A032443 -id:A032443 - cp's OEIS frontend

A062344 Triangle of binomial(2*n, k) with n >= k.

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

Offset: 0

Henry Bottomley, Jul 06 2001

From Wolfdieter Lang, Sep 19 2012: (Start)
The triangle a(n,k) appears in the formula F(2*l+1)^(2*n) = (sum(a(n,k)*L(2*(n-k)*(2*l+1)),k=0..n-1) + a(n,n))/5^n, n>=0, l>=0, with F=A000045 (Fibonacci) and L=A000032 (Lucas).
The signed triangle as(n,k):=a(n,k)*(-1)^k appears in the formula F(2*l)^(2*n) = (sum(as(n,k)*L(4*(n-k)*l),k=0..n-1) + as(n,n))/5^n, n>=0, l>=0. Proof with the Binet-de Moivre formula for F and L and the binomial formula. (End)

			Rows start
  (1),
  (1,2),
  (1,4,6),
  (1,6,15,20)
  etc.
Row n=2, (1,4,6):
F(2*l+1)^4 = (1*L(4*(2*l+1)) + 4*L(2*(2*l+1)) + 6)/25,
F(2*l)^4 = (1*L(8*l) - 4*L(4*l) + 6)/25, l>=0, F=A000045, L=A000032. See a comment above. - _Wolfdieter Lang_, Sep 19 2012

G. C. Greubel, Rows n = 0..100 of triangle, flattened
E. H. M. Brietzke, An identity of Andrews and a new method for the Riordan array proof of combinatorial identities, Discrete Math., 308 (2008), 4246-4262.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
Index entries for triangles and arrays related to Pascal's triangle

Columns include (sometimes truncated) A000012, A005843, A000384, A002492, A053134 etc. Right hand side includes A000984, A001791, A002694, A002696 etc. Row sums are A032443. Row alternate differences (e.g., 6-4+1=3 or 20-15+6-1=10) are A001700.
Cf. A122366.
a(2*n,n) gives A005810.

[[Binomial(2*n, k): k in [0..n]]: n in [0..20]]; // G. C. Greubel, Jun 28 2018

Flatten[Table[Binomial[2 n, k], {n, 0, 20}, {k, 0, n}]] (* G. C. Greubel, Jun 28 2018 *)

create_list(binomial(2*n,k),n,0,9,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

for(n=0, 20, for(k=0, n, print1(binomial(2*n, k), ", "))) \\ G. C. Greubel, Jun 28 2018

a(n,k) = a(n,k-1)*((2*n+1)/k-1) with a(n,0)=1.

G.f.: 1/((1-sqrt(1-4*x*y))^4/(16*x*y^2) + sqrt(1-4*x*y) - x). - Vladimir Kruchinin, Jan 26 2021

A100193 a(n) = Sum_{k=0..n} binomial(2n,n+k)3^k.

Original entry on oeis.org

1, 5, 27, 146, 787, 4230, 22686, 121476, 649731, 3472382, 18546922, 99023292, 528535726, 2820451964, 15048601308, 80283276936, 428271193827, 2284478396334, 12185310873138, 64993897108236, 346655914156602, 1848916875734004, 9861224376230628, 52594507923308856

Offset: 0

Paul Barry, Nov 08 2004

A transform of 3^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 4^n under the mapping g(x)->(1/(c(x)*sqrt(1-4*x)))*g(x*c(x)).

Hankel transform is A127357. In general, the Hankel transform of Sum_{k=0..n} C(2n,k)*r^(n-k) is the sequence with g.f. 1/(1-2x+r^2*x^2). - Paul Barry, Jan 11 2007

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A032443, A100192, A000108, A127357.

