A001700 -id:A001700 - cp's OEIS frontend

A069720 a(n) = 2^(n-1)binomial(2n-1, n).

1, 6, 40, 280, 2016, 14784, 109824, 823680, 6223360, 47297536, 361181184, 2769055744, 21300428800, 164317593600, 1270722723840, 9848101109760, 76467608616960, 594748067020800, 4632774416793600, 36135640450990080, 282202144474398720, 2206307674981662720, 17266755717247795200

Offset: 1

Valery A. Liskovets, Apr 07 2002

Number of rooted unicursal planar maps with n edges (unicursal means that exactly two nodes are of odd valency; there is an Eulerian path).

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Harlan J. Brothers, Pascal's Prism: Supplementary Material.
Valery A. Liskovets and Timothy R. S. Walsh, Enumeration of Eulerian and unicursal planar maps, Discr. Math., 282 (2004), 209-221.
Sheng-Liang Yang, Yan-Ni Dong, and Tian-Xiao He, Some matrix identities on colored Motzkin paths, Discrete Mathematics 340.12 (2017): 3081-3091.

First superdiagonal of number array A082137.

Cf. A000079, A001700, A003584, A069723, A069724, A082143.

a069720 n = (a000079 $ n - 1) * (a001700 $ n - 1)
-- Reinhard Zumkeller, Jan 15 2015

[2^(n-2)*Binomial(2*n, n): n in [1..25]]; // Vincenzo Librandi, Apr 14 2018

Z:=(1-sqrt(1-2*z))*4^(n-1)/sqrt(1-2*z): Zser:=series(Z, z=0, 32): seq(coeff(Zser, z, n), n=1..20); # Zerinvary Lajos, Jan 01 2007

Table[2^(n-1) Binomial[2n-1,n],{n,20}] (* Harvey P. Dale, Jan 20 2013 *)

a(n) = binomial(2*n-1,n)<<(n-1) \\ Charles R Greathouse IV, Feb 06 2017

def A069720(n): return 2^(n-2)*binomial(2*n, n)
print([A069720(n) for n in range(1,31)]) # G. C. Greubel, Jan 18 2025

a(n) = 2^(n-2)*binomial(2*n, n).
G.f.: (1-sqrt(1-8*x))/(4*x*sqrt(1-8*x)) = 2/(sqrt(1-8*x)*(1-sqrt(1-8*x))) - 1/(2*x). - Paul Barry, Sep 06 2004
D-finite with recurrence: n*a(n) - 4*(2*n-1)*a(n-1) = 0. - R. J. Mathar, Apr 01 2012
E.g.f.: a(n) = n! * [x^n] (exp(4*x)*BesselI(0, 4*x) - 1)/4. - Peter Luschny, Aug 25 2012
From Reinhard Zumkeller, Jan 15 2015: (Start)
a(n) = A000079(n-1) * A001700(n-1); for n > 1:
a(n) = 2*A082143(n-1). (End)
From Amiram Eldar, Jan 16 2024: (Start)
Sum_{n>=1} 1/a(n) = 4/7 + 32*arcsin(1/(2*sqrt(2)))/(7*sqrt(7)).
Sum_{n>=1} (-1)^(n+1)/a(n) = 4/9 + 16*log(2)/27. (End)
a(n) = ((2*n)!/4) * [x^n] (BesselI(0, 2*sqrt(2*x)) - 1). - G. C. Greubel, Jan 18 2025

A091867 Triangle read by rows: T(n,k) = number of Dyck paths of semilength n having k peaks at odd height.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 3, 4, 6, 0, 1, 6, 15, 10, 10, 0, 1, 15, 36, 45, 20, 15, 0, 1, 36, 105, 126, 105, 35, 21, 0, 1, 91, 288, 420, 336, 210, 56, 28, 0, 1, 232, 819, 1296, 1260, 756, 378, 84, 36, 0, 1, 603, 2320, 4095, 4320, 3150, 1512, 630, 120, 45, 0, 1, 1585, 6633, 12760, 15015, 11880, 6930, 2772, 990, 165, 55, 0, 1

