A097613 -id:A097613 - cp's OEIS frontend

A051960 a(n) = C(n)*(3n+2) where C(n) = Catalan numbers = A000108.

2, 5, 16, 55, 196, 714, 2640, 9867, 37180, 140998, 537472, 2057510, 7904456, 30458900, 117675360, 455657715, 1767883500, 6871173870, 26747767200, 104268528210, 406975466040, 1590307356300, 6220814327520, 24357232569150, 95452906901976, 374369872911804

Offset: 0

Barry E. Williams, Jan 05 2000

If Y is a fixed 2-subset of a 2n-set X then a(n-1) is the number of n-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007

a(n-1) is the number of vertices in the n-dimensional halohedron (or equivalently, n-cycle cubeahedron). - Vincent Pilaud, May 12 2020

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Robert Israel, Table of n, a(n) for n = 0..1600
Moa Apagodu and Doron Zeilberger, Using the "Freshman's Dream" to Prove Combinatorial Congruences, arXiv:1606.03351 [math.CO], 2016. Also Amer. Math. Monthly. 124 (2017), 597-608.
Satyan L. Devadoss, Timothy Heath, and Cid Vipismakul, Deformations of bordered Riemann surfaces and associahedral polytopes, arXiv:1002.1676 [math.AG], 2010.
S. L. Devadoss, T. Heath, and W. Vipismakul, Deformations of bordered surfaces and convex polytopes, Notices Amer. Math. Soc. 58 (2011), no. 4, 530-541.
S. B. Ekhad and M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017).
Milan Janjic, Two Enumerative Functions.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.

Cf. A000108 and A051924.
Cf. A024482 and A097613.
Half A028283.

[Catalan(n)*(3*n+2): n in [0..30]]; // Vincenzo Librandi, Oct 01 2015

a := n -> 4^n*(2+3*n)*GAMMA(1/2+n)/(sqrt(Pi)*GAMMA(2+n)):
seq(a(n), n=0..25); # Peter Luschny, Dec 14 2015

Table[CatalanNumber[n] (3n+2), {n,0,30}] (* Michael De Vlieger, Sep 30 2015 *)

a(n):=sum(binomial(n-k+1,k)*2^(n-2*k+1)*binomial(n,k),k,0,(n+1)/2); /* Vladimir Kruchinin, Sep 30 2015 */

a(n) = (3*n+2)*binomial(2*n, n)/(n+1);
vector(30, n, a(n-1)) \\ Altug Alkan, Sep 30 2015

(n+1)*a(n) - 2*(n+2)*a(n-1) - 4*(2*n-3)*a(n-2) = 0. - conjectured by R. J. Mathar, Oct 02 2014, verified by Robert Israel, Sep 30 2015
G.f.: (1 + 2*x)/(2*x*sqrt(1-4*x)) - 1/(2*x). - Vladimir Kruchinin, Sep 30 2015.
a(n) = Sum_{k=0..(n+1)/2} (binomial(n-k+1,k)*2^(n-2*k+1)*binomial(n,k)). - Vladimir Kruchinin, Sep 30 2015.
a(n) = 4^n*(2+3*n)*Gamma(n + 1/2)/(sqrt(Pi)*Gamma(n+2)). - Peter Luschny, Dec 14 2015
a(n - 1) = A051924(n) + A000108(n - 1). - F. Chapoton, Mar 05 2022
Sum_{n>=0} a(n)/8^n = 5*sqrt(2) - 4. - Amiram Eldar, May 06 2023
E.g.f.: exp(2*x)*(2*BesselI(0,2*x) + BesselI(1,2*x)). - Stefano Spezia, May 14 2025
a(n) = 2*binomial(2*n, n) + binomial(2*n, n-1) = 2*A000984(n) + A001791(n). - Peter Bala, Aug 23 2025

A260878 Number of set partitions of {1, 2, ..., 2*n} with sizes in {[n, n], [2n]}.

Original entry on oeis.org

2, 2, 4, 11, 36, 127, 463, 1717, 6436, 24311, 92379, 352717, 1352079, 5200301, 20058301, 77558761, 300540196, 1166803111, 4537567651, 17672631901, 68923264411, 269128937221, 1052049481861, 4116715363801, 16123801841551, 63205303218877, 247959266474053

Offset: 0

Peter Luschny, Aug 02 2015

Third column in A260876.

