A005810 -id:A005810 - cp's OEIS frontend

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

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

A094527 Triangle T(n,k), read by rows, defined by T(n,k) = binomial(2*n,n-k).

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 20, 15, 6, 1, 70, 56, 28, 8, 1, 252, 210, 120, 45, 10, 1, 924, 792, 495, 220, 66, 12, 1, 3432, 3003, 2002, 1001, 364, 91, 14, 1, 12870, 11440, 8008, 4368, 1820, 560, 120, 16, 1, 48620, 43758, 31824, 18564, 8568, 3060, 816, 153, 18, 1, 184756, 167960

Offset: 0

Paul Barry, May 07 2004

Right-hand side of even-numbered rows of Pascal's triangle.
Row sums are A032443. Reverse of A062344. Right-hand side of A034870. Binomial transform of trinomial triangle A094531.
Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = 2*T(n-1,0) + 2*T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Mar 14 2007
Central coefficients T(2n,n) are binomial(4n,n) (A005810).
The A- and Z-sequence for this Riordan triangle is [1,2,1] and [2,2], respectively. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. See also the Philippe Deléham comment above. - Wolfdieter Lang, Nov 22 2012

			The triangle T(n,k) begins:
  n\k      0      1      2     3     4     5    6    7   8  9 10
  0:       1
  1:       2      1
  2:       6      4      1
  3:      20     15      6     1
  4:      70     56     28     8     1
  5:     252    210    120    45    10     1
  6:     924    792    495   220    66    12    1
  7:    3432   3003   2002  1001   364    91   14    1
  8:   12870  11440   8008  4368  1820   560  120   16   1
  9:   48620  43758  31824 18564  8568  3060  816  153  18  1
  10: 184756 167960 125970 77520 38760 15504 4845 1140 190 20  1
  ... Reformatted ad extended by _Wolfdieter Lang_, Nov 22 2012
From _Paul Barry_, Sep 07 2009: (Start)
Production array is
  2, 1,
  2, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Nov 22 2012: (Start)
Recurrence from the Riordan A-sequence [1,2,1]: T(4,1) = 56 = 1*T(3,0) + 2*T(3,1) + 1*T(3,2) = 1*20 + 2*15 + 1*6.
Recurrence from the Riordan Z-sequence [2,2]: T(7,0) = 3432 = 2*T(6,0) + 2*T(6,1) = 2*924 + 2*792. See the _Philippe Deléham_ comment above. (End)

Indranil Ghosh, Rows 0..100 of triangle, flattened
Paul Barry, On the Connection Coefficients of the Chebyshev-Boubaker polynomials, The Scientific World Journal, Volume 2013 (2013), Article ID 657806, 10 pages.
Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 19.
A. Luzón, D. Merlini, M. A. Morón, R. Sprugnoli, Complementary Riordan arrays, Discrete Applied Mathematics, 172 (2014) 75-87.
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Article 12.3.3, 2012. - From _N. J. A. Sloane_, Sep 16 2012
P. Peart and W.-J. Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263
T. M. Richardson, The Reciprocal Pascal Matrix, arXiv preprint arXiv:1405.6315 [math.CO], 2014.

Cf. A007318, A032443, A039599, A039598.

A094527 := proc(n,k)
    binomial(2*n,n-k) ;
end proc: # R. J. Mathar, Jun 04 2013

T[n_, k_] := Binomial[2*n, n - k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2017 *)

Riordan array (1/sqrt(1-4*x), (1-2*x-sqrt(1-4*x))/(2*x)). Column k has e.g.f. exp(2*x)*Bessel_I(k, 2*x). - Paul Barry, Jul 14 2005
Product of Riordan arrays (1/(1-x), x/(1-x)) (Pascal's triangle, A007318) and (1/sqrt(1-2x-3x^2), (1-x-sqrt(1-2x-3x^2))/(2x)) (A094531). Inverse is A110162. - Paul Barry, Jul 14 2005
T(n,k) = Sum_{j=0..n} C(n,j)*C(n,j-k). - Paul Barry, Mar 07 2006
T(n,k) = Sum_{h>=k} A039599(n,h). Sum_{k=0..n} T(n,k) = A032443(n). - Philippe Deléham, May 01 2006
Sum_{k=0..n} T(n,k)^2 = A036910(n). - Philippe Deléham, May 07 2006
Sum_{k=0..n} T(n,k)*(-1)^k = A088218(n). - Philippe Deléham, Mar 14 2007
From Wolfdieter Lang, Nov 22 2012: (Start)
The o.g.f. for the row polynomials P(n,x) := Sum_{k=0..n} T(n,k)*x^k is G(z,x) = (-x + (1+x)*z + x*z*c(z))/(sqrt(1-4*z)*((1+x)^2*z -x)) with c the o.g.f. of A000108 (Catalan). This follows from the Riordan property.
The o.g.f. for column no. k is (c(x)-1)^k/sqrt(1-4*x) (from the Riordan property). (End)
From Peter Bala, Jun 29 2015: (Start)
Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = ( 1 - 2*x - sqrt(1 - 4*x) )/(2*x) and so belongs to the hitting time subgroup of the Riordan group (see Peart and Woan, Example 5.1).
T(n,k) = [x^(n-k)] f(x)^n with f(x) = (1 + x)^2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)
From Peter Bala, Jul 21 2015: (Start)
n-th row polynomial R(n,t) = [x^n] ( (1 + (1 + t)*x)^2/(1 + t*x) )^n.
exp ( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (2 + t)*x + (5 + 4*t + t^2)*x^2 + ... is the o.g.f. for A039598. (End)

