cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A216795 a(n) = sum_{k=0..n} binomial(n,k)^4 * 3^k.

Original entry on oeis.org

1, 4, 58, 1000, 19426, 412744, 9195796, 212836432, 5062716850, 123033947464, 3041489363188, 76243484446672, 1933564156575364, 49518970223489680, 1278877982692134568, 33269141292429734560, 870987510534775369810, 22930499187530338390600, 606700679139764282611540
Offset: 0

Views

Author

Vaclav Kotesovec, Sep 16 2012

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k]^4*3^k, {k, 0, n}], {n, 0, 25}]

Formula

Recurrence: -(x-1)^4*(n+1)^3*(n+2)*(16*(4*x^2 + 17*x + 4)*n^4 + 184*(4*x^2 + 17*x + 4)*n^3 + (3137*x^2 + 13351*x + 3137)*n^2 + (5867*x^2 + 25051*x + 5867)*n + 4061*x^2 + 17438*x + 4061)*a(n) + (n+2)*(64*(4*x^5 + 141*x^4 + 655*x^3 + 655*x^2 + 141*x + 4)*n^7 + 1024*(4*x^5 + 141*x^4 + 655*x^3 + 655*x^2 + 141*x + 4)*n^6 + 4*(6857*x^5 + 242368*x^4 + 1126775*x^3 + 1126775*x^2 + 242368*x + 6857)*n^5 + 8*(12439*x^5 + 442336*x^4 + 2059985*x^3 + 2059985*x^2 + 442336*x + 12439)*n^4 + (211031*x^5 + 7579744*x^4 + 35400065*x^3 + 35400065*x^2 + 7579744*x + 211031)*n^3 + (261344*x^5 + 9524206*x^4 + 44667470*x^3 + 44667470*x^2 + 9524206*x + 261344)*n^2 + (174888*x^5 + 6498997*x^4 + 30655175*x^3 + 30655175*x^2 + 6498997*x + 174888)*n + 15*(3251*x^5 + 123835*x^4 + 588594*x^3 + 588594*x^2 + 123835*x + 3251))*a(n+1) -(32*(12*x^4-197*x^3-1030*x^2-197*x + 12)*n^8 + 624*(12*x^4-197*x^3-1030*x^2-197*x + 12)*n^7 + 10*(6295*x^4-103673*x^3-542004*x^2-103673*x + 6295)*n^6 + 4*(74418*x^4-1233343*x^3-6448430*x^2-1233343*x + 74418)*n^5 + (864893*x^4-14467663*x^3-75685660*x^2-14467663*x + 864893)*n^4 + 20*(78938*x^4-1336491*x^3-7002327*x^2-1336491*x + 78938)*n^3 + (1764932*x^4-30321697*x^3-159367410*x^2-30321697*x + 1764932)*n^2 + (1102551*x^4-19262286*x^3-101826010*x^2-19262286*x + 1102551)*n + 10*(29405*x^4-523232*x^3-2793306*x^2-523232*x + 29405))*a(n+2) + (n+3)*(64*(4*x^3 + 21*x^2 + 21*x + 4)*n^7 + 1152*(4*x^3 + 21*x^2 + 21*x + 4)*n^6 + 12*(2899*x^3 + 15226*x^2 + 15226*x + 2899)*n^5 + 4*(35609*x^3 + 187226*x^2 + 187226*x + 35609)*n^4 + (340693*x^3 + 1795162*x^2 + 1795162*x + 340693)*n^3 + (474743*x^3 + 2511132*x^2 + 2511132*x + 474743)*n^2 + (355831*x^3 + 1894439*x^2 + 1894439*x + 355831)*n + 10*(11039*x^3 + 59401*x^2 + 59401*x + 11039))*a(n+3) -(n+3)*(n+4)^3*(16*(4*x^2 + 17*x + 4)*n^4 + 120*(4*x^2 + 17*x + 4)*n^3 + (1313*x^2 + 5599*x + 1313)*n^2 + 15*(103*x^2 + 443*x + 103)*n + 659*x^2 + 2882*x + 659)*a(n+4) = 0, this is case x=3.
a(n) ~ (1+3^(1/4))^3/(4*sqrt(2)*3^(3/8)*Pi^(3/2)) * (1+3^(1/4))^(4*n)/n^(3/2). - Vaclav Kotesovec, Sep 19 2012
Generally, Sum_{k=0..n} binomial(n,k)^p*x^k is asymptotic a(n) ~ (1+x^(1/p))^(p*n+p-1)/sqrt((2*pi*n)^(p-1)*p*x^(1-1/p)). This is case p=4, x=3. - Vaclav Kotesovec, Sep 19 2012

Extensions

Minor edits by Vaclav Kotesovec, Mar 31 2014

A328808 Constant term in the expansion of (3 + x + y + z + 1/x + 1/y + 1/z + x*y + y*z + z*x + 1/(x*y) + 1/(y*z) + 1/(z*x) + x*y*z + 1/(x*y*z))^n.

