A002457 -id:A002457 - cp's OEIS frontend

A167870 a(n) = 16^n * Sum_{k=0..n} binomial(2*k,k)^3 / 16^k.

1, 24, 600, 17600, 624600, 25996608, 1204834752, 59701593600, 3086972400600, 164324590337600, 8935798773354816, 494019944564058624, 27678350810730366400, 1567912312203901862400, 89647910047704725798400

Offset: 0

Alexander Adamchuk, Nov 14 2009

nonn

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), A167713 (B=16).

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

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

Cf. A079727, A167867, A167868, A167869, A167870, A167872.

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

Table[16^n Sum[Binomial[2k,k]^3/16^k,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Jan 21 2012 *)

a(n) = 16^n * Sum_{k=0..n} binomial(2*k,k)^3 / 16^k.
Recurrence: n^3*a(n) = 8*(10*n^3 - 12*n^2 + 6*n - 1)*a(n-1) - 128*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+2)/(3*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

More terms from Sean A. Irvine, Apr 27 2010

A167871 a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.

Original entry on oeis.org

1, 72, 4824, 316736, 20614104, 1335305664, 86248451520, 5560325134848, 357992555533272, 23026456586057408, 1479999826835627328, 95071036081670530560, 6104320340924619384256, 391801560518407856592384

Offset: 0

Alexander Adamchuk, Nov 14 2009

nonn

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), A167713 (B=16).
The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).
p^2 divides all a(n) from n = (p-1)/2 to n = p-1 for prime p of the form p = 4k+3, p = {3,7,11,19,23,31,43,47,59,...} = A002145.
p^2 divides all a(n) from n = (2p-1 - (p-1)/2) to n = 2p-1 for prime p of the form p = 4k+3.
p^2 divides all a(n) from n = (3p-1 - (p-1)/2) to n = 3p-1 for prime p of the form p = 4k+3.
p^2 divides all a(n) from n = (p^2-1)/2 to n = p^2-1 for prime p of the form p = 4k+3.

W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd ed., Wiley, 1968, p. 361.

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

Cf. A079727, A167867, A167868, A167869, A167870, A167872.

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859, A002145.

Table[64^n Sum[Binomial[2k,k]^3/64^k,{k,0,n}],{n,0,20}] (* Vincenzo Librandi, Mar 26 2012 *)

a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.
Recurrence: n^3*a(n) = 8*(4*n-1)*(4*n^2 - 2*n + 1)*a(n-1) - 512*(2*n-1)^3 *a(n-2). - Vaclav Kotesovec, Aug 14 2013
a(n) ~ 64^n*(Pi/GAMMA(3/4)^4 - 2/(Pi^(3/2)*sqrt(n))). - Vaclav Kotesovec, Aug 14 2013

More terms from Sean A. Irvine, Apr 25 2010

A189765 Triangle in which row n has the n(n+1)/2 elements of the lower triangular part of the inverse of the n-th order Hilbert matrix.

Original entry on oeis.org

1, 4, -6, 12, 9, -36, 192, 30, -180, 180, 16, -120, 1200, 240, -2700, 6480, -140, 1680, -4200, 2800, 25, -300, 4800, 1050, -18900, 79380, -1400, 26880, -117600, 179200, 630, -12600, 56700, -88200, 44100, 36, -630, 14700, 3360, -88200, 564480, -7560, 211680

Offset: 1

T. D. Noe, May 02 2011

The n-th order Hilbert matrix has elements h(i,j) = 1/(i+j-1) for 1 <= i,j <=n. Only the lower triangular matrix is shown because the Hilbert matrix and its inverse are symmetric. The n-th row begins with n^2 and ends with A000515(n+1).
The sums of select rows of the inverse matrix are sequences A002457, A002736, A002738, A007531, and A054559.
The largest magnitude in the matrix is A210356(n). - T. D. Noe, Mar 28 2012
The sum of the elements of the n-th matrix is n^2. - T. D. Noe, Apr 02 2012

			Row 3 is 9, -36, 192, 30, -180, 180 which corresponds to the inverse
  9  -36   30
-36  192 -180
30 -180  180

