A027306 -id:A027306 - cp's OEIS frontend

A034869 Right half of Pascal's triangle.

1, 1, 2, 1, 3, 1, 6, 4, 1, 10, 5, 1, 20, 15, 6, 1, 35, 21, 7, 1, 70, 56, 28, 8, 1, 126, 84, 36, 9, 1, 252, 210, 120, 45, 10, 1, 462, 330, 165, 55, 11, 1, 924, 792, 495, 220, 66, 12, 1, 1716, 1287, 715, 286, 78, 13, 1, 3432, 3003, 2002, 1001, 364, 91, 14, 1

Offset: 0

N. J. A. Sloane

From R. J. Mathar, May 13 2006: (Start)
Also flattened table of the expansion coefficients of x^n in Chebyshev Polynomials T_k(x) of the first kind:
x^n is 2^(1-n) multiplied by the sum of floor(1+n/2) terms using only terms T_k(x) with even k if n even, only terms T_k(x) with odd k if n is odd and halving the coefficient a(..) in front of any T_0(x):
x^0=2^(1-0) a(0)/2 T_0(x)
x^1=2^(1-1) a(1) T_1(x)
x^2=2^(1-2) [a(2)/2 T_0(x)+a(3) T_2(x)]
x^3=2^(1-3) [a(4) T_1(x)+a(5) T_3(x)]
x^4=2^(1-4) [a(6)/2 T_0(x)+a(7) T_2(x) +a(8) T_4(x)]
x^5=2^(1-5) [a(9) T_1(x)+a(10) T_3(x) +a(11) T_5(x)]
x^6=2^(1-6) [a(12)/2 T_0(x)+a(13) T_2(x) +a(14) T_4(x) +a(15) T_6(x)]
x^7=2^(1-7) [a(16) T_1(x)+a(17) T_3(x) +a(18) T_5(x) +a(19) T_7(x)]" (End)
T(n,k) = A034868(n,floor(n/2)-k), k = 0..floor(n/2). - Reinhard Zumkeller, Jul 27 2012
Rows are binomial(r-1,(2r+1-(-1)^r)\4 -n ) where r is the row and n is the term. Columns are binomial(2m+c-3,m-1) where c is the column and m is the term. - Anthony Browne, May 17 2016

			The table starts:
  1
  1
  2 1
  3 1
  6 4 1
  ...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened

Cf. A007318, A008619 (row lengths).
Cf. A110654.
Cf. A034868 (left half), A014413, A014462, A027306 (row sums).
Columns k=0-1-2-3-4 give: A001405, A037955, A037956, A037957, A037958.

a034869 n k = a034869_tabf !! n !! k
a034869_row n = a034869_tabf !! n
a034869_tabf = [1] : f 0 [1] where
   f 0 us'@(_:us) = ys : f 1 ys where
                    ys = zipWith (+) us' (us ++ [0])
   f 1 vs@(v:_) = ys : f 0 ys where
                  ys = zipWith (+) (vs ++ [0]) ([v] ++ vs)
-- Reinhard Zumkeller, improved Dec 21 2015, Jul 27 2012

for n from 0 to 60 do for j from n mod 2 to n by 2 do print( binomial(n,(n-j)/2) ); od; od; # R. J. Mathar, May 13 2006
# Second program:
egf:= k-> BesselI(2*k, 2*x) + BesselI(2*k+1, 2*x):
A034869:= (n, k)-> n! * coeff(series(egf(k), x, n+1), x, n):
seq(print(seq(A034869(n, k), k=0..iquo(n, 2))), n=0..14); # Mélika Tebni, Sep 05 2024

Table[Binomial[n, k], {n, 0, 14}, {k, Ceiling[n/2], n}] // Flatten (* Michael De Vlieger, May 19 2016 *)

for(n=0, 14, for(k=ceil(n/2), n, print1(binomial(n, k),", ");); print();) \\ Indranil Ghosh, Mar 31 2017

import math
from sympy import binomial
for n in range(15):
    print([binomial(n, k) for k in range(math.ceil(n/2), n + 1)]) # Indranil Ghosh, Mar 31 2017

E.g.f. of column k: BesselI(2*k,2*x) + BesselI(2*k+1,2*x). - Mélika Tebni, Sep 05 2024

Keyword fixed and example added by Franklin T. Adams-Watters, May 27 2010

A045723 Number of configurations, excluding reflections and black-white interchanges, of n black and n white beads on a string.