Original entry on oeis.org

1, 3, 23, 225, 2583, 32133, 422069, 5757699, 80790775, 1158593589, 16905540753, 250185539079, 3746205581589, 56652844671855, 864032059578879, 13274539401672345, 205252378269637815, 3191578469685269925, 49876569284504593505, 782943268394316187815
Offset: 0

Views

Author

Seiichi Manyama, Oct 28 2019

Keywords

Crossrefs

Column k=4 of A328807.

Programs

  • Mathematica
    Table[Sum[Binomial[n, i]*Sum[Binomial[i, j]^4, {j, 0, i}], {i, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 28 2019 *)
  • PARI
    {a(n) = polcoef(polcoef(polcoef((1+(1+x)*(1+y)*(1+z)+(1+1/x)*(1+1/y)*(1+1/z))^n, 0), 0), 0)}
    
  • PARI
    {a(n) = sum(i=0, n, binomial(n, i)*sum(j=0, i, binomial(i, j)^4))}

Formula

a(n) = Sum_{i=0..n} binomial(n,i)*Sum_{j=0..i} binomial(i,j)^4.
From Vaclav Kotesovec, Oct 28 2019: (Start)
Recurrence: n^3*a(n) = (2*n - 1)*(8*n^2 - 8*n + 3)*a(n-1) + (n-1)*(22*n^2 - 44*n + 13)*a(n-2) - 44*(n-2)*(n-1)*(2*n - 3)*a(n-3) + 51*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ sqrt(2) * 17^(n + 3/2) / (64 * Pi^(3/2) * n^(3/2)). (End)

A337332 a(n) = Sum_{k=0..n}C(n,k)*C(n+k,k)*C(2k,k)*C(2n-2k,n-k)*(-8)^(n-k).

Original entry on oeis.org

1, -12, 228, -3504, 44580, -298032, 1407504, -275772096, 21324125988, -966349948080, 32198201397648, -831808446595776, 16275197594916624, -210881419152530112, 1110165241205298240, -28746364298042321664, 4877709692143697517348, -323151109677783574203312, 13976671241536620108719376
Offset: 0

Views

Author

Zhi-Wei Sun, Aug 23 2020

Keywords

Comments

(-1)^n*a(n) > 0, and Sum_{k=0..n} C(n,k)*C(n+k,k)*C(2k,k)*C(2n-2k,n-k)*(-1)^(n-k) = Sum_{k=0..n}C(n,k)^4.
Conjecture 1: Sum_{k>=0}(4k+1) a(k)/(-48)^k = sqrt(72+42*sqrt(3))/Pi.
Conjecture 2: For each n > 0, the number (Sum_{k=0..n-1} (-1)^k*(4k+1)*48^(n-1-k)*a(k))/n is a positive integer.
Conjecture 3: For any prime p > 3, the square of (Sum_{k=0..p-1} (4k+1)a(k)/(-48)^k)/p is congruent to 14*(3/p)-(p/3)-12 modulo p, where (a/p) is the Legendre symbol.
Conjecture 4: Let p > 3 be a prime, and let S(p) = Sum_{k=0..p-1} a(k)/(-48)^k. If p == 1 (mod 4) and p = x^2 + 4y^2 with x and y integers, then S(p) == 4x^2-2p (mod p^2). If p == 3 (mod 4), then S(p) == 0 (mod p^2).

Examples

			a(1) = C(1,0)*C(1,0)*C(0,0)*C(2,1)*(-8) + C(1,1)*C(2,1)*C(2,1)*C(0,0) = -16 + 4 = -12.
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[Binomial[n,k]Binomial[n+k,k]Binomial[2k,k]Binomial[2(n-k),n-k](-8)^(n-k),{k,0,n}];
    Table[a[n],{n,0,18}]

Formula

a(n) = (-8)^n*binomial(2*n, n)*hypergeom([1/2, -n, -n, n + 1], [1, 1, 1/2 - n], 1/8). - Peter Luschny, Aug 24 2020

A357603 a(n) is the number of different pairs of shortest paths in an n X n lattice going between opposite corners in opposite directions and not meeting at their middle point.