T. D. Noe, Rows n = 1..25, flattened
Eric W. Weisstein, MathWorld: Hilbert Matrix

Cf. A002457, A002736, A002738, A005249 (determinant), A007531, A054559, A189766 (trace).

lowerTri[m_List] := Module[{n = Length[m]}, Flatten[Table[Take[m[[i]], i], {i, n}]]]; Flatten[Table[lowerTri[Inverse[HilbertMatrix[n]]], {n, 6}]]

a(n,i,j) = (-1)^(i+j) (i+j-1) binomial(n+i-1, n-j) binomial(n+j-1, n-i) binomial(i+j-2, i-1)^2 is the (i,j) element of the inverse of the n-th Hilbert matrix.

A331514 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/(1 - 2kx + ((k-2)*x)^2)^(3/2).

Original entry on oeis.org

1, 1, 0, 1, 3, -6, 1, 6, 6, 0, 1, 9, 30, 10, 30, 1, 12, 66, 140, 15, 0, 1, 15, 114, 450, 630, 21, -140, 1, 18, 174, 1000, 2955, 2772, 28, 0, 1, 21, 246, 1850, 8430, 18963, 12012, 36, 630, 1, 24, 330, 3060, 18855, 69384, 119812, 51480, 45, 0

Offset: 0

Seiichi Manyama, Jan 19 2020

			Square array begins:
    1,  1,    1,     1,     1,      1, ...
    0,  3,    6,     9,    12,     15, ...
   -6,  6,   30,    66,   114,    174, ...
    0, 10,  140,   450,  1000,   1850, ...
   30, 15,  630,  2955,  8430,  18855, ...
    0, 21, 2772, 18963, 69384, 187425, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=1..5 give A000217(n+1), A002457, A002695(n+1), A331515, A331516.

Cf. A307883, A331511.

T[n_, k_] = 1/2 * Sum[If[k == 2 && n == j - 1, 1, (k - 2)^(n + 1 - j)] * j * Binomial[n + 1, j] * Binomial[n + 1 + j, j], {j, 1, n + 1}]; Table[Table[T[n, k - n], {n, 0, k}], {k, 0, 9}] //Flatten (* Amiram Eldar, Jan 20 2020 *)

T(n,k) = (1/2)*sum(j=1,n+1,(k-2)^(n+1-j)*j*binomial(n+1,j)*binomial(n+1+j,j));
matrix(7, 7, n, k, T(n-1, k-1)) \\ Michel Marcus, Jan 20 2020

T(n,k) = (1/2) * Sum_{j=1..n+1} (k-2)^(n+1-j) * j * binomial(n+1,j) * binomial(n+1+j,j).
n * T(n,k) = k * (2*n+1) * T(n-1,k) - (k-2)^2 * (n+1) * T(n-2,k) for n > 1.
T(n,k) = ((n+2)/2) * Sum_{j=0..n} (k-1)^j * binomial(n+1,j) * binomial(n+1,j+1).
T(n,k) = Sum_{j=0..n} (k/2)^j * (-(k-2)^2/(2*k))^(n-j) * (2*j+1) * binomial(2*j,j) * binomial(j,n-j) for k > 0. - Seiichi Manyama, Aug 20 2025

A336828 a(n) = Sum_{k=0..n} binomial(n,k)^2 * k^n.

Original entry on oeis.org

1, 1, 8, 108, 2144, 56250, 1836792, 71799504, 3269445888, 169974711630, 9934458411800, 644825382429096, 46022332032100800, 3582265183110626740, 302002255041807372080, 27413749834141448520000, 2665789990569658618398720, 276477318687585566522176470

Offset: 0

Ilya Gutkovskiy, Aug 05 2020

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..343

Cf. A000984, A002457, A037966, A037972, A072034, A074334, A187021, A329444, A329913, A336214, A341815.

Join[{1}, Table[Sum[Binomial[n, k]^2 k^n, {k, 0, n}], {n, 1, 17}]]

a(n) = sum(k=0, n, binomial(n, k)^2*k^n); \\ Michel Marcus, Aug 05 2020

a(n) ~ c * d^n * (n-1)!, where d = (1 + 2*LambertW(exp(-1/2)/2)) / (4*LambertW(exp(-1/2)/2)^2) = 6.476217542109791521947605963458797355564... and c = 0.21617818094152997942246965143216887599763501682724844713834495... - Vaclav Kotesovec, Feb 20 2021

A337369 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1-2(k+4)x+((k-4)x)^2) (1+(k-4)x+sqrt(1-2(k+4)x+((k-4)x)^2)) )).

