A199033 -id:A199033 - cp's OEIS frontend

A219197 Self-convolution equals A199033.

1, 2, 9, 46, 253, 1452, 8570, 51594, 315225, 1948010, 12147881, 76316508, 482392198, 3064987460, 19560379470, 125309993974, 805458510441, 5192500350906, 33561539356277, 217429403317006, 1411572472199649, 9181398851046632, 59821825063376124, 390382132833183204

Offset: 0

Paul D. Hanna, Nov 14 2012

nonn

Conjecture: a(n) is never congruent to 3 modulo 4.

			G.f.: A(x) = 1 + 2*x + 9*x^2 + 46*x^3 + 253*x^4 + 1452*x^5 +...
where A(x)^2 = 1 + 4*x + 22*x^2 + 128*x^3 + 771*x^4 + 4744*x^5 +...+ A199033(n)*x^n +...
Also, the g.f. A(x) satisfies: A(x) = G(x) * F(x*G(x)^2) where
F(x) = 1 + x + 3*x^2 + 7*x^3 + 19*x^4 + 51*x^5 + 141*x^6 +...+ A002426(n)*x^n +...
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...+ A001764(n)*x^n +...

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

Cf. A199033, A183161, A002426.

A002426[n_] := Sum[Binomial[n, 2*k]*Binomial[2*k, k], {k, 0, Floor[n/2]}]; Table[Sum[A002426[k]*Binomial[3*n - k + 1, n - k]*(2*k + 1)/(3*n - k + 1), {k, 0, n}], {n, 0, 50} ] (* G. C. Greubel, Mar 06 2017 *)

{a(n)=local(A2=sum(m=0, n, sum(k=0, m, binomial(m+k+1, m-k)*binomial(2*m-k+1, k))*x^m+x*O(x^n))); polcoeff(A2^(1/2), n)}
for(n=0, 30, print1(a(n), ", "))

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

{A002426(n)=sum(k=0, n\2, binomial(n, 2*k)*binomial(2*k, k))}
{a(n)=if(n==0, 1, sum(k=0, n, A002426(k)*binomial(3*n-k+1, n-k)*(2*k+1)/(3*n-k+1)))}
for(n=0, 30, print1(a(n), ", "))

{A097893(n)=sum(m=0, n, sum(k=0, m\2, binomial(m, 2*k)*binomial(2*k, k)))}
{a(n)=if(n==0, 1, sum(k=0, n, A097893(k)*binomial(3*n-k, n-k)*2*k/(3*n-k)))}
for(n=0, 30, print1(a(n), ", "))

Sum_{k=0..n} a(n-k)*a(k) = Sum_{k=0..n} C(n+k+1,n-k)*C(2*n-k+1,k) = A199033(n).
G.f.: A(x) = G(x) / sqrt(1 - 2*x*G(x)^2 - 3*x^2*G(x)^4), where G(x) = 1 + x*G(x)^3 = g.f. of A001764.
G.f.: A(x) = Sum_{n>=0} A002426(n) * x^n * G(x)^(2*n+1), where A002426 are the central trinomial coefficients and G(x) = 1 + x*G(x)^3 = g.f. of A001764.
a(n) = Sum_{k=0..n} A002426(k) * C(3*n-k+1,n-k) * (2*k+1)/(3*n-k+1) for n>0, where A002426 are the central trinomial coefficients.
From Vaclav Kotesovec, Oct 05 2020: (Start)
Recurrence: 32*(n-1)*n*(2*n + 1)*(49*n^2 - 210*n + 222)*a(n) = 4*(n-1)*(10388*n^4 - 55104*n^3 + 96925*n^2 - 64446*n + 15006)*a(n-1) - 6*(22050*n^5 - 173439*n^4 + 536588*n^3 - 814340*n^2 + 604331*n - 174702)*a(n-2) - 81*(n-2)*(3*n - 7)*(3*n - 5)*(49*n^2 - 112*n + 61)*a(n-3).
a(n) ~ 3^(3*n + 7/4) / (Gamma(1/4) * n^(3/4) * 2^(2*n + 5/2)). (End)

A183160 a(n) = Sum_{k=0..n} C(n+k,n-k)C(2n-k,k).