			The set partitions counted by a(3) = 11 are: {{1, 2, 3, 4, 5, 6}},
{{1, 2, 4}, {3, 5, 6}}, {{1, 2, 3}, {4, 5, 6}}, {{1, 3, 4}, {2, 5, 6}},
{{1, 3, 5}, {2, 4, 6}}, {{1, 4, 5}, {2, 3, 6}}, {{1, 5, 6}, {2, 3, 4}},
{{1, 4, 6}, {2, 3, 5}}, {{1, 3, 6}, {2, 4, 5}}, {{1, 2, 6}, {3, 4, 5}},
{{1, 2, 5}, {3, 4, 6}}.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Hans Salié, Über die Koeffizienten der Blasiusschen Reihe, Math. Nachr. 1955, (14, 4-6), 241--248.

a(n) = A112849(n) for n >= 2. - Alois P. Heinz, Aug 06 2015
a(n) = A052473(n+2) - 1.
a(n) = A088218(n) + 1.
a(n) = (-1)^n*A110556(n) + 1.
a(n+1) - a(n) = A097613(n+1) for n > 0.
Cf. A260876, A001700.
Cf. A323230 (d=0), this sequence (d=1), A323229 (d=2).

a := proc(n) option remember;
if n < 2 then [2, 2][n+1] else ((4*n - 2)*a(n-1) - 3*n + 2)/n fi end:
seq(a(n), n=0..26); # Or:
egf := n -> exp(exp(x)*(1 - (GAMMA(n,x)/GAMMA(n)))):
a := n -> `if`(n<2, 2, (2*n)!*coeff(series(egf(n), x, 2*n+1), x, 2*n)):
seq(a(n), n=0..26); # Peter Luschny, Aug 02 2019

Table[Binomial[2 n - 1, n] + 1, {n, 0, 26}] (* or *)
CoefficientList[Series[(4 x^2 - 13 x + 3 + Sqrt[(1 - 4 x) (x - 1)^2])/(2 (4 x - 1) (x - 1)), {x, 0, 26}], x] (* Michael De Vlieger, Feb 26 2017 *)

Sage
```
print([A260876(n,2) for n in (0..30)])
```

# Alternative:
def A260878():
    a, f, s, n = 2, 2, 1, 1
    yield a
    while True:
        yield a
        f += 4; s += 3; n += 1
        a = (f*a - s)/n
a = A260878()
print([next(a) for n in range(27)]) # Peter Luschny, Aug 02 2019

G.f.: (4*x^2 - 13*x + 3 + sqrt((1 - 4*x)*(x - 1)^2))/(2*(4*x - 1)*(x - 1)). - Alois P. Heinz, Aug 06 2015
a(n) = Binomial(2*n-1, n) + 1. - Vladimir Kruchinin, Feb 26 2017
The generating function G(x) satisfies the differential equation x^3 + 2*x = (4*x^4 - 9*x^3 + 6*x^2 - x)*diff(G(x), x) + (2*x^3 - 4*x^2 + 2*x)*G(x). - Peter Luschny, Feb 12 2019
From Peter Luschny, Aug 02 2019: (Start)
a(n) = ((4*n - 2)*a(n-1) - 3*n + 2)/n for n >= 2.
a(n) = (2*n)! * [x^(2*n)] exp(exp(x)*(1 - (Gamma(n,x)/Gamma(n)))) for n >= 2.
a(n) ~ 4^n/sqrt(4*Pi*n). More precise asymptotic estimates are:
1 + (4^n/sqrt(n*Pi)) * (1/2 - 1/(16*n) * (1 - 1/(16*n))), and
1 + 4^n*(2 - 2/N^2 + 21/N^4 - 671/N^6) / sqrt(2*N*Pi) with N = 8*n + 2.
Let b(n) = binomial(2*(n-1), n-1) + 1 = A323230(n) for n >= 0. Then by Salié:
p divides a(p+k) - b(k+1) if p is a prime > k and 0 <= k <= 4.
Conjecture: p divides a(p+5) - b(6) if p is a prime > b(6).
If p is a prime divisor of n then a(n) == a(n/p) (mod p) (by Salié, theorem 2).
(End)
From Peter Bala, Apr 20 2024: (Start)
a(n) = Sum_{k = 0..n} (-1)^k * 3*n/(2*n + k) * binomial(2*n+k, n-k) for n >= 1.
a(n) = Sum_{k = 0..n} (-1)^k * 3*n/(n + 2*k) * binomial(2*n+k-1, n-k) for n >= 1.
(-1)^n * a(n) equals the n-th order Taylor polynomial (centered at 0) of 1/c(x)^(3*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108. (End)

A359720 T(n,k) = coefficient of x^ny^k in A(x,y) such that: x = Sum_{n=-oo..+oo} (-1)^n x^n * (y + x^n)^n * A(x,y)^n.