Offset: 0

Emeric Deutsch, Mar 10 2004

Number of ordered trees with n edges having k leaves at odd height. Row sums are the Catalan numbers (A000108). T(n,0)=A005043(n). Sum_{k=0..n} k*T(n,k) = binomial(2n-2,n-1).
T(n,k)=number of Dyck paths of semilength n and having k ascents of length 1 (an ascent is a maximal string of consecutive up steps). Example: T(4,2)=6 because we have UdUduud, UduuddUd, uuddUdUd, uudUdUdd, UduudUdd and uudUddUd (the ascents of length 1 are indicated by U instead of u).
T(n,k) is the number of Łukasiewicz paths of length n having k level steps (i.e., (1,0)). A Łukasiewicz path of length n is a path in the first quadrant from (0,0) to (n,0) using rise steps (1,k) for any positive integer k, level steps (1,0) and fall steps (1,-1) (see R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 223, Exercise 6.19w; the integers are the slopes of the steps). Example: T(4,1)=4 because we have HU(2)DD, U(2)HDD, U(2)DHD and U(2)DDH, where H=(1,0), U(1,1), U(2)=(1,2) and D=(1,-1). - Emeric Deutsch, Jan 06 2005
T(n,k) = number of noncrossing partitions of [n] containing k singleton blocks. Also, T(n,k) = number of noncrossing partitions of [n] containing k adjacencies. An adjacency is an occurrence of 2 consecutive integers in the same block (here 1 and n are considered consecutive). In fact, the statistics # singletons and # adjacencies have a symmetric joint distribution.
Exponential Riordan array [e^x*(Bessel_I(0,2x)-Bessel_I(1,2x)),x]. - Paul Barry, Mar 03 2011
T(n,k) is the number of ordered trees having n edges and exactly k nodes with one child. - Geoffrey Critzer, Feb 25 2013
From Tom Copeland, Nov 04 2014: (Start)
Summing the coeff. of the partitions in A134264 for a Lagrange inversion formula (see also A249548) containing (h_1)^k = (1')^k gives this triangle, so this array's o.g.f. H(x,t) = x + t * x^2 + (1 + t^2) * x^3 ... is the inverse of the o.g.f. of A104597 with a sign change, i.e., H^(-1)(x,t) = (x-x^2) / [1 + (t-1)(x-x^2)] = Cinv(x)/[1 + (t-1)Cinv(x)] = P[Cinv(x),t-1] where Cinv(x)= x * (1-x) is the inverse of C(x) = [1-sqrt(1-4*x)]/2, an o.g.f. for the Catalan numbers A000108, and P(x,t) = x/(1+t*x) with inverse Pinv(x,t) = -P(-x,t) = x/(1-t*x). Therefore,
O.g.f.: H(x,t) = C[Pinv(x,t-1)] = C[P(x,1-t)] = C[x/(1-(t-1)x)] = {1-sqrt[1-4*x/(1-(t-1)x)]}/2 (for A091867). Reprising,
Inverse O.g.f.: H^(-1)(x,t) = x*(1-x) / [1 + (t-1)x(1-x)] = P[Cinv(x),t-1].
From general arguments in A134264, the row polynomials are an Appell sequence with lowering operator d/dt, having the umbral property (p(.,t)+a)^n=p(n,t+a) with e.g.f. = e^(x*t)/w(x), where 1/w(x)= e.g.f. of first column for the Motzkin numbers in A005043. (Mislabeled argument corrected on Jan 31 2016.)
Cf. A124644 (t-shifted polynomials), A026378 (t=-4), A001700 (t=-3), A005773 (t=-2), A126930 (t=-1) and A210736 (t=-1, a(0)=0, unsigned), A005043 (t=0), A000108 (t=1), A007317 (t=2), A064613 (t=3), A104455 (t=4), A030528 (for inverses).
(End)
The sequence of binomial transforms A126930, A005043, A000108, ... in the above comment appears in A126930 and the link therein to a paper by F. Fite et al. on page 42. - Tom Copeland, Jul 23 2016

			T(4,2)=6 because we have (ud)uu(ud)dd, uu(ud)dd(ud), uu(ud)(ud)dd, (ud)(ud)uudd, (ud)uudd(ud) and uudd(ud)(ud) (here u=(1,1), d=(1,-1) and the peaks at odd height are shown between parentheses).
Triangle begins:
   1;
   0,   1;
   1,   0,   1;
   1,   3,   0,   1;
   3,   4,   6,   0,  1;
   6,  15,  10,  10,  0,  1;
  15,  36,  45,  20, 15,  0, 1;
  36, 105, 126, 105, 35, 21, 0, 1;
  ...

R. Sedgewick and P. Flajolet, Analysis of Algorithms, Addison and Wesley, 1996, page 254 (first edition)

Alois P. Heinz, Rows n = 0..200, flattened
David Callan, On conjugates for set partitions and integer compositions , arXiv:math/0508052 [math.CO], 2005.
A. Sapounakis, I. Tasoulas and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
F. Yano and H. Yoshida, Some set partition statistics in non-crossing partitions and generating functions, Discr. Math., 307 (2007), 3147-3160.

Cf. A000108, A001700, A005043, A005773, A007317, A026378, A030528, A064613, A104455, A104597, A124644, A126930, A210736.

T := proc(n,k) if k>n then 0 elif k=n then 1 else (binomial(n+1,k)/(n+1))*sum(binomial(n+1-k,j)*binomial(n-k-j-1,j-1),j=1..floor((n-k)/2)) fi end: seq(seq(T(n,k),k=0..n),n=0..12);
T := (n,k) -> (-1)^(n+k)*binomial(n,k)*hypergeom([-n+k,1/2],[2],4): seq(seq(simplify(T(n, k)), k=0..n), n=0..10); # Peter Luschny, Jul 27 2016
# alternative Maple program:
b:= proc(x, y, t) option remember; expand(`if`(x=0, 1,
      `if`(y>0, b(x-1, y-1, 0)*z^irem(t*y, 2), 0)+
      `if`(y (p-> seq(coeff(p, z, i), i=0..n))(b(2*n, 0$2)):
seq(T(n), n=0..16);  # Alois P. Heinz, May 12 2017

nn=10;cy = ( 1 + x - x y - ( -4x(1+x-x y) + (-1 -x + x y)^2)^(1/2))/(2(1+x-x y)); Drop[CoefficientList[Series[cy,{x,0,nn}],{x,y}],1]//Grid  (* Geoffrey Critzer, Feb 25 2013 *)
Table[Which[k == n, 1, k > n, 0, True, (Binomial[n + 1, k]/(n + 1)) Sum[Binomial[n + 1 - k, j] Binomial[n - k - j - 1, j - 1], {j, Floor[(n - k)/2]}]], {n, 0, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 25 2016 *)