Table[Binomial[2*n,n]*Hypergeometric2F1[1,-n,1+n,-3],{n,0,20}] (* Vaclav Kotesovec, Feb 03 2014 *)

G.f.: (sqrt(1-4x)+1)/(sqrt(1-4x)*(4*sqrt(1-4x)-2)).
G.f.: sqrt(1-4x)*(3*sqrt(1-4x)-8x+3)/((1-4x)(6-32x)).
a(n) = Sum_{k=0..n} binomial(2n, n-k)*3^k.
a(n) = (Sum_{k=0..n} binomial(2n, n-k))*(Sum_{j=0..n} binomial(n, j)*(-1)^(n-j)*4^j).
a(n) = Sum_{k=0..n} C(n+k-1,k)*4^(n-k). - Paul Barry, Sep 28 2007
Conjecture: 9*n*a(n) + 6*(11-18*n)*a(n-1) + 16*(26*n-37)*a(n-2) + 256*(5-2*n)*a(n-3) = 0. - R. J. Mathar, Nov 09 2012
a(n) ~ (16/3)^n. - Vaclav Kotesovec, Feb 03 2014
a(n) = [x^n] 1/((1 - x)^n*(1 - 4*x)). - Ilya Gutkovskiy, Oct 12 2017

A300192 Triangle read by rows: row n consists of the coefficients of the expansion of the polynomial (x^2 + 2x + 1)^n + (x^2 - 1)(x + 1)^n.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 1, 0, 2, 6, 6, 2, 0, 3, 13, 22, 18, 7, 1, 0, 4, 23, 56, 75, 60, 29, 8, 1, 0, 5, 36, 115, 215, 261, 215, 121, 45, 10, 1, 0, 6, 52, 206, 495, 806, 938, 798, 496, 220, 66, 12, 1, 0, 7, 71, 336, 987, 2016, 3031, 3452, 3010, 2003, 1001, 364, 91

Offset: 0

Franck Maminirina Ramaharo, Feb 28 2018

			The triangle T(n, k) begins:
n\k  0  1   2    3    4     5     6     7     8     9    10   11  12  13 14
0:   0  0   1
1:   0  1   2    1
2:   0  2   6    6    2
3:   0  3  13   22   18     7     1
4:   0  4  23   56   75    60    29     8     1
5:   0  5  36  115  215   261   215   121    45    10     1
6:   0  6  52  206  495   806   938   798   496   220    66   12   1
7:   0  7  71  336  987  2016  3031  3452  3010  2003  1001  364  91  14  1

M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972.
J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996.

Paul Barry, On the Connection Coefficients of the Chebyshev-Boubaker polynomials, The Scientific World Journal, Volume 2013 (2013), Article ID 657806, 10 pages.
Milan Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
A. M. Mathai and P. N. Rathie, Enumeration of almost cubic maps, Journal of Combinatorial Theory, Series B, Vol 13 (1972), 83-90.
Franck Ramaharo, A one-variable bracket polynomial for some Turk's head knots, arXiv:1807.05256 [math.CO], 2018.
Franck Ramaharo, A bracket polynomial for 2-tangle shadows, arXiv:2002.06672 [math.CO], 2020.

Row sums: A000302 (powers of 4).

Cf. A034870, A034871, A032443.

T := (n, k) -> binomial(2*n, k) + binomial(n, k - 2) - binomial(n, k);
for n from 0 to 10 do seq(T(n, k), k = 0 .. max(2*n, n + 2)) od;

T(n, k) := binomial(2*n, k) + binomial(n, k - 2) - binomial(n, k)$
a : []$
for n:0 thru 10 do
  a : append(a, makelist(T(n, k), k, 0, max(2*n, n + 2)))$
a;

row(n) = Vecrev((x^2 + 2*x + 1)^n + (x^2 - 1)*(x + 1)^n); \\ Michel Marcus, Nov 12 2022

T(n,k) = binomial(2*n,k) + binomial(n,k-2) - binomial(n,k).
T(n,k) = T(n-1,k-1)+ T(n-1,k) + A034871(n-1,k-1), with T(n,0) = T(0,1) = 0 and T(0,2) = 1
T(n,1) = A001477(n).
T(n,2) = A143689(n).
T(n,3) = n + A002492(n-1) - A000292(n-2).
T(n,n) = A247493(n+1,n).
T(n,n+1) = n + A001791(n).
T(n,n+2) = 1 + A002694(n), n >= 2.
T(n,n+k) = binomial(2*n, n-k) = A094527(n,k), for k >= 3 and n>=k.
G.f.: 1/(1 - y*(x^2 + 2*x + 1)) + (x^2 - 1)/(1 - y*(x + 1)).