Original entry on oeis.org

1, 2, 11, 62, 367, 2232, 13820, 86662, 548591, 3498146, 22436251, 144583496, 935394436, 6071718512, 39523955552, 257913792342, 1686627623151, 11050540084902, 72522925038257, 476669316338542, 3137209052543927, 20672732229560032, 136374124374593072, 900541325129687272

Offset: 0

Paul D. Hanna, Dec 27 2010

nonn

			G.f.: A(x) = 1 + 2*x + 11*x^2 + 62*x^3 + 367*x^4 + 2232*x^5 +...
A(x)^(1/2) = 1 + x + 5*x^2 + 26*x^3 + 145*x^4 + 841*x^5 + 5006*x^6 +...+ A183161(n)*x^n +...
Given triangle A085478(n,k) = C(n+k,n-k), which begins:
  1;
  1,  1;
  1,  3,  1;
  1,  6,  5,  1;
  1, 10, 15,  7, 1;
  1, 15, 35, 28, 9, 1; ...
ILLUSTRATE formula a(n) = Sum_{k=0..n} A085478(n,k)*A085478(n,n-k):
a(2) = 11 = 1*1 + 3*3 + 1*1;
a(3) = 62 = 1*1 + 6*5 + 5*6 + 1*1;
a(4) = 367 = 1*1 + 10*7 + 15*15 + 7*10 + 1*1;
a(5) = 2232 = 1*1 + 15*9 + 35*28 + 28*35 + 9*15 + 1*1; ...

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

Cf. A001764, A085478, A183161, A199033, A218622.
Cf. A026641, A147855, A371753.
Cf. A005809, A006256, A160906, A226751.
Cf. A171822.

[(&+[Binomial(n+k, 2*k)*Binomial(2*n-k, k): k in [0..n]]): n in [0..25]]; // G. C. Greubel, Feb 22 2021

Table[Sum[Binomial[n+k,n-k]Binomial[2n-k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Jul 19 2011 *)
Table[HypergeometricPFQ[{-n, -n, 1/2 -n, n+1}, {1/2, 1, -2*n}, 1], {n, 0, 25}] (* G. C. Greubel, Feb 22 2021 *)

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

{a(n)=local(G=1); for(i=0, n, G=1+x*G^3+O(x^(n+1))); polcoeff(1/(1-2*x*G^2-3*x^2*G^4), n)} \\ Paul D. Hanna, Nov 03 2012

{a(n)=local(G=1); for(i=0, n, G=1+x*G^3+O(x^(n+1))); polcoeff(1/(1+3*x*G-5*x*G^2), n)} \\ Paul D. Hanna, Jun 16 2013
for(n=0, 30, print1(a(n), ", "))

a = lambda n: binomial(3*n+1,n)*hypergeometric([1,-n],[2*n+2],2)
[simplify(a(n)) for n in range(26)] # Peter Luschny, May 19 2015

a(n) = Sum_{k=0..n} A085478(n,k)*A085478(n,n-k).
Self-convolution of A183161 (an integer sequence):
a(n) = Sum_{k=0..n} A183161(k)*A183161(n-k).
a(n) = Sum_{k=0..n} binomial(2*n+k,k) * cos((n+k)*Pi). - Arkadiusz Wesolowski, Apr 02 2012
Recurrence: 320*n*(2*n-1)*a(n) = 8*(346*n^2 + 79*n - 327)*a(n-1) + 6*(1688*n^2-6241*n+5981)*a(n-2) + 261*(3*n-7)*(3*n-5)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 3^(3*n+3/2)/(2^(2*n+3)*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 20 2012
...
G.f.: A(x) = 1/(1 - 2*x*G(x)^2 - 3*x^2*G(x)^4), where G(x) = 1 + x*G(x)^3 = g.f. of A001764. - Paul D. Hanna, Nov 03 2012
G.f.: A(x) = 1 + x*d/dx { log( G(x)^5/(1+x*G(x)^2) )/2 }, where G(x) = 1 + x*G(x)^3 = g.f. of A001764. - Paul D. Hanna, Nov 04 2012
G.f.: A(x) = 1/(1 + 3*x*G(x) - 5*x*G(x)^2), where G(x) = 1 + x*G(x)^3 = g.f. of A001764. - Paul D. Hanna, Jun 16 2013
a(n) = C(3*n+1,n)*Hyper2F1([1,-n],[2*n+2],2). - Peter Luschny, May 19 2015
a(n) = [x^n] 1/((1 - x^2)*(1 - x)^(2*n)). - Ilya Gutkovskiy, Oct 25 2017
From G. C. Greubel, Feb 22 2021: (Start)
a(n) = Sum_{k=0..n} A171822(n, k).
a(n) = Hypergeometric 4F3([-n, -n, 1/2 -n, n+1], [1/2, 1, -2*n], 1). (End)
a(n) = Sum_{k=0..floor(n/2)} binomial(3*n-2*k-1,n-2*k). - Seiichi Manyama, Apr 05 2024
a(n) = Sum_{k=0..n} (-2)^(n-k) * binomial(3*n+1,k). - Seiichi Manyama, Aug 03 2025
From Seiichi Manyama, Aug 14 2025: (Start)
a(n) = Sum_{k=0..n} (-1)^k * 2^(n-k) * binomial(3*n+1,k) * binomial(3*n-k,n-k).
G.f.: g^2/((-1+2*g) * (3-2*g)) where g = 1+x*g^3 is the g.f. of A001764. (End)
G.f.: B(x)^2/(1 + 4*(B(x)-1)/3), where B(x) is the g.f. of A005809. - Seiichi Manyama, Aug 15 2025

