A032443 -id:A032443 - cp's OEIS frontend

A345915 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum <= 0.

0, 3, 6, 10, 12, 13, 15, 20, 24, 25, 27, 30, 36, 40, 41, 43, 46, 48, 49, 50, 51, 53, 54, 55, 58, 60, 61, 63, 72, 80, 81, 83, 86, 92, 96, 97, 98, 99, 101, 102, 103, 106, 108, 109, 111, 116, 120, 121, 123, 126, 136, 144, 145, 147, 150, 156, 160, 161, 162, 163

Offset: 1

Gus Wiseman, Jul 08 2021

nonn

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

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 sequence of terms together with the corresponding compositions begins:
     0: ()
     3: (1,1)
     6: (1,2)
    10: (2,2)
    12: (1,3)
    13: (1,2,1)
    15: (1,1,1,1)
    20: (2,3)
    24: (1,4)
    25: (1,3,1)
    27: (1,2,1,1)
    30: (1,1,1,2)
    36: (3,3)
    40: (2,4)
    41: (2,3,1)

The version for Heinz numbers of partitions is A028260 (counted by A027187).
These compositions are counted by A058622.
These are the positions of terms <= 0 in A124754.
The reverse-alternating version is A345916.
The opposite (k >= 0) version is A345917.
The strictly negative (k < 0) version is A345919.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A236913 counts partitions of 2n with reverse-alternating sum <= 0.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000097, A000346, A008549, A025047, A032443, A114121, A163493, A344607, A344609, A344610.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[stc[#]]<=0&]

A345916 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum <= 0.

Original entry on oeis.org

0, 3, 5, 9, 10, 13, 15, 17, 18, 23, 25, 29, 33, 34, 36, 39, 41, 43, 45, 46, 49, 50, 53, 55, 57, 58, 61, 63, 65, 66, 68, 71, 75, 77, 78, 81, 85, 89, 90, 95, 97, 98, 103, 105, 109, 113, 114, 119, 121, 125, 129, 130, 132, 135, 136, 139, 141, 142, 145, 147, 149

Offset: 1

Gus Wiseman, Jul 08 2021

nonn

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

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 sequence of terms together with the corresponding compositions begins:
     0: ()
     3: (1,1)
     5: (2,1)
     9: (3,1)
    10: (2,2)
    13: (1,2,1)
    15: (1,1,1,1)
    17: (4,1)
    18: (3,2)
    23: (2,1,1,1)
    25: (1,3,1)
    29: (1,1,2,1)
    33: (5,1)
    34: (4,2)
    36: (3,3)

The version for Heinz numbers of partitions is A000290.
These compositions are counted by A058622.
These are the positions of terms <= 0 in A344618.
The opposite (k >= 0) version is A345914.
The version for unreversed alternating sum is A345915.
The strictly negative (k < 0) version is A345920.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A236913 counts partitions of 2n with reverse-alternating sum <= 0.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000346, A008549, A025047, A027187, A028260, A032443, A114121, A163493, A344607, A344610, A345908.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Select[Range[0,100],sats[stc[#]]<=0&]

A066381 a(n) = Sum_{k=0..n} binomial(4*n,k).

Original entry on oeis.org

1, 5, 37, 299, 2517, 21700, 190051, 1683218, 15033173, 135142796, 1221246132, 11083374659, 100946732307, 922205369324, 8446802334994, 77542088287444, 713250450657109, 6572130378649468, 60652194138406780, 560522209086365852

Offset: 0

N. J. A. Sloane, Dec 23 2001

nonn

Harry J. Smith, Table of n, a(n) for n = 0..150

Cf. A002293, A032443, A066380, A371739.

ogf := eval(s/((s-2)*(3*s-4)), s = RootOf(1-s+x*s^4, s));
series(ogf, x=0, 25); # Mark van Hoeij, May 05 2013