A345926 Number of distinct possible alternating sums of permutations of the multiset of prime indices of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 2, 2, 2, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 4, 1, 2, 2, 1, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 1, 4, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 2, 1, 2, 2, 3

Offset: 1

Gus Wiseman, Jul 14 2021

nonn

First differs from A096825 at a(90) = 3, A096825(90) = 4.
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.
The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. Of course, the alternating sum of prime indices is also the reverse-alternating sum of reversed prime indices.
Also the number of possible values of A056239(d) where d is a divisor of n with half as many prime factors (rounded up) as n.

			Grouping the 12 permutations of {1,2,2,3} by alternating sum k gives:
  k = -2: (1223) (1322) (2213) (2312)
  k =  0: (1232) (2123) (2321) (3212)
  k =  2: (2132) (2231) (3122) (3221)
so a(90) = 3.

The version for prime factors instead of indices is A343943.
A000005 counts divisors.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A001414 adds up prime factors, row sums of A027746.
A056239 adds up prime indices,  row sums of A112798.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A345197 counts compositions by length and alternating sum.
A344610 counts partitions by sum and positive reverse-alternating sum.
Cf. A008549, A032443, A083399, A096825, A239830, A344607, A344609, A344651, A345957, A345960, A345961.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Table[Length[Union[ats/@Permutations[primeMS[n]]]],{n,100}]

from sympy import factorint, primepi
from sympy.utilities.iterables import multiset_combinations
def A345926(n):
    fs = dict((primepi(a),b) for (a,b) in factorint(n).items())
    return len(set(sum(d) for d in multiset_combinations(fs, (sum(fs.values())+1)//2))) # Chai Wah Wu, Aug 23 2021

A371739 a(n) = Sum_{k=0..n} binomial(5*n,k).

Original entry on oeis.org

1, 6, 56, 576, 6196, 68406, 768212, 8731848, 100146724, 1156626990, 13432735556, 156713948672, 1835237017324, 21560768699762, 253994850228896, 2999267652451776, 35490014668470052, 420718526924212654, 4995548847105422048, 59402743684137281920

Offset: 0

Seiichi Manyama, Apr 05 2024

Cf. A032443, A066380, A066381.

Cf. A002294, A163455, A386699.

Table[32^n - Binomial[5*n, 1+n] * Hypergeometric2F1[1, 1 - 4*n, 2+n, -1], {n, 0, 20}] (* Vaclav Kotesovec, Apr 05 2024 *)

PARI
```
a(n) = sum(k=0, n, binomial(5*n, k));
```

a(n) = [x^n] 1/((1-2*x) * (1-x)^(4*n)).
a(n) ~ 5^(5*n + 1/2) / (3*sqrt(Pi*n) * 2^(8*n - 1/2)). - Vaclav Kotesovec, Apr 05 2024
a(n) = Sum_{k=0..floor(n/2)} binomial(5*n+1,n-2*k). - Seiichi Manyama, Apr 09 2024
a(n) = binomial(1+5*n, n)*hypergeom([1, (1-n)/2, -n/2], [1+2*n, 3/2+2*n], 1). - Stefano Spezia, Apr 09 2024
a(n) = Sum_{k=0..n} 2^(n-k) * binomial(4*n+k-1,k). - Seiichi Manyama, Jul 30 2025
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(5*n,k) * binomial(5*n-k-1,n-k). - Seiichi Manyama, Aug 08 2025
G.f.: g/((2-g) * (5-4*g)) where g = 1+x*g^5 is the g.f. of A002294. - Seiichi Manyama, Aug 12 2025
G.f.: 1/(1 - x*g^4*(10-4*g)) where g = 1+x*g^5 is the g.f. of A002294. - Seiichi Manyama, Aug 17 2025

A293574 a(n) = Sum_{k=0..n} n^(n-k)*binomial(n+k-1,k).

Original entry on oeis.org

1, 2, 11, 82, 787, 9476, 139134, 2422218, 48824675, 1118286172, 28679699578, 814027423892, 25330145185646, 857375286365768, 31360145331198428, 1232586016712594010, 51805909208539809315, 2318588202311267591852, 110085368092924083334626, 5526615354023679440754396, 292501304641192746350100410

Offset: 0

Ilya Gutkovskiy, Oct 12 2017

nonn

a(n) is the n-th term of the main diagonal of iterated partial sums array of powers of n (see example).