Original entry on oeis.org

1, 1, 3, 7, 23, 71, 252, 890, 3299, 12283, 46508, 176870, 677294, 2602198, 10034104, 38787572, 150289699, 583434323, 2268861516, 8836447022, 34461940538, 134564992898, 526025965864, 2058359779052, 8061905791118, 31602659998046

Offset: 0

Wouter Meeussen

nonn

This sequence (with offset 0) equals the probable number of inequivalent classes of permutations acting on an n-party state under the trace norm in the context of permutation criteria for separability. - Lieven Clarisse, Apr 28 2006
Number of connected components of an undirected graph where the nodes are the n-subsets of {1,...,2n} and an edge (A,B) appears if B = {1,...,2n} \ A or B = {2n + 1 - i: i in A}. See Mathematics Magazine link. - Rob Pratt, Aug 10 2015
Number of distinct staircase walks connecting opposite corners of a square grid of side n > 1. - Christian Barrientos, Nov 25 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Eddie Cheng and Jerrold W. Grossman, Problem 1959, Mathematics Magazine 87 (Dec. 2014), p. 396.
L. Clarisse and P. Wocjan, On independent permutation separability criteria, Quant. Inf. Comp. 6 277-288, 2006, arXiv:quant-ph/0504160, 2005.

Cf. A042942.

Cf. A027306, A042971.

Table[ 1/4 (2^n + Binomial[ 2 n, n ] + 2 Binomial[ -1 + n, 1/2 (-2 + n) ]*Mod[ 1 + n, 2 ]), {n, 0, 24} ]

a(n) = (1/4)*(2^n + binomial(2*n, n) + if ((n+1)%2, 2*binomial(n-1, (1/2)*(n-2)))); \\ Michel Marcus, Nov 25 2018

a(n) = (1/4)*(2^n + C(2*n, n) + 2*C(n-1, (1/2)*(n-2))*((n+1) mod 2)).

a(n) = A042971(n) + A027306(n). - Michel Marcus, Nov 26 2018

A092429 a(n) = n! * Sum_{i,j,k,l >= 0, i+j+k+l = n} 1/(i!j!k!*l!).

Original entry on oeis.org

1, 1, 3, 10, 47, 126, 522, 1821, 8143, 26326, 109958, 396111, 1737122, 5998955, 24949277, 91979985, 397402223, 1418993350, 5881338702, 22010456331, 94022106862, 342803313261, 1416758002487, 5356198979731, 22685035586290, 83911052895151, 345921828889367

Offset: 0

Benoit Cloitre, Mar 22 2004

nonn

a(n) is even iff n is a sum of 2 distinct powers of 2.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Vaclav Kotesovec, Recurrence (of order 11)

Cf. A018900, A027306, A092255.

Column k=4 of A226873. - Alois P. Heinz, Jun 21 2013

b:= proc(n, i, t) option remember;
      `if`(t=1, 1/n!, add(b(n-j, j, t-1)/j!, j=i..n/t))
    end:
a:= n-> n!*b(n, 0, 4):
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 21 2017

Table[Sum[Sum[Sum[Sum[If[i+j+k+l==n,n!/i!/j!/k!/l!,0],{l,0,k}],{k,0,j}],{j,0,i}],{i,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jul 01 2013 *)
CoefficientList[Series[(HypergeometricPFQ[{},{},x]^4 +6*HypergeometricPFQ[{},{},x]^2 *HypergeometricPFQ[{},{1},x^2] +8*HypergeometricPFQ[{},{},x] *HypergeometricPFQ[{},{1,1},x^3] +3*HypergeometricPFQ[{},{1},x^2]^2 +6*HypergeometricPFQ[{},{1,1,1},x^4])/24, {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec after Vladeta Jovovic, Jul 01 2013 *)

a(n)=sum(i=0,n,sum(j=0,i,sum(k=0,j,sum(l=0,k,if(i+j+k+l-n,0,n!/i!/j!/k!/l!)))))