Original entry on oeis.org

1, 1, 1, 2, 4, 5, 1, 7, 21, 9, 20, 51, 49, 7, 43, 170, 179, 66, 2, 110, 454, 711, 381, 54, 262, 1367, 2390, 1894, 523, 25, 674, 3776, 8361, 8070, 3496, 469, 5, 1684, 11062, 27082, 33093, 19129, 4602, 269, 4397, 31054, 89389, 125983, 93908, 33211, 4325, 91, 11320, 89935, 283170, 470439, 421762, 200449, 43062, 2846, 14

Offset: 0

Paul D. Hanna, Jan 13 2023

Related identity: 0 = Sum_{n=-oo..+oo} x^n * (y - x^n)^n, which holds formally for all y.
The terms in row n start with index k = 0 to k = floor(2*n/3), for n >= 0.
A359721(n) = Sum_{k=0..floor(2*n/3)} T(n,k), for n >= 0 (row sums).
A357797(n) = Sum_{k=0..floor(2*n/3)} T(n,k)*2^k, for n >= 0.
A359723(n) = Sum_{k=0..floor(2*n/3)} T(n,k)*3^k, for n >= 0.
A359724(n) = Sum_{k=0..floor(2*n/3)} T(n,k)*4^k, for n >= 0.
A355357(n) = T(n,0), for n >= 0.
A359725(n) = T(n+2,1), for n >= 0.
A359726(n) = T(n+3,2), for n >= 0.
A000108(n) = T(3*n,2*n), for n >= 0.
A359722(n) = T(3*n+1,2*n), for n >= 0.
A097613(n+2) = T(3*n+2,2*n+1), for n >= 0.

			G.f.: A(x,y) = 1 + x*(1) + x^2*(1 + 2*y) + x^3*(4 + 5*y + y^2) + x^4*(7 + 21*y + 9*y^2) + x^5*(20 + 51*y + 49*y^2 + 7*y^3) + x^6*(43 + 170*y + 179*y^2 + 66*y^3 + 2*y^4) + x^7*(110 + 454*y + 711*y^2 + 381*y^3 + 54*y^4) + x^8*(262 + 1367*y + 2390*y^2 + 1894*y^3 + 523*y^4 + 25*y^5) + x^9*(674 + 3776*y + 8361*y^2 + 8070*y^3 + 3496*y^4 + 469*y^5 + 5*y^6) + x^10*(1684 + 11062*y + 27082*y^2 + 33093*y^3 + 19129*y^4 + 4602*y^5 + 269*y^6) + x^11*(4397 + 31054*y + 89389*y^2 + 125983*y^3 + 93908*y^4 + 33211*y^5 + 4325*y^6 + 91*y^7) + x^12*(11320 + 89935*y + 283170*y^2 + 470439*y^3 + 421762*y^4 + 200449*y^5 + 43062*y^6 + 2846*y^7 + 14*y^8) + x^13*(29938 + 254654*y + 905307*y^2 + 1683683*y^3 + 1798279*y^4 + 1072012*y^5 + 329533*y^6 + 41858*y^7 + 1254*y^8) + x^14*(78641 + 733725*y + 2825245*y^2 + 5954300*y^3 + 7287245*y^4 + 5277807*y^5 + 2131517*y^6 + 421554*y^7 + 30194*y^8 + 336*y^9 ) + x^15*(210044 + 2088612*y + 8854116*y^2 + 20499318*y^3 + 28639206*y^4 + 24326336*y^5 + 12274991*y^6 + 3370105*y^7 + 420102*y^8 + 15745*y^9 + 42*y^10) + ...
This irregular triangle of coefficients T(n,k) of x^n*y^k, for n >= 0, k = 0..[2*n/3], in g.f. A(x,y) begin:
n = 0: [1],
n = 1: [1],
n = 2: [1, 2],
n = 3: [4, 5, 1],
n = 4: [7, 21, 9],
n = 5: [20, 51, 49, 7],
n = 6: [43, 170, 179, 66, 2],
n = 7: [110, 454, 711, 381, 54],
n = 8: [262, 1367, 2390, 1894, 523, 25],
n = 9: [674, 3776, 8361, 8070, 3496, 469, 5],
n = 10: [1684, 11062, 27082, 33093, 19129, 4602, 269],
n = 11: [4397, 31054, 89389, 125983, 93908, 33211, 4325, 91],
n = 12: [11320, 89935, 283170, 470439, 421762, 200449, 43062, 2846, 14],
n = 13: [29938, 254654, 905307, 1683683, 1798279, 1072012, 329533, 41858, 1254],
n = 14: [78641, 733725, 2825245, 5954300, 7287245, 5277807, 2131517, 421554, 30194, 336],
n = 15: [210044, 2088612, 8854116, 20499318, 28639206, 24326336, 12274991, 3370105, 420102, 15745, 42], ...
...
in which various sequences are found along columns and diagonals:
T(n,0) = A355357(n) = [1, 1, 1, 4, 7, 20, 43, 110, 262, 674, 1684, ...],
T(n+2,1) = A359725(n) = [2, 5, 21, 51, 170, 454, 1367, 3776, 11062, ...],
T(n+3,2) = A359726(n) = [1, 9, 49, 179, 711, 2390, 8361, 27082, 89389, ...],
T(3*n,2*n) = A000108(n) = [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, ...],
T(3*n+1,2*n) = A359722(n) = [1, 9, 54, 269, 1254, 5642, 24828, 107613, ...],
T(3*n+2,2*n+1) = A097613(n+2) = [2, 7, 25, 91, 336, 1254, 4719, 17875, ...].