			For n = 2 we have:
----------------------------
0   1   [2]   3    4     5
----------------------------
1,  2,   4,   8,  16,   32, ... A000079 (powers of 2)
1,  3,   7,  15,  31,   63, ... A126646 (partial sums of A000079)
1,  4, [11], 26,  57,  120, ... A000295 (partial sums of A126646)
----------------------------
therefore a(2) = 11.

Cf. A000312, A001700, A031973, A032443, A100192, A100193.

Join[{1}, Table[Sum[n^(n - k) Binomial[n + k - 1, k], {k, 0, n}], {n, 1, 20}]]
Table[SeriesCoefficient[1/((1 - x)^n (1 - n x)), {x, 0, n}], {n, 0, 20}]
Join[{1, 2}, Table[n^(2 n)/(n - 1)^n - Binomial[2 n, n + 1] Hypergeometric2F1[1, 2 n + 1, n + 2, 1/n]/n, {n, 2, 20}]]

a(n) = sum(k=0, n, n^(n-k)*binomial(n+k-1,k)); \\ Michel Marcus, Oct 12 2017

a(n) = [x^n] 1/((1 - x)^n*(1 - n*x)).

a(n) ~ exp(1) * n^n. - Vaclav Kotesovec, Oct 16 2017

A110166 Row sums of Riordan array A110165.

Original entry on oeis.org

1, 4, 18, 85, 410, 1999, 9807, 48304, 238570, 1180615, 5851253, 29033074, 144190943, 716652070, 3564079250, 17734184365, 88280673770, 439625873215, 2189988826125, 10912480440850, 54389237971285, 271142650382080

Offset: 0

Paul Barry, Jul 14 2005

Number of 5-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
From Peter Bala, Jan 09 2022: (Start)
Conjectures: for k >= 2, the number of k-ary words of length n such that the number of 1's <= the number of 0's is equal to the coefficient of x^n in the expansion of ( k*x + 1/(1 + x) )^n, and satisfies the recurrence u(0) = 1, u(1) = k-1 and n*u(n) = (k-2)*(2*n-1)*u(n-1) - k*(k-4)*(n-1)* u(n-2) + k^(n-1) for n >= 2.
For cases see A027306 (k = 2), A027914 (k = 3) and A032443 (k = 4). (End)

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

Row sums A110165. Cf. A026375, A032443, A246437, A242586.

seq( (1/2)*(5^n + add(binomial(n,k)*binomial(2*k,k), k = 0..n)), n = 0..30); # Peter Bala, Jan 08 2022

Table[Sum[Sum[Binomial[n,j]Binomial[2j,j+k],{j,0,n}],{k,0,n}],{n,0,25}] (* Harvey P. Dale, Dec 16 2011 *)

G.f.: (1/sqrt(1-6*x+5*x^2))/(1-(1-3*x-sqrt(1-6*x+5*x^2))/(2*x)).
a(n) = Sum_{k = 0..n} Sum_{j = 0..n} C(n, j)*C(2*j, j+k).
Recurrence: n*a(n) = (11*n-8)*a(n-1) - 5*(7*n-10)*a(n-2) + 25*(n-2)*a(n-3). - Vaclav Kotesovec, Oct 18 2012
a(n) ~ 5^n/2*(1+sqrt(5)/(2*sqrt(Pi*n))). - Vaclav Kotesovec, Oct 18 2012
From Peter Bala, Jan 08 2022: (Start)
a(n) = (1/2)*(5^n + A026375(n)) = (1/2)*(5^n + Sum_{k = 0..n} binomial(n,k) *binomial(2*k,k)).
a(n) = (1/2)*(5^n)*(1 + Sum_{k = 0..n} binomial(n,k)*binomial(2*k,k)*(-1/5)^k).
a(n) = [x^n] ( 5*x + 1/(1 + x) )^n.
a(0) = 1, a(1) = 4 and n*a(n) = 3*(2*n-1)*a(n-1) - 5*(n-1)*a(n-2) + 5^(n-1) for n >= 2.
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.
Binomial transform of A032443. (End)

A356267 a(n) = Sum_{k=0..n} binomial(2n, k) p(k), where p(k) is the partition function A000041.

