A006134 -id:A006134 - cp's OEIS frontend

A054108 a(n) = (-1)^(n+1)Sum_{k=0..n+1}(-1)^kbinomial(2*k,k).

1, 5, 15, 55, 197, 727, 2705, 10165, 38455, 146301, 559131, 2145025, 8255575, 31861025, 123256495, 477823895, 1855782325, 7219352975, 28125910825, 109720617995, 428537256445, 1675561707275, 6557869020325, 25689734662775

Offset: 0

Clark Kimberling

nonn

p divides a((p-3)/2) for p in A045468 (primes congruent to {1, 4} mod 5). - Alexander Adamchuk, Jul 05 2006
The sequence 1,1,5,15,55,... has general term sum{k=0..n, (-1)^(n-k)*C(2k,k)}. Its Hankel transform is A082761. - Paul Barry, Apr 10 2007
From Paul Barry, Mar 29 2010: (Start)
The sequence 1,1,5,15,... has g.f. 1/((1+x)*sqrt(1-4x)).
The doubled sequence 1,1,1,1,5,5,... has e.g.f. dif(int((sin(x-t)+cos(x-t))*Bessel_I(0,2t),t,0,x),x). (End)

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

T(2n, n), array T as in A054106.

Cf. A066796, A000984, A054109, A006134, A045468.

Table[Sum[(-1)^(k+n)*((2k)!/(k!)^2),{k,0,n}], {n,1,50}] (* Alexander Adamchuk, Jul 05 2006 *)
CoefficientList[Series[(1/Sqrt[1-4*x]/(1+x)-1)/x, {x, 0, 20}], x]
(* or *)
Table[(-1)^(n+1)*Sum[(-1)^k*Binomial[2*k, k], {k, 0, n+1}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 06 2012 *)
Round@Table[Binomial[2 (n + 2), n + 2] Hypergeometric2F1[1, n + 5/2, n + 3, -4] - (-1)^n/Sqrt[5], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)

a(n)=(-1)^(n+1)*sum(k=0,n+1,(-1)^k*binomial(2*k,k))

from math import comb
def A054108(n): return (1 if n % 2 else -1)*sum((-1 if k % 2 else 1)*comb(2*k,k) for k in range(n+2)) # Chai Wah Wu, Jan 19 2022

a(n) = C(2n, n) - a(n-1) with a(0)=1. - Labos Elemer, Apr 26 2003
C(2n,n) - C(2n-2,n-1) + ... +(-1)^(k+n)*C(2k,k)+ ... + (-1)^(1+n)*C(2,1) + (-1)^n*C(0,0), where C(2k,k)=(2k)!/(k!)^2 - central binomial coefficients A000984[k]. - Alexander Adamchuk, Jul 05 2006
a(n) = Sum_{k=0..n} (-1)^(k+n)*((2k)!/(k!)^2). - Alexander Adamchuk, Jul 05 2006
G.f.: (1/sqrt(1-4*x)/(1+x)-1)/x = (-1 + 2/(U(0)-2*x))/(1+x) where U(k)= 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 27 2012
a(n) ~ 2^(2*n+4)/(5*sqrt(Pi*n)). - Vaclav Kotesovec, Nov 06 2012
Recurrence: (n+1)*a(n) = (3*n+1)*a(n-1) + 2*(2*n+1)*a(n-2). - Vaclav Kotesovec, Nov 06 2012

Formula from Benoit Cloitre, Sep 29 2002

Definition corrected by Vaclav Kotesovec, Nov 06 2012

A070925 Number of subsets of A = {1,2,...,n} that have the same center of gravity as A, i.e., (n+1)/2.