Table[Sum[Binomial[4*n,k],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 03 2015 *)

a[0]:1$ a[1]:5$ a[n]:=8*((3784*n^6-18764*n^5+34432*n^4 -28138*n^3+9028*n^2-24*n-315)*a[n-1]+16*(3-2*n)*(4*n-5)*(4*n-7)*(44*n^3-34*n^2-2*n+3)*a[n-2])/(3*n*(3*n-1)*(3*n-2)*(44*n^3-166*n^2 +198*n-73))$ makelist(a[n],n,0,1000); /* Tani Akinari, Sep 02 2014 */

{ for (n=0, 150, a=0; for (k=0, n, a+=binomial(4*n, k)); write("b066381.txt", n, " ", a) ) } \\ Harry J. Smith, Feb 12 2010

G.f.: s/((s-2)*(3*s-4)) where s = o.g.f.(A002293) which satisfies 1-s+x*s^4 = 0. - Mark van Hoeij, May 05 2013
a(0) = 1, a(n) = 16*a(n-1)-2*(44*n^3-34*n^2-2*n+3)*(4*n-4)!/(n!*(3*n-1)!). - Tani Akinari, Sep 02 2014
a(n) are special values of the hypergeometric function 2F1, in Maple notation: a(n)=16^n-binomial(4*n,n+1)*hypergeom([1,-3*n+1],[n+2],-1), n=0,1,... . - Karol A. Penson, Jun 03 2015
a(n) ~ (256/27)^n * sqrt(3/(2*Pi*n)). - Vaclav Kotesovec, Jun 03 2015
a(n) = [x^n] 1/((1 - 2*x)*(1 - x)^(3*n)). - Ilya Gutkovskiy, Oct 25 2017
a(n) = Sum_{k=0..floor(n/2)} binomial(4*n+1,n-2*k). - Seiichi Manyama, Apr 09 2024
a(n) = Sum_{k=0..n} 2^(n-k) * binomial(3*n+k-1,k). - Seiichi Manyama, Jul 30 2025
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n,k) * binomial(4*n-k-1,n-k). - Seiichi Manyama, Aug 08 2025
G.f.: 1/(1 - x*g^3*(8-3*g)) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 17 2025

A027914 T(n,0) + T(n,1) + ... + T(n,n), T given by A027907.

Original entry on oeis.org

1, 2, 6, 17, 50, 147, 435, 1290, 3834, 11411, 34001, 101400, 302615, 903632, 2699598, 8068257, 24121674, 72137547, 215786649, 645629160, 1932081885, 5782851966, 17311097568, 51828203475, 155188936431, 464732722872

Offset: 0

Clark Kimberling

Let b(n)=a(n) mod 2; then b(n)=1/2+(-1)^n*(1/2-A010060(floor(n/2))). - Benoit Cloitre, Mar 23 2004
Binomial transform of A027306. Inverse binomial transform of = A032443. Hankel transform is {1, 2, 3, 4, ..., n, ...}. - Philippe Deléham, Jul 20 2005
Sums of rows of the triangle in A111808. - Reinhard Zumkeller, Aug 17 2005
Number of 3-ary words of length n in which the number of 1's does not exceed the number of 0's. - David Scambler, Aug 14 2012
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 07 2022

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Moa Apagodu and Ji-Cai Liu, Congruence properties for the trinomial coefficients, arXiv:1907.13547 [math.NT], 2019.
Yassine Otmani, The 2-Pascal Triangle and a Related Riordan Array, J. Int. Seq. (2025) Vol. 28, Issue 3, Art. No. 25.3.5. See p. 24.

Cf. A025191, A027915, A081673, A092255, A055217, A005773.

a027914 n = sum $ take (n + 1) $ a027907_row n
-- Reinhard Zumkeller, Jan 22 2013

a := n -> simplify((3^n + GegenbauerC(n,-n,-1/2))/2):
seq(a(n), n=0..25); # Peter Luschny, May 12 2016

CoefficientList[ Series[ (1 + x + Sqrt[1 - 2x - 3x^2])/(2 - 4x - 6x^2), {x, 0, 26}], x] (* Robert G. Wilson v, Jul 21 2015 *)
Table[(3^n + Hypergeometric2F1[1/2 - n/2, -n/2, 1, 4])/2, {n, 0, 20}] (* Vladimir Reshetnikov, May 07 2016 *)
f[n_] := Plus @@ Take[ CoefficientList[ Sum[x^k, {k, 0, 2}]^n, x], n +1]; Array[f, 26, 0] (* Robert G. Wilson v, Jan 30 2017 *)

a(n)=sum(i=0,n,polcoeff((1+x+x^2)^n,i,x))

a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,j,if(i+j+k-n,0,(n!/i!/j!/k!)))))

