A094527 -id:A094527 - cp's OEIS frontend

A243585 Expansion of xlog'(C(C(x)-1)-1), C(x) = (1-sqrt(1-4x))/(2*x).

1, 4, 20, 106, 580, 3244, 18446, 106250, 618340, 3628600, 21438820, 127377980, 760346350, 4556473276, 27396081950, 165189725326, 998492094244, 6048338850560, 36706629690824, 223139239595840, 1358475322091620

Offset: 0

Vladimir Kruchinin, Jun 07 2014

nonn

G.-S. Cheon, H. Kim, L. W. Shapiro, Mutation effects in ordered trees, arXiv preprint arXiv:1410.1249 [math.CO], 2014 (see page 6).

Cf. A000108, A000984, A094527, A153294.

CoefficientList[Series[1/(Sqrt[(1-4*x)*(2*Sqrt[1-4*x]+5*x-2)/x]), {x, 0, 20}], x] (* Vaclav Kotesovec, Jun 08 2014 *)
A243585[n_] := Binomial[2 n, n] Hypergeometric2F1[1/2, -n, n + 1, -4];
Table[A243585[n], {n, 0, 20}] (* Peter Luschny, Aug 04 2019 *)

a(n):=sum(binomial(2*k,k)*binomial(2*n,n-k),k,0,n);

a(n) = Sum_{k=0..n} binomial(2*k,k)*binomial(2*n,n-k).
a(n) ~ 5^(2*n+1/2) / (4^n * sqrt(3*Pi*n)). - Vaclav Kotesovec, Jun 08 2014
First column of A094527^2. 1 + x*exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 4*x^2 + 18*x^3 + 86*x^4 + ... is the o.g.f. for A153294. - Peter Bala, Jul 21 2015
Conjecture D-finite with recurrence: 2*n*(2*n-1)*(3*n-5)*a(n) +(-123*n^3+328*n^2-249*n+60)*a(n-1) +50*(n-1)*(2*n-3)*(3*n-2)*a(n-2)=0. - R. J. Mathar, Jun 14 2016
a(n) = binomial(2*n, n)*hypergeom([1/2, -n], [n + 1], -4). - Peter Luschny, Aug 04 2019

A226197 Numbers of vectors with 2*n integers such that each element is either 1 or -1, and their sum > n.

Original entry on oeis.org

1, 1, 7, 9, 56, 79, 470, 697, 4048, 6196, 35443, 55455, 313912, 499178, 2804012, 4514873, 25211936, 40999516, 227881004, 373585604, 2068564064, 3414035527, 18844224462, 31278197839, 172186125456, 287191809724, 1577401391626, 2642070371194, 14483100716176, 24347999094724

Offset: 1

Alex Ratushnyak, May 31 2013

nonn

			With n=3 there are 7 vectors with sum bigger than 3:
{1, 1, 1, 1, 1, 1}
{-1, 1, 1, 1, 1, 1}
{1, -1, 1, 1, 1, 1}
{1, 1, -1, 1, 1, 1}
{1, 1, 1, -1, 1, 1}
{1, 1, 1, 1, -1, 1}
{1, 1, 1, 1, 1, -1}
So a(3) = 7.

#include 
long long count, n;
void addOne(long long sum, long long added) {
  if (added==n*2) {
    if (sum>n) ++count;
    return;
  }
  ++added;
  addOne(sum+1, added);
  addOne(sum-1, added);
}
int main() {
  for (n=1; n<99; n++) {
    count = 0;
    addOne(0, 0);
    printf("%llu, ", count);
  }
  return 0;
}

A226197 := proc(n)
    add( A094527(n,k),k=1+floor(n/2)..n) ;
end proc: # R. J. Mathar, Jun 04 2013

a[n_] := Sum[(2*n)!/((n-k)!*(n+k)!), {k, 1 + Floor[n/2], n}]; Array[a,30] (* Giovanni Resta, May 31 2013 *)

a(n) = sum_{k = 1+floor(n/2)...n} binomial(2n,n-k), sum of the n/2 rightmost elements of row n of A094527. - Giovanni Resta, May 31 2013

A269956 Triangle read by rows, T(n,k) = binomial(3*n,n+k) for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 3, 3, 15, 20, 15, 84, 126, 126, 84, 495, 792, 924, 792, 495, 3003, 5005, 6435, 6435, 5005, 3003, 18564, 31824, 43758, 48620, 43758, 31824, 18564, 116280, 203490, 293930, 352716, 352716, 293930, 203490, 116280, 735471, 1307504, 1961256, 2496144, 2704156, 2496144, 1961256, 1307504, 735471

Offset: 0

Peter Luschny, Mar 28 2016

			Triangle starts:
1,
3,    3,
15,   20,   15,
84,   126,  126,  84,
495,  792,  924,  792,  495,
3003, 5005, 6435, 6435, 5005, 3003

Felix Fröhlich, Table of n, a(n) for n = 0..10000

Cf. binomial(m*n,n+k): A000007 (m=0), A010054 (m=1), A094527 (m=2).