Paul D. Hanna, Table of n, a(n) for n = 0..4920

Cf. A359721 (row sums), A357797 (y=2), A359723 (y=3), A359724 (y=4).
Cf. A355357 (T(n,0)), A359725 (T(n+2,1)), A359726 (T(n+3,2)).
Cf. A000108 (T(3*n,2*n)), A097613 (T(3*n+2,2*n+1)), A359722 (T(3*n+1,2*n)).

/* Print this irregular triangle */
{T(n,k) = my(A=[1]); for(i=1, n, A=concat(A, 0);
A[#A] = polcoeff(x - sum(n=-#A-1, #A+1, (-1)^n * x^n * (y + x^n +x*O(x^#A) )^n * Ser(A)^n ), #A-1) );
polcoeff(A[n+1],k,y)}
for(n=0, 15, for(k=0, (2*n)\3, print1(T(n,k), ", "));print(""))

G.f.: A(x,y) = Sum_{n>=0} Sum_{k=0..floor(2*n/3)} T(n,k)*x^n*y^k may be described by the following.
(1) x = Sum_{n=-oo..+oo} (-1)^n * x^n * (y + x^n)^n * A(x,y)^n.
(2) x = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n-1)) / ( (1 + y*x^n)^n * A(x,y)^n ).
T(3*n,2*n) = binomial(2*n+1,n)/(2*n+1) = A000108(n), n >= 0 (Catalan numbers).
T(3*n+2,2*n+1) = binomial(2*n+1,n+1) + binomial(2*n+2,n) = A097613(n+2), for n >= 0.

A024482 a(n) = (1/2)*(binomial(2n, n) - binomial(2n-2, n-1)).

Original entry on oeis.org

2, 7, 25, 91, 336, 1254, 4719, 17875, 68068, 260338, 999362, 3848222, 14858000, 57500460, 222981435, 866262915, 3370764540, 13135064250, 51250632510, 200205672810, 782920544640, 3064665881940, 12007086477750, 47081501377326, 184753963255176, 725510446350004

Offset: 2

Clark Kimberling

nonn

Apparently the number of sawtooth patterns in all Dyck paths of semilength n, ([0,1],2,7,25,...). A sawtooth pattern is of the form (UD)^k, k >= 1. More generally, the number of sawtooth patterns of length > t in all Dyck paths with semilength (n+t), t >= 0. - David Scambler, Apr 23 2013

			The path udUududD has two sawtooth patterns, shown in lower case.

Stefano Spezia, Table of n, a(n) for n = 2..1600
Toufik Mansour and I. L. Ramirez, Enumerations of polyominoes determined by Fuss-Catalan words, Australas. J. Combin. 81 (3) (2021) 447-457, table 2.

Cf. A000108, A097613, A225015.

