A257633 -id:A257633 - cp's OEIS frontend

A262977 a(n) = binomial(4*n-1,n).

1, 3, 21, 165, 1365, 11628, 100947, 888030, 7888725, 70607460, 635745396, 5752004349, 52251400851, 476260169700, 4353548972850, 39895566894540, 366395202809685, 3371363686069236, 31074067324187580, 286845713747883300, 2651487106659130740, 24539426037817994160

Offset: 0

Vladimir Kruchinin, Oct 06 2015

From Gus Wiseman, Sep 28 2022: (Start)
Also the number of integer compositions of 4n with alternating sum 2n, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A348614. The a(12) = 21 compositions are:
  (6,2)  (1,2,5)  (1,1,5,1)  (1,1,1,1,4)
         (2,2,4)  (2,1,4,1)  (1,1,2,1,3)
         (3,2,3)  (3,1,3,1)  (1,1,3,1,2)
         (4,2,2)  (4,1,2,1)  (1,1,4,1,1)
         (5,2,1)  (5,1,1,1)  (2,1,1,1,3)
                             (2,1,2,1,2)
                             (2,1,3,1,1)
                             (3,1,1,1,2)
                             (3,1,2,1,1)
                             (4,1,1,1,1)
The following pertain to this interpretation:
- The case of partitions is A000712, reverse A006330.
- Allowing any alternating sum gives A013777 (compositions of 4n).
- A011782 counts compositions of n.
- A034871 counts compositions of 2n with alternating sum 2k.
- A097805 counts compositions by alternating (or reverse-alternating) sum.
- A103919 counts partitions by sum and alternating sum (reverse: A344612).
- A345197 counts compositions by length and alternating sum.
(End)

V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A006632, A002293, A004331, A005810, A052203, A224274, A257633.
Cf. A000346, A000984, A001700, A002458, A025047, A058622, A081294, A294175.
Cf. A088218, A163455, A165817.

[Binomial(4*n-1,n): n in [0..20]]; // Vincenzo Librandi, Oct 06 2015

Table[Binomial[4 n - 1, n], {n, 0, 40}] (* Vincenzo Librandi, Oct 06 2015 *)

B(x):=sum(binomial(4*n-1,n-1)*3/(4*n-1)*x^n,n,1,30);
taylor(x*diff(B(x),x,1)/B(x),x,0,20);

a(n) = binomial(4*n-1,n); \\ Michel Marcus, Oct 06 2015

G.f.: A(x)=x*B'(x)/B(x), where B(x) if g.f. of A006632.
a(n) = Sum_{k=0..n}(binomial(n-1,n-k)*binomial(3*n,k)).
a(n) = 3*A224274(n), for n > 0. - Michel Marcus, Oct 12 2015
From Peter Bala, Nov 04 2015: (Start)
The o.g.f. equals f(x)/g(x), where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A005810 (k = 0), A052203 (k = 1), A257633 (k = 2), A224274 (k = 3) and A004331 (k = 4). (End)
a(n) = [x^n] 1/(1 - x)^(3*n). - Ilya Gutkovskiy, Oct 03 2017
a(n) = A071919(3n-1,n+1) = A097805(4n,n+1). - Gus Wiseman, Sep 28 2022
From Peter Bala, Feb 14 2024: (Start)
a(n) = (-1)^n * binomial(-3*n, n).
a(n) = hypergeom([1 - 3*n, -n], [1], 1).
The g.f. A(x) satisfies A(x/(1 + x)^4) = 1/(1 - 3*x). (End)
a(n) = Sum_{k = 0..n} binomial(2*n+k-1, k)*binomial(2*n-k-1, n-k). - Peter Bala, Sep 16 2024
G.f.: 1/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 17 2025

A224274 a(n) = binomial(4*n,n)/4.

