A002674 -id:A002674 - cp's OEIS frontend

A268647 G.f.: C(x,y)^2 - S(x,y)^2 = Sum_{n>=0} x^(2n)y/[Sum_{k=0..2n+1} T(n,k)y^k], where C(x,y) = Sum_{n>=0} x^(2n) / Product_{k=1..2n} (k + y) and S(x,y) = Sum_{n>=0} x^(2n+1) / Product_{k=1..2n+1} (k + y).

0, 1, 2, 5, 4, 1, 48, 124, 120, 55, 12, 1, 2160, 6012, 6636, 3829, 1260, 238, 24, 1, 161280, 478656, 582080, 387260, 157080, 40593, 6720, 690, 40, 1, 18144000, 56772000, 74396520, 54801076, 25494150, 7927205, 1690920, 248523, 24750, 1595, 60, 1, 2874009600, 9397658880, 13075800192, 10415648880, 5357255904, 1893627736, 476011536, 86550035, 11423412, 1084083, 72072, 3185, 84, 1, 610248038400, 2071437822720, 3028563232128, 2569081620624, 1429040500160, 556365173000, 157528627256, 33179499353, 5260335080, 629597540, 56560504, 3753022, 178360, 5740, 112, 1

Offset: 0

Paul D. Hanna, Mar 01 2016

This triangle illustrates the following identity.
Given
C(x,y) = Sum_{n>=0} x^(2*n) / Product_{k=1..2*n} (k + y)
S(x,y) = Sum_{n>=0} x^(2*n+1) / Product_{k=1..2*n+1} (k + y)
then
C(x,y)^2 - S(x,y)^2 = Sum_{n>=0} x^(2*n) * y / ((n + y) * Product_{k=1..2*n} (k + y)).

			Define C(x,y) by the series:
C(x,y) = 1 + x^2/((1+y)*(2+y)) + x^4/((1+y)*(2+y)*(3+y)*(4+y)) + x^6/((1+y)*(2+y)*(3+y)*(4+y)*(5+y)*(6+y)) + x^8/((1+y)*(2+y)*(3+y)*(4+y)*(5+y)*(6+y)*(7+y)*(8+y)) +...
and define S(x,y) by the series:
S(x,y) = x/(1+y) + x^3/((1+y)*(2+y)*(3+y)) + x^5/((1+y)*(2+y)*(3+y)*(4+y)*(5+y)) + x^7/((1+y)*(2+y)*(3+y)*(4+y)*(5+y)*(6+y)*(7+y)) + x^9/((1+y)*(2+y)*(3+y)*(4+y)*(5+y)*(6+y)*(7+y)*(8+y)*(9+y)) +...
then the g.f. of this triangle begins:
C(x,y)^2 - S(x,y)^2 = 1 + x^2*y/((1+y) * (1+y)*(2+y)) + x^4*y/((2+y) * (1+y)*(2+y)*(3+y)*(4+y)) + x^6*y/((3+y) * (1+y)*(2+y)*(3+y)*(4+y)*(5+y)*(6+y)) + x^8*y/((4+y) * (1+y)*(2+y)*(3+y)*(4+y)*(5+y)*(6+y)*(7+y)*(8+y)) +...
where the rows of this triangle are formed from the coefficients in the denominators of coefficients of x^(2*n) in C(x,y)^2 - S(x,y)^2, as more clearly seen in the expansion:
C(x,y)^2 - S(x,y)^2 = y/(0 + y) + x^2 * y/(2 + 5*y + 4*y^2 + y^3) +
x^4 * y/(48 + 124*y + 120*y^2 + 55*y^3 + 12*y^4 + y^5) +
x^6 * y/(2160 + 6012*y + 6636*y^2 + 3829*y^3 + 1260*y^4 + 238*y^5 + 24*y^6 + y^7) +
x^8 * y/(161280 + 478656*y + 582080*y^2 + 387260*y^3 + 157080*y^4 + 40593*y^5 + 6720*y^6 + 690*y^7 + 40*y^8 + y^9) +...
This triangle begins:
0, 1;
2, 5, 4, 1;
48, 124, 120, 55, 12, 1;
2160, 6012, 6636, 3829, 1260, 238, 24, 1;
161280, 478656, 582080, 387260, 157080, 40593, 6720, 690, 40, 1;
18144000, 56772000, 74396520, 54801076, 25494150, 7927205, 1690920, 248523, 24750, 1595, 60, 1;
2874009600, 9397658880, 13075800192, 10415648880, 5357255904, 1893627736, 476011536, 86550035, 11423412, 1084083, 72072, 3185, 84, 1;
610248038400, 2071437822720, 3028563232128, 2569081620624, 1429040500160, 556365173000, 157528627256, 33179499353, 5260335080, 629597540, 56560504, 3753022, 178360, 5740, 112, 1; ...