E.g.f.: (t(1)^4 + 6*t(1)^2*t(2) + 8*t(1)*t(3) + 3*t(2)^2 + 6*t(4))/24 where t(1) = hypergeom([],[],x), t(2) = hypergeom([],[1],x^2), t(3) = hypergeom([],[1,1],x^3) and t(4) = hypergeom([],[1,1,1],x^4). - Vladeta Jovovic, Sep 22 2007, typo corrected by Vaclav Kotesovec, Jul 01 2013

Conjecture: a(n) ~ 4^n/4!. - Vaclav Kotesovec, Mar 07 2014

A100067 a(n) = Sum_{k=0..floor(n/2)} binomial(n,k)2^(n-2k).

Original entry on oeis.org

1, 2, 6, 14, 38, 92, 240, 590, 1510, 3740, 9476, 23564, 59372, 147968, 371636, 927374, 2324870, 5805740, 14538660, 36322340, 90898228, 227153192, 568235696, 1420236524, 3551943388, 8878506392, 22201466280, 55498465400, 138766221800, 346895496200, 867316299260, 2168213189390

Offset: 0

Paul Barry, Nov 02 2004

An inverse Chebyshev transform of x/(1-2*x), where the Chebyshev transform of g(x) is ((1-x^2)/(1+x^2))*g(x/(1+x^2)) and the inverse transform maps a g.f. A(x) to (1/sqrt(1-4*x^2))*A(x*c(x^2)) where c(x) is the g.f. of the Catalan numbers A000108. In general, Sum_{k=0..floor(n/2)} binomial(n,k) * r^(n-2*k) has g.f. 2*x/(sqrt(1-4*x^2)*(r*sqrt(1-4*x^2) + 2*x - r)). - corrected by Vaclav Kotesovec, Dec 06 2012
Generally (for r>1), a(n) ~ (r + 1/r)^n. - Vaclav Kotesovec, Dec 06 2012
Hankel transform is A088138(n+1). - Paul Barry, Jun 16 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A027306, A100068, A100069.

m:=2; [(&+[Binomial(n,k)*m^(n-2*k): k in [0..Floor(n/2)]]): n in [0..40]]; // G. C. Greubel, Jun 08 2022

CoefficientList[Series[x/(Sqrt[1-4*x^2]*(Sqrt[1-4*x^2]+x-1)), {x, 0, 20}], x] (* Vaclav Kotesovec, Dec 06 2012 *)

my(x='x+O('x^66)); Vec(x/(sqrt(1-4*x^2)*(sqrt(1-4*x^2)+x-1))) \\ Joerg Arndt, May 12 2013

m=2; [sum(binomial(n,k)*m^(n-2*k) for k in (0..n//2)) for n in (0..40)] # G. C. Greubel, Jun 08 2022

G.f.: x/(sqrt(1-4*x^2)*(sqrt(1-4*x^2)+x-1)).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*2^(n-2*k).
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1 + (-1)^(n-k))*2^k/2.
Recurrence: 2*n*(3*n-7)*a(n) = (15*n^2 - 35*n + 8)*a(n-1) + 4*(6*n^2 - 20*n + 11)*a(n-2) - 20*(n-2)*(3*n-4)*a(n-3). - Vaclav Kotesovec, Dec 06 2012
a(n) ~ 5^n/2^n. - Vaclav Kotesovec, Dec 06 2012

A345907 Triangle giving the main antidiagonals of the matrices counting integer compositions by length and alternating sum (A345197).

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 2, 2, 1, 1, 0, 0, 4, 3, 1, 1, 0, 0, 3, 6, 4, 1, 1, 0, 0, 6, 9, 8, 5, 1, 1, 0, 0, 0, 18, 18, 10, 6, 1, 1, 0, 0, 0, 10, 36, 30, 12, 7, 1, 1, 0, 0, 0, 20, 40, 60, 45, 14, 8, 1, 1, 0, 0, 0, 0, 80, 100, 90, 63, 16, 9, 1, 1

Offset: 0

Gus Wiseman, Jul 26 2021

The matrices (A345197) count the integer compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2. Here, the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

Problem: What are the column sums? They appear to match A239201, but it is not clear why.