Original entry on oeis.org

1, 7, 55, 455, 3876, 33649, 296010, 2629575, 23535820, 211915132, 1917334783, 17417133617, 158753389900, 1451182990950, 13298522298180, 122131734269895, 1123787895356412, 10358022441395860, 95615237915961100, 883829035553043580, 8179808679272664720, 75788358475481302185

Offset: 1

Gary Detlefs, Apr 02 2013

In general, binomial(k*n,n)/k = binomial(k*n-1,n-1).
Sequences in the OEIS related to this identity are:
. C(2n,n) = A000984, C(2n,n)/2 = A001700;
. C(3n,n) = A005809, C(3n,n)/3 = A025174;
. C(4n,n) = A005810, C(4n,n)/4 = a(n);
. C(5n,n) = A001449, C(5n,n)/5 = A163456;
. C(6n,n) = A004355, C(6n,n)/6 is not in the OEIS.
Conjecture: a(n) == 1 (mod n^3) iff n is an odd prime.
It is known that a(p) == 1(mod p^3) for prime p >= 3. See Mestrovic, Section 3. - Peter Bala, Oct 09 2015

			For n=2, binomial(4*n,n) = binomial(8,2) = 8*7/2 = 28, so a(2) = 28/4 = 7. - _Michael B. Porter_, Jul 12 2016

G. C. Greubel, Table of n, a(n) for n = 1..1000
Dmitry Kruchinin and Vladimir Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequence, Vol. 18 (2015), article 15.4.6.
Romeo Meštrović, Lucas’ theorem: its generalizations, extensions and applications (1878-2014), arXiv:1409.3820v1 [math.NT], 2014.
N. J. Wildberger and Dean Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly (2025). See section 12.

Cf. A001700, A001764, A025174, A163456, A002293, A069271, A004331, A005810, A052203, A257633, A262977, A227726.

[Binomial(4*n,n) div 4: n in [1..25]]; // Vincenzo Librandi, Jun 03 2015

Maple
```
seq(binomial(4*n,n)/4, n=1..17);
```

Table[Binomial[4 n, n]/4, {n, 30}] (* Vincenzo Librandi, Jun 03 2015 *)

a(n) = binomial(4*n,n)/4; /* Joerg Arndt, Apr 02 2013 */

a(n) = binomial(4*n,n)/4 = A005810(n)/4.
a(n) = binomial(4*n-1,n-1).
G.f.: A(x) = B'(x)/B(x), where B(x) = 1 + x*B(x)^4 is g.f. of A002293. - Vladimir Kruchinin, Aug 13 2015
From Peter Bala, Oct 08 2015: (Start)
a(n) = 1/2*[x^n] (C(x)^2)^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A163456.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 4*x^2 + 22*x^3 + ... is the o.g.f. for A002293.
exp( 2*Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 9*x^2 + 52*x^3 + ... is the o.g.f. for A069271. (End)
From Peter Bala, Nov 04 2015: (Start)
With an offset of 1, the o.g.f. equals f(x)*g(x)^3, where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A262977 (k = -1), A005810 (k = 0), A052203 (k = 1), A257633 (k = 2) and A004331 (k = 4). (End)
a(n) = 1/5*[x^n] (1 + x)/(1 - x)^(3*n + 1) = 1/5*[x^n]( 1/C(-x) )^(5*n), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A227726. - Peter Bala, Jul 12 2016
a(n) ~ 2^(8*n-3/2)*3^(-3*n-1/2)*n^(-1/2)/sqrt(Pi). - Ilya Gutkovskiy, Jul 12 2016
O.g.f.: A(x) = f(x)/(1 - 3*f(x)), where f(x) = series reversion (x/(1 + x)^4) = x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + ... is the o.g.f. of A002293 with the initial term omitted. Cf. A025174. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/3)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+3)*n+k-1,k) = C(4*n,n)/4 and (1/4)*Sum_{k = 0..n} (-1)^k*C(x*n,n-k)*C((x-4)*n+k-1,k) = C(4*n,n)/4, both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
Right-hand side of the identity (1/3)*Sum_{k = 0..2*n} (-1)^k*binomial(5*n-k-1,2*n-k)*binomial(3*n+k-1,k) = binomial(4*n,n)/4. - Peter Bala, Mar 09 2022
a(n) = [x^n] G(x)^n, where G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... is the g.f. of A001764. - Peter Bala, Oct 17 2024

A052203 a(n) = (4n+1)*binomial(4n,n)/(3n+1).

