A001700 -id:A001700 - cp's OEIS frontend

A000097 Number of partitions of n if there are two kinds of 1's and two kinds of 2's.

1, 2, 5, 9, 17, 28, 47, 73, 114, 170, 253, 365, 525, 738, 1033, 1422, 1948, 2634, 3545, 4721, 6259, 8227, 10767, 13990, 18105, 23286, 29837, 38028, 48297, 61053, 76926, 96524, 120746, 150487, 187019, 231643, 286152, 352413, 432937, 530383, 648245

Offset: 0

N. J. A. Sloane

Also number of partitions of 2*n with exactly 2 odd parts (offset 1). - Vladeta Jovovic, Jan 12 2005
Also number of transitions from one partition of n+2 to another, where a transition consists of replacing any two parts with their sum. Remove all 1' and 2' from the partition, replacing them with ((number of 2') + 1) and ((number of 1') + (number of 2') + 1); these are the two parts being summed. Number of partitions of n into parts of 2 kinds with at most 2 parts of the second kind, or of n+2 into parts of 2 kinds with exactly 2 parts of the second kind. - Franklin T. Adams-Watters, Mar 20 2006
From Christian Gutschwager (gutschwager(AT)math.uni-hannover.de), Feb 10 2010: (Start)
a(n) is also the number of pairs of partitions of n+2 which differ by only one box (for bijection see the first Gutschwager link).
a(n) is also the number of partitions of n with two parts marked.
a(n) is also the number of partitions of n+1 with two different parts marked. (End)
Convolution of A000041 and A008619. - Vaclav Kotesovec, Aug 18 2015
a(n) = P(/2,n), a particular case of P(/k,n) defined as follows: P(/0,n) = A000041(n) and P(/k,n) = P(/k-1, n) + P(/k-1,n-k) + P(/k-1, n-2k) + ... Also, P(/k,n) = the convolution of A000041 and the partitions of n with exactly k parts, and g.f. P(/k,n) = (g.f. for P(n)) * 1/(1-x)...(1-x^k). - Gregory L. Simay, Mar 22 2018
a(n) is also the sum of binomial(D(p),2) in partitions p of (n+3), where D(p)= number of different sizes of parts in p. - Emily Anible, Apr 03 2018
Also partitions of 2*(n+1) with alternating sum 2. Also partitions of 2*(n+1) with reverse-alternating sum -2 or 2. - Gus Wiseman, Jun 21 2021
Define the distance graph of the partitions of n using the distance function in A366156 as follows: two vertices (partitions) share an edge if and only if the distance between the vertices is 2. Then a(n) is the number of edges in the distance graph of the partitions of n. - Clark Kimberling, Oct 12 2023

			a(3) = 9 because we have 3, 2+1, 2+1', 2'+1, 2'+1', 1+1+1, 1+1+1', 1+1'+1' and 1'+1'+1'.
From _Gus Wiseman_, Jun 22 2021: (Start)
The a(0) = 1 through a(4) = 9 partitions of 2*(n+1) with exactly 2 odd parts:
  (1,1)  (3,1)    (3,3)      (5,3)
         (2,1,1)  (5,1)      (7,1)
                  (3,2,1)    (3,3,2)
                  (4,1,1)    (4,3,1)
                  (2,2,1,1)  (5,2,1)
                             (6,1,1)
                             (3,2,2,1)
                             (4,2,1,1)
                             (2,2,2,1,1)
The a(0) = 1 through a(4) = 9 partitions of 2*(n+1) with alternating sum 2:
  (2)  (3,1)    (4,2)        (5,3)
       (2,1,1)  (2,2,2)      (3,3,2)
                (3,2,1)      (4,3,1)
                (3,1,1,1)    (3,2,2,1)
                (2,1,1,1,1)  (4,2,1,1)
                             (2,2,2,1,1)
                             (3,2,1,1,1)
                             (3,1,1,1,1,1)
                             (2,1,1,1,1,1,1)
(End)

H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 90.
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe and Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Christian Gutschwager, The skew diagram poset and components of skew characters, arXiv:1104.0008 [math.CO], 2011.
Christian Gutschwager, Reduced Kronecker products which are multiplicity free or contain only a few components, Eur. J. Combinat. 31 (2010) 1996-2005. doi:10.1016/j.ejc.2010.05.008.
J. P. Robinson, Edges in the poset of partitions of an integer, J. Combin. Theory Ser. A, 48 (1988), 236-238.
N. J. A. Sloane, Transforms

First differences are in A024786.
Cf. A000098, A000710.
Third column of Riordan triangle A008951 and of triangle A103923.
The case of reverse-alternating sum 1 or alternating sum 0 is A000041.
The case of reverse-alternating sum -1 or alternating sum 1 is A000070.
The normal case appears to be A004526 or A065033.
The strict case is A096914.
The case of reverse-alternating sum 2 is A120452.
The case of reverse-alternating sum -2 is A344741.
A001700 counts compositions with alternating sum 2.
A035363 counts partitions into even parts.
A058696 counts partitions of 2n.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A124754 gives alternating sums of standard compositions (reverse: A344618).
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
Cf. A006330, A027187, A239830, A306145, A343941, A344607, A344608, A344619, A344650, A344651, A344740.
Shift of A093695.

with(numtheory): etr:= proc(p) local b; b:=proc(n) option remember; local d,j; if n=0 then 1 else add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n)/n fi end end: a:= etr(n->`if`(n<3,2,1)): seq(a(n), n=0..40); # Alois P. Heinz, Sep 08 2008

CoefficientList[Series[1/((1 - x) (1 - x^2) Product[1 - x^k, {k, 1, 100}]), {x, 0, 100}], x] (* Ben Branman, Mar 07 2012 *)
etr[p_] := Module[{b}, b[n_] := b[n] = If[n == 0, 1, Sum[Sum[d*p[d], {d, Divisors[j]}]*b[n - j], {j, 1, n}]/n]; b]; a = etr[If[# < 3, 2, 1]&]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Apr 09 2014, after Alois P. Heinz *)
(1/((1 - x) (1 - x^2) QPochhammer[x]) + O[x]^50)[[3]] (* Vladimir Reshetnikov, Nov 22 2016 *)
Table[Length@IntegerPartitions[n,All,Join[{1,2},Range[n]]],{n,0,15}] (* Robert Price, Jul 28 2020 and Jun 21 2021 *)
T[n_, 0] := PartitionsP[n];
T[n_, m_] /; (n >= m (m + 1)/2) := T[n, m] = T[n - m, m - 1] + T[n - m, m];
T[, ] = 0;
a[n_] := T[n + 3, 2];
Table[a[n], {n, 0, 60}] (* Jean-François Alcover, May 30 2021 *)
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];Table[Length[Select[IntegerPartitions[n],ats[#]==2&]],{n,0,30,2}] (* Gus Wiseman, Jun 21 2021 *)

my(x = 'x + O('x^66)); Vec( 1/((1-x)*(1-x^2)*eta(x)) ) \\ Joerg Arndt, Apr 29 2013

Euler transform of 2 2 1 1 1 1 1...
G.f.: 1/( (1-x) * (1-x^2) * Product_{k>=1} (1-x^k) ).
a(n) = Sum_{j=0..floor(n/2)} A000070(n-2*j), n>=0.
a(n) = A014153(n)/2 + A087787(n)/4 + A000070(n)/4. - Vaclav Kotesovec, Nov 05 2016
a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n/3)) / (4*Pi^2) * (1 + 35*Pi/(24*sqrt(6*n))). - Vaclav Kotesovec, Aug 18 2015, extended Nov 05 2016
a(n) = A120452(n) + A344741(n). - Gus Wiseman, Jun 21 2021

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 04 2004
Edited by Emeric Deutsch, Mar 23 2005
More terms from Franklin T. Adams-Watters, Mar 20 2006
Edited by Charles R Greathouse IV, Apr 20 2010

A027306 a(n) = 2^(n-1) + ((1 + (-1)^n)/4)*binomial(n, n/2).

Original entry on oeis.org

1, 1, 3, 4, 11, 16, 42, 64, 163, 256, 638, 1024, 2510, 4096, 9908, 16384, 39203, 65536, 155382, 262144, 616666, 1048576, 2449868, 4194304, 9740686, 16777216, 38754732, 67108864, 154276028, 268435456, 614429672, 1073741824, 2448023843

Offset: 0

Clark Kimberling

Inverse binomial transform of A027914. Hankel transform (see A001906 for definition) is {1, 2, 3, 4, ..., n, ...}. - Philippe Deléham, Jul 21 2005
Number of walks of length n on a line that starts at the origin and ends at or above 0. - Benjamin Phillabaum, Mar 05 2011
Number of binary integers (i.e., with a leading 1 bit) of length n+1 which have a majority of 1-bits. E.g., for n+1=4: (1011, 1101, 1110, 1111) a(3)=4. - Toby Gottfried, Dec 11 2011
Number of distinct symmetric staircase walks connecting opposite corners of a square grid of side n > 1. - Christian Barrientos, Nov 25 2018
From Gus Wiseman, Aug 20 2021: (Start)
Also the number of integer compositions of n + 1 with alternating sum > 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A345917. For example, the a(0) = 1 through a(4) = 11 compositions are:
  (1)  (2)  (3)    (4)    (5)
            (21)   (31)   (32)
            (111)  (112)  (41)
                   (211)  (113)
                          (122)
                          (212)
                          (221)
                          (311)
                          (1121)
                          (2111)
                          (11111)
The following relate to these compositions:
- The unordered version is A027193.
- The complement is counted by A058622.
- The reverse unordered version is A086543.
- The version for alternating sum >= 0 is A116406.
- The version for alternating sum < 0 is A294175.
- Ranked by A345917. (End)
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 07 2022

			From _Gus Wiseman_, Aug 20 2021: (Start)
The a(0) = 1 through a(4) = 11 binary numbers with a majority of 1-bits (Gottfried's comment) are:
  1   11   101   1011   10011
           110   1101   10101
           111   1110   10110
                 1111   10111
                        11001
                        11010
                        11011
                        11100
                        11101
                        11110
                        11111
The version allowing an initial zero is A058622.
(End)

A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992, Eq. (4.2.1.6)

Alois P. Heinz, Table of n, a(n) for n = 0..1000
F. Disanto, A. Frosini, and S. Rinaldi, Square involutions, J. Int. Seq. 14 (2011) # 11.3.5.
Zachary Hamaker and Eric Marberg, Atoms for signed permutations, arXiv:1802.09805 [math.CO], 2018.
Donatella Merlini and Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.

a(n) = Sum{(k+1)T(n, m-k)}, 0<=k<=[ (n+1)/2 ], T given by A008315.
Column k=2 of A226873. - Alois P. Heinz, Jun 21 2013
Cf. A008949, A248574, A001405, A246437.
The even bisection is A000302.
The odd bisection appears to be A032443.
Cf. A000984, A001700, A001791, A008549, A011782, A088218, A097805, A163493, A182616, A345197.