T(n, k) = [binomial(n+1, k)/(n+1)]*Sum_{j=1..floor((n-k)/2)} binomial(n+1-k, j)*binomial(n-k-j-1, j-1) for kn. G.f.=G=G(t, z) satisfies z(1+z-tz)G^2-(1+z-tz)G+1=0. T(n, k)=r(n-k)*binomial(n, k), where r(n)=A005043(n) are the Riordan numbers.
G.f.: 1/(1-xy-x^2/(1-x-xy-x^2/(1-x-xy-x^2/(1-x-xy-x^2/(1-... (continued fraction). - Paul Barry, Aug 03 2009
Sum_{k=0..n} T(n,k)*x^k = A126930(n), A005043(n), A000108(n), A007317(n), A064613(n), A104455(n) for x = -1,0,1,2,3,4 respectively. - Philippe Deléham, Dec 03 2009
Sum_{k=0..n} (-1)^(n-k)*T(n,k)*x^k = A168491(n), A099323(n+1), A001405(n), A005773(n+1), A001700(n), A026378(n+1), A005573(n), A122898(n) for x = -1, 0, 1, 2, 3, 4, 5, 6 respectively. - Philippe Deléham, Dec 03 2009
E.g.f.: e^(x+xy)*(Bessel_I(0,2x)-Bessel_I(1,2x)). - Paul Barry, Mar 10 2010
From Tom Copeland, Nov 06 2014: (Start)
O.g.f.: H(x,t) = {1-sqrt[1-4x/(1-(t-1)x)]}/2 (shifted index, as given in Copeland's comment, see comp. inverse there).
H(x,t)= x / [1-(C.+(t-1))x] = Sum_{n>=1} (C.+ (t-1))^(n-1)*x^n umbrally, e.g., (a.+b.)^2 = a_0*b_2 + 2 a_1*b1_+ a_0*b_2, where (C.)^n = C_n are the Catalan numbers (1,1,2,5,14,..) of A000108.
This shows directly that the lowering operator for the polynomials is D=d/dt, i.e., D p(n,t)= D(C. + (t-1))^n = n * (C. + (t-1))^(n-1) = n*p(n-1,t), so that the polynomials form an Appell sequence, and that p(n,0) gives a Motzkin sum, or Riordan, number A005043.
(End)
T(n,k) = (-1)^(n+k)*binomial(n,k)*hypergeom([k-n,1/2],[2],4). - Peter Luschny, Jul 27 2016

A349057 Numbers k such that the k-th composition in standard order is not weakly alternating.

Original entry on oeis.org

37, 46, 52, 53, 69, 75, 78, 92, 93, 101, 104, 105, 107, 110, 116, 117, 133, 137, 139, 142, 150, 151, 156, 157, 165, 174, 180, 181, 184, 185, 186, 187, 190, 197, 200, 201, 203, 206, 208, 209, 210, 211, 214, 215, 220, 221, 229, 232, 233, 235, 238, 244, 245, 261

Offset: 1

Gus Wiseman, Dec 04 2021

nonn

We define a sequence to be weakly alternating if it is alternately weakly increasing and weakly decreasing, starting with either.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The terms and corresponding compositions begin:
   37: (3,2,1)
   46: (2,1,1,2)
   52: (1,2,3)
   53: (1,2,2,1)
   69: (4,2,1)
   75: (3,2,1,1)
   78: (3,1,1,2)
   92: (2,1,1,3)
   93: (2,1,1,2,1)
  101: (1,3,2,1)
  104: (1,2,4)
  105: (1,2,3,1)
  107: (1,2,2,1,1)
  110: (1,2,1,1,2)
  116: (1,1,2,3)
  117: (1,1,2,2,1)

The strong case is A345168, complement A345167, counted by A345192.
The strong anti-run case is A345169, counted by A345195.
Including all non-anti-runs gives A348612, complement A333489.
These compositions are counted by A349053, complement A349052.
The directed cases are counted by A129852 (incr.) and A129853 (decr.).
The complement for patterns is A349058, strong A345194.
The complement for ordered factorizations is A349059, strong A348610.
Partitions of this type are counted by A349061, complement A349060.
Partitions of this type are ranked by A349794.
Non-strict partitions of this type are counted by A349796.
Permutations of prime indices of this type are counted by A349797.
A001250 counts alternating permutations, complement A348615.
A003242 counts Carlitz (anti-run) compositions, complement A261983.
A011782 counts compositions.
A025047 counts alternating/wiggly compositions, directed A025048, A025049.
A345164 counts alternating permutations of prime indices, weak A349056.
A345165 counts partitions w/o an alternating permutation, ranked by A345171.
A345170 counts partitions w/ an alternating permutation, ranked by A345172.
A349054 counts strict alternating compositions.
Cf. A001700, A096441, A128761, A344615, A344654, A345173, A348613, A349051, A349794, A349795, A349799.

stc[n_]:=Differences[Prepend[Join@@Position[ Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
whkQ[y_]:=And@@Table[If[EvenQ[m], y[[m]]<=y[[m+1]],y[[m]]>=y[[m+1]]],{m,1,Length[y]-1}];
Select[Range[0,100],!whkQ[stc[#]]&&!whkQ[-stc[#]]&]

A103371 Number triangle T(n,k) = C(n,n-k)*C(n+1,n-k).