Original entry on oeis.org

1, 1, 6, 1, 7, 30, 1, 8, 51, 140, 1, 9, 74, 393, 630, 1, 10, 99, 736, 3139, 2772, 1, 11, 126, 1175, 7606, 25653, 12012, 1, 12, 155, 1716, 14499, 80464, 212941, 51480, 1, 13, 186, 2365, 24310, 183195, 864772, 1787607, 218790, 1, 14, 219, 3128, 37555, 352716, 2351805, 9400192, 15134931, 923780

Offset: 0

Seiichi Manyama, Aug 25 2020

			Square array begins:
     1,     1,     1,      1,      1,      1, ...
     6,     7,     8,      9,     10,     11, ...
    30,    51,    74,     99,    126,    155, ...
   140,   393,   736,   1175,   1716,   2365, ...
   630,  3139,  7606,  14499,  24310,  37555, ...
  2772, 25653, 80464, 183195, 352716, 610897, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..5 give A002457, A273055, A337370, A245927, A002458, A243947.
Main diagonal gives A337387.
Cf. A337389, A337464.

T[n_, k_] := Sum[If[k == 0, Boole[n == j], k^(n - j)] * Binomial[2*j, j] * Binomial[2*n + 1, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 25 2020 *)

{T(n, k) = sum(j=0, n, k^(n-j)*binomial(2*j, j)*binomial(2*n+1, 2*j))}

T(n,k) = Sum_{j=0..n} k^(n-j) * binomial(2*j,j) * binomial(2*n+1,2*j).
T(0,k) = 1, T(1,k) = k+6 and n * (2*n+1) * (4*n-3) * T(n,k) = (4*n-1) * (4*(k+4)*n^2-2*(k+4)*n-k-2) * T(n-1,k) - (k-4)^2 * (n-1) * (2*n-1) * (4*n+1) * T(n-2,k) for n > 1. - Seiichi Manyama, Aug 29 2020
For fixed k > 0, T(n,k) ~ (2 + sqrt(k))^(2*n + 3/2) / sqrt(8*k*Pi*n). - Vaclav Kotesovec, Aug 31 2020

A342982 Triangle read by rows: T(n,k) is the number of tree-rooted planar maps with n edges and k+1 faces, n >= 0, k = 0..n.

Original entry on oeis.org

1, 1, 1, 2, 6, 2, 5, 30, 30, 5, 14, 140, 280, 140, 14, 42, 630, 2100, 2100, 630, 42, 132, 2772, 13860, 23100, 13860, 2772, 132, 429, 12012, 84084, 210210, 210210, 84084, 12012, 429, 1430, 51480, 480480, 1681680, 2522520, 1681680, 480480, 51480, 1430

Offset: 0

Andrew Howroyd, Apr 03 2021

The number of vertices is n + 1 - k.

A tree-rooted planar map is a planar map with a distinguished spanning tree.

			Triangle begins:
    1;
    1,     1;
    2,     6,     2;
    5,    30,    30,      5;
   14,   140,   280,    140,     14;
   42,   630,  2100,   2100,    630,    42;
  132,  2772, 13860,  23100,  13860,  2772,   132;
  429, 12012, 84084, 210210, 210210, 84084, 12012, 429;
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)
R. C. Mullin, On the Enumeration of Tree-Rooted Maps, Canadian Journal of Mathematics, Volume 19, 1967, pp. 174-183.

Columns k=0..2 are A000108, A002457, 2*A002803.
Row sums are A005568.
Central coefficients are A342983.
Cf. A001263, A215288, A269920, A342984.