List([0..35],n->Sum([0..Int(n/2)],k->Binomial(n,k))); # Muniru A Asiru, Nov 27 2018

a027306 n = a008949 n (n `div` 2)  -- Reinhard Zumkeller, Nov 14 2014

[2^(n-1)+(1+(-1)^n)/4*Binomial(n, n div 2): n in [0..40]]; // Vincenzo Librandi, Jun 19 2016

a:= proc(n) add(binomial(n, j), j=0..n/2) end:
seq(a(n), n=0..32); # Zerinvary Lajos, Mar 29 2009

Table[Sum[Binomial[n, k], {k, 0, Floor[n/2]}], {n, 1, 35}]
(* Second program: *)
a[0] = a[1] = 1; a[2] = 3; a[n_] := a[n] = (2(n-1)(2a[n-2] + a[n-1]) - 8(n-2) a[n-3])/n; Array[a, 33, 0] (* Jean-François Alcover, Sep 04 2016 *)

a(n)=if(n<0,0,(2^n+if(n%2,0,binomial(n, n/2)))/2)

a(n) = Sum_{k=0..floor(n/2)} binomial(n,k).
Odd terms are 2^(n-1). Also a(2n) - 2^(2n-1) is given by A001700. a(n) = 2^n + (n mod 2)*binomial(n, (n-1)/2).
E.g.f.: (exp(2x) + I_0(2x))/2.
O.g.f.: 2*x/(1-2*x)/(1+2*x-((1+2*x)*(1-2*x))^(1/2)). - Vladeta Jovovic, Apr 27 2003
a(n) = A008949(n, floor(n/2)); a(n) + a(n-1) = A248574(n), n > 0. - Reinhard Zumkeller, Nov 14 2014
From Peter Bala, Jul 21 2015: (Start)
a(n) = [x^n]( 2*x - 1/(1 - x) )^n.
O.g.f.: (1/2)*( 1/sqrt(1 - 4*x^2) + 1/(1 - 2*x) ).
Inverse binomial transform is (-1)^n*A246437(n).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + ... is the o.g.f. for A001405. (End)
a(n) = Sum_{k=1..floor((n+1)/2)} binomial(n-1,(2n+1-(-1)^n)/4 -k). - Anthony Browne, Jun 18 2016
D-finite with recurrence: n*a(n) + 2*(-n+1)*a(n-1) + 4*(-n+1)*a(n-2) + 8*(n-2)*a(n-3) = 0. - R. J. Mathar, Aug 09 2017

Better description from Robert G. Wilson v, Aug 30 2000 and from Yong Kong (ykong(AT)curagen.com), Dec 28 2000

A057079 Periodic sequence: repeat [1,2,1,-1,-2,-1]; expansion of (1+x)/(1-x+x^2).

Original entry on oeis.org

1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1

Offset: 0

Wolfdieter Lang, Aug 04 2000

Inverse binomial transform of A057083. Binomial transform of A061347. The sums of consecutive pairs of elements give A084103. - Paul Barry, May 15 2003
Hexaperiodic sequence identical to its third differences. - Paul Curtz, Dec 13 2007
a(n+1) is the Hankel transform of A001700(n+1)-A001700(n). - Paul Barry, Apr 21 2009
Non-simple continued fraction expansion of 1 = 1+1/(2+1/(1+1/(-1+...))). - R. J. Mathar, Mar 08 2012
Pisano period lengths: 1, 3, 2, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, ... - R. J. Mathar, Aug 10 2012
Alternating row sums of Riordan triangle A111125. - Wolfdieter Lang, Oct 18 2012
Periodic sequences of this type can be also calculated by a(n) = c + floor(q/(p^m-1)*p^n) mod p, where c is a constant, q is the number representing the periodic digit pattern and m is the period length. c, p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D, min = minimum value of elements in D. Than c := min, p := max - min + 1 and q := p^m*sum_{i=1..m} (D(i)-min)/p^i. Example: D = (1, 2, 1, -1, -2, -1), c = -2, m = 6, p = 5 and q = 12276 for this sequence. - Hieronymus Fischer, Jan 04 2013

			G.f. = 1 + 2*x + x^2 - x^3 - 2*x^4 - x^5 + x^6 + 2*x^7 + x^8 - x^9 - 2*x^10 + x^11 + ...

Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
T.-X. He and L. W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic. 532 (2017) 25-41, theorem 2.5, k=3.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,-1).
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.

Cf. A049310. Apart from signs, same as A061347.

Cf. A002264, A010872.

A057079:=n->[1, 2, 1, -1, -2, -1][(n mod 6)+1]: seq(A057079(n), n=0..100); # Wesley Ivan Hurt, Mar 10 2015

a[n_] := {1, 2, 1, -1, -2, -1}[[Mod[n, 6] + 1]]; Array[a, 100, 0] (* Jean-François Alcover, Jul 05 2013 *)
CoefficientList[Series[(1 + x)/(1 - x + x^2), {x, 0, 71}], x] (* Michael De Vlieger, Jul 10 2017 *)
PadRight[{},100,{1,2,1,-1,-2,-1}] (* Harvey P. Dale, Nov 11 2024 *)

{a(n) = [1, 2, 1, -1, -2, -1][n%6 + 1]}; /* Michael Somos, Jul 14 2006 */

{a(n) = if( n<0, n = 2-n); polcoeff( (1 + x) / (1 - x + x^2) + x * O(x^n), n)}; /* Michael Somos, Jul 14 2006 */

a(n)=2^(n%3%2)*(-1)^(n\3) \\ Tani Akinari, Aug 15 2013

a(n) = S(n, 1) + S(n-1, 1) = S(2*n, sqrt(3)); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 1) = A010892(n).
a(n) = 2*cos((n-1)*Pi/3) = a(n-1) - a(n-2) = -a(n-3) = a(n-6) = (A022003(n+1)+1)*(-1)^floor(n/3). Unsigned a(n) = 4 - a(n-1) - a(n-2). - Henry Bottomley, Mar 29 2001
a(n) = (-1)^floor(n/3) + ((-1)^floor((n-1)/3) + (-1)^floor((n+1)/3))/2. - Mario Catalani (mario.catalani(AT)unito.it), Jan 07 2003
a(n) = (1/2 - sqrt(3)*i/2)^(n-1) + (1/2 + sqrt(3)*i/2)^(n-1) = cos(Pi*n/3) + sqrt(3)*sin(Pi*n/3). - Paul Barry, Mar 15 2004
The period 3 sequence (2, -1, -1, ...) has a(n) = 2*cos(2*Pi*n/3) = (-1/2 - sqrt(3)*i/2)^n + (-1/2 + sqrt(3)*i/2)^n. - Paul Barry, Mar 15 2004
Euler transform of length 6 sequence [2, -2, -1, 0, 0, 1]. - Michael Somos, Jul 14 2006
G.f.: (1 + x) / (1 - x + x^2) = (1 - x^2)^2 * (1 - x^3) / ((1 - x)^2 * (1 - x^6)). a(n) = a(2-n) for all n in Z. - Michael Somos, Jul 14 2006
a(n) = A033999(A002264(n))*(A000035(A010872(n))+1). - Hieronymus Fischer, Jun 20 2007
a(n) = (3*A033999(A002264(n)) - A033999(n))/2. - Hieronymus Fischer, Jun 20 2007
a(n) = (-1)^floor(n/3)*((n mod 3) mod 2 + 1). - Hieronymus Fischer, Jun 20 2007
a(n) = (3*(-1)^floor(n/3) - (-1)^n)/2. - Hieronymus Fischer, Jun 20 2007
a(n) = (-1)^((n-1)/3) + (-1)^((1-n)/3). - Jaume Oliver Lafont, May 13 2010
E.g.f.: E(x) = S(0), S(k) = 1 + 2*x/(6*k+1 - x*(6*k+1)/(4*(3*k+1) + x + 4*x*(3*k+1)/(6*k + 3 - x - x*(6*k+3)/(3*k + 2 + x - x*(3*k+2)/(12*k + 10 + x - x*(12*k+10)/(x - (6*k+6)/S(k+1))))))); (continued fraction). - Sergei N. Gladkovskii, Dec 14 2011
a(n) = -2 + floor((281/819)*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 04 2013
a(n) = -2 + floor((11/14)*5^(n+1)) mod 5. - Hieronymus Fischer, Jan 04 2013
a(n) = A010892(n) + A010892(n-1).
a(n) = ( (1+i*sqrt(3))^(n-1) + (1-i*sqrt(3))^(n-1) )/2^(n-1), where i=sqrt(-1). - Bruno Berselli, Dec 01 2014
a(n) = 2*sin((2n+1)*Pi/6). - Wesley Ivan Hurt, Apr 04 2015
a(n) = hypergeom([-n/2-2, -n/2-5/2], [-n-4], 4). - Peter Luschny, Dec 17 2016
G.f.: 1 / (1 - 2*x / (1 + 3*x / (2 - x))). - Michael Somos, Dec 29 2016
a(n) = (2*n+1)*(Sum_{k=0..n} ((-1)^k/(2*k+1))*binomial(n+k,2*k)) for n >= 0. - Werner Schulte, Jul 10 2017
Sum_{n>=0} (a(n)/(2*n+1))*x^(2*n+1) = arctan(x/(1-x^2)) for -1 < x < 1. - Werner Schulte, Jul 10 2017
E.g.f.: exp(x/2)*(sqrt(3)*cos(sqrt(3)*x/2) + 3*sin(sqrt(3)*x/2))/sqrt(3). - Stefano Spezia, Aug 04 2025

A008549 Number of ways of choosing at most n-1 items from a set of size 2*n+1.