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 12, 1, 5, 40, 60, 20, 1, 6, 75, 200, 150, 30, 1, 7, 126, 525, 700, 315, 42, 1, 8, 196, 1176, 2450, 1960, 588, 56, 1, 9, 288, 2352, 7056, 8820, 4704, 1008, 72, 1, 10, 405, 4320, 17640, 31752, 26460, 10080, 1620, 90, 1, 11, 550, 7425, 39600, 97020

Offset: 0

Paul Barry, Feb 03 2005

Columns include A000027, A002411, A004302, A108647, A134287. Row sums are C(2n+1,n+1) or A001700.
T(n-1,k-1) is the number of ways to put n identical objects into k of altogether n distinguishable boxes. See the partition array A035206 from which this triangle arises after summing over all entries related to partitions with fixed part number k.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - Abdullahi Umar, Oct 02 2008
The o.g.f. of the (n+1)-th diagonal is given by G(n, x) = (n+1)*Sum_{k=1..n} A001263(n, k)*x^(k-1) / (1 - x)^(2*n+1), for n >= 1 and for n = 0 it is G(0, x) = 1/(1-x). - Wolfdieter Lang, Jul 31 2017

			The triangle T(n, k) begins:
n\k  0   1    2     3     4     5     6    7  8 9 ...
0:   1
1:   2   1
2:   3   6    1
3:   4  18   12     1
4:   5  40   60    20     1
5:   6  75  200   150    30     1
6:   7 126  525   700   315    42     1
7:   8 196 1176  2450  1960   588    56    1
8:   9 288 2352  7056  8820  4704  1008   72  1
9:  10 405 4320 17640 31752 26460 10080 1620 90 1
...  reformatted. - _Wolfdieter Lang_, Jul 31 2017
From _R. J. Mathar_, Mar 29 2013: (Start)
The matrix inverse starts
       1;
      -2,       1;
       9,      -6,      1;
     -76,      54,    -12,      1;
    1055,    -760,    180,    -20,   1;
  -21906,   15825,  -3800,    450, -30,   1;
  636447, -460026, 110775, -13300, 945, -42, 1; (End)
O.g.f. of 4th diagonal [4, 40,200, ...] is G(3, x) = 4*(1 + 3*x + x^2)/(1 - x)^7, from the n = 3 row [1, 3, 1] of A001263. See a comment above. - _Wolfdieter Lang_, Jul 31 2017

Reinhard Zumkeller, Rows n = 0..125 of table, flattened
Per Alexandersson, Svante Linusson, Samu Potka, and Joakim Uhlin, Refined Catalan and Narayana cyclic sieving, arXiv:2010.11157 [math.CO], 2020.
Paul Barry and A. Hennessy, A Note on Narayana Triangles and Related Polynomials, Riordan Arrays, and MIMO Capacity Calculations, J. Int. Seq. 14 (2011), Article 11.3.8.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. II. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 16.
R. Cori and G. Hetyei, Counting genus one partitions and permutations, arXiv:1306.4628 [math.CO], 2013.
Wolfdieter Lang, On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving full transformations, Semigroup Forum 72 (2006), 51-62.

Cf. A008459, A122899, A194595, A197653.
Cf. A007318, A000894 (central terms), A132813 (mirrored).
Cf. A000027, A002411, A004302, A108647, A134287, A135278, A001263.

a103371 n k = a103371_tabl !! n !! k
a103371_row n = a103371_tabl !! n
a103371_tabl = map reverse a132813_tabl
-- Reinhard Zumkeller, Apr 04 2014

/* As triangle */ [[Binomial(n,n-k)*Binomial(n+1,n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 01 2017

A103371 := (n,k) -> binomial(n,k)^2*(n+1)/(k+1);
seq(print(seq(A103371(n, k), k=0..n)), n=0..7); # Peter Luschny, Oct 19 2011

Flatten[Table[Binomial[n,n-k]Binomial[n+1,n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, May 26 2014 *)
CoefficientList[Series[Series[E^(x(1+y))(BesselI[0,2*x*Sqrt[y]]+BesselI[1,2*x*Sqrt[y]]/Sqrt[y]),{x,0,8}],{y,0,8}],{x,y}]*Range[0,8]! (* Natalia L. Skirrow, Apr 14 2025 *)

create_list(binomial(n,k)*binomial(n+1,k+1),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

for(n=0,10, for(k=0,n, print1(binomial(n,k)*binomial(n+1,k+1), ", "))) \\ G. C. Greubel, Nov 09 2018