Table[(2 n)!/(k!*(k + 1)!*(n - k)!*(n - k + 1)!), {n, 0, 8}, {k, 0, n}] // Flatten (* Michael De Vlieger, Apr 06 2021 *)

T(n,k) = {(2*n)!/(k!*(k+1)!*(n-k)!*(n-k+1)!)}
{ for(n=0, 10, print(vector(n+1, k, T(n,k-1)))) }

T(n,k) = (2*n)!/(k!*(k+1)!*(n-k)!*(n-k+1)!).
T(n,n-k) = T(n,k).
T(n, floor(n/2)) = A215288(n).
T(n,k) = A000108(n) * A001263(n+1,k+1). - Werner Schulte, Apr 04 2021

A368753 Irregular triangle read by rows: T(n,k) is the defect of the k-th balanced string of left/right parentheses of length 2*n, where strings within a row are in reverse lexicographical order.

Original entry on oeis.org

1, 0, 2, 2, 1, 1, 0, 0, 3, 3, 3, 2, 3, 3, 2, 2, 1, 1, 2, 2, 1, 1, 0, 0, 1, 0, 0, 0, 4, 4, 4, 4, 3, 4, 4, 4, 3, 4, 4, 3, 3, 2, 2, 4, 4, 4, 3, 4, 4, 3, 3, 2, 2, 3, 3, 2, 2, 1, 1, 2, 1, 1, 1, 3, 3, 3, 2, 3, 3, 2, 2, 1, 1, 2, 2, 1, 1, 0, 0, 1, 0, 0, 0, 2, 2, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0

Offset: 1

Paolo Xausa, Jan 05 2024

See A368750 for the definition of balanced strings and atoms/co-atoms.
The defect is half the length of co-atoms or, equivalently, the number of indices where the i-th right parenthesis precedes the i-th left parenthesis (see Knuth, 2011).
Knuth reports a result by MacMahon (1909) and Chung and Feller (1949): exactly A000108(n) balanced strings of length 2*n have defect d, for 0 <= d <= n.

			Triangle begins:
  [1] 1 0;
  [2] 2 2 1 1 0 0;
  [3] 3 3 3 2 3 3 2 2 1 1 2 2 1 1 0 0 1 0 0 0;
  ...
The strings corresponding to row 2, in reverse lexicographical order, are:
  "))((" (defect 2),
  ")()(" (defect 2),
  ")(()" (defect 1),
  "())(" (defect 1),
  "()()" (defect 0) and
  "(())" (defect 0).
For the string "())((())))(()(", for example, the defect is calculated as follows:
.
  atom
  |   co-atom
  |   |   atom  co-atom
  |   |   |     |     co-atom
  |   |   |     |     |
  ()  )(  (())  ))((  )(
      *         **    *
.
  defect = length of co-atoms / 2 = 8 / 2 = 4 = number of indices where the i-th right parenthesis precedes the i-th left parenthesis (marked with asterisks).

Donald E. Knuth, The Art of Computer Programming, Vol. 4A: Combinatorial Algorithms, Part 1, Addison-Wesley, 2011, Section 7.2.1.6, exercise 60, pp. 478 and 797.

Paolo Xausa, Table of n, a(n) for n = 1..17576 (rows 1..8 of the triangle, flattened).
K. L. Chung and W. Feller, On Fluctuations in Coin-Tossing, PNAS, vol. 35, no. 10, 1949, pp. 605-608.
J. L. Hodges, Galton's Rank-Order Test, Biometrika, vol. 42, no. 1/2, 1955, pp. 261-262.
P. A. MacMahon, Memoir on the Theory of the Partitions of Numbers. Part IV: On the Probability That the Successful Candidate at an Election by Ballot May Never at Any Time Have Fewer Votes Than the One Who Is Unsuccessful; on a Generalization of This Question; and on Its Connexion with Other Questions of Partition, Permutation, and Combination, Philosophical Transactions of the Royal Society of London, Series A, Containing Papers of a Mathematical or Physical Character, vol. 209, 1909, pp. 153-175.

Cf. A000108.
Cf. A000984 (row lengths), A002457 (row sums), A362030 and A368804 (binary words).
Cf. A368750 (atoms), A368751 (co-atoms), A368752 (all atoms).

strings[n_]:=Permutations[PadLeft[PadLeft[{},n,1],2n]];
defect[s_]:=Count[Position[s,1]-Position[s,0],_?Positive,{2}];
Array[Map[defect,strings[#]]&,5]

A374497 Expansion of 1/(1 - 4x - 4x^2)^(3/2).