x='x+O('x^99); Vec((1+x+(1-2*x-3*x^2)^(1/2))/(2*(1-2*x-3*x^2))) \\ Altug Alkan, May 12 2016

a(n) = ( 3^n + A002426(n) )/2; lim n -> infinity a(n+1)/a(n) = 3; 3^n < 2*a(n) < 3^(n+1). - Benoit Cloitre, Sep 28 2002
From Benoit Cloitre, Jan 26 2003: (Start)
a(n) = (1/2) *( Sum{k = 0..n} binomial(n,k)*binomial(n-k,k) + 3^n );
a(n) = Sum_{k = 0..n} Sum_{i = 0..k} binomial(n,i)*binomial(n-i,k);
a(n) = 3^n/2*(1+c/sqrt(n)+O(n^-1/2)) where c=0.5... (End)
c = sqrt(3/Pi)/2 = 0.4886025119... - Vaclav Kotesovec, May 07 2016
a(n) = n!*Sum(i+j+k=n, 1/(i!*j!*k!)) 0<=i<=n, 0<=k<=j<=n. - Benoit Cloitre, Mar 23 2004
G.f.: (1+x+sqrt(1-2x-3x^2))/(2(1-2x-3x^2)); a(n) = Sum_{k = 0..n} floor((k+2)/2)*Sum_{i = 0..floor((n-k)/2)} C(n,i)*C(n-i,i+k)* ((k+1)/(i+k+1)). - Paul Barry, Sep 23 2005; corrected Jan 20 2008
D-finite with recurrence: n*a(n) +(-5*n+4)*a(n-1) +3*(n-2)*a(n-2) +9*(n-2)*a(n-3)=0. - R. J. Mathar, Dec 02 2012
G.f.: (1+x+1/G(0))/(2*(1-2*x-3*x^2)), where G(k)= 1 + x*(2+3*x)*(4*k+1)/(4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 30 2013
From Peter Bala, Jul 21 2015: (Start)
a(n) = [x^n]( 3*x - 1/(1 - x) )^n.
1 + x*exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 5*x^3 + 13*x^4 + 35*x^5 + ... is the o.g.f. for A005773. (End)
a(n) = (3^n + GegenbauerC(n,-n,-1/2))/2. - Peter Luschny, May 12 2016

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

A066380 a(n) = Sum_{k=0..n} binomial(3*n,k).

Original entry on oeis.org

1, 4, 22, 130, 794, 4944, 31180, 198440, 1271626, 8192524, 53009102, 344212906, 2241812648, 14637774688, 95786202688, 628002401520, 4124304597834, 27126202533252, 178651732923346, 1178005033926998, 7776048412324714

Offset: 0

N. J. A. Sloane, Dec 23 2001

R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 425.

Stefano Spezia, Table of n, a(n) for n = 0..1200 (first 151 terms from Harry J. Smith)

Cf. A001764, A032443, A066381, A371739.

A066380:=n->add(binomial(3*n,k), k=0..n): seq(A066380(n), n=0..20); # Wesley Ivan Hurt, Sep 18 2014

Table[Sum[Binomial[3 n, k], {k, 0, n}], {n, 0, 20}] (* Geoffrey Critzer, May 27 2013 *)
a[n_] := 8^n - (2*n)/(n+1)*Binomial[3*n, n]*Hypergeometric2F1[1, -2*n+1, n+2, -1]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Dec 02 2013 *)

a[0]:1$ a[n]:=8*a[n-1]-(5*n^2+n-2)*(3*n-3)!/((2*n-1)!*n!)$ makelist(a[n],n,0,200); /* Tani Akinari, Sep 02 2014 */