Entry revised by N. J. A. Sloane, Mar 23 2007

A386811 a(n) = Sum_{k=0..n} binomial(4*n+1,k).

Original entry on oeis.org

1, 6, 46, 378, 3214, 27896, 245506, 2182396, 19548046, 176142312, 1594831736, 14497410186, 132224930146, 1209397179048, 11088872706188, 101890087382168, 937973964234638, 8649109175873288, 79872298511230120, 738583466508887304, 6837944227813170424

Offset: 0

Seiichi Manyama, Aug 03 2025

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..500

Cf. A000302, A160906, A386812.
Cf. A005810, A078995, A147855, A226733, A226761, A385605.
Cf. A098430, A383716.
Cf. A066381.

[&+[Binomial(4*n+1, k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Aug 21 2025

Table[Sum[Binomial[4*n+1,k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Aug 07 2025 *)

a(n) = sum(k=0, n, binomial(4*n+1, k));

a(n) = [x^n] 1/((1-2*x) * (1-x)^(3*n+1)).
a(n) = Sum_{k=0..n} 2^(n-k) * binomial(3*n+k,k).
D-finite with recurrence +645*n*(3*n-1)*(3*n-2)*a(n) +8*(-56722*n^3+213090*n^2-305978*n+150255)*a(n-1) +128*(62908*n^3-282348*n^2+385070*n-126735)*a(n-2) +12288*(-2486*n^3+8918*n^2+758*n-18935)*a(n-3) -2949120*(2*n-7)*(4*n-13)*(4*n-11)*a(n-4)=0. - R. J. Mathar, Aug 03 2025
a(n) = 2^(4*n+1) - binomial(4*n+1, n)*(hypergeom([1, -1-3*n], [1+n], -1) - 1). - Stefano Spezia, Aug 05 2025
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n+1,k) * binomial(4*n-k,n-k). - Seiichi Manyama, Aug 07 2025
a(n) ~ 2^(8*n + 3/2) / (sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Aug 07 2025
G.f.: g^2/((2-g) * (4-3*g)) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 12 2025
G.f.: B(x)^2/(1 + (B(x)-1)/2), where B(x) is the g.f. of A005810. - Seiichi Manyama, Aug 15 2025
G.f.: 1/(1 - x*g^2*(8-2*g)) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 16 2025

A060539 Table by antidiagonals of number of ways of choosing k items from n*k.

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 20, 15, 4, 1, 70, 84, 28, 5, 1, 252, 495, 220, 45, 6, 1, 924, 3003, 1820, 455, 66, 7, 1, 3432, 18564, 15504, 4845, 816, 91, 8, 1, 12870, 116280, 134596, 53130, 10626, 1330, 120, 9, 1, 48620, 735471, 1184040, 593775, 142506, 20475, 2024, 153, 10

Offset: 1

Henry Bottomley, Apr 02 2001

			Square array A(n,k) begins:
  1,  1,    1,     1,      1,       1,        1, ...
  2,  6,   20,    70,    252,     924,     3432, ...
  3, 15,   84,   495,   3003,   18564,   116280, ...
  4, 28,  220,  1820,  15504,  134596,  1184040, ...
  5, 45,  455,  4845,  53130,  593775,  6724520, ...
  6, 66,  816, 10626, 142506, 1947792, 26978328, ...
  7, 91, 1330, 20475, 324632, 5245786, 85900584, ...

Seiichi Manyama, Antidiagonals n = 1..140, flattened (first 20 antidiagonals from Harry J. Smith)
H. J. Brothers, Pascal's Prism: Supplementary Material.