Number triangle T(n, k) = C(n, n-k)*C(n+1, n-k) = C(n, k)*C(n+1, k+1); Column k of this triangle has g.f. Sum_{j=0..k} (C(k, j)*C(k+1, j) * x^(k+j))/(1-x)^(2*k+2); coefficients of the numerators are the rows of the reverse triangle C(n, k)*C(n+1, k).
T(n,k) = C(n, k)*Sum_{j=0..(n-k)} C(n-j, k). - Paul Barry, Jan 12 2006
T(n,k) = (n+1-k)*N(n+1,k+1), with N(n,k):=A001263(n,k), the Narayana triangle (with offset [1,1]).
O.g.f.: ((1-(1-y)*x)/sqrt((1-(1+y)*x)^2-4*x^2*y) -1)/2, (from o.g.f. of A001263, Narayana triangle). - Wolfdieter Lang, Nov 13 2007
From Peter Bala, Jan 24 2008: (Start)
Matrix product of A007318 and A122899.
O.g.f. for row n: (1-x)^n*P(n,1,0,(1+x)/(1-x)) = 1/(2*x)*(1-x)^(n+1)*( Legendre_P(n+1,(1+x)/(1-x)) - Legendre_P(n,(1+x)/(1-x)) ), where P(n,a,b,x) denotes the Jacobi polynomial.
O.g.f. for column k: x^k/(1-x)^(k+2)*P(k,0,1,(1+x)/(1-x)). Compare with A008459. (End)
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,1). (Cf. A194595, A197653, A197654). - Peter Luschny, Oct 20 2011
T(n,k) = A003056(n+1,k+1)*C(n,k)^2/(k+1). - Peter Luschny, Oct 29 2011
T(n,k) = A007318(n, k)*A135278(n, k), n >= k >= 0. - Wolfdieter Lang, Jul 31 2017
From Natalia L. Skirrow, Apr 14 2025: (Start)
T(n,k) = A008459(n,k) + n*N(n,k+1).
E.g.f.: e^(x*(1+y))*(I_0(2*x*sqrt(y)) + I_1(2*x*sqrt(y))/sqrt(y)), where I_n is the modified Bessel function of the first kind. (The I_0 contributes A008459(n,k), the I_1 contributes n*N(n,k+1))
O.g.f. for row n: (n+1)*2F1(-n,-n;2;y) = (n+1)*2F1(2+n,2+n;2;y)*(1-y)^(2*(n+1)) (by Euler's hypergeometric transformation); (n+1)*2F1(2+n,2+n;2;y) is the o.g.f. for row n of (k+n+1)!^2/(k!*(k+1)!*n!*(n+1)!), which is column n+1 of A132812.
O.g.f. for column k: 2F1(1+k,2+k;1;x)*x^k = 2F1(-k,-1-k;1;x)*x^k/(1-x)^(2+2*k). 2F1(-k,-1-k;1;x) is the kth row of A132813, the reflection of the kth row of this triangle.
O.g.f. for diagonal d (beginning at a(d,0)): (d+1)*x^d*2F1(d+1,d+2;2;x*y). 2F1(d+1,d+2;2;x) = 2F1(1-d,-d;2;x)/(1-x)^(2*d+1), numerator being the o.g.f. of row d of the Narayana triangle.
These respectively yield:
T(n,k) = Sum_{i=0..n+k} C(2*(n+1),i)*(-1)^i*A132812(n+1+k-i,n+1),
T(d+k,k) = Sum_{i=0..k} C(d-i+1+2*k,d-i)*T(k,k-i),
T(d+k,k) = Sum_{i=0..d} C(k-i + 2*d,k-i)*N(d,i+1)*(d+1).
E.g.f. for column k: 1F1(2+k;1;x)*x^k/k!.
E.g.f. for diagonal d: (d+1)*x^d*1F1(d+2;2;x*y)/d!. (End)

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A046899 Triangle in which n-th row is {binomial(n+k,k), k=0..n}, n >= 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 10, 20, 1, 5, 15, 35, 70, 1, 6, 21, 56, 126, 252, 1, 7, 28, 84, 210, 462, 924, 1, 8, 36, 120, 330, 792, 1716, 3432, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 1, 11, 66, 286, 1001

Offset: 0

N. J. A. Sloane

C(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Row sums are A001700.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) of waist k (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
If T(r,c), r=0,1,2,..., c=1,2,...,(r+1), are the triangle elements, then for r > 0, T(r,c) = binomial(r+c-1,c-1) = M(r,c) is the number of monotonic mappings from an ordered set of r elements into an ordered set of c elements. For example, there are 15 monotonic mappings from an ordered set of 4 elements into an ordered set of 3 elements. For c > r+1, use the identity M(r,c) = M(c-1,r+1) = T(c-1,r+1). For example, there are 210 monotonic mappings from an ordered set of 4 elements into an ordered set of 7 elements, because M(4,7) = T(6,5) = 210. Number of monotonic endomorphisms in a set of r elements, M(r,r), therefore appear on the second diagonal of the triangle which coincides with A001700. - Stanislav Sykora, May 26 2012
Start at the origin. Flip a fair coin to determine steps of (1,0) or (0,1). Stop when you are a (perpendicular) distance of n steps from the x axis or the y axis. For k = 0,1,...,n-1, C(n-1,k)/2^(n+k) is the probability that you will stop on the point (n,k). This is equal to the probability that you will stop on the point (k,n). Hence, Sum_{k=0..n} C(n,k)/2^(n+k) = 1. - Geoffrey Critzer, May 13 2017