Original entry on oeis.org

1, 1, 3, 3, 7, 7, 19, 17, 51, 47, 151, 137, 471, 427, 1519, 1391, 5043, 4651, 17111, 15883, 59007, 55123, 206259, 193723, 729095, 688007, 2601639, 2465133, 9358943, 8899699, 33904323, 32342235, 123580883, 118215779, 452902071, 434314137, 1667837679, 1602935103

Offset: 1

Sharon Sela (sharonsela(AT)hotmail.com), May 20 2002

nonn

From Gus Wiseman, Apr 15 2023: (Start)
Also the number of nonempty subsets of {0..n} with mean n/2. The a(0) = 1 through a(5) = 7 subsets are:
  {0}  {0,1}  {1}      {0,3}      {2}          {0,5}
              {0,2}    {1,2}      {0,4}        {1,4}
              {0,1,2}  {0,1,2,3}  {1,3}        {2,3}
                                  {0,2,4}      {0,1,4,5}
                                  {1,2,3}      {0,2,3,5}
                                  {0,1,3,4}    {1,2,3,4}
                                  {0,1,2,3,4}  {0,1,2,3,4,5}
(End)

			Of the 32 (2^5) sets which can be constructed from the set A = {1,2,3,4,5} only the sets {3}, {2, 3, 4}, {2, 4}, {1, 2, 4, 5}, {1, 2, 3, 4, 5}, {1, 3, 5}, {1, 5} give an average of 3.

Fausto A. C. Cariboni, Table of n, a(n) for n = 1..40

The odd bisection is A000980(n) - 1 = 2*A047653(n) - 1.
For median instead of mean we have A100066, bisection A006134.
Including the empty set gives A222955.
The one-based version is A362046, even bisection A047653(n) - 1.
A007318 counts subsets by length.
A067538 counts partitions with integer mean, strict A102627.
A231147 counts subsets by median.
A327481 counts subsets by integer mean.
Cf. A024718, A057552, A079309, A133406, A326512, A326513, A326537, A327475, A349156, A359893.

Needs["DiscreteMath`Combinatorica`"]; f[n_] := Block[{s = Subsets[n], c = 0, k = 2}, While[k < 2^n + 1, If[ (Plus @@ s[[k]]) / Length[s[[k]]] == (n + 1)/2, c++ ]; k++ ]; c]; Table[ f[n], {n, 1, 20}]
(* second program *)
Table[Length[Select[Subsets[Range[0,n]],Mean[#]==n/2&]],{n,0,10}] (* Gus Wiseman, Apr 15 2023 *)

From Gus Wiseman, Apr 18 2023: (Start)
a(2n+1) = A000980(n) - 1.
a(n) = A222955(n) - 1.
a(n) = 2*A362046(n) + 1.
(End)

Edited by Robert G. Wilson v and John W. Layman, May 25 2002

a(34)-a(38) from Fausto A. C. Cariboni, Oct 08 2020

A079727 a(n) = 1 + C(2,1)^3 + C(4,2)^3 + ... + C(2n,n)^3.

Original entry on oeis.org

1, 9, 225, 8225, 351225, 16354233, 805243257, 41229480825, 2172976383825, 117106008311825, 6423711336265041, 357470875526646609, 20131502573232075025, 1145190201805448075025, 65706503254247744075025

Offset: 0

Benoit Cloitre, Feb 17 2003

a(n) seems to have an interesting congruence property: For p prime, a(p) == 8 (mod p) if and only if p == 3, 5, 7, or 13 (mod 14); i.e., iff p = 7 or p is in A003625.
From Peter Bala, Jul 12 2024: (Start)
Zhi-Wei Sun (2010) conjectured that if p is an odd prime such that the Legendre symbol (p/7) = -1 (i.e., if p == 3, 5, 6 (mod 7)) then a(p-1) == 0 (mod p^2). Otherwise, if (p/7) = 1 then a(p-1) == 4*x^2 - 2*p (mod p^2) where p = x^2 + 7*y^2 with x, y in Z.
The author’s twin brother Zhi_Hong Sun confirmed the conjecture in the case (p/7) = -1.
Conjectures: if prime p is in A003625 then
1) a(p^2) == 8 + p^2 (mod p^3)
2) a(p*(p-1)) == p^2 (mod p^3)
3) a((p^2-1)/2) == p^2 (mod p^4) (all checked up to p = 101).
4) if n is a product of distinct primes from A003625 then a((n-1)/2) is divisible by n^2. (End)

Seiichi Manyama, Table of n, a(n) for n = 0..556
Zhi-Hong Sun, Congruences concerning Legendre polynomials II", Theorem 3.2, arXiv:1012.3898v2 [math.NT], 2010-2012.
Zhi-Wei Sun, Open conjectures on congruences, Part A, conjecture A1, arXiv:0911.5665v59 [math.NT], 2009-2011.