Rows include A000012, A000984, A005809, A005810, A001449, A004355, A004368.
Columns include A000027, A000384, A006566, A060541.
Main diagonal is A014062.
Cf. A295772.

A:= (n, k)-> binomial(n*k, k):
seq(seq(A(n, 1+d-n), n=1..d), d=1..10);  # Alois P. Heinz, Jul 28 2023

{ i=0; for (m=1, 20, for (n=1, m, k=m - n + 1; write("b060539.txt", i++, " ", binomial(n*k, k))); ) } \\ Harry J. Smith, Jul 06 2009

A(n,k) = binomial(n*k,k) = A007318(n*k,k) = A060538(n,k)/A060538(n-1,k).

A169958 a(n) = binomial(9*n, n).

Original entry on oeis.org

1, 9, 153, 2925, 58905, 1221759, 25827165, 553270671, 11969016345, 260887834350, 5720645481903, 126050526132804, 2788629694000605, 61902409203193230, 1378095785451705375, 30756373941461374800, 687917389635036844569, 15415916972482007401455, 346051021610256116115150

Offset: 0

N. J. A. Sloane, Aug 07 2010

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..110

binomial(k*n,n): A000984 (k = 2), A005809 (k = 3), A005810 (k = 4), A001449 (k = 5), A004355 (k = 6), A004368 (k = 7), A004381 (k = 8), A169959 - A169961 (k = 10 thru 12).

[Binomial(9*n, n): n in [0..50] ]; // Vincenzo Librandi, Apr 21 2011

A169958[n_] := Binomial[9*n, n]; Array[A169958, 20, 0] (* Paolo Xausa, Aug 20 2025 *)

a(n) = C(9*n-1, n-1)*C(81*n^2, 2)/(3*n*C(9*n+1, 3)), n > 0. - Gary Detlefs, Jan 02 2014
From Peter Bala, Feb 21 2022: (Start)
The o.g.f. A(x) is algebraic: (1 - A(x))*(1 + 8*A(x))^8 +  (9^9)*x*A(x)^9 = 0.
Sum_{n >= 1} a(n)*( x*(8*x + 9)^8/(9^9*(1 + x)^9) )^n = x. (End)
D-finite with recurrence 128*n*(8*n-5) *(4*n-1) *(8*n-7) *(2*n-1) *(8*n-1) *(4*n-3) *(8*n-3)*a(n) -81*(9*n-7) *(9*n-5) *(3*n-1) *(9*n-1) *(9*n-8) *(3*n-2) *(9*n-4) *(9*n-2)*a(n-1)=0. - R. J. Mathar, Aug 19 2025
G.f.: 8F7(8/9, 7/9, 2/3, 5/9, 4/9, 1/3, 2/9 ,1/9 ; 7/8, 3/4, 5/8, 1/2, 3/8, 1/4, 1/8; 387420489/16777216*x). - R. J. Mathar, Aug 19 2025

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, 1159120046626942400, 10735998891545372445

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
a(n) ~ 2^(8*n+1/2) / (3^(3*n+3/2) * sqrt(Pi*n)). - Amiram Eldar, Sep 07 2025

A004381 Binomial coefficient C(8n,n).

Original entry on oeis.org

1, 8, 120, 2024, 35960, 658008, 12271512, 231917400, 4426165368, 85113005120, 1646492110120, 32006008361808, 624668654531480, 12233149001721760, 240260199935164200, 4730523156632595024, 93343021201262177400, 1845382436487682488000

Offset: 0

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.

T. D. Noe, Table of n, a(n) for n=0..100
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].

Row 8 of A060539.

binomial(k*n,n): A000984 (k = 2), A005809 (k = 3), A005810 (k = 4), A001449 (k = 5), A004355 (k = 6), A004368 (k = 7), A169958 - A169961 (k = 9 thru 12).

[Binomial(8*n, n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

Table[Binomial[8 n, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

from math import comb
def A004381(n): return comb(n<<3,n) # Chai Wah Wu, Aug 01 2023

a(n) = C(8*n-1,n-1)*C(64*n^2,2)/(3*n*C(8*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
From Ilya Gutkovskiy, Jan 16 2017: (Start)
O.g.f.: 7F6(1/8,1/4,3/8,1/2,5/8,3/4,7/8; 1/7,2/7,3/7,4/7,5/7,6/7; 16777216*x/823543).
E.g.f.: 7F7(1/8,1/4,3/8,1/2,5/8,3/4,7/8; 1/7,2/7,3/7,4/7,5/7,6/7,1; 16777216*x/823543).
a(n) ~ 2^(24*n+1)/(sqrt(Pi*n)*7^(7*n+1/2)). (End)
From Peter Bala, Feb 20 2022: (Start)
The o.g.f. A(x) is algebraic: (1 - A(x))*(1 + 7*A(x))^7 +  (8^8)*x*A(x)^8 = 0.
Sum_{n >= 1} a(n)*( x*(7*x + 8)^7/(8^8*(1 + x)^8) )^n = x. (End)
From Seiichi Manyama, Aug 16 2025: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(8*n+1,k).
G.f.: 1/(1 - 8*x*g^7) where g = 1+x*g^8 is the g.f. of A007556.
G.f.: g/(8-7*g) where g = 1+x*g^8 is the g.f. of A007556. (End)

A004368 Binomial coefficient C(7n,n).