Original entry on oeis.org

0, 2, 18, 236, 3090, 42252, 589932, 8383608, 120720402, 1756863020, 25789460268, 381298472568, 5671808350572, 84807208655288, 1273785187835640, 19207311526394736, 290631247129611282, 4411188317020786668, 67137528193253129484, 1024357917198436543800
Offset: 0

Views

Author

Janaka Rodrigo, Oct 05 2022

Keywords

Comments

Equivalently, a(n) is the number of different ways to interchange the positions of two men standing at opposite corners of an n X n lattice without meeting each other.
The two men start to move simultaneously at the same constant speed; one always moves to the right or upward, the other always moves to the left or downward.
All terms are even.

Examples

			Let the lattice points of a lattice of size 2 X 2 be labeled 1,2,3,4,5,6,7,8,9, and let men A and B start at points 1 and 9, respectively.
                          man B
                          starts
             7---8---9 <-- here
             |   |   |
   man A     4---5---6
   starts    |   |   |
    here --> 1---2---3
.
The lattice paths available for A are 14789, 14589, 14569, 12589, 12569, 12369 and those available for B are 98741, 98541, 98521, 96541, 96521, 96321.
A002894(2) = 36 is the number of different ways to exchange positions, that is, 6 X 6 or (C(4,2))^2.
The different ways they can meet halfway on their paths are as follows:
If A selects 14789, B must select 98741. If A selects one of 14589, 14569, 12589, 12569, B must select one of 98541, 98521, 96541, 96521. If A selects 12369, B must select 96321.
Therefore the total number of choices available is 1 + 4*4 + 1 = 18 and this is given by A005260(2) = (C(2,0))^4 + (C(2,1))^4 + (C(2,2))^4 = 18.
Therefore the total number of such different pairs is a(2) = 36 - 18 = 18.
		

Crossrefs

Formula

a(n) = A002894(n) - A005260(n).

A007804 Related to the asymptotic expansion of Sum_{k = 0..n} C(n,k)^4.

Original entry on oeis.org

1, 30, 1730, 152340, 18177750, 2742927876, 500848449300, 107365679361000, 26431620161451750, 7348772237141884500, 2277376143931016207100, 778374526612873263759000, 290867891728117751744917500
Offset: 0

Views

Author

Keywords

Crossrefs

Cf. A005260.

A334237 a(n) = 2*Sum_{k=0..n-1} binomial(n,k)^2*binomial(n,k+1)^2.

Original entry on oeis.org

2, 16, 198, 2368, 30100, 392544, 5248782, 71501056, 989177508, 13859716000, 196282985756, 2805235913088, 40408113882344, 586055349387200, 8551024115349150, 125431745952519168, 1848653992986172324, 27362153523832614432, 406546456064695351020
Offset: 1

Views

Author

Bradley Klee, Apr 19 2020

Keywords

Comments

a(n) is also the number of simultaneous walks between two walkers on an n X n grid, subject to a "social distancing" constraint. The rules are the same as in A005260, but the counting criterion is changed so that the walkers cannot meet. Instead, they must be separated by closest-approach distance of sqrt(2) after n steps. Each term a(n) is a hypergeometric single sum, so Zeilberger's algorithm applies, and a(n) must also satisfy a p-recurrence.

References

  • B. Klee and É. Angelini, "Social Distancing and A005260", [math-fun] mailing list, Apr. 19, 2020.

Crossrefs

Cf. A005260.

Programs

  • Mathematica
    RecurrenceTable[{Dot[{(n-1)^2*(n+1)^3*(5*n^2-10*n+4),
    -2*n^2*(2*n-1)*(15*n^4-30*n^3+7*n^2+8*n-8),
    -4*(n-1)^2*n*(4*n-5)*(4*n-3)*(5*n^2-1)},
    a[n-#]&/@Range[0,2]] == 0, a[0] == 0, a[1] == 2},
    a, {n, 0, 100}]
  • PARI
    a(n) = 2*sum(k=0, n-1, binomial(n,k)^2*binomial(n,k+1)^2); \\ Michel Marcus, Apr 19 2020

Formula

D-finite with recurrence (n-1)^2*(n+1)^3*(5*n^2-10*n+4)*a(n) - 2*n^2*(2*n-1)*(15*n^4-30*n^3+7*n^2+8*n-8)*a(n-1) - 4*(n-1)^2*n*(4*n-5)*(4*n-3)*(5*n^2-1)*a(n-2) = 0.
a(n) ~ 2^(4*n + 3/2) / (Pi*n)^(3/2). - Vaclav Kotesovec, Apr 20 2020
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