Paul D. Hanna, Table of n, a(n) for n = 0..991 terms in rows 0..30 of this triangle in flattened form.

Cf. A322627 (diagonal).

/* C(x,y)^2 - S(x,y)^2 = Sum_{n>=0} x^(2*n)*y/[Sum_{k=0..2*n+1} T(n,k)*y^k] */
{T(n,k) = my(C=1,S=x); C = sum(m=0,n+1, x^(2*m)/prod(k=1,2*m, k + y) +x*O(x^(2*n)));
S = sum(m=1,n+1, x^(2*m-1)/prod(k=1,2*m-1, k + y) +x*O(x^(2*n)));
polcoeff( y/polcoeff( C^2 - S^2, 2*n, x), k, y)}
for(n=0,10, for(k=0,2*n+1, print1(T(n,k),", "));print(""))

/* (n + y)*Product_{k=1..2*n} (k + y) = Sum_{k=0..2*n+1} T(n,k)*y^k */
{T(n,k) = polcoeff((n + y)*prod(k=1,2*n, k + y), k, y)}
for(n=0,10, for(k=0,2*n+1, print1(T(n,k),", "));print(""))

G.f. of row n: (n + y) * Product_{k=1..2*n} (k + y) = Sum_{k=0..2*n+1} T(n,k)*y^k, for n>=0.

Row sums equal A002674 (with offset): A002674(n+1) = (n+1)*(2*n+1)!.

A304001 Number of permutations of [n] whose up-down signature has a nonnegative total sum.

Original entry on oeis.org

1, 1, 1, 5, 12, 93, 360, 3728, 20160, 259535, 1814400, 27820524, 239500800, 4251096402, 43589145600, 877606592736, 10461394944000, 235288904377275, 3201186852864000, 79476406782222500, 1216451004088320000, 33020655481590446318, 562000363888803840000

Offset: 0

Alois P. Heinz, May 04 2018

nonn

The up-down signature has (+1) for each ascent and (-1) for each descent.

Alois P. Heinz, Table of n, a(n) for n = 0..450

Bisections give: A002674 (even part), A179457(2n+1,n+1) (odd part).

Cf. A000246 (for nonnegative partial sums), A006551 (total sums are 0 or 1), A008292, A303287.

b:= proc(u, o, t) option remember; (n->
     `if`(t>=n, n!, `if`(t<-n, 0,
      add(b(u-j, o+j-1, t-1), j=1..u)+
      add(b(u+j-1, o-j, t+1), j=1..o))))(u+o)
    end:
a:= n-> `if`(n=0, 1, add(b(j-1, n-j, 0), j=1..n)):
seq(a(n), n=0..25);
# second Maple program:
a:= n-> `if`(irem(n, 2, 'r')=0, ceil(n!/2),
         add(combinat[eulerian1](n, j), j=0..r)):
seq(a(n), n=0..25);

Eulerian1[n_, k_] := If[k == 0, 1, If[n == 0, 0, Sum[(-1)^j (k - j + 1)^n Binomial[n + 1, j], {j, 0, k + 1}]]];
a[n_] := Module[{r, m}, {r, m} = QuotientRemainder[n, 2]; If[m == 0, Ceiling[n!/2], Sum[Eulerian1[n, j], {j, 0, r}]]];
a /@ Range[0, 25] (* Jean-François Alcover, Mar 26 2021, after 2nd Maple program *)

A066991 Square array read by descending antidiagonals of number of ways of dividing n*k labeled items into k unlabeled orders with n items in each order.