			Triangle begins:
   1
   1   1
   0   1   1
   0   1   1   1
   0   2   2   1   1
   0   0   4   3   1   1
   0   0   3   6   4   1   1
   0   0   6   9   8   5   1   1
   0   0   0  18  18  10   6   1   1
   0   0   0  10  36  30  12   7   1   1
   0   0   0  20  40  60  45  14   8   1   1
   0   0   0   0  80 100  90  63  16   9   1   1
   0   0   0   0  35 200 200 126  84  18  10   1   1
   0   0   0   0  70 175 400 350 168 108  20  11   1   1
   0   0   0   0   0 350 525 700 560 216 135  22  12   1   1

Row sums are A163493.
Rows are the antidiagonals of the matrices given by A345197.
The main diagonals of A345197 are A346632, with sums A345908.
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).
Other diagonals are A008277 of A318393 and A055884 of A320808.
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 A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000346, A007318, A008549, A025047, A114121, A344610.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Table[Table[Length[Select[Join@@Permutations/@IntegerPartitions[n,{n-k}],k==(n+ats[#])/2-1&]],{k,0,n-1}],{n,0,15}]

A349155 Numbers k such that the k-th composition in standard order has sum equal to negative twice its reverse-alternating sum.

Original entry on oeis.org

0, 9, 130, 135, 141, 153, 177, 193, 225, 2052, 2059, 2062, 2069, 2074, 2079, 2089, 2098, 2103, 2109, 2129, 2146, 2151, 2157, 2169, 2209, 2242, 2247, 2253, 2265, 2289, 2369, 2434, 2439, 2445, 2457, 2481, 2529, 2561, 2689, 2818, 2823, 2829, 2841, 2865, 2913

Offset: 1

Gus Wiseman, Nov 22 2021

nonn

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 reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

			The terms and corresponding compositions begin:
     0: ()
     9: (3,1)
   130: (6,2)
   135: (5,1,1,1)
   141: (4,1,2,1)
   153: (3,1,3,1)
   177: (2,1,4,1)
   193: (1,6,1)
   225: (1,1,5,1)
  2052: (9,3)
  2059: (8,2,1,1)
  2062: (8,1,1,2)
  2069: (7,2,2,1)
  2074: (7,1,2,2)
  2079: (7,1,1,1,1,1)
  2089: (6,2,3,1)
  2098: (6,1,3,2)
  2103: (6,1,2,1,1,1)

These compositions are counted by A224274 up to 0's.
An unordered version is A348617, counted by A001523 up to 0's.
The positive version is A349153, unreversed A348614.
The unreversed version is A349154.
Positive unordered unreversed: A349159, counted by A000712 up to 0's.
A positive unordered version is A349160, counted by A006330 up to 0's.
A003242 counts Carlitz compositions.
A011782 counts compositions.
A025047 counts alternating or wiggly compositions, complement A345192.
A034871, A097805, and A345197 count compositions by alternating sum.
A103919 counts partitions by alternating sum, reverse A344612.
A116406 counts compositions with alternating sum >=0, ranked by A345913.
A138364 counts compositions with alternating sum 0, ranked by A344619.
Cf. A000070, A000346, A001250, A001700, A008549, A027306, A058622, A088218, A114121, A120452, A262977, A294175, A345917.
Statistics of standard compositions:
- The compositions themselves are the rows of A066099.
- Number of parts is given by A000120, distinct A334028.
- Sum and product of parts are given by A070939 and A124758.
- Maximum and minimum parts are given by A333766 and A333768.
- Heinz number is given by A333219.
Classes of standard compositions:
- Partitions and strict partitions are ranked by A114994 and A333256.
- Multisets and sets are ranked by A225620 and A333255.
- Strict and constant compositions are ranked by A233564 and A272919.
- Carlitz compositions are ranked by A333489, complement A348612.
- Alternating compositions are ranked by A345167, complement A345168.

stc[n_]:=Differences[Prepend[ Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Select[Range[0,1000],Total[stc[#]]==-2*sats[stc[#]]&]

A110166 Row sums of Riordan array A110165.

Original entry on oeis.org

1, 4, 18, 85, 410, 1999, 9807, 48304, 238570, 1180615, 5851253, 29033074, 144190943, 716652070, 3564079250, 17734184365, 88280673770, 439625873215, 2189988826125, 10912480440850, 54389237971285, 271142650382080

Offset: 0

Paul Barry, Jul 14 2005

Number of 5-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
From Peter Bala, Jan 09 2022: (Start)
Conjectures: for k >= 2, the number of k-ary words of length n such that the number of 1's <= the number of 0's is equal to the coefficient of x^n in the expansion of ( k*x + 1/(1 + x) )^n, and satisfies the recurrence u(0) = 1, u(1) = k-1 and n*u(n) = (k-2)*(2*n-1)*u(n-1) - k*(k-4)*(n-1)* u(n-2) + k^(n-1) for n >= 2.
For cases see A027306 (k = 2), A027914 (k = 3) and A032443 (k = 4). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Row sums A110165. Cf. A026375, A032443, A246437, A242586.