Original entry on oeis.org

1, 7, 91, 1330, 20475, 324632, 5245786, 85900584, 1420494075, 23667689815, 396704524216, 6681687099710, 112992892764570, 1917283000904460, 32626924340528840, 556608279578340080, 9516306085765295355, 163011740982048945441

Offset: 0

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.

T. D. Noe, Table of n, a(n) for n = 0..100
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].

binomial(k*n,n): A000984 (k = 2), A005809 (k = 3), A005810 (k = 4), A001449 (k = 5), A004355 (k = 6), A004381 (k = 8), A169958 - A169961 (k = 9 thru 12).

Cf. A002296.

[Binomial(7*n,n): n in [0..20]]; // Vincenzo Librandi, Oct 06 2015

Table[Binomial[7n,n],{n,0,20}] (* Harvey P. Dale, Apr 05 2014 *)

B(x):=sum(binomial(7*n,n-1)/n*x^n,n,1,30);
taylor(x*diff(B(x),x)/B(x),x,0,10); /* Vladimir Kruchinin, Oct 05 2015 */

a(n) = binomial(7*n,n) \\ Altug Alkan, Oct 05 2015

a(n) = C(7*n-1,n-1)*C(49*n^2,2)/(3*n*C(7*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
G.f.: A(x) = x*B'(x)/B(x), where B(x)+1 is g.f. of A002296. - Vladimir Kruchinin, Oct 05 2015
From Ilya Gutkovskiy, Jan 16 2017: (Start)
O.g.f.: 6F5(1/7,2/7,3/7,4/7,5/7,6/7; 1/6,1/3,1/2,2/3,5/6; 823543*x/46656).
E.g.f.: 6F6(1/7,2/7,3/7,4/7,5/7,6/7; 1/6,1/3,1/2,2/3,5/6,1; 823543*x/46656).
a(n) ~ sqrt(7/3)*7^(7*n)/(2*sqrt(Pi*n)*6^(6*n)). (End)
From Peter Bala, Feb 20 2022: (Start)
The o.g.f. A(x) is algebraic: (1 - A(x))*(1 + 6*A(x))^6 +  (7^7)*x*A(x)^7 = 0.
Sum_{n >= 1} a(n)*( x*(6*x + 7)^6/(7^7*(1 + x)^7) )^n = x. (End)

A147855 G.f.: 1 / (1 + 4xG(x)^2 - 7xG(x)^3) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.

Original entry on oeis.org

1, 3, 22, 174, 1444, 12323, 107104, 942952, 8381596, 75053100, 676017962, 6118171326, 55591175956, 506805088026, 4633571685968, 42468065811884, 390071875757852, 3589637747968964, 33089300640166360, 305476314574338648, 2823932709938708824, 26137341654281261347

Offset: 0

Paul D. Hanna, Jun 16 2013

nonn

			G.f.: A(x) = 1 + 3*x + 22*x^2 + 174*x^3 + 1444*x^4 + 12323*x^5 +...
A related series is G(x) = 1 + x*G(x)^4, where
G(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + 7084*x^6 +...
G(x)^2 = 1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + 2394*x^5 + 17710*x^6 +...
G(x)^3 = 1 + 3*x + 15*x^2 + 91*x^3 + 612*x^4 + 4389*x^5 + 32890*x^6 +...
such that A(x) = 1/(1 + 4*x*G(x)^2 - 7*x*G(x)^3).

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

Cf. A226705, A002293.
Cf. A026641, A183160, A371753.
Cf. A005810, A078995, A226733, A226761, A385605, A386811.