Original entry on oeis.org

0, 1, 6, 29, 130, 562, 2380, 9949, 41226, 169766, 695860, 2842226, 11576916, 47050564, 190876696, 773201629, 3128164186, 12642301534, 51046844836, 205954642534, 830382690556, 3345997029244, 13475470680616, 54244942336114, 218269673491780, 877940640368572

Offset: 0

N. J. A. Sloane

Area under Dyck excursions (paths ending in 0): a(n) is the sum of the areas under all Dyck excursions of length 2*n (nonnegative walks beginning and ending in 0 with jumps -1,+1).
Number of inversions in all 321-avoiding permutations of [n+1]. Example: a(2)=6 because the 321-avoiding permutations of [3], namely 123,132,312,213,231, have 0, 1, 2, 1, 2 inversions, respectively. - Emeric Deutsch, Jul 28 2003
Convolution of A001791 and A000984. - Paul Barry, Feb 16 2005
a(n) = total semilength of "longest Dyck subpath" starting at an upstep U taken over all upsteps in all Dyck paths of semilength n. - David Callan, Jul 25 2008
[1,6,29,130,562,2380,...] is convolution of A001700 with itself. - Philippe Deléham, May 19 2009
From Ran Pan, Feb 04 2016: (Start)
a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) bounce off the diagonal y = x to the right. This is related to paired pattern P_2 in Pan and Remmel's link and more details can be found in Section 3.2 in the link.
a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) horizontally cross the diagonal y = x. This is related to paired pattern P_3 in Pan and Remmel's link and more details can be found in Section 3.3 in the link.
2*a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) bounce off the diagonal y = x. This is related to paired pattern P_2 and P_4 in Pan and Remmel's link and more details can be found in Section 4.2 in the link.
2*a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) cross the diagonal y = x. This is related to paired pattern P_3 and P_4 in Pan and Remmel's link and more details can be found in Section 4.3 in the link. (End)
From Gus Wiseman, Jul 17 2021: (Start)
Also the number of integer compositions of 2*(n+1) with alternating sum < 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(3) = 29 compositions of 8 are:
  (1,7)  (1,5,2)  (1,1,1,5)  (1,1,1,4,1)  (1,1,1,1,1,3)
  (2,6)  (1,6,1)  (1,1,2,4)  (1,2,1,3,1)  (1,1,1,2,1,2)
  (3,5)  (2,5,1)  (1,2,1,4)  (1,3,1,2,1)  (1,1,1,3,1,1)
                  (1,2,2,3)  (1,4,1,1,1)  (1,2,1,1,1,2)
                  (1,3,1,3)               (1,2,1,2,1,1)
                  (1,3,2,2)               (1,3,1,1,1,1)
                  (1,4,1,2)
                  (1,4,2,1)
                  (1,5,1,1)
                  (2,1,1,4)
                  (2,2,1,3)
                  (2,3,1,2)
                  (2,4,1,1)
Also the number of integer compositions of 2*(n+1) with reverse-alternating sum < 0. For a bijection, keep the odd-length compositions and reverse the even-length ones.
Also the number of 2*(n+1)-digit binary numbers with more 0's than 1's. For example, the a(2) = 6 binary numbers are: 100000, 100001, 100010, 100100, 101000, 110000; or in decimal: 32, 33, 34, 36, 40, 48.
(End)

			a(2) = 6 because there are 6 ways to choose at most 1 item from a set of size 5: You can choose the empty set, or you can choose any of the five one-element sets.
G.f. = x + 6*x^2 + 29*x^3 + 130*x^4 + 562*x^5 + 2380*x^6 + 9949*x^7 + ...

D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.

Indranil Ghosh, Table of n, a(n) for n = 0..1500 (terms 0..200 from T. D. Noe)
José Agapito, Ângela Mestre, Maria M. Torres, and Pasquale Petrullo, On One-Parameter Catalan Arrays, Journal of Integer Sequences, 18 (2015), Article 15.5.1.
Octavio Arizmendi, Daniel Perales, and Josue Vazquez-Becerra, Finite Free Convolution: Infinitesimal Distributions, arXiv:c [math.PR], 2025. See p. 34.
Jean Christophe Aval, Adrien Boussicault, Patxi Laborde-Zubieta, and Mathias Pétréolle, Generating series of Periodic Parallelogram polyominoes, arXiv:1612.03759, 2016.
Roland Bacher, On generating series of complementary plane trees, arXiv:math/0409050 [math.CO], 2004.
Vijay Balasubramanian, Javier M. Magan, and Qingyue Wu, A Tale of Two Hungarians: Tridiagonalizing Random Matrices, arXiv:2208.08452 [hep-th], 2022.
Cyril Banderier, Analytic combinatorics of random walks and planar maps, Ph. D. Thesis, 2001. [Broken link]
Adrien Boussicault and P. Laborde-Zubieta, Periodic Parallelogram Polyominoes, arXiv preprint arXiv:1611.03766 [math.CO], 2016.
AJ Bu, Explicit Generating Functions for the Sum of the Areas Under Dyck and Motzkin Paths (and for Their Powers), arXiv:2310.17026 [math.CO], 2023.
AJ Bu and Doron Zeilberger, Using Symbolic Computation to Explore Generalized Dyck Paths and Their Areas, arXiv:2305.09030 [math.CO], 2023.
Alexander Burstein and Sergi Elizalde, Total occurrence statistics on restricted permutations, arXiv:1305.3177 [math.CO], 2013.
Robin Chapman, Moments of Dyck paths, Discrete Math., 204 (1999), 113-117.
Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, Pinnacle sets of signed permutations, arXiv:2301.02628 [math.CO], 2023.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
Niklas G. Johansson, Efficient Simulation of the Deutsch-Jozsa Algorithm, Master's Project, Department of Electrical Engineering & Department of Physics, Chemistry and Biology, Linkoping University, April, 2015.
Miles Jones, Sergey Kitaev, and Jeffrey Remmel, Frame patterns in n-cycles, arXiv preprint arXiv:1311.3332 [math.CO], 2013.
James A. Mingo and Josue Vazquez-Becerra, The Asymptotic Infinitesimal Distribution of a Real Wishart Random Matrix, arXiv:2112.15231 [math.PR], 2021.
Henri Mühle, Symmetric Chain Decompositions and the Strong Sperner Property for Noncrossing Partition Lattices, arXiv:1509.06942v1 [math.CO], 2015.
Ran Pan and Jeffrey B. Remmel, Paired patterns in lattice paths, arXiv:1601.07988 [math.CO], 2016.
Elisa Pergola, Two bijections for the area of Dyck paths, Discrete Math., 241 (2001), 435-447.
Wen-jin Woan, Area of Catalan Paths, Discrete Math., 226 (2001), 439-444.

Cf. A038608, A045894, A057571, A109196, A153338.
Odd bisection of A294175 (even is A000346).
For integer compositions of 2*(n+1) with alternating sum k < 0 we have:
- The opposite (k > 0) version is A000302.
- The weak (k <= 0) version is (also) A000302.
- The k = 0 version is A001700 or A088218.
- The reverse-alternating version is also A008549 (this sequence).
- These compositions are ranked by A053754 /\ A345919.
- The complement (k >= 0) is counted by A114121.
- The case of reversed integer partitions is A344743(n+1).
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A345197 counts compositions by length and alternating sum.
Cf. A000070, A001791, A007318, A025047, A027306, A032443, A058622, A120452, A163493, A239830, A344611.

[4^n-Binomial(2*n+1, n): n in [0..30]]; // Vincenzo Librandi, Feb 04 2016

A008549:=n->4^n-binomial(2*n+1,n): seq(A008549(n), n=0..30);

Table[4^n-Binomial[2n+1,n],{n,0,30}] (* Harvey P. Dale, May 11 2011 *)
a[ n_] := If[ n<-4, 0, 4^n - Binomial[2 n + 2, n + 1] / 2] (* Michael Somos, Jan 25 2014 *)