[(3*n-2)*Catalan(n-1)/2: n in [2..40]]; // G. C. Greubel, Apr 03 2024

Z:=(1-z-sqrt(1-4*z))/sqrt(1-4*z)/2: Zser:=series(Z, z=0, 32): seq(coeff(Zser, z, n), n=2..25); # Zerinvary Lajos, Jan 16 2007

Table[(Binomial[2n,n]-Binomial[2n-2,n-1])/2,{n,2,30}] (* Harvey P. Dale, Mar 04 2011 *)

[(3*n-2)*catalan_number(n-1)/2 for n in range(2,41)] # G. C. Greubel, Apr 03 2024

a(n) = A051924(n)/2. - Zerinvary Lajos, Jan 16 2007
From R. J. Mathar, Nov 09 2018: (Start)
D-finite with recurrence n*a(n) - (5*n-4)*a(n-1) + 2*(2*n-5)*a(n-2) = 0.
n*(3*n-5)*a(n) - 2*(3*n-2)*(2*n-3)*a(n-1) = 0. (End)
a(n) ~ 3*2^(2*n-3)/sqrt(n*Pi). - Stefano Spezia, May 09 2023
From G. C. Greubel, Apr 03 2024: (Start)
a(n) = (3*n-2)*A000108(n-1)/2.
G.f.: ((1-x)*sqrt(1-4*x) - (1+x)*(1-4*x))/(2*(1-4*x)).
E.g.f.: (1/2)*( -1 - x + exp(2*x)*( (1-x)*BesselI(0, 2*x) + x*BesselI(1, 2*x) ) ). (End)

A097070 Consider all compositions (ordered partitions) of n into n parts, allowing zeros. E.g., for n = 3 we get 300, 030, 003, 210, 120, 201, 102, 021, 012, 111. Then a(n) is the total number of 1's.

Original entry on oeis.org

1, 2, 9, 40, 175, 756, 3234, 13728, 57915, 243100, 1016158, 4232592, 17577014, 72804200, 300874500, 1240940160, 5109183315, 21002455980, 86213785350, 353452638000, 1447388552610, 5920836618840, 24197138082780, 98801168731200, 403095046038750, 1643337883690776, 6694900194799404

Offset: 1

Amy J. Kolan, Sep 15 2004

Number of compositions of n into n parts, allowing zeros = binomial(2*n-1,n) = A088218 = essentially A001700.

			The compositions for n=2 are 20, 02, 11. There are two 1's in these so a(2) = 2.
From _Robert G. Wilson v_, Sep 16 2004: (Start)
The case n = 5:
A. There are 5 combinations associated with the numbers 50000: 50000, 05000, 00500, 00050, 00005.
B. There are 20 combinations associated with the numbers 41000.
C. There are 20 combinations associated with 32000.
D. There are 30 combinations associated with 31100.
E. There are 30 combinations associated with 22100.
F. There are 20 combinations associated with 21110.
G. There is one combinations associated with 11111.
The number of 1's associated with A is 0, with B 20, with C 0, with D 60, with E 30, with F 60 and with G 5. 0 + 20 + 0 + 60 + 30 + 60 + 5 = 175.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
G.-S. Cheon, H. Kim, and L. W. Shapiro, Mutation effects in ordered trees, arXiv preprint arXiv:1410.1249 [math.CO], 2014.

Cf. A037965.
Cf. A000984, A001622, A088218.
cf. A005430, A002740, A002457, A000917, A110609, A097613, A001791, A110609.

List([1..30], n-> n*Binomial(2*n-3, n-1)); # G. C. Greubel, Jul 27 2019

[n*Binomial(2*n-3, n-1): n in [1..30]]; // Vincenzo Librandi, Jul 13 2019

A097070 := n -> ifelse(n=1, 1, 2^(n-2)*JacobiP(n-1, -1/2, -n+2, 3)):
seq(simplify(A097070(n)), n = 1..28);  # Peter Luschny, Jan 22 2025

Table[n*Binomial[2n-3, n-1], {n, 30}] (* Robert G. Wilson v, Sep 17 2004 *)