Cf. A002476.

Cf. Sum_{k = 0..n} binomial(2*k, k)^m: A006134 (m=1), A115257 (m=2), this sequence (m=3).

[&+[Binomial(2*k, k)^3: k in [0..n]]: n in [0..20]]; // Vincenzo Librandi, Nov 16 2016

Table[Sum[Binomial[2 k, k]^3, {k, 0, n}], {n, 0, 14}] (* Michael De Vlieger, Nov 15 2016 *)

makelist(sum(binomial(2*k,k)^3,k,0,n),n,0,12); /* Emanuele Munarini, Nov 15 2016 */

PARI
```
a(n)=sum(k=0,n,binomial(2*k,k)^3)
```

a(n) = Sum_{k=0..n} binomial(2*k,k)^3.
G.f.: hypergeom([1/2, 1/2, 1/2], [1, 1], 64*x)/(1-x). - Vladeta Jovovic, Feb 18 2003
G.f.: hypergeom([1/4,1/4],[1],64*x)^2/(1-x). - Mark van Hoeij, Nov 17 2011
Recurrence: (n+2)^3*a(n+2)-(5*n+8)*(13*n^2+38*n+28)*a(n+1)+8*(2n+3)^3*a(n)=0. - Emanuele Munarini, Nov 15 2016
a(n) ~ 2^(6*n+6) / (63*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Nov 16 2016

A144635 a(n) = 5^nSum_{ k=0..n } binomial(2k,k)/5^k.

Original entry on oeis.org

1, 7, 41, 225, 1195, 6227, 32059, 163727, 831505, 4206145, 21215481, 106782837, 536618341, 2693492305, 13507578125, 67693008145, 339066121115, 1697664211795, 8497396194275, 42522326235175, 212749477704695, 1064285646397915, 5323532330953295, 26625895085494075

Offset: 0

N. J. A. Sloane, Jan 21 2009

nonn

Cf. A000032, A006134, A082590, A132310, A002457.

Table[5^n Sum[Binomial[2k,k]/5^k,{k,0,n}],{n,0,30}] (* Harvey P. Dale, Aug 08 2011 *)
Round@Table[5^(n + 1/2) - 2^(n + 1) (2 n + 1)!! Hypergeometric2F1[1, n + 3/2, n + 2, 4/5]/(5 (n + 1)!), {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster - Vladimir Reshetnikov, Oct 14 2016 *)

a(n) = 5^n*sum(k=0, n, binomial(2*k,k)/5^k); \\ Michel Marcus, Oct 14 2016

From Vaclav Kotesovec, Jun 12 2013: (Start)
G.f.: 1/((1-5*x)*sqrt(1-4*x)).
Recurrence: n*a(n) = (9*n-2)*a(n-1) - 10*(2*n-1)*a(n-2).
a(n) ~ 5^(n+1/2). (End)
a(n) = 5^(n+1/2) - 2^(n+1)*(2*n+1)!!*hypergeom([1,n+3/2], [n+2], 4/5)/(5*(n+1)!). - Vladimir Reshetnikov, Oct 14 2016
a(n) = Sum_{k=0..n} C(2*n+1,n-k)*A000032(2*k+1). - Vladimir Kruchinin Jan 14 2025

A362046 Number of nonempty subsets of {1..n} with mean n/2.

Original entry on oeis.org

0, 0, 1, 1, 3, 3, 9, 8, 25, 23, 75, 68, 235, 213, 759, 695, 2521, 2325, 8555, 7941, 29503, 27561, 103129, 96861, 364547, 344003, 1300819, 1232566, 4679471, 4449849, 16952161, 16171117, 61790441, 59107889, 226451035, 217157068, 833918839, 801467551, 3084255127