{a(n)=if(n<0, 0, 4^n - binomial(2*n+1, n))} /* Michael Somos Oct 31 2006 */

{a(n) = if( n<-4, 0, n++; (4^n / 2 - binomial(2*n, n)) / 2)} /* Michael Somos, Jan 25 2014 */

import math
def C(n,r):
    f=math.factorial
    return f(n)/f(r)/f(n-r)
def A008549(n):
    return int((4**n)-C(2*n+1,n)) # Indranil Ghosh, Feb 18 2017

a(n) = 4^n - C(2*n+1, n).
a(n) = Sum_{k=1..n} Catalan(k)*4^(n-k): convolution of Catalan numbers and powers of 4.
G.f.: x*c(x)^2/(1 - 4*x), c(x) = g.f. of Catalan numbers. - Wolfdieter Lang
Note Sum_{k=0..2*n+1} binomial(2*n+1, k) = 2^(2n+1). Therefore, by the symmetry of Pascal's triangle, Sum_{k=0..n} binomial(2*n+1, k) = 2^(2*n) = 4^n. This explains why the following two expressions for a(n) are equal: Sum_{k=0..n-1} binomial(2*n+1, k) = 4^n - binomial(2*n+1, n). - Dan Velleman
G.f.: (2*x^2 - 1 + sqrt(1 - 4*x^2))/(2*(1 + 2*x)*(2*x - 1)*x^3).
a(n) = Sum_{k=0..n} C(2*k, k)*C(2*(n-k), n-k-1). - Paul Barry, Feb 16 2005
Second binomial transform of 2^n - C(n, floor(n/2)) = A045621(n). - Paul Barry, Jan 13 2006
a(n) = Sum_{0 < i <= k < n} binomial(n, k+i)*binomial(n, k-i). - Mircea Merca, Apr 05 2012
D-finite with recurrence (n+1)*a(n) + 2*(-4*n-1)*a(n-1) + 8*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 03 2012
0 = a(n) * (256*a(n+1) - 224*a(n+2) + 40*a(n+3)) + a(n+1) * (-32*a(n+1) + 56*a(n+2) - 14*a(n+3)) + a(n+2) * (-2*a(n+2) + a(n+3)) if n > -5. - Michael Somos, Jan 25 2014
Convolution square is A045894. - Michael Somos, Jan 25 2014
HANKEL transform is [0, -1, 2, -3, 4, -5, ...]. - Michael Somos, Jan 25 2014
BINOMIAL transform of [0, 0, 1, 3, 11, 35,...] (A109196) is [0, 0, 1, 6, 29, 130, ...]. - Michael Somos, Jan 25 2014
(n+1) * a(n) = A153338(n+1). - Michael Somos, Jan 25 2014
a(n) = Sum_{m = n+2..2*n+1} binomial(2*n+1,m), n >= 0. - Wolfdieter Lang, May 22 2015
E.g.f.: (exp(2*x) - BesselI(0,2*x) - BesselI(1,2*x))*exp(2*x). - Ilya Gutkovskiy, Aug 30 2016

Better description from Dan Velleman (djvelleman(AT)amherst.edu), Dec 01 2000

A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 14, 14, 6, 1, 42, 48, 27, 8, 1, 132, 165, 110, 44, 10, 1, 429, 572, 429, 208, 65, 12, 1, 1430, 2002, 1638, 910, 350, 90, 14, 1, 4862, 7072, 6188, 3808, 1700, 544, 119, 16, 1, 16796, 25194, 23256, 15504, 7752, 2907, 798, 152, 18, 1

Offset: 0

N. J. A. Sloane

T(n,k) is the number of leaves at level k+1 in all ordered trees with n+1 edges. - Emeric Deutsch, Jan 15 2005
Riordan array ((1-2x-sqrt(1-4x))/(2x^2),(1-2x-sqrt(1-4x))/(2x)). Inverse array is A053122. - Paul Barry, Mar 17 2005
T(n,k) is the number of walks of n steps, each in direction N, S, W, or E, starting at the origin, remaining in the upper half-plane and ending at height k (see the R. K. Guy reference, p. 5). Example: T(3,2)=6 because we have ENN, WNN, NEN, NWN, NNE and NNW. - Emeric Deutsch, Apr 15 2005
Triangle T(n,k), 0<=k<=n, read by rows given by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 30 2007
Number of (2n+1)-step walks from (0,0) to (2n+1,2k+1) and consisting of steps u=(1,1) and d=(1,-1) in which the path stays in the nonnegative quadrant. Examples: T(2,0)=5 because we have uuudd, uudud, uuddu, uduud, ududu; T(2,1)=4 because we have uuuud, uuudu, uuduu, uduuu; T(2,2)=1 because we have uuuuu. - Philippe Deléham, Apr 16 2007, Apr 18 2007
Triangle read by rows: T(n,k)=number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and two types of steps H=(1,0); example: T(3,1)=14 because we have UDU, UUD, 4 HHU paths, 4 HUH paths and 4 UHH paths. - Philippe Deléham, Sep 25 2007
This triangle belongs to the family of triangles defined by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = x*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
With offset [1,1] this is the (ordinary) convolution triangle a(n,m) with o.g.f. of column m given by (c(x)-1)^m, where c(x) is the o.g.f. for Catalan numbers A000108. See the Riordan comment by Paul Barry.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) with exactly k fixed points. - Abdullahi Umar, Oct 02 2008
T(n,k)/2^(2n+1) = coefficients of the maximally flat lowpass digital differentiator of the order N=2n+3. - Pavel Holoborodko (pavel(AT)holoborodko.com), Dec 19 2008
The signed triangle S(n,k) := (-1)^(n-k)*T(n,k) provides the transformation matrix between f(n,l) := L(2*l)*5^n*F(2*l)^(2*n+1) (F=Fibonacci numbers A000045, L=Lucas numbers A000032) and F(4*l*(k+1)), k = 0, ..., n, for each l>=0: f(n,l) = Sum_{k=0..n} S(n,k)*F(4*l*(k+1)), n>=0, l>=0. Proof: the o.g.f. of the l.h.s., G(l;x) := Sum_{n>=0} f(n,l)*x^n = F(4*l)/(1 - 5*F(2*l)^2*x) is shown to match the o.g.f. of the r.h.s.: after an interchange of the n- and k-summation, the Riordan property of S = (C(x)/x,C(x)) (compare with the above comments by Paul Barry), with C(x) := 1 - c(-x), with the o.g.f. c(x) of A000108 (Catalan numbers), is used, to obtain, after an index shift, first Sum_{k>=0} F(4*l*(k))*GS(k;x), with the o.g.f of column k of triangle S which is GS(k;x) := Sum_{n>=k} S(n,k)*x^n = C(x)^(k+1)/x. The result is GF(l;C(x))/x with the o.g.f. GF(l,x) := Sum_{k>=0} F(4*l*k)*x^k = x*F(4*l)/(1-L(4*l)*x+x^2) (see a comment on A049670, and A028412). If one uses then the identity L(4*n) - 5*F(2*n)^2 = 2 (in Koshy's book [reference under A065563] this is No. 15, p. 88, attributed to Lucas, 1876), the proof that one recovers the o.g.f. of the l.h.s. from above boils down to a trivial identity on the Catalan o.g.f., namely 1/c^2(-x) = 1 + 2*x - (x*c(-x))^2. - Wolfdieter Lang, Aug 27 2012
O.g.f. for row polynomials R(x) := Sum_{k=0..n} a(n,k)*x^k:
  ((1+x) - C(z))/(x - (1+x)^2*z) with C the o.g.f. of A000108 (Catalan numbers). From Riordan ((C(x)-1)/x,C(x)-1), compare with a Paul Barry comment above. This coincides with the o.g.f. given by Emeric Deutsch in the formula section. - Wolfdieter Lang, Nov 13 2012
The A-sequence for this Riordan triangle is [1,2,1] and the Z-sequence is [2,1]. See a W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 13 2012
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n+1, 2*k+1). T(n, k) appears in the formula for the (2*n+1)-th power of the algebraic number rho(N) := 2*cos(Pi/N) = R(N, 2) in terms of the even-indexed diagonal/side length ratios R(N, 2*(k+1)) = S(2*k+1, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310): rho(N)^(2*n+1) = Sum_{k=0..n} T(n, k)*R(N, 2*(k+1)), n >= 0, identical in N >= 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears. For the even powers of rho(n) see A039599. (End)
The tridiagonal Toeplitz production matrix P in the Example section corresponds to the unsigned Cartan matrix for the simple Lie algebra A_n as n tends to infinity (cf. Damianou ref. in A053122). -  Tom Copeland, Dec 11 2015 (revised Dec 28 2015)
T(n,k) is the number of pairs of non-intersecting walks of n steps, each in direction N or E, starting at the origin, and such that the end points of the two paths are separated by a horizontal distance of k. See Shapiro 1976. - Peter Bala, Apr 12 2017
Also the convolution triangle of the Catalan numbers A000108. - Peter Luschny, Oct 07 2022

			Triangle T(n,k) starts:
n\k     0      1      2      3      4     5    6    7   8  9 10
0:      1
1:      2      1
2:      5      4      1
3:     14     14      6      1
4:     42     48     27      8      1
5:    132    165    110     44     10     1
6:    429    572    429    208     65    12    1
7:   1430   2002   1638    910    350    90   14    1
8:   4862   7072   6188   3808   1700   544  119   16   1
9:  16796  25194  23256  15504   7752  2907  798  152  18  1
10: 58786  90440  87210  62016  33915 14364 4655 1120 189 20  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 13 2012.
Production matrix begins:
2, 1
1, 2, 1
0, 1, 2, 1
0, 0, 1, 2, 1
0, 0, 0, 1, 2, 1
0, 0, 0, 0, 1, 2, 1
0, 0, 0, 0, 0, 1, 2, 1
0, 0, 0, 0, 0, 0, 1, 2, 1
- _Philippe Deléham_, Nov 07 2011
From _Wolfdieter Lang_, Nov 13 2012: (Start)
Recurrence: T(5,1) = 165 = 1*42 + 2*48 +1*27. The Riordan A-sequence is [1,2,1].
Recurrence from Riordan Z-sequence [2,1]: T(5,0) = 132 = 2*42 + 1*48. (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
  Example for rho(N) = 2*cos(Pi/N) powers:
  n=2: rho(N)^5 = 5*R(N, 2) + 4*R(N, 4) + 1*R(N, 6) = 5*S(1, rho(N)) + 4*S(3, rho(N)) + 1*S(5, rho(N)), identical in N >= 1. For N=5 (the pentagon with only one distinct diagonal) the degree delta(5) = 2, hence R(5, 4) and R(5, 6) can be reduced, namely to R(5, 1) = 1 and R(5, 6) = -R(5,1) = -1, respectively. Thus rho(5)^5 = 5*R(N, 2) + 4*1  + 1*(-1) = 3 + 5*R(N, 2) = 3 + 5*rho(5), with the golden section rho(5). (End)

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L. W. Shapiro, A Catalan triangle, Discrete Math. 14 (1976), no. 1, 83-90. [Annotated scanned copy]
Yidong Sun and Fei Ma, Minors of a Class of Riordan Arrays Related to Weighted Partial Motzkin Paths, arXiv preprint arXiv:1305.2015 [math.CO], 2013.
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.
Paweł J. Szabłowski, Yet another way of calculating moments of the Kesten's distribution and its consequences, arXiv:2106.10461 [math.CO], 2021.
Charles Zhao-Chen Wang and Yi Wang, Total positivity of Catalan triangle, Discrete Math. 338 (2015), no. 4, 566--568. MR3300743.
W.-J. Woan, L. Shapiro, and D. G. Rogers, The Catalan numbers, the Lebesgue integral and 4^{n-2}, Amer. Math. Monthly, 104 (1997), 926-931.
Sheng-Liang Yang, Yan-Ni Dong, and Tian-Xiao He, Some matrix identities on colored Motzkin paths, Discrete Mathematics 340.12 (2017): 3081-3091.