A201635 Triangle formed by T(n,n) = (-1)^n*Sum_{j=0..n} C(-n,j), T(n,k) = Sum_{j=0..k} T(n-1,j) for k=0..n-1, and n>=0, read by rows.

Original entry on oeis.org

1, 1, 0, 1, 1, 2, 1, 2, 4, 6, 1, 3, 7, 13, 22, 1, 4, 11, 24, 46, 80, 1, 5, 16, 40, 86, 166, 296, 1, 6, 22, 62, 148, 314, 610, 1106, 1, 7, 29, 91, 239, 553, 1163, 2269, 4166, 1, 8, 37, 128, 367, 920, 2083, 4352, 8518, 15792, 1, 9, 46, 174, 541, 1461, 3544, 7896

Offset: 0

Peter Luschny, Nov 14 2012

Notation: If a sequence id is starred then the offset and/or some terms are different. Starred terms indicate the variance.
Row sums: [A026641 ] [1,  1,  4,  13,  46,  166,   610]
--
T(j+2, 2) [A000124*] [1*, 2 , 4,   7,  11,  16,     22]
T(j+3, 3) [A003600*] [1*, 2*, 6,  13,  24,  40,     62]
--
T(j  , j) [A072547 ] [1,  0,  2,   6,  22,  80,    296]
T(j+1, j) [A026641 ] [1,  1,  4,  13,  46,  166,   610]
T(j+2, j) [A014300 ] [1,  2,  7,  24,  86,  314,  1163]
T(j+3, j) [A014301*] [1,  3, 11,  40, 148,  553,  2083]
T(j+4, j) [A172025 ] [1,  4, 16,  62, 239,  920,  3544]
T(j+5, j) [A172061 ] [1,  5, 22,  91, 367, 1461,  5776]
T(j+6, j) [A172062 ] [1,  6, 29, 128, 541, 2232,  9076]
T(j+7, j) [A172063 ] [1,  7, 37, 174, 771, 3300, 13820]
--
T(2j  ,j) [Central ] [1,  1,  7,  40, 239, 1461,  9076]
T(2j+1,j) [A183160 ] [1,  2, 11,  62, 367, 2232, 13820]
T(2j+2,j) [        ] [1,  3, 16,  91, 541, 3300, 20476]
T(2j+3,j) [A199033*] [1,  4, 22, 128, 771, 4744, 29618]

			Triangle begins as:
[n]|k->
[0] 1
[1] 1, 0
[2] 1, 1,  2
[3] 1, 2,  4,  6
[4] 1, 3,  7, 13,  22
[5] 1, 4, 11, 24,  46,  80
[6] 1, 5, 16, 40,  86, 166, 296
[7] 1, 6, 22, 62, 148, 314, 610, 1106.

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

A201635 := proc(n,k) option remember; local j;
if n=k then (-1)^n*add(binomial(-n,j), j=0..n)
else add(A201635(n-1,j), j=0..k) fi end:
for n from 0 to 7 do seq(A(n,k), k=0..n) od;

T[n_, k_]:= T[n, k]= If[k==n, (-1)^n*Sum[Binomial[-n, j], {j, 0, n}], Sum[T[n-1, j], {j, 0, k}]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 27 2019 *)

{T(n,k) = if(k==n, (-1)^n*sum(j=0,n, binomial(-n,j)), sum(j=0,k, T(n-1,j)))};
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 27 2019

@CachedFunction
def A201635(n, k):
    if n==k: return (-1)^n*add(binomial(-n, j) for j in (0..n))
    return add(A201635(n-1, j) for j in (0..k))
for n in (0..7) : [A201635(n, k) for k in (0..n)]

cp's OEIS Frontend

A219197 Self-convolution equals A199033.

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A183160 a(n) = Sum_{k=0..n} C(n+k,n-k)*C(2*n-k,k).

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A201635 Triangle formed by T(n,n) = (-1)^n*Sum_{j=0..n} C(-n,j), T(n,k) = Sum_{j=0..k} T(n-1,j) for k=0..n-1, and n>=0, read by rows.

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A183160 a(n) = Sum_{k=0..n} C(n+k,n-k)C(2n-k,k).