Original entry on oeis.org

1, 5, 36, 286, 2380, 20349, 177100, 1560780, 13884156, 124403620, 1121099408, 10150595910, 92263734836, 841392966470, 7694644696200, 70539987842520, 648045936942300, 5964720367660956, 54991682779774384, 507749884105448600, 4694436188839116720

Offset: 0

Barry E. Williams, Jan 28 2000

Central terms of the triangles in A122366 and A111418. - Reinhard Zumkeller, Aug 30 2006 and Mar 14 2014

a(n) is the number of paths from (0,0) to (4n,n), taking north and east steps while avoiding exactly 2 consecutive north steps. - Shanzhen Gao, Apr 15 2010

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Cf. A002293, A051944, A004331, A005810, A224274, A257633, A262977.

a052203 n = a122366 (2 * n) n  -- Reinhard Zumkeller, Mar 14 2014

[Binomial(4*n+1, n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

Table[Binomial[4 n + 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

vector(30, n, n--; (4*n+1)*binomial(4*n,n)/(3*n+1)) \\ Altug Alkan, Nov 05 2015

a(n) = C(4n+1, n); a(n) is asymptotic to c/sqrt(n)*(256/27)^n with c=0.614... - Benoit Cloitre, Jan 27 2003 [c = 2^(5/2)/(3^(3/2)*sqrt(Pi)) = 0.61421182128... - Vaclav Kotesovec, Feb 14 2019]
G.f.: g^2/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
G.f.: hypergeom([1/2, 3/4, 5/4], [2/3, 4/3], (256/27)*x). - Robert Israel, Aug 07 2014
D-finite with recurrence 3*n*(3*n-1)*(3*n+1)*a(n) - 8*(4*n+1)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Nov 26 2012
From Peter Bala, Nov 04 2015: (Start)
The o.g.f. equals f(x)*g(x), where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293.
More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A262977 (k = -1), A005810 (k = 0), A257633 (k = 2), A224274 (k = 3) and A004331 (k = 4). (End)
a(n) = [x^n] 1/(1 - x)^(3*n+2). - Ilya Gutkovskiy, Oct 03 2017
a(n) = Sum_{k = 0..n} binomial(2*n+k+1, k)*binomial(2*n-k-1, n-k). - Peter Bala, Sep 17 2024

More terms from James Sellers, Jan 31 2000

A004331 Binomial coefficient C(4n,n-1).

Original entry on oeis.org

1, 8, 66, 560, 4845, 42504, 376740, 3365856, 30260340, 273438880, 2481256778, 22595200368, 206379406870, 1889912732400, 17345898649800, 159518999862720, 1469568786235308, 13559593014190944, 125288932441604200

Offset: 1

N. J. A. Sloane

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.

Seiichi Manyama, Table of n, a(n) for n = 1..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Cf. A002293, A005810, A052203, A224274, A257633, A262977.

Cf. A001791, A004319, A004343, A004356, A004369, A004382.

#A004331
seq(binomial(4*n - 1,n), n = 0..20);

a[n_] := Binomial[4*n, n - 1]; Array[a, 19] (* Amiram Eldar, May 09 2020 *)

vector(30, n, binomial(4*n, n-1)) \\ Altug Alkan, Nov 05 2015

G.f.: (g^2-g)/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
With an offset of 0, the o.g.f. equals f(x)*g(x)^4, where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A262977 (k = -1), A005810 (k = 0), A052203 (k = 1), A257633 (k = 2) and A224274 (k = 3). - Peter Bala, Nov 04 2015
D-finite with recurrence 3*(n-1)*(3*n-1)*(3*n+1)*a(n) -8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Mar 19 2025

A264773 Triangle T(n,k) = binomial(4n - 3k, 3n - 2k), 0 <= k <= n.