Original entry on oeis.org

1, 1, 2, 1, 12, 6, 1, 120, 360, 24, 1, 1680, 60480, 20160, 120, 1, 30240, 19958400, 79833600, 1814400, 720, 1, 665280, 10897286400, 871782912000, 217945728000, 239500800, 5040, 1, 17297280, 8892185702400, 20274183401472000

Offset: 1

Henry Bottomley, Feb 01 2002

T(p,k) = (pk)!/k! is divisible by p^k but not p^(k+1) for p prime; e.g., T(3,4) = 3^4*11*10*8*7*5*4*2*1 = 19958400.

			The array begins:
  n\k|   1        2             3                   4  ...
  --------------------------------------------------------
   1 |   1,       1,            1,                  1, ...
   2 |   2,      12,          120,               1680, ...
   3 |   6,     360,        60480,           19958400, ...
   4 |  24,   20160,     79833600,       871782912000, ...
   5 | 120, 1814400, 217945728000, 101370917007360000, ...
  ...

Paolo Xausa, Table of n, a(n) for n = 1..820 (antidiagonals 1..40 of the array, flattened).

Rows include A000012, A001813, A064350.
Columns include A000142, A002674, A065961.
Cf. A060538, A060539, A060540.

Table[((n-k+1)*k)!/k!, {n, 10}, {k, n, 1, -1}] (* Paolo Xausa, Feb 19 2024 *)

T(n,k) = (n*k)!/k!.

Edited by Paolo Xausa, Feb 19 2024

A143844 Triangle T(n,k) = k^2 read by rows.

Original entry on oeis.org

0, 0, 1, 0, 1, 4, 0, 1, 4, 9, 0, 1, 4, 9, 16, 0, 1, 4, 9, 16, 25, 0, 1, 4, 9, 16, 25, 36, 0, 1, 4, 9, 16, 25, 36, 49, 0, 1, 4, 9, 16, 25, 36, 49, 64, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121

Offset: 0

Paul Curtz, Sep 03 2008

This is triangle A133819 with an additional leading column of zeros.

There is a family of even integer-valued polynomials p_n(x) = product_{k=0..n} (x^2 - T(n,k))/ A002674(n+1). We find p_0(x) in A000290, p_1(x) in A002415, p_2(x) essentially in A040977, p_3(x) in A053347 and p_4(x) in A054334. - Paul Curtz, Jun 10 2011

Cf. A000290, A002262, A002415, A002674, A040977, A053347, A054334, A133819.

Table[Range[0,n]^2,{n,0,15}]//Flatten (* Harvey P. Dale, Sep 08 2017 *)

for(n=0,9,for(k=0,n,print1(k^2", "))) \\ Charles R Greathouse IV, Jun 10 2011

T(n,k) = (A002262(n,k))^2.

G.f.: x*y*(1 + x*y)/((1 - x)*(1 - x*y)^3). - Stefano Spezia, Feb 21 2024

Definition simplified by R. J. Mathar, Sep 07 2009

A226731 a(n) = (2n - 1)!/(2n).

Original entry on oeis.org

20, 630, 36288, 3326400, 444787200, 81729648000, 19760412672000, 6082255020441600, 2322315553259520000, 1077167364120207360000, 596585001666576384000000, 388888194657798291456000000

Offset: 3

Wesley Ivan Hurt, Jun 15 2013

For n < 3, the formula does not produce an integer.