Offset: 0

Gus Wiseman, Apr 12 2023

nonn

			The a(2) = 1 through a(7) = 8 subsets:
  {1}  {1,2}  {2}      {1,4}      {3}          {1,6}
              {1,3}    {2,3}      {1,5}        {2,5}
              {1,2,3}  {1,2,3,4}  {2,4}        {3,4}
                                  {1,2,6}      {1,2,4,7}
                                  {1,3,5}      {1,2,5,6}
                                  {2,3,4}      {1,3,4,6}
                                  {1,2,3,6}    {2,3,4,5}
                                  {1,2,4,5}    {1,2,3,4,5,6}
                                  {1,2,3,4,5}

Using range 0..n gives A070925.
Including the empty set gives A133406.
Even bisection is A212352.
For median instead of mean we have A361801, the doubling of A079309.
A version for partitions is A361853, for median A361849.
A000980 counts nonempty subsets of {1..2n-1} with mean n.
A007318 counts subsets by length.
A067538 counts partitions with integer mean, strict A102627.
A231147 appears to count subsets by median, full-steps A013580.
A327475 counts subsets with integer mean, A000975 integer median.
A327481 counts subsets by integer mean.
Cf. A006134, A024718, A057552, A349156, A359893, A361654, A361864.

Table[Length[Select[Subsets[Range[n]],Mean[#]==n/2&]],{n,0,15}]

a(n) = (A070925(n) - 1)/2.
a(n) = A133406(n) - 1.
a(2n) = A212352(n) = A000980(n)/2 - 1.

A371965 a(n) is the sum of all peaks in the set of Catalan words of length n.

Original entry on oeis.org

0, 0, 0, 1, 6, 27, 111, 441, 1728, 6733, 26181, 101763, 395693, 1539759, 5997159, 23381019, 91244934, 356427459, 1393585779, 5453514729, 21358883439, 83718027429, 328380697629, 1288947615849, 5062603365999, 19896501060225, 78239857877649, 307831771279549, 1211767933187601

Offset: 0

Stefano Spezia, Apr 14 2024

nonn

			a(3) = 1 because there is 1 Catalan word of length 3 with one peak: 010.
a(4) = 6 because there are 6 Catalan words of length 4 with one peak: 0010, 0100, 0101, 0110, 0120, and 0121 (see Figure 10 at p. 19 in Baril et al.).

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Jean-Luc Baril, Pamela E. Harris, Kimberly J. Harry, Matt McClinton, and José L. Ramírez, Enumerating runs, valleys, and peaks in Catalan words, arXiv:2404.05672 [math.CO], 2024. See Corollary 4.7, p. 19.

Cf. A371963, A371964.

Cf. A002054, A275289.

a:= proc(n) option remember; `if`(n<3, 0,
      a(n-1)+binomial(2*n-3, n-3))
    end:
seq(a(n), n=0..28);  # Alois P. Heinz, Apr 15 2024
# Second Maple program:
A371965 := series((exp(2*x)*BesselI(0,2*x)-1)/2-exp(x)*(int(BesselI(0,2*x)*exp(x), x)), x = 0, 29):
seq(n!*coeff(A371965, x, n), n = 0 .. 28); # Mélika Tebni, Jun 15 2024

CoefficientList[Series[(1-3x-(1-x)Sqrt[1-4x])/(2(1-x) Sqrt[1-4x]),{x,0,28}],x]

from math import comb
def A371965(n): return sum(comb((n-i<<1)-3,n-i-3) for i in range(n-2)) # Chai Wah Wu, Apr 15 2024

G.f.: (1 - 3*x - (1 - x)*sqrt(1 - 4*x))/(2*(1 - x)*sqrt(1 - 4*x)).
a(n) = Sum_{i=1..n-1} binomial(2*(n-i)-1,n-i-2).
a(n) ~ 2^(2*n)/(6*sqrt(Pi*n)).
a(n)/A371963(n) ~ 1.
a(n) - a(n-1) = A002054(n-2).
From Mélika Tebni, Jun 15 2024: (Start)
E.g.f.: (exp(2*x)*BesselI(0,2*x)-1)/2 - exp(x)*Integral_{x=-oo..oo} BesselI(0,2*x)*exp(x) dx.
a(n) = binomial(2*n,n)*(1/2 + hypergeom([1,n+1/2],[n+1],4)) + i/sqrt(3) - 0^n/2.
a(n) = (3*A106191(n) + A006134(n) + 4*0^n) / 8.
a(n) = A281593(n) - (A000984(n) + 0^n) / 2. (End)
Binomial transform of A275289. - Alois P. Heinz, Jun 20 2025

A094639 Partial sums of squares of Catalan numbers (A000108).