Table[Sum[Binomial[2*n+k,n-k]*Binomial[2*n-k,k],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 16 2013 *)

{a(n)=sum(k=0, n, binomial(2*n+k, n-k)*binomial(2*n-k, k))}
for(n=0, 30, print1(a(n), ", "))

{a(n)=sum(k=0, n, binomial(k, n-k)*binomial(4*n-k, k))}
for(n=0, 30, print1(a(n), ", "))

{a(n)=local(G=1+x); for(i=0, n,G=1+x*G^4+x*O(x^n)); polcoeff(1/(1+4*x*G^2-7*x*G^3), n)}
for(n=0, 30, print1(a(n), ", "))

{a(n)=local(G=1+x); for(i=0, n,G=1+x*G^4+x*O(x^n)); polcoeff(1/(1-3*x*G^2-7*x^2*G^6), n)}
for(n=0, 30, print1(a(n), ", "))

a(n) = Sum_{k=0..n} C(k, n-k) * C(4*n-k, k).
a(n) = Sum_{k=0..n} C(n+k, n-k) * C(3*n-k, k).
a(n) = Sum_{k=0..n} C(2*n+k, n-k) * C(2*n-k, k).
a(n) = Sum_{k=0..n} C(3*n+k, n-k) * C(n-k, k).
a(n) = Sum_{k=0..n} C(4*n+k, n-k) * C(-k, k).
G.f.: 1 / (1 - 3*x*G(x)^2 - 7*x^2*G(x)^6) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.
a(n) ~ 2^(8*n+5/2)/(5*sqrt(Pi*n)*3^(3*n+1/2)). - Vaclav Kotesovec, Jun 16 2013
From Seiichi Manyama, Apr 05 2024: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(4*n-2*k-1,n-2*k).
a(n) = [x^n] 1/((1-x^2) * (1-x)^(3*n)). (End)
From Seiichi Manyama, Aug 05 2025: (Start)
a(n) = Sum_{k=0..n} (-2)^(n-k) * binomial(4*n+1,k).
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(3*n+k,k). (End)
From Seiichi Manyama, Aug 14 2025: (Start)
a(n) = Sum_{k=0..n} (-1)^k * 2^(n-k) * binomial(4*n+1,k) * binomial(4*n-k,n-k).
G.f.: G(x)^2/((-1+2*G(x)) * (4-3*G(x))) where G(x) = 1+x*G(x)^4 is the g.f. of A002293. (End)
G.f.: B(x)^2/(1 + 5*(B(x)-1)/4), where B(x) is the g.f. of A005810. - Seiichi Manyama, Aug 15 2025

A348614 Numbers k such that the k-th composition in standard order has sum equal to twice its alternating sum.

Original entry on oeis.org

0, 9, 11, 14, 130, 133, 135, 138, 141, 143, 148, 153, 155, 158, 168, 177, 179, 182, 188, 208, 225, 227, 230, 236, 248, 2052, 2057, 2059, 2062, 2066, 2069, 2071, 2074, 2077, 2079, 2084, 2089, 2091, 2094, 2098, 2101, 2103, 2106, 2109, 2111, 2120, 2129, 2131

Offset: 1

Gus Wiseman, Oct 29 2021

nonn

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			The terms together with their binary indices begin:
    0: ()
    9: (3,1)
   11: (2,1,1)
   14: (1,1,2)
  130: (6,2)
  133: (5,2,1)
  135: (5,1,1,1)
  138: (4,2,2)
  141: (4,1,2,1)
  143: (4,1,1,1,1)
  148: (3,2,3)
  153: (3,1,3,1)
  155: (3,1,2,1,1)
  158: (3,1,1,1,2)

Gus Wiseman, Statistics, classes, and transformations of standard compositions

The unordered case (partitions) is counted by A000712, reverse A006330.
These compositions are counted by A262977.
Except for 0, a subset of A345917 (which is itself a subset of A345913).
A000346 = even-length compositions with alt sum != 0, complement A001700.
A011782 counts compositions.
A025047 counts wiggly compositions, ranked by A345167.
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).
A116406 counts compositions with alternating sum >=0, ranked by A345913.
A138364 counts compositions with alternating sum 0, ranked by A344619.
A345197 counts compositions by length and alternating sum.
Cf. A000984, A002458, A004331, A005810, A224274.
Cf. A008549, A013777, A027306, A058622, A088218, A114121, A120452, A126869, A163493, A294175, A344604.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Select[Range[0,1000],Total[stc[#]]==2*ats[stc[#]]&]

cp's OEIS Frontend

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

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A094527 Triangle T(n,k), read by rows, defined by T(n,k) = binomial(2*n,n-k).

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A386811 a(n) = Sum_{k=0..n} binomial(4*n+1,k).

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A060539 Table by antidiagonals of number of ways of choosing k items from n*k.

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A004381 Binomial coefficient C(8n,n).

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A147855 G.f.: 1 / (1 + 4xG(x)^2 - 7xG(x)^3) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.