a(n) = n*binomial(2*n-3, n-1); \\ Joerg Arndt, Feb 17 2015

[n*binomial(2*n-3, n-1) for n in (1..30)] # G. C. Greubel, Jul 27 2019

a(n) = n*binomial(2*n-3, n-1).
More generally, total number of k's (k>=0) in all ordered partitions of n into n parts, allowing zeros, is n*binomial(2*n-k-2, n-2) if n >= k, 0 otherwise.
Total number of 0's is given by A005430.
From Vladeta Jovovic, Sep 17 2004: (Start)
a(n) = Sum_{k=0..n} k*binomial(n, k)*binomial(n-2, k-2).
G.f.: x*(1 -2*x +(1-4*x)^(3/2))/(2*(1-4*x)^(3/2)).
E.g.f.: (x/2)*(exp(2*x)*BesselI(0, 2*x)+1). (End)
a(n) = A014107(n)*A000108(n-2). - Philippe Deléham, Apr 12 2007
a(n) = n*A088218(n-1) for n > 0. - Werner Schulte, Jan 22 2017
From Bruce J. Nicholson, Jul 11 2019: (Start)
a(n) = A002740(n) + A097613(n).
a(n) = A110609(n-1) - A002457(n-2) + A097613(n).
a(n) = A005430(n-1) - A000917(n-3) for n > 1.
a(n) = A002457(n-1) - A037965(n) - A000917(n-3) for n > 1.
a(n) = A037965(n)/2.
a(n) = A001700(n-2)*n.
a(n) = A001791(n-2)*n + A000984(n-2)*n for n > 1. (End)
From Amiram Eldar, May 16 2022: (Start)
Sum_{n>=1} 1/a(n) = 4*Pi/(3*sqrt(3)) - Pi^2/9.
Sum_{n>=1} (-1)^(n+1)/a(n) = 8*log(phi)/sqrt(5) - 4*log(phi)^2, where phi is the golden ratio (A001622). (End)
a(n) = 2^(n-2)*JacobiP(n-1, -1/2, -n+2, 3) for n > 1. - Peter Luschny, Jan 22 2025

Formula, more terms and comments from Vladeta Jovovic, Sep 15 2004

A209561 Triangle of coefficients of polynomials u(n,x) jointly generated with A209562; see the Formula section.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 4, 3, 1, 4, 7, 7, 4, 1, 5, 11, 14, 11, 5, 1, 6, 16, 25, 25, 16, 6, 1, 7, 22, 41, 50, 41, 22, 7, 1, 8, 29, 63, 91, 91, 63, 29, 8, 1, 9, 37, 92, 154, 182, 154, 92, 37, 9, 1, 10, 46, 129, 246, 336, 336, 246, 129, 46, 10, 1, 11, 56, 175, 375, 582, 672

Offset: 1

Clark Kimberling, Mar 10 2012

Alternating row sums: 1,0,1,1,1,1,1,1,1,1,1,1,1,1,...

For a discussion and guide to related arrays, see A208510.

			First five rows:
1
1...1
2...2...1
3...4...3...1
4...7...7...4...1
First three polynomials v(n,x): 1, 1 + x, 2 + 2x + x^2.

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened

Cf. A209562, A208510.

Cf. A083329 (row sums), A097613 (central terms).

a209561 n k = a209561_tabl !! (n-1) !! (k-1)
a209561_row n = a209561_tabl !! (n-1)
a209561_tabl = [1] : iterate
               (\row -> zipWith (+) ([1] ++ row) (row ++ [0])) [1,1]
-- Reinhard Zumkeller, Dec 26 2012

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := x*u[n - 1, x] + v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A209561 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A209562 *)

u(n,x)=x*u(n-1,x)+v(n-1,x),
v(n,x)=x*u(n-1,x)+v(n-1,x) +1,
where u(1,x)=1, v(1,x)=1.
T(n,n) = 1; T(n,k) = A051597(n-2,k-1), 1 <= k < n. - Reinhard Zumkeller, Dec 26 2012

A265612 a(n) = CatalanNumber(n+1)n(1+3n)/(6+2n).

Original entry on oeis.org

0, 1, 7, 35, 156, 660, 2717, 11011, 44200, 176358, 700910, 2778446, 10994920, 43459650, 171655785, 677688675, 2674776720, 10555815270, 41656918050, 164401379610, 648887951400, 2561511781920, 10113397410402, 39937416869070, 157743149913776, 623178050662300

Offset: 0

Peter Luschny, Dec 15 2015

nonn

This is row n=7 in the array A(n,k) = (rf(k+n-2,k-1)-(k-1)*(k-2)*rf(k+n-2, k-3))/ (k-1)! if n>=3 and A(n,0)=0, A(n,1)=1, A(n,2)=n; rf(n,k) denotes the rising factorial. See the cross-references for other values of n and the table in A264357.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A007946, A097613, A051960, A029651, A051924, A129869, A220101, A264357, A265613.