			The triangle is the lower triangular part of the square array:
  1|  1,  1,   1,   1,    1,    1,     1,     1,     1, ...
  1,  2|  3,   4,   5,    6,    7,     8,     9,    10, ...
  1,  3,  6|  10,  15,   21,   28,    36,    45,    55, ...
  1,  4, 10,  20|  35,   56,   84,   120,   165,   220, ...
  1,  5, 15,  35,  70|  126,  210,   330,   495,   715, ...
  1,  6, 21,  56, 126,  252|  462,   792,  1287,  2002, ...
  1,  7, 28,  84, 210,  462,  924|  1716,  3003,  5005, ...
  1,  8, 36, 120, 330,  792, 1716,  3432|  6435, 11440, ...
  1,  9, 45, 165, 495, 1287, 3003,  6435, 12870| 24310, ...
  1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620| ...
The array read by antidiagonals gives the binomial triangle.
From _Reinhard Zumkeller_, Jul 27 2012: (Start)
Take the first n elements of the n-th diagonal (NW to SE) of left half of Pascal's triangle and write it as n-th row on the triangle on the right side, see above
  0:                 1                    1
  1:               1   _                  1  2
  2:             1   2  __                1  3  6
  3:           1   3  __  __              1  4 10 20
  4:         1   4   6  __  __            1  5 15 35 70
  5:       1   5  10  __  __  __          1  6 21 56 .. ..
  6:     1   6  15  20  __  __  __        1  7 28 .. .. .. ..
  7:   1   7  21  35  __  __  __  __      1  8 .. .. .. .. .. ..
  8: 1   8  28  56  70  __  __  __  __    1 .. .. .. .. .. .. .. .. (End)

H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83.

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Karl Dilcher and Maciej Ulas, Arithmetic properties of polynomial solutions of the Diophantine equation P(x)x^(n+1)+Q(x)(x+1)^(n+1)=1, arXiv:1909.11222 [math.NT], 2019. See Qn(x) Table 1 p. 2.
H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83. (Annotated scanned copy)
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving full transformations, Semigroup Forum 72 (2006), 51-62.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A000108, A046900, A001700, A007318, A034868, A092392, A239103, A178300.

import Data.List (transpose)
a046899 n k = a046899_tabl !! n !! k
a046899_row n = a046899_tabl !! n
a046899_tabl = zipWith take [1..] $ transpose a007318_tabl
-- Reinhard Zumkeller, Jul 27 2012

/* As triangle */ [[Binomial(n+k, n): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 18 2015

for n from 0 to 10 do seq( binomial(n+m,n), m = 0 .. n) od; # Zerinvary Lajos, Dec 09 2007

t[n_, k_] := Binomial[n + k, n]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 12 2013 *)

/* same as in A092566 but use */
steps=[[1, 0], [1, 0] ];
/* Joerg Arndt, Jul 01 2011 */

for n in (0..9):
    print([multinomial(n, k) for k in (0..n)]) # Peter Luschny, Dec 24 2020

T(n,k) = A092392(n,n-k), k = 0..n. - Reinhard Zumkeller, Jul 27 2012
T(n,k) = A178300(n,k), n>0, k = 1..n. - L. Edson Jeffery, Jul 23 2014
T(n,k) = (n + 1)*hypergeom([-n, 1 - k], [2], 1). - Peter Luschny, Jan 09 2022
T(n,k) = hypergeom([-n, -k], [1], 1). - Peter Luschny, Mar 21 2024
G.f.: 1/((1-2x*y*C(x*y))*(1-x*C(x*y))), where C(x) is the g.f. for A000108, the Catalan numbers. - Michael D. Weiner, Jul 31 2024

More terms from James Sellers

A357623 Skew-alternating sum of the n-th composition in standard order.

Original entry on oeis.org

0, 1, 2, 0, 3, 1, -1, -1, 4, 2, 0, 0, -2, -2, -2, 0, 5, 3, 1, 1, -1, -1, -1, 1, -3, -3, -3, -1, -3, -1, 1, 1, 6, 4, 2, 2, 0, 0, 0, 2, -2, -2, -2, 0, -2, 0, 2, 2, -4, -4, -4, -2, -4, -2, 0, 0, -4, -2, 0, 0, 2, 2, 2, 0, 7, 5, 3, 3, 1, 1, 1, 3, -1, -1, -1, 1, -1

Offset: 0

Gus Wiseman, Oct 08 2022

sign

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ....

			The 358-th composition is (2,1,3,1,2) so a(358) = 2 - 1 - 3 + 1 + 2 = 1.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

See link for sequences related to standard compositions.
Positions of positive firsts appear to be A029744.
The half-alternating form is A357621, reverse A357622.
The reverse version is A357624.
Positions of zeros are A357627, reverse A357628.
The version for prime indices is A357630.
The version for Heinz numbers of partitions is A357634.
A124754 gives alternating sum of standard compositions, reverse A344618.
A357637 counts partitions by half-alternating sum, skew A357638.
A357641 counts comps w/ half-alt sum 0, partitions A357639, even A357642.
Cf. A001700, A001511, A053251, A344619, A357136, A357182, A357183, A357184, A357185, A357625, A357626.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
Table[skats[stc[n]],{n,0,100}]

A100100 Triangle T(n,k) = binomial(2*n-k-1, n-k) read by rows.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 10, 6, 3, 1, 35, 20, 10, 4, 1, 126, 70, 35, 15, 5, 1, 462, 252, 126, 56, 21, 6, 1, 1716, 924, 462, 210, 84, 28, 7, 1, 6435, 3432, 1716, 792, 330, 120, 36, 8, 1, 24310, 12870, 6435, 3003, 1287, 495, 165, 45, 9, 1, 92378, 48620, 24310, 11440, 5005, 2002