Original entry on oeis.org

1, 6, 36, 200, 1080, 5712, 29792, 153792, 787680, 4009280, 20304768, 102405888, 514678528, 2579028480, 12890311680, 64283809792, 319954540032, 1589720712192, 7886437652480, 39069462835200, 193307835764736, 955361266917376, 4716674314223616, 23264437702656000

Offset: 0

Seiichi Manyama, Jul 09 2024

nonn

Cf. A001788, A002457, A025163.
Cf. A006139, A374511, A374513.
Cf. A071356.

a[n_]:= Sum[(2*k+1)*Binomial[2*k,k]*Binomial[k,n-k],{k,0,n}]; Array[a,24,0] (* Stefano Spezia, May 08 2025 *)

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

a(0) = 1, a(1) = 6; a(n) = (2*(2*n+1)*a(n-1) + 4*(n+1)*a(n-2))/n.
a(n) = binomial(n+2,2) * A071356(n).
a(n) = Sum_{k=0..n} (2*k+1) * binomial(2*k,k) * binomial(k,n-k). - Seiichi Manyama, Oct 19 2024
a(n) = ((n+2)/2) * Sum_{k=0..floor(n/2)} 2^(n-k) * binomial(n+1,n-2*k) * binomial(2*k+1,k). - Seiichi Manyama, Aug 20 2025

A382514 Expansion of 1/(1 - x/(1 - 4*x)^(3/2)).

Original entry on oeis.org

1, 1, 7, 43, 255, 1493, 8695, 50517, 293163, 1700335, 9859019, 57156631, 331332423, 1920621431, 11132911939, 64531189379, 374047777319, 2168115796941, 12567146992975, 72843402779669, 422224417571347, 2447350774345341, 14185640454054279, 82224565359415849

Offset: 0

Seiichi Manyama, Mar 30 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Cf. A026671, A382515.

Cf. A002457.

R := PowerSeriesRing(Rationals(), 40); f := 1/(1 - x/(1 - 4*x)^(3/2)); seq := [ Coefficient(f, n) : n in [0..30] ]; seq; // Vincenzo Librandi, Apr 09 2025

Table[Sum[4^(n-k)*Binomial[n+k/2-1,n-k],{k,0,n}],{n,0,35}] (* Vincenzo Librandi, Apr 09 2025 *)

a(n) = sum(k=0, n, 4^(n-k)*binomial(n+k/2-1, n-k));

a(n) = Sum_{k=0..n} 4^(n-k) * binomial(n+k/2-1,n-k).

D-finite with recurrence (-n+1)*a(n) +2*(8*n-13)*a(n-1) +5*(-19*n+43)*a(n-2) +2*(126*n-361)*a(n-3) +128*(-2*n+7)*a(n-4)=0. - R. J. Mathar, Mar 31 2025

cp's OEIS Frontend

A167870 a(n) = 16^n * Sum_{k=0..n} binomial(2*k,k)^3 / 16^k.

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A167871 a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.

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A189765 Triangle in which row n has the n(n+1)/2 elements of the lower triangular part of the inverse of the n-th order Hilbert matrix.

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A331514 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/(1 - 2*k*x + ((k-2)*x)^2)^(3/2).

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A336828 a(n) = Sum_{k=0..n} binomial(n,k)^2 * k^n.

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A337369 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1-2*(k+4)*x+((k-4)*x)^2) * (1+(k-4)*x+sqrt(1-2*(k+4)*x+((k-4)*x)^2)) )).

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A342982 Triangle read by rows: T(n,k) is the number of tree-rooted planar maps with n edges and k+1 faces, n >= 0, k = 0..n.

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A368753 Irregular triangle read by rows: T(n,k) is the defect of the k-th balanced string of left/right parentheses of length 2*n, where strings within a row are in reverse lexicographical order.

A331514 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/(1 - 2kx + ((k-2)*x)^2)^(3/2).

A337369 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1-2(k+4)x+((k-4)x)^2) (1+(k-4)x+sqrt(1-2(k+4)x+((k-4)x)^2)) )).

A374497 Expansion of 1/(1 - 4x - 4x^2)^(3/2).