{ for (n=0, 150, a=0; for (k=0, n, a+=binomial(3*n, k)); write("b066380.txt", n, " ", a) ) } \\ Harry J. Smith, Feb 12 2010

a(n) ~ C(3n, n)(2 - 4/n + O(1/n^2)).
G.f.: (1-g)/((3*g-1)*(2*g-1)) where g*(1-g)^2 = x. - Mark van Hoeij, Nov 10 2011
G.f.: x*(d/dx)log((F(x)-1)/(2-F(x))), where F(x) is g.f. of A001764. - Vladimir Kruchinin, Jun 13 2014
a(0)=1, a(n) = 8*a(n-1) - (5*n^2+n-2)*(3*n-3)!/((2*n-1)!*n!). - Tani Akinari, Sep 02 2014
a(n) = [x^n] 1/((1 - 2*x)*(1 - x)^(2*n)). - Ilya Gutkovskiy, Oct 25 2017
a(n) = Sum_{k=0..floor(n/2)} binomial(3*n+1,n-2*k). - Seiichi Manyama, Apr 09 2024
a(n) = binomial(1+3*n, n)*hypergeom([1, (1-n)/2, -n/2], [1+n, 3/2+n], 1). - Stefano Spezia, Apr 09 2024
a(n) = Sum_{k=0..n} 2^(n-k) * binomial(2*n+k-1,k). - Seiichi Manyama, Jul 30 2025
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(3*n,k) * binomial(3*n-k-1,n-k). - Seiichi Manyama, Aug 08 2025
From Seiichi Manyama, Aug 17 2025: (Start)
G.f.: 1/(1 - x*g^2*(6-2*g)) where g = 1+x*g^3 is the g.f. of A001764.
G.f.: g/((2-g) * (3-2*g)) where g = 1+x*g^3 is the g.f. of A001764. (End)

A054143 Triangular array T given by T(n,k) = Sum_{0 <= j <= i-n+k, n-k <= i <= n} C(i,j) for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 4, 7, 1, 5, 11, 15, 1, 6, 16, 26, 31, 1, 7, 22, 42, 57, 63, 1, 8, 29, 64, 99, 120, 127, 1, 9, 37, 93, 163, 219, 247, 255, 1, 10, 46, 130, 256, 382, 466, 502, 511, 1, 11, 56, 176, 386, 638, 848, 968, 1013, 1023, 1, 12, 67, 232, 562, 1024, 1486, 1816, 1981, 2036, 2047

Offset: 0

Clark Kimberling, Mar 18 2000

Row sums given by A001787.
T(n, n) = -1 + 2^(n+1).
T(2*n, n) = 4^n.
T(2*n+1, n) = A000346(n).
T(2*n-1, n) = A032443(n).
A054143 is the fission of the polynomial sequence ((x+1)^n) by the polynomial sequence (q(n,x)) given by q(n,x) = x^n + x^(n-1) + ... + x + 1. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,  3;
  1,  4,  7;
  1,  5, 11, 15;
  1,  6, 16, 26, 31;
  1,  7, 22, 42, 57, 63;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A000346, A001787, A032443.

Diagonal sums give A005672. - Paul Barry, Feb 07 2003

Flat(List([0..12], n-> List([0..n], k-> Sum([n-k..n], i-> Sum([0..i-n+k], j-> Binomial(i,j) ))))); # G. C. Greubel, Aug 01 2019

T:= func< n,k | (&+[ (&+[ Binomial(i,j): j in [0..i-n+k]]): i in [n-k..n]]) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019

A054143_row := proc(n) add(add(binomial(n,n-i)*x^(k+1),i=0..k),k=0..n-1); coeffs(sort(%)) end; seq(print(A054143_row(n)),n=1..6); # Peter Luschny, Sep 29 2011