Cf. A005809.

/* As triangle */ [[Binomial(3*n, n+k): k in [0..n]]: n in [0.. 7]]; // Vincenzo Librandi, Mar 28 2016

for n from 0 to 6 do seq(binomial(3*n,n+k), k=0..n) od;

Table[Binomial[3 n, n + k], {n, 0, 10}, {k, 0, n}]//Flatten (* Vincenzo Librandi, Mar 28 2016 *)

t(n, k) = binomial(3*n, n+k)
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* The following function call prints the upper six rows of the triangle */
trianglerows(6) \\ Felix Fröhlich, Mar 28 2016

T(n,0) = T(n,n) = A005809(n).

A380113 Triangle read by rows: The inverse matrix of the central factorials A370707, row n normalized by (-1)^(n - k)*A370707(n, n).

Original entry on oeis.org

1, 1, 1, 3, 4, 1, 10, 15, 6, 1, 35, 56, 28, 8, 1, 126, 210, 120, 45, 10, 1, 462, 792, 495, 220, 66, 12, 1, 1716, 3003, 2002, 1001, 364, 91, 14, 1, 6435, 11440, 8008, 4368, 1820, 560, 120, 16, 1, 24310, 43758, 31824, 18564, 8568, 3060, 816, 153, 18, 1

Offset: 0

Peter Luschny, Jan 12 2025

The inverse matrix of A370707 is a rational matrix and the normalization serves to make it a matrix over the integers. Note that the normalization factor A370707(n, n) = FallingFactorial(n, n) * RisingFactorial(n, n) extends A002674 to n = 0.

			Triangle starts:
  [0] [    1]
  [1] [    1,     1]
  [2] [    3,     4,     1]
  [3] [   10,    15,     6,     1]
  [4] [   35,    56,    28,     8,    1]
  [5] [  126,   210,   120,    45,   10,    1]
  [6] [  462,   792,   495,   220,   66,   12,   1]
  [7] [ 1716,  3003,  2002,  1001,  364,   91,  14,   1]
  [8] [ 6435, 11440,  8008,  4368, 1820,  560, 120,  16,  1]
  [9] [24310, 43758, 31824, 18564, 8568, 3060, 816, 153, 18, 1]
.
Row 3 of the matrix inverse of the central factorials is [-1/36, 1/24, -1/60, 1/360]. Normalized with (-1)^(n-k)*360 gives row 3 of T.

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).

Variant: A094527.

Cf. A370707, A002674, A008311, A088218 and A110556 (column 0), A081294 (row sums), A000007 (alternating row sums), A005810 (central terms).

T := (n, k) -> if n = k then 1 elif k = 0 then binomial(2*n, n - k)/2 else binomial(2*n, n - k) fi: seq(seq(T(n, k), k = 0..n), n = 0..9);

A380113[n_, k_] := Binomial[2*n, n - k]/(Boole[k == 0 && n > 0] + 1);
Table[A380113[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Jan 13 2025 *)

def Trow(n):
    def cf(n, k): return falling_factorial(n, k)*rising_factorial(n, k)
    def w(n): return factorial(n)*rising_factorial(n, n)
    m = matrix(QQ, n + 1, lambda x, y: cf(x, y)).inverse()
    return [(-1)^(n-k)*w(n)*m[n, k] for k in range(n+1)]
for n in range(10): print(Trow(n))

T(n, k) = (-1)^(n - k) * ff(n, n) * rf(n, n) * M^(-1)(ff(n, k) * rf(n, k)) where ff denotes the falling factorial, rf the rising factorial and M^(-1)(t(n, k)) the matrix inverse to the matrix with entries t(n, k).
T(n, k) = binomial(2*n, n - k) for 0 < k < n. T(n, n) = 1; T(n, 0) = (-1)^n*binomial(-n, n).
Sum_{k=0..n} T(n, k)*cos(k*x) = 2^(n-1)*(cos(x)+1)^n. (After Philippe Deléham in A008311).