A265612 := n -> 2*4^n*GAMMA(3/2+n)*n*(1+3*n)/(sqrt(Pi)*GAMMA(4+n)):
seq(simplify(A265612(n)), n=0..25);

Table[SeriesCoefficient[(5 x + (I (x - 1) (7 x - 2))/Sqrt[4 x - 1] - 2 - x^2)/(2 x^3), {x, 0, n}], {n, 0, 25}] (* or *)
Table[2*4^n Gamma[3/2 + n] n (1 + 3 n)/(Sqrt[Pi] Gamma[4 + n]), {n, 0, 25}] (* or *)
Table[CatalanNumber[n + 1] n ((1 + 3 n)/(6 + 2 n)), {n, 0, 25}] (* Michael De Vlieger, Dec 15 2015 *)

for(n=0,25, print1(round(2*4^n*gamma(3/2+n)*n*(1+3*n)/(sqrt(Pi)*gamma(4+n))), ", ")) \\ G. C. Greubel, Feb 06 2017

a = lambda n: catalan_number(n+1)*n*(1+3*n)/(6+2*n)
[a(n) for n in range(26)]

G.f.: (5*x+(I*(x-1)*(7*x-2))/sqrt(4*x-1)-2-x^2)/(2*x^3).
a(n) = 2*4^n*Gamma(3/2+n)*n*(1+3*n)/(sqrt(Pi)*Gamma(4+n)).
a(n) = (rf(5+n, n-1)-(n-1)*(n-2)*rf(5+n, n-3))/(n-1)! for n>=3, rf(n,k) the rising factorial.
a(n) = a(n-1)*(2*n*(1+3*n)*(1+2*n)/((n-1)*(3*n-2)*(3+n))) for n>=2.
a(n) ~  4^n*(6-(127/4)/n+(7995/64)/n^2-(223405/512)/n^3+(23501457/16384)/n^4-...) /sqrt(n*Pi).
a(n) = [x^n] x*(1 + x)/(1 - x)^(n+4). - Ilya Gutkovskiy, Oct 09 2017

A300789 Heinz numbers of integer partitions whose Young diagram can be tiled by dominos.

Original entry on oeis.org

1, 3, 4, 7, 9, 10, 12, 13, 16, 19, 21, 22, 25, 27, 28, 29, 34, 36, 37, 39, 40, 43, 46, 48, 49, 52, 53, 55, 57, 61, 62, 63, 64, 70, 71, 75, 76, 79, 81, 82, 84, 85, 87, 88, 89, 90, 91, 94, 100, 101, 107, 108, 111, 112, 113, 115, 116, 117, 118, 121, 129, 130, 131

Offset: 1

Gus Wiseman, Mar 12 2018

nonn

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

This sequence is conjectured to be the Heinz numbers of integer partitions in which the odd parts appear as many times in even as in odd positions.

			Sequence of integer partitions whose Young diagram can be tiled by dominos begins: (), (2), (11), (4), (22), (31), (211), (6), (1111), (8), (42), (51), (33), (222), (411).

Alois P. Heinz, Table of n, a(n) for n = 1..20000
Solomon W. Golomb, Tiling with polyominoes, Journal of Combinatorial Theory, 1-2 (1966), 280-296.
Wikipedia, Domino tiling

Cf. A000712, A000898, A001405, A004003, A045931, A097613, A099390, A299926, A300056, A300060, A300787, A300788, A304662.

a:= proc(n) option remember; local k; for k from 1+
      `if`(n=1, 0, a(n-1)) while (l-> add(`if`(l[i]::odd,
       (-1)^i, 0), i=1..nops(l))<>0)(sort(map(i->
       numtheory[pi](i[1])$i[2], ifactors(k)[2]))) do od; k
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, May 22 2018

primeMS[n_]:=If[n===1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
Select[Range[100],Total[(-1)^Flatten[Position[primeMS[#],_?OddQ]]]===0&] (* Conjectured *)

A300787 Number of integer partitions of n in which the even parts appear as often at even positions as at odd positions.

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 6, 8, 12, 15, 21, 27, 38, 47, 63, 79, 106, 130, 170, 209, 272, 330, 422, 512, 653, 784, 986, 1183, 1482, 1765, 2191, 2604, 3218, 3804, 4666, 5504, 6726, 7898, 9592, 11240, 13602, 15880, 19122, 22277, 26733, 31048, 37102, 43003, 51232, 59220

Offset: 0

Gus Wiseman, Mar 12 2018

nonn

			The a(7) = 8 partitions: (7), (511), (421), (331), (322), (31111), (22111), (1111111). Missing are: (61), (52), (43), (4111), (3211), (2221), (211111).

Even- and odd-indexed terms are A006330 and A001523 respectively, which add up to A000712.