seq( (1/2)*(5^n + add(binomial(n,k)*binomial(2*k,k), k = 0..n)), n = 0..30); # Peter Bala, Jan 08 2022

Table[Sum[Sum[Binomial[n,j]Binomial[2j,j+k],{j,0,n}],{k,0,n}],{n,0,25}] (* Harvey P. Dale, Dec 16 2011 *)

G.f.: (1/sqrt(1-6*x+5*x^2))/(1-(1-3*x-sqrt(1-6*x+5*x^2))/(2*x)).
a(n) = Sum_{k = 0..n} Sum_{j = 0..n} C(n, j)*C(2*j, j+k).
Recurrence: n*a(n) = (11*n-8)*a(n-1) - 5*(7*n-10)*a(n-2) + 25*(n-2)*a(n-3). - Vaclav Kotesovec, Oct 18 2012
a(n) ~ 5^n/2*(1+sqrt(5)/(2*sqrt(Pi*n))). - Vaclav Kotesovec, Oct 18 2012
From Peter Bala, Jan 08 2022: (Start)
a(n) = (1/2)*(5^n + A026375(n)) = (1/2)*(5^n + Sum_{k = 0..n} binomial(n,k) *binomial(2*k,k)).
a(n) = (1/2)*(5^n)*(1 + Sum_{k = 0..n} binomial(n,k)*binomial(2*k,k)*(-1/5)^k).
a(n) = [x^n] ( 5*x + 1/(1 + x) )^n.
a(0) = 1, a(1) = 4 and n*a(n) = 3*(2*n-1)*a(n-1) - 5*(n-1)*a(n-2) + 5^(n-1) for n >= 2.
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.
Binomial transform of A032443. (End)

A226875 Number of n-length words w over a 5-ary alphabet {a1,a2,...,a5} such that #(w,a1) >= #(w,a2) >= ... >= #(w,a5) >= 0, where #(w,x) counts the letters x in word w.

Original entry on oeis.org

1, 1, 3, 10, 47, 246, 882, 3921, 18223, 84790, 432518, 1863951, 8892842, 42656147, 204204353, 1025014815, 4728033983, 22948258742, 111605089014, 541696830843, 2708218059022, 12861557284425, 62938669549583, 308273057334413, 1508708926286914, 7533652902408071

Offset: 0

Alois P. Heinz, Jun 21 2013

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Column k=5 of A226873.

Cf. A027306, A092255, A092429.

b:= proc(n, i, t) option remember;
      `if`(t=1, 1/n!, add(b(n-j, j, t-1)/j!, j=i..n/t))
    end:
a:= n-> n!*b(n, 0, 5):
seq(a(n), n=0..30);

Table[Sum[Sum[Sum[Sum[Sum[If[i+j+k+l+m==n,n!/i!/j!/k!/l!/m!,0],{m,0,l}],{l,0,k}],{k,0,j}],{j,0,i}],{i,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jul 01 2013 *)
CoefficientList[Series[(HypergeometricPFQ[{},{},x]^5 + 10*HypergeometricPFQ[{},{},x]^3*HypergeometricPFQ[{},{1},x^2] + 20*HypergeometricPFQ[{},{},x]^2*HypergeometricPFQ[{},{1,1},x^3] + 20*HypergeometricPFQ[{},{1},x^2]*HypergeometricPFQ[{},{1,1},x^3] + 15*HypergeometricPFQ[{},{1},x^2]^2*HypergeometricPFQ[{},{},x] + 30*HypergeometricPFQ[{},{1,1,1},x^4]*HypergeometricPFQ[{},{},x] + 24*HypergeometricPFQ[{},{1,1,1,1},x^5])/5!,{x,0,20}],x]*Range[0,20]! (* more efficient, Vaclav Kotesovec, Jul 01 2013 *)

Conjecture: a(n) ~ 5^n/5!. - Vaclav Kotesovec, Mar 07 2014

A226881 Number of n-length binary words w with #(w,0) >= #(w,1) >= 1, where #(w,x) gives the number of digits x in w.