Offset: 0

Paul Barry, Nov 08 2004

Sometimes called a Catalan triangle, although there are many other triangles that carry that name - see A009766, A008315, A028364, A033184, A053121, A059365, A062103.
Number of nodes of outdegree k in all ordered trees with n edges. Equivalently, number of ascents of length k in all Dyck paths of semilength n. Example: T(3,2) = 3 because the Dyck paths of semilength 3 are UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and UUUDDD, where U = (1,1), D = (1,-1), the ascents of length 2 being shown between parentheses. - Emeric Deutsch, Nov 19 2006
Riordan array (f(x), x*g(x)) where f(x) is the g.f. of A088218 and g(x) is the g.f. of A000108. - Philippe Deléham, Jan 23 2010
T(n,k) is the number of nonnegative paths of upsteps U = (1,1) and downsteps D = (1,-1) of length 2*n with k returns to  ground level, the horizontal line through the initial vertex. Example: T(2,1) = 2 counts UDUU, UUDD. Also, T(n,k) = number of these paths whose last descent has length k, that is, k downsteps follow the last upstep. Example: T(2,1) = 2 counts UUUD, UDUD. - David Callan, Nov 21 2011
Belongs to the hitting-time subgroup of the Riordan group. Multiplying this triangle by the square Pascal matrix gives A092392 read as a square array. See the example below. - Peter Bala, Nov 03 2015

			From _Paul Barry_, Mar 15 2010: (Start)
Triangle begins in row n=0 with columns 0<=k<=n as:
    1;
    1,   1;
    3,   2,   1;
   10,   6,   3,  1;
   35,  20,  10,  4,  1;
  126,  70,  35, 15,  5, 1;
  462, 252, 126, 56, 21, 6, 1;
Production matrix begins
  1, 1;
  2, 1, 1;
  3, 1, 1, 1;
  4, 1, 1, 1, 1;
  5, 1, 1, 1, 1, 1;
  6, 1, 1, 1, 1, 1, 1;
  7, 1, 1, 1, 1, 1, 1, 1;
(End)
A092392 as a square array = A100100 * square Pascal matrix:
/1   1  1  1 ...\   / 1          \/1 1  1  1 ...\
|2   3  4  5 ...|   | 1 1        ||1 2  3  4 ...|
|6  10 15 21 ...| = | 3 2 1      ||1 3  6 10 ...|
|20 35 56 84 ...|   |10 6 3 1    ||1 4 10 20 ...|
|70 ...         |   |35 ...      ||1 ...        |
- _Peter Bala_, Nov 03 2015

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Gi-Sang Cheon, Hana Kim, and Louis W. Shapiro, An algebraic structure for Faber polynomials, Lin. Alg. Applic. 433 (2010) 1170-1179.
Tian-Xiao He and Renzo Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math. 309 (2009), no. 12, 3962-3974.

Row sums are A000984. Equivalent to A092392, to which A088218 has been added as a first column. Columns include A088218, A000984, A001700, A001791, A002054, A002694. Diagonal sums are A100217. Matrix inverse is A100218.

Cf. A059481 (mirrored). Cf. A033184, A094527, A113955.

a100100 n k = a100100_tabl !! n !! n
a100100_row n = a100100_tabl !! n
a100100_tabl = [1] : f a092392_tabl where
   f (us : wss'@(vs : wss)) = (vs !! 1 : us) : f wss'
-- Reinhard Zumkeller, Jan 15 2014

/* As triangle */ [[Binomial(2*n - k - 1, n - k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Nov 21 2018

A100100 := proc(n,k)
    binomial(2*n-k-1,n-1) ;
end proc:
seq(seq(A100100(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 06 2015

Flatten[Table[Binomial[2 n - k - 1, n - k], {n, 0, 11}, {k, 0, n}]] (* Vincenzo Librandi, Nov 21 2018 *)

T(n,k)=binomial(2*n-k-1,n-k) \\ Charles R Greathouse IV, Jan 16 2012

From Peter Bala, Sep 06 2015: (Start)
Matrix product A094527 * P^(-1) = A113955 * P^(-2), where P denotes Pascal's triangle A007318.
Essentially, the logarithmic derivative of A033184. (End)
Let column(k) = [T(n, k), n >= k], then the generating function for column(k) is (2/(sqrt(1-4*x)+1))^(k-1)/sqrt(1-4*x). - Peter Luschny, Mar 19 2021
O.g.f. row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, i.e. o.g.f. of the triangle, G(z,x) = 1/((2 - c(z))*(1 - x*z*c(z))), with c the o.g.f. of A000108 (Catalan). See the Riordan coomment by Philippe Deléham above. - Wolfdieter Lang, Apr 06 2021

A183134 Square array A(n,k) by antidiagonals. A(n,k) is the number of length 2n k-ary words (n,k>=0), either empty or beginning with the first character of the alphabet, that can be built by repeatedly inserting doublets into the initially empty word.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 3, 1, 0, 1, 1, 5, 10, 1, 0, 1, 1, 7, 29, 35, 1, 0, 1, 1, 9, 58, 181, 126, 1, 0, 1, 1, 11, 97, 523, 1181, 462, 1, 0, 1, 1, 13, 146, 1145, 4966, 7941, 1716, 1, 0, 1, 1, 15, 205, 2131, 14289, 48838, 54573, 6435, 1, 0

Offset: 0

Alois P. Heinz, Dec 26 2010

Column k > 2 is asymptotic to 2^(2*n) * (k-1)^(n+1) / ((k-2)^2 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Sep 07 2014