Mirror image of A050166. Row sums are A001700.

Cf. A000108, A008313, A039599, A183134, A094527, A033184, A035324, A053122, A172417.

/* As triangle: */ [[Binomial(2*n,n-k) - Binomial(2*n,n-k-2): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2015

T:=(n,k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); # N. J. A. Sloane, Aug 26 2013
# Uses function PMatrix from A357368. Adds row and column above and to the left.
PMatrix(10, n -> binomial(2*n, n) / (n + 1)); # Peter Luschny, Oct 07 2022

Flatten[Table[Binomial[2n, n-k] - Binomial[2n, n-k-2], {n,0,9}, {k,0,n}]] (* Jean-François Alcover, May 03 2011 *)

T(n,k)=binomial(2*n,n-k) - binomial(2*n,n-k-2) \\ Charles R Greathouse IV, Nov 07 2016

# Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle.
def A039598_triangle(n) :
    D = [0]*(n+2); D[1] = 1
    b = True; h = 1
    for i in range(2*n) :
        if b :
            for k in range(h,0,-1) : D[k] += D[k-1]
            h += 1
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        b = not b
        if b : print([D[z] for z in (1..h-1) ])
A039598_triangle(10)  # Peter Luschny, May 01 2012

Row n: C(2n, n-k) - C(2n, n-k-2).
a(n, k) = C(2n+1, n-k)*2*(k+1)/(n+k+2) = A050166(n, n-k) = a(n-1, k-1) + 2*a(n-1, k)+ a (n-1, k+1) [with a(0, 0) = 1 and a(n, k) = 0 if n<0 or nHenry Bottomley, Sep 24 2001
From Philippe Deléham, Feb 14 2004: (Start)
T(n, 0) = A000108(n+1), T(n, k) = 0 if n0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+2) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
Sum_{k>=0} T(m, k)*T(n, k) = A000108(m+n+1). (End)
T(n, k) = A009766(n+k+1, n-k) = A033184(n+k+2, 2k+2). - Philippe Deléham, Feb 14 2004
Sum_{j>=0} T(k, j)*A039599(n-k, j) = A028364(n, k). - Philippe Deléham, Mar 04 2004
Antidiagonal Sum_{k=0..n} T(n-k, k) = A000957(n+3). - Gerald McGarvey, Jun 05 2005
The triangle may also be generated from M^n * [1,0,0,0,...], where M = an infinite tridiagonal matrix with 1's in the super- and subdiagonals and [2,2,2,...] in the main diagonal. - Gary W. Adamson, Dec 17 2006
G.f.: G(t,x) = C^2/(1-txC^2), where C = (1-sqrt(1-4x))/(2x) is the Catalan function. From here G(-1,x)=C, i.e., the alternating row sums are the Catalan numbers (A000108). - Emeric Deutsch, Jan 20 2007
Sum_{k=0..n} T(n,k)*x^k = A000957(n+1), A000108(n), A000108(n+1), A001700(n), A049027(n+1), A076025(n+1), A076026(n+1) for x=-2,-1,0,1,2,3,4 respectively (see square array in A067345). - Philippe Deléham, Mar 21 2007, Nov 04 2011
Sum_{k=0..n} T(n,k)*(k+1) = 4^n. - Philippe Deléham, Mar 30 2007
Sum_{j>=0} T(n,j)*binomial(j,k) = A035324(n,k), A035324 with offset 0 (0 <= k <= n). - Philippe Deléham, Mar 30 2007
T(n,k) = A053121(2*n+1,2*k+1). - Philippe Deléham, Apr 16 2007, Apr 18 2007
T(n,k) = A039599(n,k) + A039599(n,k+1). - Philippe Deléham, Sep 11 2007
Sum_{k=0..n+1} T(n+1,k)*k^2 = A029760(n). - Philippe Deléham, Dec 16 2007
Sum_{k=0..n} T(n,k)*A059841(k) = A000984(n). - Philippe Deléham, Nov 12 2008
G.f.: 1/(1-xy-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-.... (continued fraction).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A001700(n), A194723(n+1), A194724(n+1), A194725(n+1), A194726(n+1), A194727(n+1), A194728(n+1), A194729(n+1), A194730(n+1) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Nov 03 2011
From Peter Bala, Dec 21 2014: (Start)
This triangle factorizes in the Riordan group as ( C(x), x*C(x) ) * ( 1/(1 - x), x/(1 - x) ) = A033184 * A007318, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
Let U denote the lower unit triangular array with 1's on or below the main diagonal and zeros elsewhere. For k = 0,1,2,... define U(k) to be the lower unit triangular block array
/I_k 0\
\ 0  U/ having the k X k identity matrix I_k as the upper left block; in particular, U(0) = U. Then this array equals the bi-infinite product (...*U(2)*U(1)*U(0))*(U(0)*U(1)*U(2)*...). (End)
From Peter Bala, Jul 21 2015: (Start)
O.g.f. G(x,t) = (1/x) * series reversion of ( x/f(x,t) ), where f(x,t) = ( 1 + (1 + t)*x )^2/( 1 + t*x ).
1 + x*d/dx(G(x,t))/G(x,t) = 1 + (2 + t)*x + (6 + 4*t + t^2)*x^2 + ... is the o.g.f for A094527. (End)
Conjecture: Sum_{k=0..n} T(n,k)/(k+1)^2 = H(n+1)*A000108(n)*(2*n+1)/(n+1), where H(n+1) = Sum_{k=0..n} 1/(k+1). - Werner Schulte, Jul 23 2015
From Werner Schulte, Jul 25 2015: (Start)
Sum_{k=0..n} T(n,k)*(k+1)^2 = (2*n+1)*binomial(2*n,n). (A002457)
Sum_{k=0..n} T(n,k)*(k+1)^3 = 4^n*(3*n+2)/2.
Sum_{k=0..n} T(n,k)*(k+1)^4 = (2*n+1)^2*binomial(2*n,n).
Sum_{k=0..n} T(n,k)*(k+1)^5 = 4^n*(15*n^2+15*n+4)/4. (End)
The o.g.f. G(x,t) is such that G(x,t+1) is the o.g.f. for A035324, but with an offset of 0, and G(x,t-1) is the o.g.f. for A033184, again with an offset of 0. - Peter Bala, Sep 20 2015
Denote this lower triangular array by L; then L * transpose(L) is the Cholesky factorization of the Hankel matrix ( 1/(i+j)*binomial(2*i+2*j-2, i+j-1) )A172417%20read%20as%20a%20square%20array.%20See%20Chamberland,%20p.%201669.%20-%20_Peter%20Bala">i,j >= 1 = A172417 read as a square array. See Chamberland, p. 1669. - _Peter Bala, Oct 15 2023

Typo in one entry corrected by Philippe Deléham, Dec 16 2007

A032443 a(n) = Sum_{i=0..n} binomial(2*n, i).

Original entry on oeis.org

1, 3, 11, 42, 163, 638, 2510, 9908, 39203, 155382, 616666, 2449868, 9740686, 38754732, 154276028, 614429672, 2448023843, 9756737702, 38897306018, 155111585372, 618679078298, 2468152192772, 9848142504068, 39301087452632, 156861290196878, 626155256640188

Offset: 0

J. H. Conway, Dec 11 1999

Array interpretation: first row is filled with 1's, first column with powers of 2, b(i,j) = b(i-1,j) + b(i,j-1); then a(n) = b(n,n). - Benoit Cloitre, Apr 01 2002
   1   1   1   1   1   1   1 ...
   2   3   4   5   6   7   8 ...
   4   7  11  16  22 ...
   8  15  26  42  64 ...
  16  31  ..  99 163 ...
From Gary W. Adamson, Dec 27 2008: (Start)
This is an analog of the Catalan sequence: Let M denote an infinite Cartan matrix (-1's in the super and subdiagonals and (2,2,2,...) in the main diagonal which we modify to (1,2,2,2,...)). Then A000108 can be generated by accessing the leftmost term in M^n * [1,0,0,0,...]. Change the operation to M^n * [1,2,3,...] to get this sequence. Or, take iterates M * [1,2,3,...] -> M * ANS, -> M * ANS,...; accessing the leftmost term. (End)
Convolved with the Catalan sequence, A000108: (1, 1, 2, 5, 14, ...) = powers of 4, A000302: (1, 4, 16, 64, ...). - Gary W. Adamson, May 15 2009
Row sums of A094527. - Paul Barry, Sep 07 2009
Hankel transform of the aeration of this sequence is C(n+2, 2). - Paul Barry, Sep 26 2009
Number of 4-ary words of length n in which the number of 1's does not exceed the number of 0's. - David Scambler, Aug 14 2012
Number of options available to a voter who has up to n (0-n) votes to allot among 2n candidates. - Lorraine Lee, Apr 27 2019
2*a(n-1) is the number of all triangulations that can be obtained from a Möbius strip with n chosen points on its edge. See Bazier-Matte et al. - Michel Marcus, Sep 15 2020

			G.f. = 1 + 3*x + 11*x^2 + 42*x^3 + 163*x^4 + 638*x^5 + 2510*x^6 + 9908*x^7 + ...
According to the second formula, we see the fourth row of Pascal's triangle has the terms 1,4,6,4,1 and the partial sums are 1,5,11,15,16. Using these we get 1*1 + 4*5 + 6*11 + 4*15*1*16 = 1 + 20 + 66 + 60 + 16 = 163 = a(4). - _J. M. Bergot_, Apr 29 2014

D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Véronique Bazier-Matte, Ruiyan Huang, and Hanyi Luo, Number of Triangulations of a Möbius Strip, arXiv:2009.05785 [math.CO], 2020.
A. Bernini, F. Disanto, R. Pinzani and S. Rinaldi, Permutations defining convex permutominoes, J. Int. Seq. 10 (2007) # 07.9.7.
Celeste Damiani, Paul Martin, and Eric C. Rowell, Generalisations of Hecke algebras from Loop Braid Groups, arXiv:2008.04840 [math.GT], 2020.
M. Klazar, Twelve countings with rooted plane trees, European Journal of Combinatorics 18 (1997), 195-210; Addendum, 18 (1997), 739-740.
Mircea Merca, A Note on Cosine Power Sums J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.