The ratio of the product of the partition parts of 2n into exactly two parts to the sum of the partition parts of 2n into exactly two parts. For example, a(3) = 20, and 2*3 = 6 has 3 partitions into exactly two parts: (5,1), (4,2), (3,3). Forming the ratio of product to sum (of parts), we have (5*1*4*2*3*3)/(5+1+4+2+3+3) = 360/18 = 20. - Wesley Ivan Hurt, Jun 24 2013

			a(3) = (2*3 - 1)!/(2*3) = 5!/6 = 120/6 = 20.

Seiichi Manyama, Table of n, a(n) for n = 3..225
Index entries for sequences related to partitions.

Cf. A001105, A002674, A005843, A009445, A093353, A143623, A211374.

seq((2*k-1)!/(2*k),k=1..20); # Wesley Ivan Hurt, Jun 24 2013

Table[(2n-1)!/(2n),{n,3,20}] (* Harvey P. Dale, Jun 19 2013 *)

a(n) = (2*n-1)!/(2*n); \\ Michel Marcus, Nov 01 2016

a(n) = A009445(n-1)/A005843(n) = A002674(n)/A001105(n). - Wesley Ivan Hurt, Jun 24 2013
a(n) ~ sqrt(Pi)*2^(2*n-1)*n^(2*n-3/2)/exp(2*n). - Ilya Gutkovskiy, Nov 01 2016
From Amiram Eldar, Mar 10 2022: (Start)
Sum_{n>=3} 1/a(n) = e - 8/3.
Sum_{n>=3} (-1)^(n+1)/a(n) = cos(1) + sin(1) - 4/3. (End)

A370379 Number of compositions of n where there are (2*k)!/2 sorts of part k.

Original entry on oeis.org

1, 1, 13, 385, 21061, 1864921, 243833533, 44133789745, 10556951897461, 3223557261840841, 1223184443268467053, 564530822421956927905, 311384269987431969105061, 202282520358685311116600761, 152856358784713560205903602973

Offset: 0

Seiichi Manyama, Feb 17 2024

Cf. A002674, A051295, A370378.

my(N=20, x='x+O('x^N)); Vec(1/(1-sum(k=1, N, (2*k)!/2*x^k)))

G.f.: 1 / (1 - Sum_{k>=1} (2*k)!/2 * x^k).

a(0) = 1; a(n) = Sum_{k=1..n} (2*k)!/2 * a(n-k).

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).

A382527 a(n) = Sum_{j = 1..n} (-1)^(n+j) * j^(2n+4) binomial(2*n, n-j).

Original entry on oeis.org

1, 252, 52920, 12640320, 3632428800, 1264085222400, 529085049292800, 263564384219136000, 154550100069421056000, 105562401683780321280000, 83178863857362412339200000, 74951718050379657373286400000, 76628603945744083606044672000000, 88258468221509704910254374912000000

Offset: 1

Peter Bala, Mar 30 2025

Compare with the identities Sum_{j = 1..n} (-1)^(n+j) * j^n * binomial(n, n-j) = n! and Sum_{j = 1..n} (-1)^(n+j) * j^(2*n+2) * binomial(2*n, n-j) = n*(n+1)*(2*n+1)/6 * (2*n)!/2 = A000330(n) * (2*n)!/2. (Campbell, Eq. 17).

Paolo Xausa, Table of n, a(n) for n = 1..200
John Campbell, Applications of Gosper’s nonlocal derangement identity, Maple Transactions, 5, 1, Article 16724, Feb. 2025.

Cf. A002674, A060493.

seq(add((-1)^(n+j) * j^(2*n+4) * binomial(2*n, n-j), j = 1..n), n = 1..20);

A382527[n_] := n*(5*n - 1)*(2*n + 4)!/2880; Array[A382527, 15] (* Paolo Xausa, Apr 03 2025 *)

a(n) = n*(n+1)*(n+2)*(2*n+3)*(5*n-1)*(2*n+1)!/6!.
a(n) = (2*n)!/2 * (Sum_{1 <= i <= j <= n} i^2*j^2) = (2*n)!/2 * A060493(n).
a(n) = 2*n*(n+2)*(2*n+3)*(5*n-1)/((n-1)*(5*n-6)) * a(n-1) with a(1) = 1.