Original entry on oeis.org

1, 2, 6, 31, 227, 1991, 19415, 203456, 2248356, 25887400, 307993016, 3763786812, 47032778956, 598933188956, 7751562502556, 101741582076581, 1351906409905481, 18159677984049581, 246298405721739581

Offset: 0

André F. Labossière, May 27 2004

Koshy and Salmassi give an elementary proof that the only prime Catalan numbers are A000108(2) = 2 and A000108(3) = 5. Franklin T. Adams-Watters showed that the only semiprime Catalan number is A000108(4) = 14. The subsequence of primes in the partial sum of squares of Catalan numbers begins: 2, 31, 227, 101741582076581. [Jonathan Vos Post, May 27 2010]

Conjecture: For any positive integer n, the polynomial P_n(x) = sum_{k = 0}^n(C_k)^2*x^k (with C_k = binomial(2k, k)/(k+1)) is irreducible over the field of rational numbers. [Zhi-Wei Sun, Mar 23 2013]

Michael De Vlieger, Table of n, a(n) for n = 0..838
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Joel E. Cohen, Variance Functions of Asymptotically Exponentially Increasing Integer Sequences Go Beyond Taylor's Law, J. Int. Seq., Vol. 25 (2022), Article 22.9.3.

Cf. A000108, A094638, A014137, A001246, A033536, A000984, A006134, A082894, A002897, A079727.

Accumulate[CatalanNumber[Range[0,20]]^2] (* Harvey P. Dale, May 01 2011 *)

a(n) = Sum_{k=0..n} ((2k)!/(k!)^2/(k+1))^2. - Alexander Adamchuk, Feb 16 2008
Sum_{i=1..n} [c(i)]^2 = Sum_{i=1..n} [C(2*i-2, i-1)/i]^2 = (1/(n-1)!)^2 * [ n^C(2*n-4, 1) + {2*C(n-1, 2)}*n^(2*n-5) + {C(n-2, 0) + 4*C(n-2, 1) + 13*C(n-2, 2) + 22*C(n-2, 3) + 12*C(n-2, 4)}*n^C(2*n-6, 1) + {12*C(n-3, 1) + 152*C(n-3, 2) + 458*C(n-3, 3) + 640*C(n-3, 4) + 440*C(n-3, 5) + 120*C(n-3, 6)}*n^(2*n-7) + {40*C(n-4, 0) + 313*C(n-4, 1) + 2332*C(n-4, 2) + 9536*C(n-4, 3) + 21409*C(n-4, 4) + 28068*C(n-4, 5) + 21700*C(n-4, 6) + 9240*C(n-4, 7) + 1680*C(n-4, 8) + ... + C(n-3, 0)*((n-1)!)^2 ].
Recurrence: (n+1)^2*a(n) = (17*n^2 - 14*n + 5)*a(n-1) - 4*(2*n - 1)^2*a(n-2). - Vaclav Kotesovec, Jul 01 2016
a(n) ~ 2^(4*n+4) /(15*Pi*n^3). - Vaclav Kotesovec, Jul 01 2016

A100066 Expansion of x/((1-x)sqrt(1-4x^2)).

Original entry on oeis.org

0, 1, 1, 3, 3, 9, 9, 29, 29, 99, 99, 351, 351, 1275, 1275, 4707, 4707, 17577, 17577, 66197, 66197, 250953, 250953, 956385, 956385, 3660541, 3660541, 14061141, 14061141, 54177741, 54177741, 209295261, 209295261, 810375651, 810375651

Offset: 0

Paul Barry, Nov 02 2004

An inverse Chebyshev transform of x/(1-x+x^2), 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.

Hankel transform of a(n+1) is A120582. The Hankel transform of a(n) is (-1)*[x^n] x/(1+2*x-4*x^2). - Paul Barry, Mar 29 2010

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi).

