176, 756, 252, 924, 3234, 5880, 7056, 5880, 3234, 924, 3432, 13728, 28512, 39600, 39600, 28512, 13728, 3432, 12870, 57915, 135135, 212355, 245025, 212355, 135135, 57915, 12870

Offset: 0

Peter Luschny, Mar 21 2024

The main diagonal and column 0 of the triangle are the central binomial coefficients, which are the sums of the squares of Pascal's triangle entries. This sum representation can be generalized, and all terms can be seen as sums of coefficients of some polynomials. (See the Example section.)

To see this, consider T(n, k) as the value of the polynomials P(n, k)(x) at x = 1, where P(n, k)(x) = H([-n, -k], [1], x)*H([-n, -n + k], [1], x) and H denotes the hypergeometric sum 2F1. For instance column 0 is given by the row sums of A008459, and column 1 by the row sums of A371401.

			Triangle starts:
[0]    1;
[1]    2,     2;
[2]    6,     9,     6;
[3]   20,    40,    40,    20;
[4]   70,   175,   225,   175,    70;
[5]  252,   756,  1176,  1176,   756,   252;
[6]  924,  3234,  5880,  7056,  5880,  3234,   924;
[7] 3432, 13728, 28512, 39600, 39600, 28512, 13728, 3432;
.
Because of the symmetry, only the sum representation of terms with k <= n/2 are shown.
0:                 [1]
1:               [1+1]
2:             [1+4+1],               [1+4+4]
3:           [1+9+9+1],            [1+9+21+9]
4:      [1+16+36+16+1],       [1+16+66+76+16],        [1+16+76+96+36]
5: [1+25+100+100+25+1], [1+25+160+340+205+25], [1+25+190+460+400+100]

Column 0 and main diagonal are A000984.
Column 1 and subdiagonal are A097070.
Row sums are A045721.
The even bisection of the alternating row sums is A005809.
The central terms are A188662.
Cf. A046899, A092392, A371395, A244038, A371399.
Cf. A008459, A371401.

T := (n, k) -> binomial(k + n, k) * binomial(2*n - k, n):
seq(print(seq(T(n, k), k = 0..n)), n = 0..8);

T[n_, k_] := Hypergeometric2F1[-n, -k, 1, 1] Hypergeometric2F1[-n, -n +k, 1, 1];
Table[T[n, k], {n, 0, 7}, {k, 0, n}]

T(n, k) = A046899(n, k) * A092392(n, k).
T(n, k) = A046899(n, k) * A046899(n, n - k).
T(n, k) = A092392(n, k) * A092392(n, n - k).
T(n, k) = A371395(n, k) * (n + 1).
T(n, k) = hypergeom([-n, -k], [1], 1) * hypergeom([-n, -n + k], [1], 1).
2^n*Sum_{k=0..n} T(n, k)*(1/2)^k = A244038(n).
2^n*Sum_{k=0..n} T(n, k)*(-1/2)^k = A371399(n).

cp's OEIS Frontend

A088218 Total number of leaves in all rooted ordered trees with n edges.

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A014107 a(n) = n*(2*n-3).

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A189911 Row sums of the extended Catalan triangle A189231.

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A371400 Triangle read by rows: T(n, k) = binomial(k + n, k)*binomial(2*n - k, n).

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A275329 a(n) = (2+[n/2])*n!/((1+[n/2])*[n/2]!^2).

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A371401 Triangle read by rows: T(n, k) = [x^k] (n*x + 1)*Hypergeometric([-n, -n + 1], [1], x).

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A014107 a(n) = n(2n-3).

A371400 Triangle read by rows: T(n, k) = binomial(k + n, k)binomial(2n - k, n).

A275329 a(n) = (2+[n/2])n!/((1+[n/2])[n/2]!^2).

A371401 Triangle read by rows: T(n, k) = [x^k] (nx + 1)Hypergeometric([-n, -n + 1], [1], x).