Binomial transform of A027914. Hankel transform is {1, 2, 3, 4, ..., n, ...}.

Cf. A000108, A000302, A110166.

[(4^n+Binomial(2*n, n))/2:n in [0.. 30]]; // Vincenzo Librandi, Oct 18 2014

A032443:=n->(4^n + binomial(2*n, n))/2; seq(A032443(n), n=0..30); # Wesley Ivan Hurt, Apr 15 2014

a[ n_] := If[ n<-3, 0, (4^n + Binomial[2 n, n]) / 2]; Table[a[n], {n, 0, 30}] (* Michael Somos, Jan 25 2014 *)

A032443 = n->sum(i=0,n,binomial(2*n,i)) \\ M. F. Hasler, Jan 02 2014

A032443 = n->binomial(2*n-1,n)+4^n\2 \\ M. F. Hasler, Jan 02 2014

{a(n) = if( n<-3, 0, (4^n + binomial(2*n, n)) / 2)} /* Michael Somos, Jan 25 2014 */

a(n) = (4^n+binomial(2*n, n))/2. - David W. Wilson
a(n) = Sum_{0 <= i_1 <= i_2 <= n} binomial(n, i_2) * binomial(n, i_1+i_2). - Benoit Cloitre, Oct 14 2004
Sequence with interpolated zeros has a(n) = Sum_{k=0..floor(n/2)} (if((n-2k) mod 2)=0, C(n, k), 0). - Paul Barry, Jan 14 2005
a(n) = Sum_{k=0..n} C(n+k-1, k)*2^(n-k). - Paul Barry, Sep 28 2007
E.g.f.: exp(2*x)*(exp(2*x) + BesselI(0,2*x))/2. For BesselI see Abramowitz-Stegun (reference and link under A008277), p. 375, eq. 9.6.10. See also A000984 for its e.g.f. - Wolfdieter Lang, Jan 16 2012
From Sergei N. Gladkovskii, Aug 13 2012: (Start)
G.f.: (1/sqrt(1-4*x) + 1/(1-4*x))/2 = G(0)/2 where G(k) = 1 + ((2*k)!)/(k!)^2/(4^k - 4*x*(16^k)/( 4*x*(4^k) + ((2*k)!)/(k!)^2/G(k+1))); (continued fraction).
E.g.f.: G(0)/2 where G(k) = 1 + ((2*k)!)/(k!)^2/(4^k - 4*x*(16^k)/( 4*x*(4^k) + (k+1)*((2*k)!)/(k!)^2/G(k+1))); (continued fraction).
(End)
O.g.f.: (1 - x*(2 + c(x)))/(1 - 4*x)^(3/2), with c the o.g.f. of A000108 (Catalan). - Wolfdieter Lang, Nov 22 2012
D-finite with recurrence: n*a(n) + 2*(-4*n+3)*a(n-1) + 8*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Dec 04 2012
a(n) = binomial(2*n-1,n) + floor(4^n/2), or a(n+1) = A001700(n) + A004171(n), for all n >= 0. See A000346 for the difference. - M. F. Hasler, Jan 02 2014
0 = a(n) * (256*a(n+1) - 224*a(n+2) + 40*a(n+3)) + a(n+1) * (-32*a(n+1) + 56*a(n+2) - 14*a(n+3)) + a(n+2) * (-2*a(n+2) + a(n+3)) if n > -4. - Michael Somos, Jan 25 2014
a(n) = coefficient of x^n in (4*x + 1 / (1 + x))^n. - Michael Somos, Jan 25 2014
Binomial transform is A110166. - Michael Somos, Jan 25 2014
Asymptotics: a(n) ~ 2^(2*n-1)*(1+1/sqrt(Pi*n)). - Fung Lam, Apr 13 2014
Self-convolution is A240879. - Fung Lam, Apr 13 2014
a(0) = 1, a(n+1) = A001700(n) + 2^(2n+1). - Philippe Deléham, Oct 11 2014
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 10*x^2 + 35*x^3 + 126*x^4 + ... is the o.g.f. for A001700. - Peter Bala, Jul 21 2015
a(n) = 4*a(n-1) - A000108(n-1). - Bob Selcoe, Apr 02 2017
a(n) = [x^n] 1/((1 - x)^n*(1 - 2*x)). - Ilya Gutkovskiy, Oct 12 2017
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 08 2022
O.g.f.: ( hypergeom([1/4, 3/4], [1/2], 4*x) )^2. - Peter Bala, Mar 04 2022
a(n) = binomial(2*n, n) * hypergeom([1, -n], [n + 1], -1). - Peter Luschny, Oct 06 2023

A116406 Expansion of ((1 + x - 2x^2) + (1+x)*sqrt(1-4x^2))/(2(1-4x^2)).

Original entry on oeis.org

1, 1, 2, 3, 7, 11, 26, 42, 99, 163, 382, 638, 1486, 2510, 5812, 9908, 22819, 39203, 89846, 155382, 354522, 616666, 1401292, 2449868, 5546382, 9740686, 21977516, 38754732, 87167164, 154276028, 345994216, 614429672, 1374282019, 2448023843

Offset: 0

Paul Barry, Feb 13 2006

Interleaving of A114121 and A032443. Row sums of A116405. Binomial transform is A116409.
Appears to be the number of n-digit binary numbers not having more zeros than ones; equivalently, the number of unrestricted Dyck paths of length n not going below the axis. - Ralf Stephan, Mar 25 2008
From Gus Wiseman, Jun 20 2021: (Start)
Also the number compositions of n with alternating sum >= 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. The a(0) = 1 through a(5) = 11 compositions are:
  ()  (1)  (2)   (3)    (4)     (5)
           (11)  (21)   (22)    (32)
                 (111)  (31)    (41)
                        (112)   (113)
                        (121)   (122)
                        (211)   (212)
                        (1111)  (221)
                                (311)
                                (1121)
                                (2111)
                                (11111)
(End)
From J. Stauduhar, Jan 14 2022: (Start)
Also, for n >= 2, first differences of partial row sums of Pascal's triangle. The first ceiling(n/2)+1 elements of rows n=0 to n=4 in Pascal's triangle are:
     1
    1 1
   1 2
  1 3 3
1 4 6
...
The cumulative sums of these partial rows form the sequence 1,3,6,13,24,..., and its first differences are a(2),a(3),a(4),... in this sequence.
(End)

The alternating sum = 0 case is A001700 or A088218.
The alternating sum > 0 case appears to be A027306.
The bisections are  A032443 (odd) and A114121 (even).
The alternating sum <= 0 version is A058622.
The alternating sum < 0 version is A294175.
The restriction to reversed partitions is A344607.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A124754 gives the alternating sum of standard compositions.
A344610 counts partitions by sum and positive reverse-alternating sum.
A344616 lists the alternating sums of partitions by Heinz number.
Cf. A000041, A000070, A000097, A003242, A006330, A028260, A058696, A119899, A239830, A344605, A344611, A344650, A344739.

CoefficientList[Series[((1+x-2x^2)+(1+x)Sqrt[1-4x^2])/(2(1-4x^2)),{x,0,40}],x] (* Harvey P. Dale, Aug 16 2012 *)
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],ats[#]>=0&]],{n,0,15}] (* Gus Wiseman, Jun 20 2021 *)

a(n) = A114121(n/2)*(1+(-1)^n)/2 + A032443((n-1)/2)*(1-(-1)^n)/2.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-1,k). - Paul Barry, Oct 06 2007
Conjecture: n*(n-3)*a(n) +2*(-n^2+4*n-2)*a(n-1) -4*(n-2)^2*a(n-2) +8*(n-2)*(n-3)*a(n-3)=0. - R. J. Mathar, Nov 28 2014
a(n) ~ 2^(n-2) * (1 + (3+(-1)^n)/sqrt(2*Pi*n)). - Vaclav Kotesovec, May 30 2016
a(n) = 2^(n-1) - A294175(n). - Gus Wiseman, Jun 27 2021

A058622 a(n) = 2^(n-1) - ((1+(-1)^n)/4)*binomial(n, n/2).