Cf. A000108, A006134, A120582.

R:=PowerSeriesRing(Rationals(), 40);
[0] cat Coefficients(R!( x/((1-x)*Sqrt(1-4*x^2)) )); // G. C. Greubel, Mar 17 2025

a:=n->sum(binomial(2*j,j), j=0..n): seq(a(n/2), n=-1..40); # Zerinvary Lajos, Apr 30 2007

CoefficientList[Series[x/((1-x)*Sqrt[1-4*x^2]), {x, 0, 40}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

my(x='x+O('x^41)); concat(0, Vec(x/((1-x)*sqrt(1-4*x^2)))) \\ G. C. Greubel, Jan 30 2017; Mar 17 2025

def b(n): return sum(binomial(2*k,k) for k in range(n+1))
def A100066(n): return b((n-1)//2)
print([A100066(n) for n in range(41)]) # G. C. Greubel, Mar 17 2025

a(n) = Sum_{k=0..n} if(mod(n-k, 2)=0, binomial(n, (n-k)/2) * 2*sin(Pi*k/3) / sqrt(3)).
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1+(-1)^(n-k))*sin(Pi*k/3)/sqrt(3).
a(n) = Sum_{k=0..n} binomial(n-1, (n-1)/2)*(1-(-1)^k)/2.
a(n+1) = Sum_{k=0..floor(n/2)} binomial(2k, k) = Sum{k=0..n} binomial(k, k/2)*(1+(-1)^k)/2.
a(2n-1) = a(2n) = A006134(n-1) = Sum_{k=0..n}( (2*k)!/(k!)^2 ) for n > 0. - Alexander Adamchuk, Feb 23 2007
From Paul Barry, Mar 29 2010: (Start)
G.f.: x*(1+x)/((1-x^2)*sqrt(1-4*x^2)) = x/((1-x)*sqrt(1-4*x^2)).
E.g.f.: Integral_{t=0..x} exp(x-t)*Bessel_I(0,2t). (End)
D-finite with recurrence: (n-1)*a(n) - (n-1)*a(n-1) - 4*(n-2)*a(n-2) + 4*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 24 2012
a(n) ~ (3-(-1)^n) * 2^(n+1/2) / (6*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 12 2014

A167713 a(n) = 16^n * Sum_{k=0..n} binomial(2*k, k) / 16^k.

Original entry on oeis.org

1, 18, 294, 4724, 75654, 1210716, 19372380, 309961512, 4959397062, 79350401612, 1269606610548, 20313706474200, 325019306291356, 5200308911062296, 83204942617113336, 1331279082028930896, 21300465313063974726, 340807445011357201836, 5452919120190790364676

Offset: 0

Alexander Adamchuk, Nov 10 2009

nonn

p^2 divides a((p-3)/2) for prime p of the form p = 6k + 1 (A002476).

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), a(n) = A167713 (B=16).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Eric Weisstein's World of Mathematics, Central Binomial Coefficient.
Eric Weisstein's World of Mathematics, Binomial Sums.

Cf. A000984, A002476, A066796, A006134, A082590, A132310, A002457, A144635.

A167713 := proc(n) coeftayl( 1/(1-16*x)/sqrt(1-4*x),x=0,n) ; end proc: seq(A167713(n),n=0..40) ; # R. J. Mathar, Nov 13 2009

Table[ 16^n * Sum[ (2k)!/(k!)^2 / 16^k, {k,0,n} ], {n,0,20} ]
CoefficientList[Series[1 / ((1 - 16 x) Sqrt[1 - 4 x]), {x, 0, 20}], x] (* Vincenzo Librandi, May 27 2013 *)

a(n) = 16^n * Sum_{k=0..n} ((2k)!/(k!)^2) / 16^k.
a(n) = 16^n * Sum_{k=0..n} binomial(2k,k) / 16^k.
G.f.: 1/((1-16*x)*sqrt(1-4*x)). - R. J. Mathar, Nov 13 2009
From Vaclav Kotesovec, Oct 20 2012: (Start)
Recurrence: n*a(n) = 2*(10*n-1)*a(n-1) - 32*(2*n-1)*a(n-2).
a(n) ~ 2^(4*n+1)/sqrt(3). (End)