A383929 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n, k) (n-k)^(3*n).

Original entry on oeis.org

1, 1, 60, 16626, 12640320, 20421928750, 60233972198400, 293230314199497444, 2192804991244707840000, 23869875368184417393486678, 362747302615636095725568000000, 7442995512384107947406685870219196, 200637069747857913587015560318156800000, 6945549555749361962465324588957867814958924

Offset: 0

Vaclav Kotesovec, May 15 2025

nonn

Cf. A002674, A298851, A383916, A383930.

Join[{1}, Table[Sum[(-1)^(n-k)*Binomial[2*n, n-k]*k^(3*n), {k, 0, n}], {n, 1, 15}]]

a(n) ~ 2^(2*n + 1/2) * r^(3*n + 1) * n^(3*n) / (sqrt(3 - r^2) * exp(3*n) * (r^2 - 1)^n), where r = 1.1647414545521878292908344008181647954486720209245020743652... is the root of the equation (1 + r)/(1 - r) = -exp(3/r).

A383930 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n, k) (n-k)^(5*n).

Original entry on oeis.org

1, 1, 1020, 14152314, 1071646712640, 286802348769420190, 209974096349134108992000, 355016116241074708829385321492, 1228958111984894631846657261766656000, 7960240318398277162915923478914410838135990, 89961580311571094335785117669395413813764096000000

Offset: 0

Vaclav Kotesovec, May 15 2025

nonn

In general, for m>2, Sum_{k=0..n} (-1)^(n-k) * binomial(2*n, n-k) * k^(m*n) ~ 2^(2*n + 1/2) * r^(m*n + 1) * n^(m*n) / (sqrt(m + (2-m)*r^2) * exp(m*n) * (r^2 - 1)^n), where r is the root of the equation (1 + r)/(1 - r) = -exp(m/r).

Cf. A002674 (m=2), A383929 (m=3), A298851*A002674 (m=4).

Cf. A383917.

Join[{1}, Table[Sum[(-1)^(n-k)*Binomial[2*n, n-k]*k^(5*n), {k, 0, n}], {n, 1, 12}]]

a(n) ~ 2^(2*n + 1/2) * r^(5*n + 1) * n^(5*n) / (sqrt(5 - 3*r^2) * exp(5*n) * (r^2 - 1)^n), where r = 1.0145858159274292356581282820876562174881159476120290450838... is the root of the equation (1 + r)/(1 - r) = -exp(5/r).

cp's OEIS Frontend

A268647 G.f.: C(x,y)^2 - S(x,y)^2 = Sum_{n>=0} x^(2*n)*y/[Sum_{k=0..2*n+1} T(n,k)*y^k], where C(x,y) = Sum_{n>=0} x^(2*n) / Product_{k=1..2*n} (k + y) and S(x,y) = Sum_{n>=0} x^(2*n+1) / Product_{k=1..2*n+1} (k + y).

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A304001 Number of permutations of [n] whose up-down signature has a nonnegative total sum.

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Mathematica

A066991 Square array read by descending antidiagonals of number of ways of dividing n*k labeled items into k unlabeled orders with n items in each order.

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Mathematica

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A143844 Triangle T(n,k) = k^2 read by rows.

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PARI

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A226731 a(n) = (2n - 1)!/(2n).

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A370379 Number of compositions of n where there are (2*k)!/2 sorts of part k.

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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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A268647 G.f.: C(x,y)^2 - S(x,y)^2 = Sum_{n>=0} x^(2n)y/[Sum_{k=0..2n+1} T(n,k)y^k], where C(x,y) = Sum_{n>=0} x^(2n) / Product_{k=1..2n} (k + y) and S(x,y) = Sum_{n>=0} x^(2n+1) / Product_{k=1..2n+1} (k + y).

A382527 a(n) = Sum_{j = 1..n} (-1)^(n+j) * j^(2n+4) binomial(2*n, n-j).

A383929 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n, k) (n-k)^(3*n).

A383930 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n, k) (n-k)^(5*n).