Original entry on oeis.org

0, 1, 1, 4, 5, 16, 22, 64, 93, 256, 386, 1024, 1586, 4096, 6476, 16384, 26333, 65536, 106762, 262144, 431910, 1048576, 1744436, 4194304, 7036530, 16777216, 28354132, 67108864, 114159428, 268435456, 459312152, 1073741824, 1846943453

Offset: 0

Yong Kong (ykong(AT)curagen.com), Dec 29 2000

a(n) is the number of n-digit binary sequences that have more 1's than 0's. - Geoffrey Critzer, Jul 16 2009
Maps to the number of walks that end above 0 on the number line with steps being 1 or -1. - Benjamin Phillabaum, Mar 06 2011
Chris Godsil observes that a(n) is the independence number of the (n+1)-folded cube graph; proof is by a Cvetkovic's eigenvalue bound to establish an upper bound and a direct construction of the independent set by looking at vertices at an odd (resp., even) distance from a fixed vertex when n is odd (resp., even). - Stan Wagon, Jan 29 2013
Also the number of subsets of {1,2,...,n} that contain more odd than even numbers.  For example, for n=4, a(4)=5 and the 5 subsets are {1}, {3}, {1,3}, {1,2,3}, {1,3,4}. See A014495 when same number of even and odd numbers. - Enrique Navarrete, Feb 10 2018
Also half the number of length-n binary sequences with a different number of zeros than ones. This is also the number of integer compositions of n with nonzero alternating sum, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. Also the number of integer compositions of n+1 with alternating sum <= 0, ranked by A345915 (reverse: A345916). - Gus Wiseman, Jul 19 2021

			G.f. = x + x^2 + 4*x^3 + 5*x^4 + 16*x^5 + 22*x^6 + 64*x^7 + 93*x^8 + ...
From _Gus Wiseman_, Jul 19 2021: (Start)
The a(1) = 1 through a(5) = 16 compositions with nonzero alternating sum:
  (1)  (2)  (3)      (4)      (5)
            (1,2)    (1,3)    (1,4)
            (2,1)    (3,1)    (2,3)
            (1,1,1)  (1,1,2)  (3,2)
                     (2,1,1)  (4,1)
                              (1,1,3)
                              (1,2,2)
                              (1,3,1)
                              (2,1,2)
                              (2,2,1)
                              (3,1,1)
                              (1,1,1,2)
                              (1,1,2,1)
                              (1,2,1,1)
                              (2,1,1,1)
                              (1,1,1,1,1)
(End)

A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992, Eq. (4.2.1.7)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Folded Cube Graph
Eric Weisstein's World of Mathematics, Independence Number

The odd bisection is A000302.
The even bisection is A000346.
The following relate to compositions with nonzero alternating sum:
- The complement is counted by A001700 or A138364.
- The version for alternating sum > 0 is A027306.
- The unordered version is A086543 (even bisection: A182616).
- The version for alternating sum < 0 is A294175.
- These compositions are ranked by A345921.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A345197 counts compositions by length and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, 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, 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 A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000097, A007318, A008549, A034871, A114121, A120452, A163493, A210736, A239830, A344611.

[(2^n -(1+(-1)^n)*Binomial(n, Floor(n/2))/2)/2: n in [0..40]]; // G. C. Greubel, Aug 08 2022

Table[Sum[Binomial[n, Floor[n/2 + i]], {i, 1, n}], {n, 0, 32}] (* Geoffrey Critzer, Jul 16 2009 *)
a[n_] := If[n < 0, 0, (2^n - Boole[EvenQ @ n] Binomial[n, Quotient[n, 2]])/2]; (* Michael Somos, Nov 22 2014 *)
a[n_] := If[n < 0, 0, n! SeriesCoefficient[(Exp[2 x] - BesselI[0, 2 x])/2, {x, 0, n}]]; (* Michael Somos, Nov 22 2014 *)
Table[2^(n - 1) - (1 + (-1)^n) Binomial[n, n/2]/4, {n, 0, 40}] (* Eric W. Weisstein, Mar 21 2018 *)
CoefficientList[Series[2 x/((1-2x)(1 + 2x + Sqrt[(1+2x)(1-2x)])), {x, 0, 40}], x] (* Eric W. Weisstein, Mar 21 2018 *)
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],ats[#]!=0&]],{n,0,15}] (* Gus Wiseman, Jul 19 2021 *)

a(n) = 2^(n-1) - ((1+(-1)^n)/4)*binomial(n, n\2); \\ Michel Marcus, Dec 30 2015

my(x='x+O('x^100)); concat(0, Vec(2*x/((1-2*x)*(1+2*x+((1+2*x)*(1-2*x))^(1/2))))) \\ Altug Alkan, Dec 30 2015

from math import comb
def A058622(n): return (1<>1)>>1) if n else 0 # Chai Wah Wu, Aug 25 2025

[(2^n - binomial(n, n//2)*((n+1)%2))/2 for n in (0..40)] # G. C. Greubel, Aug 08 2022

a(n) = 2^(n-1) - ((1+(-1)^n)/4)*binomial(n, n/2).
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n, i).
G.f.: 2*x/((1-2*x)*(1+2*x+((1+2*x)*(1-2*x))^(1/2))). - Vladeta Jovovic, Apr 27 2003
E.g.f: (e^(2x)-I_0(2x))/2 where I_n is the Modified Bessel Function. - Benjamin Phillabaum, Mar 06 2011
Logarithmic derivative of the g.f. of A210736 is a(n+1). - Michael Somos, Nov 22 2014
Even index: a(2n) = 2^(n-1) - A088218(n). Odd index: a(2n+1) = 2^(2n). - Gus Wiseman, Jul 19 2021
D-finite with recurrence n*a(n) +2*(-n+1)*a(n-1) +4*(-n+1)*a(n-2) +8*(n-2)*a(n-3)=0. - R. J. Mathar, Sep 23 2021
a(n) = 2^n-A027306(n). - R. J. Mathar, Sep 23 2021
A027306(n) - a(n) = A126869(n). - R. J. Mathar, Sep 23 2021

A119620 Number of partitions of floor(3n/2) into n parts each from {1,2,...,n}.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 3, 5, 5, 7, 7, 11, 11, 15, 15, 22, 22, 30, 30, 42, 42, 56, 56, 77, 77, 101, 101, 135, 135, 176, 176, 231, 231, 297, 297, 385, 385, 490, 490, 627, 627, 792, 792, 1002, 1002, 1255, 1255, 1575, 1575, 1958, 1958, 2436, 2436, 3010, 3010, 3718, 3718

Offset: 0

John W. Layman, Jun 07 2006

The bisection {1,1,2,3,5,7,11,15,22,...} agrees with the initial terms of A008641, Number of partitions of n into at most 12 parts and also A008635, Molien series for A_12.
a(2n+1)=a(2n) for all n>0. If the partition {...,1} is a member of a(2n) then the partition {...,1,1} is a member of a(2n+1). - Robert G. Wilson v, Jun 09 2006
Number of partitions of n where all parts (except for possibly the first part) are even; see example. - Joerg Arndt, Apr 22 2013
For n >= 2, a(n) = number of partitions p of n such that floor(n/2) is a part of p.  For n >= 1, a(n) = number of partitions p of n such that ceiling(n/2) is a part of p. - Clark Kimberling, Feb 28 2014
From Gus Wiseman, Oct 28 2021: (Start)
If we insert zeros every three terms, this counts partitions of n such that n = floor(3*k/2), where k is the number of parts. This counts by sum rather than length. These partitions are ranked by A347452.
Also the number of integer partitions of n with alternating product 1, where the alternating product of a sequence (y_1,...,y_k) is Product_i y_i^((-1)^(i-1)). These are the conjugates of the partitions (ranked by A336119) described in Arndt's comment above. For example, the a(2) = 1 through a(10) = 7 partitions are:
  11  111  22    221    33      331      44        441        55
           1111  11111  2211    22111    2222      22221      3322
                        111111  1111111  3311      33111      4411
                                         221111    2211111    222211
                                         11111111  111111111  331111
                                                              22111111
                                                              1111111111
These partitions are ranked by A028982. The odd-length case is A035363 (shifted), which is also the version for sum instead of product. The multiplicative version (factorizations) is A347438.
(End)

			For n=8, floor(3*n/2) is 12 and there are five partitions of 12 into 8 parts each in the range 1-8 inclusive, namely: {5,1,1,1,1,1,1,1}, {4,2,1,1,1,1,1,1}, {3,3,1,1,1,1,1,1}, {3,2,2,1,1,1,1,1} and {2,2,2,2,1,1,1,1}. Thus a(8)=5.
From _Joerg Arndt_, Apr 22 2013: (Start)
a(8) = a(9) = 5, counting the following partitions where all parts (except for possibly the first part) are even:
01:  [ 2 2 2 2 ]
02:  [ 4 2 2 ]
03:  [ 4 4 ]
04:  [ 6 2 ]
05:  [ 8 ]
and
01:  [ 3 2 2 2 ]
02:  [ 5 2 2 ]
03:  [ 5 4 ]
04:  [ 7 2 ]
05:  [ 9 ]
(End)
G.f. = 1 + x + x^2 + x^3 + 2*x^4 + 2*x^5 + 3*x^6 + 3*x^7 + 5*x^8 + 5*x^9 + 7*x^10 + ...

Cf. A008641, A008635.
Both bisections are A000041.
An adjoint version is A108711.
A027187 counts partitions of even length.
A027193 counts partitions of odd length.
A325534 counts separable partitions.
A325535 counts inseparable partitions.
Cf. A000070, A001700, A028983, A086543, A182616, A236559, A236914, A344654, A347444, A347452.

# Using the function EULER from Transforms (see link at the bottom of the page).
[1, op(EULER([1,0,seq(irem(n,2),n=2..55)]))]; # Peter Luschny, Aug 19 2020

(* first do *) Needs["DiscreteMath`Combinatorica`"] (* then *) f[n_] := f[n] = Length@ Select[ Partitions[ Floor[3n/2], n], Length@# == n &]; Table[ If[n > 1, f[2Floor[n/2]], f[n]], {n, 57}] (* Robert G. Wilson v, Jun 09 2006 *)
Table[ PartitionsP[ Floor[n/2]], {n, 57}] (* Robert G. Wilson v, Jun 09 2006 *)
Table[Count[IntegerPartitions[n], p_ /; MemberQ[p, Ceiling[n/2]]], {n, 50}] (* Clark Kimberling, Feb 28 2014 *)
a[ n_] := SeriesCoefficient[ (1 + x) / QPochhammer[x^2], {x, 0, n}]; (* Michael Somos, Mar 01 2014 *)

a(n)=numbpart(n\2); \\ Joerg Arndt, Apr 22 2013

a(n) = A000041(floor(n/2)). - Vladeta Jovovic, Jun 10 2006

G.f.: (Sum_{n>=0} x^(4*n) / Product_{k=1..n} (1-x^(2*k))) / (1 - x). - Michael Somos, Mar 01 2014 [corrected by Jason Yuen, Jan 24 2025]

More terms from Robert G. Wilson v, Jun 09 2006

Added a(0)=1. - Michael Somos, Mar 01 2014

A001448 a(n) = binomial(4n,2n) or (4n)!/((2n)!(2n)!).