			A(3,2) = 10, because 10 words of length 6 beginning with the first character of the 2-letter alphabet {a, b} can be built by repeatedly inserting doublets (words with two equal letters) into the initially empty word: aaaaaa, aaaabb, aaabba, aabaab, aabbaa, aabbbb, abaaba, abbaaa, abbabb, abbbba.
Square array A(n,k) begins:
  1,  1,   1,    1,    1,     1,  ...
  0,  1,   1,    1,    1,     1,  ...
  0,  1,   3,    5,    7,     9,  ...
  0,  1,  10,   29,   58,    97,  ...
  0,  1,  35,  181,  523,  1145,  ...
  0,  1, 126, 1181, 4966, 14289,  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
C. Kassel and C. Reutenauer, Algebraicity of the zeta function associated to a matrix over a free group algebra, arXiv preprint arXiv:1303.3481, 2013
A. Lakshminarayan, Z. Puchala, K. Zyczkowski, Diagonal unitary entangling gates and contradiagonal quantum states, arXiv preprint arXiv:1407.1169, 2014

Rows 0-10 give: A000012, A057427, A004273, A079273(k) for k>0, A194716, A194717, A194718, A194719, A194720, A194721, A194722.
Columns 0-10 give: A000007, A000012, A001700(n-1) for n>0, A194723, A194724, A194725, A194726, A194727, A194728, A194729, A194730.
Main diagonal gives A248828.
Coefficients of row polynomials for k>0 in k, (k+1) are given by A050166, A157491.
Cf. A007318, A183135.

A:= proc(n, k)
      local j;
      if n=0  then 1
    elif k<=1 then k
              else add(binomial(2*n,j)*(n-j)*(k-1)^j, j=0..n-1)/n
      fi
    end:
seq(seq(A(n, d-n), n=0..d), d=0..10);

a[n_, k_] := If[ n == 0, 1 , If[ k <= 1, k, Sum [Binomial[2*n, j]*(n-j)*(k-1)^j, {j, 0, n-1}] / n ] ]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Dec 09 2013, translated from Maple *)

A(n,k) = 1 if n=0, A(n,k) = k if n>0 and k<=1, and A(n,k) = 1/n * Sum_{j=0..n-1} C(2*n,j)*(n-j)*(k-1)^j else.
A(n,k) = A183135(n,k) if n=0 or k<2, A(n,k) = A183135(n,k)/k else.
G.f. of column k: 1/(1-k*x) if k<2, (1-1/k) * (1 + 2 / (k-2 + k * sqrt (1-(4*k-4)*x))) else.

A344741 Number of integer partitions of 2n with reverse-alternating sum -2.

Original entry on oeis.org

0, 0, 1, 2, 4, 8, 14, 24, 39, 62, 95, 144, 212, 309, 442, 626, 873, 1209, 1653, 2245, 3019, 4035, 5348, 7051, 9229, 12022, 15565, 20063, 25722, 32847, 41746, 52862, 66657, 83768, 104873, 130889, 162797, 201902, 249620, 307789, 378428, 464122, 567721, 692828, 843448

Offset: 0

Gus Wiseman, Jun 08 2021

nonn

The reverse-alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i. This is equal to (-1)^(r-1) times the number of odd parts, where r is the greatest part, so a(n) is the number of integer partitions of 2n with exactly two odd parts, neither of which is the greatest.

Also the number of reversed integer partitions of 2n with alternating sum -2.

			The a(2) = 1 through a(6) = 14 partitions:
  (31)  (42)    (53)      (64)        (75)
        (3111)  (3221)    (3331)      (4332)
                (4211)    (4222)      (4431)
                (311111)  (4321)      (5322)
                          (5311)      (5421)
                          (322111)    (6411)
                          (421111)    (322221)
                          (31111111)  (333111)
                                      (422211)
                                      (432111)
                                      (531111)
                                      (32211111)
                                      (42111111)
                                      (3111111111)

The version for -1 instead of -2 is A000070.
The non-reversed negative version is A000097.
The ordered version appears to be A001700.
The version for 1 instead of -2 is A035363.
The whole set of partitions of 2n is counted by A058696.
The strict case appears to be A065033.
The version for -1 instead of -2 is A306145.
The version for 2 instead of -2 is A344613.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
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. A001250, A003242, A006330, A027187, A028260, A344604, A344607, A344608, A344650, A344651, A344654, A344739.

sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Table[Length[Select[IntegerPartitions[n],sats[#]==-2&]],{n,0,30,2}]
- or -
Table[Length[Select[IntegerPartitions[n],EvenQ[Max[#]]&&Count[#,_?OddQ]==2&]],{n,0,30,2}]

More terms from Bert Dobbelaere, Jun 12 2021

cp's OEIS Frontend

A069720 a(n) = 2^(n-1)*binomial(2*n-1, n).

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A091867 Triangle read by rows: T(n,k) = number of Dyck paths of semilength n having k peaks at odd height.

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A349057 Numbers k such that the k-th composition in standard order is not weakly alternating.

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A103371 Number triangle T(n,k) = C(n,n-k)*C(n+1,n-k).

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A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

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A046899 Triangle in which n-th row is {binomial(n+k,k), k=0..n}, n >= 0.

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A069720 a(n) = 2^(n-1)binomial(2n-1, n).