A386825 Triangle read by rows: T(n,k) = 3^(n-k)C(2n,n-k).

Original entry on oeis.org

1, 6, 1, 54, 12, 1, 540, 135, 18, 1, 5670, 1512, 252, 24, 1, 61236, 17010, 3240, 405, 30, 1, 673596, 192456, 40095, 5940, 594, 36, 1, 7505784, 2189187, 486486, 81081, 9828, 819, 42, 1, 84440070, 25019280, 5837832, 1061424, 147420, 15120, 1080, 48, 1, 956987460, 287096238

Offset: 0

Enrique Navarrete, Aug 04 2025

Row sums are A386826.

			Triangle begins:
        1;
        6,       1;
       54,      12,      1;
      540,     135,     18,     1;
     5670,    1512,    252,    24,    1;
    61236,   17010,   3240,   405,   30,   1;
   673596,  192456,  40095,  5940,  594,  36,  1;
  7505784, 2189187, 486486, 81081, 9828, 819, 42, 1;
  ...

Cf. A094527, A128417, A386826.

Flatten[Table[3^(n-k) Binomial[2n, n-k], {n, 0, 9}, {k, 0, n}]]

T(n,k) = 3^(n-k)*A094527(n,k).

A111505 Right half of Pascal's triangle (A007318) with zeros.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 0, 3, 1, 0, 0, 6, 4, 1, 0, 0, 0, 10, 5, 1, 0, 0, 0, 20, 15, 6, 1, 0, 0, 0, 0, 35, 21, 7, 1, 0, 0, 0, 0, 70, 56, 28, 8, 1, 0, 0, 0, 0, 0, 126, 84, 36, 9, 1, 0, 0, 0, 0, 0, 252, 210, 120, 45, 10, 1, 0, 0, 0, 0, 0, 0, 462, 330, 165

Offset: 0

Philippe Deléham, Nov 16 2005

A034869 is the version without zeros.

			Triangle begins:
1;
0, 1;
0, 2, 1;
0, 0, 3, 1;
0, 0, 6, 4, 1;
0, 0, 0, 10, 5, 1;
0, 0, 0, 20, 15, 6, 1;
0, 0, 0, 0, 35, 21, 7, 1;
0, 0, 0, 0, 70, 56, 28, 8, 1;
0, 0, 0, 0, 0, 126, 84, 36, 9, 1;
0, 0, 0, 0, 0, 252, 210, 120, 45, 10, 1;
0, 0, 0, 0, 0, 0, 462, 330, 165, 55, 11, 1;
0, 0, 0, 0, 0, 0, 924, 792, 495, 220, 66, 12, 1;
0, 0, 0, 0, 0, 0, 0, 1716, 1287, 715, 286, 78, 13, 1;
0, 0, 0, 0, 0, 0, 0, 3432, 3003, 2002, 1001, 364, 91, 14, 1;

Cf. A007318, A034869, A094527, A111418.

Sum_{n, n>=k} T(n, k) = A001700(k).
Sum_{k =0..2*n} T(2*n, k) = A032443(n).
Sum_{k=0..2*n+1} T(2*n+1, k) = 4^n = A000302(n).
Sum_{k=0..2*n} T(2*n, k)^2 = A036910(n).
Sum_{k=0..2*n+1} T(2*n+1, k)^2 = C(4*n+1, 2*n) = A002458(n) . Paul D. Hanna

cp's OEIS Frontend

A243585 Expansion of x*log'(C(C(x)-1)-1), C(x) = (1-sqrt(1-4*x))/(2*x).

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A226197 Numbers of vectors with 2*n integers such that each element is either 1 or -1, and their sum > n.

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A269956 Triangle read by rows, T(n,k) = binomial(3*n,n+k) for n>=0 and 0<=k<=n.

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A380113 Triangle read by rows: The inverse matrix of the central factorials A370707, row n normalized by (-1)^(n - k)*A370707(n, n).

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A386825 Triangle read by rows: T(n,k) = 3^(n-k)*C(2*n,n-k).

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A111505 Right half of Pascal's triangle (A007318) with zeros.

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A243585 Expansion of xlog'(C(C(x)-1)-1), C(x) = (1-sqrt(1-4x))/(2*x).

A386825 Triangle read by rows: T(n,k) = 3^(n-k)C(2n,n-k).