Original entry on oeis.org

1, 6, 70, 924, 12870, 184756, 2704156, 40116600, 601080390, 9075135300, 137846528820, 2104098963720, 32247603683100, 495918532948104, 7648690600760440, 118264581564861424, 1832624140942590534, 28453041475240576740, 442512540276836779204, 6892620648693261354600

Offset: 0

N. J. A. Sloane

Corollary 8 in Chapman et alia says: "For n>=1, there are binomial(4n,2n) binary sequences of length 4n+1 with the property that for all j, the j-th occurrence of 10 appears in positions 4j+1 and 4j+2 or later (if it exists at all)." - Peter Luschny, Nov 21 2011

Sequence terms are given by [x^n] ( (1 + x)^(k+2)/(1 - x)^k )^n for k = 2. See the cross references for related sequences obtained from other values of k. - Peter Bala, Sep 29 2015

			a(n) = (1/Pi)*Integral_{x=0..4} x^(2n)/sqrt(4-(x-2)^2) dx. - _Paul Barry_, Sep 17 2010
G.f. = 1 + 6*x + 70*x^2 + 924*x^3 + 12870*x^4 + 184756*x^5 + 2704156*x^6 + ...

T. D. Noe, Table of n, a(n) for n = 0..100
R. J. Chapman, T. Y. Chowa, A. Khetana, D. P. Moulton and R. J. Waters, Simple formulas for lattice paths avoiding certain periodic staircase boundaries, Journal of Combinatorial Theory, Series A 116 (2009) 205-214.
Maciej Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv preprint arXiv:1410.5747 [math.CO], 2014.
Quy Nhan, A finite sum of binomial coefficients, MathStackExchange, 2025-06-21.
Tito Piezas III, Fermat primes and Binomial sums, tpiezas, A Collection of Algebraic IdentitiesJune 3, 2012.
K. H. Pilehrood and T. H. Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, J. Int. Seq. 18 (2015) 15.11.7.
Ricardo A. Podestá, New identities for binary Krawtchouk polynomials, binomial coefficients and Catalan numbers, arXiv:1603.09156 [math.CO], 2016.
Renzo Sprugnoli, Sums of reciprocals of the central binomial coefficients, INTEGERS 6 (2006) #A27.

Bisection of A000984. Cf. A002458, A066357, A000984 (k = 0), A091527 (k = 1), A262732 (k = 3), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6).

Cf. A370100, A370101.

[Factorial(4*n)/(Factorial(2*n)*Factorial(2*n)): n in [0..20]]; // Vincenzo Librandi, Sep 13 2011

Maple
```
A001448 := n-> binomial(4*n,2*n) ;
```

Table[Binomial[4n,2n],{n,0,20}] (* Harvey P. Dale, Apr 26 2014 *)
a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {-2 n, -2 n}, {1}, 1]]; (* Michael Somos, Oct 22 2014 *)

a(n)=binomial(4*n,2*n) \\ Charles R Greathouse IV, Sep 13 2011

from math import comb
def A001448(n): return comb(n<<2,n<<1) # Chai Wah Wu, Aug 10 2023

a(n) = A000984(2*n).
Using Stirling's formula in sequence A000142 it is easy to get the asymptotic expression a(n) ~ 16^n / sqrt(2 * Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
From Wolfdieter Lang, Dec 13 2001: (Start)
a(n) = 2*A001700(2*n-1) = (2*n+1)*C(2*n), n >= 1, C(n) := A000108(n) (Catalan).
G.f.: (1-y*((1+4*y)*c(y)-(1-4*y)*c(-y)))/(1-(4*y)^2) with y^2=x, c(y) = g.f. for A000108 (Catalan). (End)
a(n) ~ 2^(-1/2)*Pi^(-1/2)*n^(-1/2)*2^(4*n)*{1 - (1/16)*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
a(n) = (1/Pi)*Integral_{x=-2..2} (2+x)^(2*n)/sqrt((2-x)*(2+x)) dx. Peter Luschny, Sep 12 2011
G.f.: (1/2) * (1/sqrt(1+4*sqrt(x)) + 1/sqrt(1-4*sqrt(x))). - Mark van Hoeij, Oct 25 2011
Sum_{n>=1} 1/a(n) = 16/15 + Pi*sqrt(3)/27 - 2*sqrt(5)*log(phi)/25, [T. Trif, Fib Quart 38 (2000) 79] with phi=A001622. - R. J. Mathar, Jul 18 2012
D-finite with recurrence n*(2*n-1)*a(n) -2*(4*n-1)*(4*n-3)*a(n-1)=0. - R. J. Mathar, Dec 02 2012
G.f.: sqrt((1 + sqrt(1-16*x))/(2*(1-16*x))) = 1 + 6*x/(G(0)-6*x), where G(k) = 2*x*(4*k+3)*(4*k+1) + (2*k+1)*(k+1) - 2*x*(k+1)*(2*k+1)*(4*k+5)*(4*k+7)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jun 30 2013
a(n) = hypergeom([1-2*n,-2*n],[2],1)*(2*n+1). - Peter Luschny, Sep 22 2014
From Michael Somos, Oct 22 2014: (Start)
0 = a(n)*(+65536*a(n+2) - 16896*a(n+3) + 858*a(n+4)) + a(n+1)*(-3584*a(n+2) + 1176*a(n+3) - 66*a(n+4)) + a(n+2)*(+14*a(n+2) - 14*a(n+3) + a(n+4)) for all n in Z.
0 = a(n)^2*(+196608*a(n+1)^2 - 40960*a(n+1)*a(n+2) + 2100*a(n+2)^2) + a(n)*a(n+1)*(-12288*a(n+1)^2 + 2840*a(n+1)*a(n+2) - 160*a(n+2)^2) + a(n+1)^2*(+180*a(n+1)^2 - 48*a(n+1)*a(n+2) + 3*a(n+2)^2) for all n in Z. (End)
a(n) = [x^n] ( (1 + x)^4/(1 - x)^2 )^n; exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 6*x + 53*x^2 + 554*x^3 + ... = Sum_{n >= 0} A066357(n+1)*x^n. - Peter Bala, Jun 23 2015
a(n) = Sum_{i = 0..n} binomial(4*n,i) * binomial(3*n-i-1,n-i). - Peter Bala, Sep 29 2015
a(n) = A000984(n)*Product_{j=0..n} (2^j/(j!*(2*j-1)!!))*A068424(n, j)^2, with A068424 the falling factorial. See (5.4) in Podestá link. - Michel Marcus, Mar 31 2016
a(n) = GegenbauerC(2*n, -2*n, -1). - Peter Luschny, May 07 2016
a(n) = [x^n] 1/sqrt(1 - 4*x)^(2*n+1). - Ilya Gutkovskiy, Oct 10 2017
a(n) is the n-th moment of the positive weight function w(x) on (0,16), i.e. a(n) = Integral_{x=0..16} x^n*w(x) dx, n = 0,1,..., where w(x) = (1/(2*Pi))/(sqrt(4 - sqrt(x))*x^(3/4)). The function w(x) is the solution of the Hausdorff moment problem and is unique. - Karol A. Penson, Mar 06 2018
a(n) = (16^n*(Beta(2*n - 1/2, 1/2) - Beta(2*n - 1/2, 3/2)))/Pi. - Peter Luschny, Mar 06 2018
E.g.f.: hypergeom([1/4,3/4],[1/2,1],16*x). - Karol A. Penson, Mar 08 2018
From Peter Bala, Feb 16 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k.
a(n) = [(x*y)^(2*n)] (1 + x + y)^(4*n). (End)
a(n) = (2^n/n!)*Product_{k = n..2*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
a(n) = Sum_{k = 0..2*n} binomial(2*n+k-1, k). - Peter Bala, Nov 02 2024
Sum_{n>=0} (-1)^n/a(n) = 16/17 + 4*sqrt(34)*(sqrt(17)-2)*arctan(sqrt(2/(sqrt(17)-1)))/(289*sqrt(sqrt(17)-1)) + 2*sqrt(34)*(sqrt(17)+2)*log((sqrt(sqrt(17)+1)-sqrt(2))/(sqrt(sqrt(17)+1)+sqrt(2)))/(289*sqrt(sqrt(17)+1)) (Sprugnoli, 2006, Theorem 3.8, p. 11; Piezas, 2012). - Amiram Eldar, Nov 03 2024
For n >= 1, a(n) = Sum_{k=1}^n a(n-k) * A337350(n) = Sum_{k=1}^n a(n-k) * a(k) * (8k + 1) / (8k^2 + 2k - 1).  For proof, see the Quy Nhan link. - Lucas A. Brown, Jun 26 2025
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(2*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(2*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((4*n-1)/2,n).
a(n) = [x^n] (1+4*x)^((4*n-1)/2). (End)

cp's OEIS Frontend

A000097 Number of partitions of n if there are two kinds of 1's and two kinds of 2's.

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A027306 a(n) = 2^(n-1) + ((1 + (-1)^n)/4)*binomial(n, n/2).

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A057079 Periodic sequence: repeat [1,2,1,-1,-2,-1]; expansion of (1+x)/(1-x+x^2).

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A008549 Number of ways of choosing at most n-1 items from a set of size 2*n+1.

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A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

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A001448 a(n) = binomial(4n,2n) or (4n)!/((2n)!(2n)!).