Original entry on oeis.org

1, 4, 1, 28, 5, 1, 220, 36, 6, 1, 1820, 286, 45, 7, 1, 15504, 2380, 364, 55, 8, 1, 134596, 20349, 3060, 455, 66, 9, 1, 1184040, 177100, 26334, 3876, 560, 78, 10, 1, 10518300, 1560780, 230230, 33649, 4845, 680, 91, 11, 1, 94143280, 13884156, 2035800, 296010, 42504, 5985, 816, 105, 12, 1

Offset: 0

Peter Bala, Nov 30 2015

Riordan array (f(x),x*g(x)), where g(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + ... is the o.g.f. for A002293 and f(x) = g(x)/(4 - 3*g(x)) = 1 + 4*x + 28*x^2 + 220*x^3 + 1820*x^4 + ... is the o.g.f. for A005810.

More generally, if (R(n,k))n,k>=0 is a proper Riordan array and m is a nonnegative integer and a > b are integers then the array with (n,k)-th element R((m + 1)*n - a*k, m*n - b*k) is also a Riordan array (not necessarily proper). Here we take R as Pascal's triangle and m = a = 3 and b = 2. See A092392, A264772, A264774 and A113139 for further examples.

			Triangle begins
  n\k |       0      1     2    3   4   5   6   7
------+-----------------------------------------------
   0  |       1
   1  |       4      1
   2  |      28      5     1
   3  |     220     36     6    1
   4  |    1820    286    45    7   1
   5  |   15504   2380   364   55   8   1
   6  |  134596  20349  3060  455  66   9   1
   7  | 1184040 177100 26334 3876 560  78  10   1
...

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of the triangle, flattened).
Peter Bala, A 4-parameter family of embedded Riordan arrays
E. Lebensztayn, On the asymptotic enumeration of accessible automata, Discrete Mathematics and Theoretical Computer Science, Vol. 12, No. 3, 2010, 75-80, Section 2.
R. Sprugnoli, An Introduction to Mathematical Methods in Combinatorics CreateSpace Independent Publishing Platform 2006, Section 5.6, ISBN-13: 978-1502925244.

A005810 (column 0), A052203 (column 1), A257633 (column 2), A224274 (column 3), A004331 (column 4). Cf. A002293, A007318, A092392 (C(2n-k,n)), A119301 (C(3n-k,n-k)), A264772, A264774.

/* As triangle: */ [[Binomial(4*n-3*k, 3*n-2*k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Dec 02 2015

A264773:= proc(n,k) binomial(4*n - 3*k, 3*n - 2*k); end proc:
seq(seq(A264773(n,k), k = 0..n), n = 0..10);

A264773[n_,k_] := Binomial[4*n - 3*k, n - k];
Table[A264773[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Feb 06 2024 *)

T(n,k) = binomial(4*n - 3*k, n - k).

O.g.f.: f(x)/(1 - t*x*g(x)), where f(x) = Sum_{n >= 0} binomial(4*n,n)*x^n and g(x) = Sum_{n >= 0} 1/(3*n + 1)*binomial(4*n,n)*x^n.

cp's OEIS Frontend

A262977 a(n) = binomial(4*n-1,n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

Formula

A224274 a(n) = binomial(4*n,n)/4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A052203 a(n) = (4n+1)*binomial(4n,n)/(3n+1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

PARI

Formula

Extensions

A004331 Binomial coefficient C(4n,n-1).

Views

Author

Keywords

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A264773 Triangle T(n,k) = binomial(4*n - 3*k, 3*n - 2*k), 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A264773 Triangle T(n,k) = binomial(4n - 3k, 3n - 2k), 0 <= k <= n.