Original entry on oeis.org

1, 3, 17, 97, 583, 3275, 18988, 104821, 584441, 3180889, 17295626, 92225785, 492811733, 2590911097, 13591889993, 70605682273, 365601169939, 1876312271003, 9605682510676, 48809295651049, 247315330613099, 1245888505795725, 6256686417801919, 31260996876796579

Offset: 0

Vaclav Kotesovec, Aug 01 2022

nonn

Cf. A000041, A032443, A218481, A286955, A356268.

Table[Sum[Binomial[2*n, k] * PartitionsP[k], {k, 0, n}], {n, 0, 30}]

a(n) ~ erfc(Pi/(2*sqrt(6))) * 2^(2*n - 3) * exp(Pi*sqrt(2*n/3) + Pi^2/24) / (sqrt(3)*n).

A356268 a(n) = Sum_{k=0..n} binomial(2n, k) q(k), where q(k) is the number of partitions into distinct parts (A000009).

Original entry on oeis.org

1, 3, 11, 62, 289, 1472, 7581, 38014, 184453, 918512, 4548393, 22077762, 107423503, 516720332, 2483445404, 11959145079, 57022343425, 270173627092, 1282971321633, 6047971597490, 28446033085527, 133714464665108, 625893086713686, 2919093380089383, 13596052503945537

Offset: 0

Vaclav Kotesovec, Aug 01 2022

nonn

Cf. A000009, A032443, A266232, A307496, A356267.

Table[Sum[Binomial[2*n, k] * PartitionsQ[k], {k, 0, n}], {n, 0, 30}]

a(n) ~ erfc(Pi/(4*sqrt(3))) * 2^(2*n - 3) * exp(Pi*sqrt(n/3) + Pi^2/48) / (3^(1/4) * n^(3/4)).

A356280 a(n) = Sum_{k=0..n} binomial(2n, n-k) p(k), where p(k) is the partition function A000041.

Original entry on oeis.org

1, 3, 12, 50, 211, 894, 3791, 16068, 68032, 287675, 1214761, 5122428, 21571028, 90718913, 381050570, 1598645263, 6699355413, 28044720813, 117281866330, 489999068614, 2045341248508, 8530263939665, 35547083083270, 148015639243691, 615870619714675, 2560734764460360

Offset: 0

Vaclav Kotesovec, Aug 01 2022

nonn

Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, VI.26. Catalan sums, p.417.

Cf. A000041, A032443, A218481, A286955, A356267, A356281.

Table[Sum[PartitionsP[k]*Binomial[2*n, n-k], {k, 0, n}], {n, 0, 30}]
nmax = 30; CoefficientList[Series[Sum[PartitionsP[k]*((1-2*x-Sqrt[1-4*x])/(2*x))^k / Sqrt[1-4*x], {k, 0, nmax}], {x, 0, nmax}], x]

a(n) ~ 2^(2*n - 1/2) * exp(3^(1/3) * Pi^(4/3) * n^(1/3) / 4) / (3*Pi*n)^(2/3).

cp's OEIS Frontend

A062344 Triangle of binomial(2*n, k) with n >= k.

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A100193 a(n) = Sum_{k=0..n} binomial(2*n,n+k)*3^k.

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A300192 Triangle read by rows: row n consists of the coefficients of the expansion of the polynomial (x^2 + 2*x + 1)^n + (x^2 - 1)*(x + 1)^n.

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A345926 Number of distinct possible alternating sums of permutations of the multiset of prime indices of n.

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A371739 a(n) = Sum_{k=0..n} binomial(5*n,k).

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A293574 a(n) = Sum_{k=0..n} n^(n-k)*binomial(n+k-1,k).

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A110166 Row sums of Riordan array A110165.

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A356267 a(n) = Sum_{k=0..n} binomial(2*n, k) * p(k), where p(k) is the partition function A000041.

A100193 a(n) = Sum_{k=0..n} binomial(2n,n+k)3^k.

A300192 Triangle read by rows: row n consists of the coefficients of the expansion of the polynomial (x^2 + 2x + 1)^n + (x^2 - 1)(x + 1)^n.

A356267 a(n) = Sum_{k=0..n} binomial(2n, k) p(k), where p(k) is the partition function A000041.

A356268 a(n) = Sum_{k=0..n} binomial(2n, k) q(k), where q(k) is the number of partitions into distinct parts (A000009).

A356280 a(n) = Sum_{k=0..n} binomial(2n, n-k) p(k), where p(k) is the partition function A000041.