Cf. A000898, A001405, A026010, A045931, A058696, A063886, A097613, A130780, A171966, A239241, A300788, A300789.

cobal[y_]:=Sum[(-1)^x,{x,Join@@Position[y,_?EvenQ]}];
Table[Length[Select[IntegerPartitions[n],cobal[#]===0&]],{n,0,50}]

A115241 Square array read by antidiagonals: T(n,p) is the number of linearly independent, homogeneous harmonic polynomials of degree n in p variables (n,p>=1).

Original entry on oeis.org

1, 2, 0, 3, 2, 0, 4, 5, 2, 0, 5, 9, 7, 2, 0, 6, 14, 16, 9, 2, 0, 7, 20, 30, 25, 11, 2, 0, 8, 27, 50, 55, 36, 13, 2, 0, 9, 35, 77, 105, 91, 49, 15, 2, 0, 10, 44, 112, 182, 196, 140, 64, 17, 2, 0, 11, 54, 156, 294, 378, 336, 204, 81, 19, 2, 0, 12, 65, 210, 450, 672, 714, 540, 285, 100

Offset: 1

Roger L. Bagula, Mar 04 2006

A115241 is jointly generated with A209688 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=x*u(n-1,x)+x*v(n-1) and v(n,x)=u(n-1,x)+v(n-1,x)+1. See the Mathematica section at A209688. - Clark Kimberling, Mar 12 2012

Clark Kimberling's recurrence in the previous comment yields an additional zero: 1, 0, 2, 0, 3, 2, ... - Georg Fischer, Sep 04 2021

			T(1,1)=1 corresponds to the polynomial x.
T(n,1)=0 for n>=2 because no polynomial in x of degree >=2 is harmonic.
T(1,2)=2 because we can take, for example, x and y.
T(2,2)=2 because we can take, for example, x^2-y^2 and xy.
T(3,3)=7 because we can take, for example, x^3-3xy^2, x^3-3xz^2, y^3-3yx^2, y^3-3yz^2, z^3-3zx^2, z^3-3zy^2 and xyz.
The square array starts:
  1, 2,  3,  4,   5,   6,    7, ...
  0, 2,  5,  9,  14,  20,   27, ...
  0, 2,  7, 16,  30,  50,   77, ...
  0, 2,  9, 25,  55, 105,  182, ...
  0, 2, 11, 36,  91, 196,  378, ...
  0, 2, 13, 49, 140, 336,  714, ...
  0, 2, 15, 64, 204, 540, 1254, ...
  ...

Harry Hochstadt, The Functions of Mathematical Physics, Wiley, New York (1971), p. 170; also Dover, New York (1986), p. 170.

Cf. A097613 (diagonal terms), A209688.

T:=(n,p)->(2*n+p-2)*binomial(n+p-3,n-1)/n: for n from 1 to 10 do seq(T(n,p),p=1..10) od; # yields the 10 by 10 upper left corner of the square array
seq(seq(T(n,p+1-n), n=1..p), p=1..12); # yields the sequence; Georg Fischer, Sep 04 2021

T[n_, m_] := Binomial[n + m - 3, n - 1]*(2*n + m - 2)/n; Table[Table[T[n, p+1-n],{n,1,p}], {p,1,12}] // Flatten (* amended by Georg Fischer, Sep 04 2021 *)

T(n,p) = (2n+p-2) * binomial(n+p-3,n-1)/n for n>=1, p>=1.

Edited by N. J. A. Sloane, Mar 07 2006

cp's OEIS Frontend

A051960 a(n) = C(n)*(3n+2) where C(n) = Catalan numbers = A000108.

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A260878 Number of set partitions of {1, 2, ..., 2*n} with sizes in {[n, n], [2n]}.

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A359720 T(n,k) = coefficient of x^n*y^k in A(x,y) such that: x = Sum_{n=-oo..+oo} (-1)^n * x^n * (y + x^n)^n * A(x,y)^n.

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A024482 a(n) = (1/2)*(binomial(2n, n) - binomial(2n-2, n-1)).

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A097070 Consider all compositions (ordered partitions) of n into n parts, allowing zeros. E.g., for n = 3 we get 300, 030, 003, 210, 120, 201, 102, 021, 012, 111. Then a(n) is the total number of 1's.

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A209561 Triangle of coefficients of polynomials u(n,x) jointly generated with A209562; see the Formula section.

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A359720 T(n,k) = coefficient of x^ny^k in A(x,y) such that: x = Sum_{n=-oo..+oo} (-1)^n x^n * (y + x^n)^n * A(x,y)^n.

A265612 a(n) = CatalanNumber(n+1)n(1+3n)/(6+2n).