(* First program *)
z=10;
p[n_,x_]:=(x+1)^n;
q[0,x_]:=1;q[n_,x_]:=x*q[n-1,x]+1;
p1[n_,k_]:=Coefficient[p[n,x],x^k];p1[n_,0]:=p[n,x]/.x->0;
d[n_,x_]:=Sum[p1[n,k]*q[n-1-k,x],{k,0,n-1}]
h[n_]:=CoefficientList[d[n,x],{x}]
TableForm[Table[Reverse[h[n]],{n,0,z}]]
Flatten[Table[Reverse[h[n]],{n,-1,z}]] (* A054143 *)
TableForm[Table[h[n],{n,0,z}]]
Flatten[Table[h[n],{n,-1,z}]] (* A104709 *)
(* Second program *)
Table[Sum[Binomial[i, j], {i, n-k, n}, {j,0,i-n+k}], {n,0,12}, {k,0,n}]// Flatten (* G. C. Greubel, Aug 01 2019 *)

T(n,k) = sum(i=n-k,n, sum(j=0,i-n+k, binomial(i,j)));
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Aug 01 2019

def T(n, k): return sum(sum( binomial(i,j) for j in (0..i-n+k)) for i in (n-k..n))
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019

T(n,k) = Sum_{0 <= j <= i-n+k, n-k <= i <= n} binomial(i,j).
T(n,k) = T(n-1,k) + 3*T(n-1,k-1) - 2*T(n-2,k-1) - 2*T(n-2,k-2), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 30 2013
From Petros Hadjicostas, Jun 05 2020: (Start)
Bivariate o.g.f.: Sum_{n,k >= 0} T(n,k)*x^n*y^k = 1/(1 - x - 3*x*y + 2*x^2*y + 2*x^2*y^2) = 1/((1 - 2*x*y)*(1 - x*(y+1))).
n-th row o.g.f.: ((1 + y)^(n+1) - (2*y)^(n+1))/(1 - y). (End)

Name edited by Petros Hadjicostas, Jun 04 2020

A212402 T(n,k)=Number of binary arrays of length n+2*k-1 with no more than k ones in any length 2k subsequence (=50% duty cycle).

Original entry on oeis.org

3, 11, 5, 42, 19, 8, 163, 74, 33, 13, 638, 291, 132, 57, 21, 2510, 1150, 527, 236, 97, 34, 9908, 4558, 2104, 959, 421, 166, 55, 39203, 18100, 8402, 3872, 1747, 747, 285, 89, 155382, 71971, 33560, 15586, 7143, 3179, 1314, 489, 144, 616666, 286454, 134075

Offset: 1

R. H. Hardin May 14 2012

Table starts
..3..11...42...163...638...2510...9908...39203...155382...616666...2449868
..5..19...74...291..1150...4558..18100...71971...286454..1140954...4547020
..8..33..132...527..2104...8402..33560..134075...535728..2140910...8556568
.13..57..236...959..3872..15586..62632..251419..1008536..4043582..16206152
.21..97..421..1747..7143..29002.117290..473171..1905675..7665886..30810054
.34.166..747..3179.13185..54042.220054..892387..3609005.14567294..58714842
.55.285.1314..5769.24322.100736.413220.1685039..6844362.27724036.112072540
.89.489.2318.10425.44794.187696.776116.3183631.12990818.52815156.214150732

			Some solutions for n=3 k=4
..0....0....0....1....0....0....0....0....1....1....1....0....1....0....1....1
..1....0....1....1....0....0....1....0....1....1....0....1....1....1....0....0
..1....1....1....0....1....0....1....0....1....0....1....0....0....0....0....0
..0....1....1....0....1....0....1....0....0....1....0....1....1....0....1....0
..1....0....0....0....1....1....1....1....0....0....1....0....0....1....0....0
..0....0....0....1....0....1....0....0....0....0....0....0....0....0....1....0
..0....0....0....0....0....0....0....0....0....0....0....0....0....1....1....1
..0....1....1....0....0....1....0....1....0....1....1....0....1....0....0....1
..0....0....0....1....0....1....0....1....1....1....1....1....0....0....0....1
..1....0....0....1....1....0....0....0....1....1....0....1....0....0....1....1

R. H. Hardin, Table of n, a(n) for n = 1..4360

Column 1 is A000045(n+3)
Column 2 is A118647(n+3)
Column 3 is A133551(n+5)
Row 1 is A032443

A100192 a(n) = Sum_{k=0..n} binomial(2n,n+k)2^k.