Original entry on oeis.org

0, 0, 2, 3, 10, 15, 41, 63, 162, 255, 637, 1023, 2509, 4095, 9907, 16383, 39202, 65535, 155381, 262143, 616665, 1048575, 2449867, 4194303, 9740685, 16777215, 38754731, 67108863, 154276027, 268435455, 614429671, 1073741823, 2448023842, 4294967295, 9756737701

Offset: 0

Alois P. Heinz, Jun 21 2013

a(n) is the number of nonempty subsets of {1,2,...,n} that contain either more even than odd numbers or the same number of even and odd numbers. For example, for n=5, a(5)=15 and the 15 subsets are {2}, {4}, {1,2}, {1,4}, {2,3}, {2,4}, {2,5}, {3,4}, {4,5}, {1,2,4}, {2,3,4}, {2,4,5}, {1,2,3,4}, {1,2,4,5}, {2,3,4,5}. - Enrique Navarrete, Dec 15 2019

			a(4) = 10: 0001, 0010, 0011, 0100, 0101, 0110, 1000, 1001, 1010, 1100.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Column k=2 of A226874.

Cf. A027306.

a:= proc(n) option remember;
      `if`(n<4, n*(n-1)*(4-n)/2, (9*(n-1)*(n-4) *a(n-1)
      +(12-32*n+6*n^2) *a(n-2) -36*(n-2)*(n-4) *a(n-3)
      +8*(n-3)*(3*n-10) *a(n-4))/ (n*(3*n-13)))
    end:
seq(a(n), n=0..40);

Table[Sum[Binomial[n, i], {i, Floor[n/2]}], {n, 0, 30}] (* Wesley Ivan Hurt, Mar 14 2015 *)

a(n) = sum(i=1, n\2, binomial(n,i)); \\ Michel Marcus, Jul 15 2022

G.f.: (3*x-1)/(2*(x-1)*(2*x-1)) + 1/(2*sqrt((1+2*x)*(1-2*x))).
a(n) = Sum_{i=1..floor(n/2)} binomial(n,i). - Wesley Ivan Hurt, Mar 14 2015
a(n) = A027306(n)-1 = 2^(n-1)-1+((1+(-1)^n)/4)*binomial(n,n/2). - Alois P. Heinz, Dec 15 2019

A322428 Sum T(n,k) of k-th largest parts of all compositions of n; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 3, 1, 8, 3, 1, 19, 8, 4, 1, 43, 20, 11, 5, 1, 94, 48, 27, 16, 6, 1, 202, 110, 64, 42, 22, 7, 1, 428, 245, 149, 100, 64, 29, 8, 1, 899, 533, 341, 228, 163, 93, 37, 9, 1, 1875, 1142, 765, 512, 383, 256, 130, 46, 10, 1, 3890, 2420, 1683, 1144, 859, 638, 386, 176, 56, 11, 1

Offset: 1

Alois P. Heinz, Dec 07 2018

			The 4 compositions of 3 are: 111, 12, 21, 3.  The sums of k-th largest parts for k=1..3 give: 1+2+2+3 = 8, 1+1+1+0 = 3, 1+0+0+0 = 1.
Triangle T(n,k) begins:
     1;
     3,    1;
     8,    3,   1;
    19,    8,   4,   1;
    43,   20,  11,   5,   1;
    94,   48,  27,  16,   6,   1;
   202,  110,  64,  42,  22,   7,   1;
   428,  245, 149, 100,  64,  29,   8,  1;
   899,  533, 341, 228, 163,  93,  37,  9,  1;
  1875, 1142, 765, 512, 383, 256, 130, 46, 10, 1;
  ...

Alois P. Heinz, Rows n = 1..50, flattened

Column k=1 gives A102712.
Row sums give A001787.
T(n+1,1+ceiling(n/2)) gives A027306.
Cf. A322427.

b:= proc(n, l) option remember; `if`(n=0, add(l[-i]*x^i,
      i=1..nops(l)), add(b(n-j, sort([l[], j])), j=1..n))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n, [])):
seq(T(n), n=1..12);

b[n_, l_] := b[n, l] = If[n == 0, Sum[l[[-i]] x^i, {i, 1, Length[l]}], Sum[b[n - j, Sort[Append[l, j]]], {j, 1, n}]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, Exponent[p, x]}]][ b[n, {}]];
Array[T, 12] // Flatten (* Jean-François Alcover, Dec 29 2018, after Alois P. Heinz *)

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