Extended by R. J. Mathar, Nov 13 2009

A361654 Triangle read by rows where T(n,k) is the number of nonempty subsets of {1,...,2n-1} with median n and minimum k.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 15, 9, 4, 1, 50, 29, 14, 5, 1, 176, 99, 49, 20, 6, 1, 638, 351, 175, 76, 27, 7, 1, 2354, 1275, 637, 286, 111, 35, 8, 1, 8789, 4707, 2353, 1078, 441, 155, 44, 9, 1, 33099, 17577, 8788, 4081, 1728, 650, 209, 54, 10, 1

Offset: 1

Gus Wiseman, Mar 23 2023

The median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length).

			Triangle begins:
     1
     2     1
     5     3     1
    15     9     4     1
    50    29    14     5     1
   176    99    49    20     6     1
   638   351   175    76    27     7     1
  2354  1275   637   286   111    35     8     1
  8789  4707  2353  1078   441   155    44     9     1
Row n = 4 counts the following subsets:
  {1,7}            {2,6}        {3,5}    {4}
  {1,4,5}          {2,4,5}      {3,4,5}
  {1,4,6}          {2,4,6}      {3,4,6}
  {1,4,7}          {2,4,7}      {3,4,7}
  {1,2,6,7}        {2,3,5,6}
  {1,3,5,6}        {2,3,5,7}
  {1,3,5,7}        {2,3,4,5,6}
  {1,2,4,5,6}      {2,3,4,5,7}
  {1,2,4,5,7}      {2,3,4,6,7}
  {1,2,4,6,7}
  {1,3,4,5,6}
  {1,3,4,5,7}
  {1,3,4,6,7}
  {1,2,3,5,6,7}
  {1,2,3,4,5,6,7}

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)
Paul Barry, A Riordan array family for some integrable lattice models, arXiv:2409.09547 [math.CO], 2024. See p. 7.
Paul Barry, Extensions of Riordan Arrays and Their Applications, Mathematics (2025) Vol. 13, No. 2, 242. See p. 12.

Row sums appear to be A006134.
Column k = 1 appears to be A024718.
Column k = 2 appears to be A006134.
Column k = 3 appears to be A079309.
A000975 counts subsets with integer median, mean A327475.
A007318 counts subsets by length.
A231147 counts subsets by median, full steps A013580, by mean A327481.
A359893 and A359901 count partitions by median.
A360005(n)/2 gives the median statistic.
Cf. A006134, A057552, A067659, A325347, A359907, A361849.

Table[Length[Select[Subsets[Range[2n-1]],Min@@#==k&&Median[#]==n&]],{n,6},{k,n}]

T(n,k) = sum(j=0, n-k, binomial(2*j+k-2, j)) \\ Andrew Howroyd, Apr 09 2023

T(n,k) = 1 + Sum_{j=1..n-k} binomial(2*j+k-2, j). - Andrew Howroyd, Apr 09 2023

cp's OEIS Frontend

A054108 a(n) = (-1)^(n+1)*Sum_{k=0..n+1}(-1)^k*binomial(2*k,k).

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A070925 Number of subsets of A = {1,2,...,n} that have the same center of gravity as A, i.e., (n+1)/2.

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A079727 a(n) = 1 + C(2,1)^3 + C(4,2)^3 + ... + C(2n,n)^3.

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Magma

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Maxima

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A144635 a(n) = 5^n*Sum_{ k=0..n } binomial(2*k,k)/5^k.

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A362046 Number of nonempty subsets of {1..n} with mean n/2.

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A371965 a(n) is the sum of all peaks in the set of Catalan words of length n.

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Maple

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A094639 Partial sums of squares of Catalan numbers (A000108).

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A100066 Expansion of x/((1-x)*sqrt(1-4*x^2)).

A054108 a(n) = (-1)^(n+1)Sum_{k=0..n+1}(-1)^kbinomial(2*k,k).

A144635 a(n) = 5^nSum_{ k=0..n } binomial(2k,k)/5^k.

A100066 Expansion of x/((1-x)sqrt(1-4x^2)).