Original entry on oeis.org

1, 4, 18, 82, 374, 1704, 7752, 35214, 159750, 723880, 3276908, 14821668, 66991436, 302605528, 1366182276, 6165204102, 27811282374, 125415953208, 565408947756, 2548400193852, 11483706241044, 51739037228688, 233070330199296, 1049777052815052, 4727770393417884

Offset: 0

Paul Barry, Nov 08 2004

A transform of 2^n under the mapping g(x)->(1/sqrt(1-4*x))*g(x*c(x)^2), where c(x) is the g.f. of the Catalan numbers A000108. A transform of 3^n under the mapping g(x)->(1/(c(x)*sqrt(1-4*x)))*g(x*c(x)).

Hankel transform is A088138(n+1). - Paul Barry, Jan 11 2007

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

Cf. A032443.

CoefficientList[Series[Sqrt[1-4*x]*(Sqrt[1-4*x]-3*x+1)/((1-4*x)*(2-9*x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

G.f.: (sqrt(1-4*x)+1)/(sqrt(1-4*x)*(3*sqrt(1-4*x)-1)).
G.f.: sqrt(1-4*x)*(sqrt(1-4*x)-3*x+1)/((1-4*x)*(2-9*x)).
a(n) = Sum_{k=0..n} binomial(2*n, n-k)*2^k.
a(n) = Sum_{k=0..n} C(2*n,k)*2^(n-k). - Paul Barry, Jan 11 2007
a(n) = Sum_{k=0..n} C(n+k-1,k)*3^(n-k). - Paul Barry, Sep 28 2007
D-finite with recurrence 2*n*a(n) +(-23*n+16)*a(n-1) +3*(29*n-44)*a(n-2) +54*(-2*n+5)*a(n-3)=0. - R. J. Mathar, Nov 24 2012
a(n) ~ (9/2)^n. - Vaclav Kotesovec, Feb 12 2014
a(n) = [x^n] 1/((1 - x)^n*(1 - 3*x)). - Ilya Gutkovskiy, Oct 12 2017

A371777 a(n) = Sum_{k=0..floor(n/3)} binomial(2n+2,n-3k).

Original entry on oeis.org

1, 4, 15, 57, 220, 858, 3368, 13276, 52479, 207861, 824527, 3274395, 13015081, 51769813, 206045841, 820475513, 3268499356, 13025237058, 51922543076, 207034128448, 825713206746, 3293865399518, 13142007903586, 52443095356218, 209304385553096, 835459642193284

Offset: 0

Seiichi Manyama, Apr 05 2024

nonn

Cf. A371778, A371779, A371780.
Cf. A144904, A371758, A371773.
Cf. A032443.

a(n) = sum(k=0, n\3, binomial(2*n+2, n-3*k));

a(n) = [x^n] 1/(((1-x)^3-x^3) * (1-x)^n).
a(n) = binomial(2*(n+1), n)*hypergeom([1, (1-n)/3, (2-n)/3, -n/3], [1+n/3, (4+n)/3, (5+n)/3], -1). - Stefano Spezia, Apr 06 2024
From Vaclav Kotesovec, Apr 08 2024: (Start)
Recurrence: n*a(n) = 3*(3*n-2)*a(n-1) - 6*(4*n-5)*a(n-2) + 8*(2*n-3)*a(n-3).
G.f.: (1 + sqrt(1-4*x))/(2*(1-x)*(1-4*x)).
a(n) ~ 2^(2*n+1)/3.  (End)

cp's OEIS Frontend

A345915 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum <= 0.

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A345916 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum <= 0.

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A027914 T(n,0) + T(n,1) + ... + T(n,n), T given by A027907.

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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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A066380 a(n) = Sum_{k=0..n} binomial(3*n,k).

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A054143 Triangular array T given by T(n,k) = Sum_{0 <= j <= i-n+k, n-k <= i <= n} C(i,j) for n >= 0 and 0 <= k <= n.

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A100192 a(n) = Sum_{k=0..n} binomial(2n,n+k)2^k.

A371777 a(n) = Sum_{k=0..floor(n/3)} binomial(2n+2,n-3k).