A201805 -id:A201805 - cp's OEIS frontend

A150500 Number of walks within N^3 (the first octant of Z^3) starting at (0,0,0) and consisting of n steps taken from {(-1, -1, -1), (-1, -1, 1), (0, 1, 0), (1, 1, -1), (1, 1, 1)}.

1, 2, 7, 25, 101, 416, 1787, 7792, 34645, 155722, 707795, 3242515, 14963665, 69458000, 324102287, 1519028843, 7147771981, 33750528146, 159860887355, 759295147045, 3615520821281, 17255165910632, 82521746019487, 395404081034830, 1897886817388201, 9124229781131546, 43930513066698367, 211803668881914847

Offset: 0

Manuel Kauers, Nov 18 2008

Alin Bostan and Manuel Kauers, Automatic Classification of Restricted Lattice Walks, arXiv:0811.2899 [math.CO], 2008-2009.
Nachum Dershowitz, Touchard's Drunkard, Journal of Integer Sequences, Vol. 20 (2017), #17.1.5.

Cf. A201805.

aux[i_Integer, j_Integer, k_Integer, n_Integer] := Which[Min[i, j, k, n] < 0 || Max[i, j, k] > n, 0, n == 0, KroneckerDelta[i, j, k, n], True, aux[i, j, k, n] = aux[-1 + i, -1 + j, -1 + k, -1 + n] + aux[-1 + i, -1 + j, 1 + k, -1 + n] + aux[i, -1 + j, k, -1 + n] + aux[1 + i, 1 + j, -1 + k, -1 + n] + aux[1 + i, 1 + j, 1 + k, -1 + n]]; Table[Sum[aux[i, j, k, n], {i, 0, n}, {j, 0, n}, {k, 0, n}], {n, 0, 10}]

a(n) = (A201805(n+1) + 3*A201805(n))/4. - Mark van Hoeij, Nov 29 2024

A328718 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(n,k) is the constant term in the expansion of (1 + x_1 + x_2 + ... + x_n + 1/x_1 + 1/x_2 + ... + 1/x_n)^k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 7, 5, 1, 1, 1, 19, 13, 7, 1, 1, 1, 51, 61, 19, 9, 1, 1, 1, 141, 221, 127, 25, 11, 1, 1, 1, 393, 1001, 511, 217, 31, 13, 1, 1, 1, 1107, 4145, 3301, 921, 331, 37, 15, 1, 1, 1, 3139, 18733, 16297, 7761, 1451, 469, 43, 17, 1, 1

Offset: 0

Seiichi Manyama, Oct 26 2019

T(n,k) is the number of k-step closed walks (from origin to origin) in n-dimensional lattice where each step changes at most one component by -1 or by +1. - Alois P. Heinz, Oct 26 2019

Conjecture: Row r is asymptotic to (2*r+1)^(n + r/2) / (2^r * (Pi*n)^(r/2)). - Vaclav Kotesovec, Oct 27 2019

			Square array begins:
   1, 1,  1,  1,   1,    1,     1,      1, ...
   1, 1,  3,  7,  19,   51,   141,    393, ...
   1, 1,  5, 13,  61,  221,  1001,   4145, ...
   1, 1,  7, 19, 127,  511,  3301,  16297, ...
   1, 1,  9, 25, 217,  921,  7761,  41889, ...
   1, 1, 11, 31, 331, 1451, 15101,  85961, ...
   1, 1, 13, 37, 469, 2101, 26041, 153553, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Rows n=0-5 give A000012, A002426, A201805, A328713, A328714, A328715.

Main diagonal is A328716.

From Vaclav Kotesovec, Oct 30 2019: (Start)
Columns:
T(n,2) = 2*n + 1.
T(n,3) = 6*n + 1.
T(n,4) = 12*n^2 + 6*n + 1.
T(n,5) = 60*n^2 - 10*n + 1.
T(n,6) = 120*n^3 + 20*n + 1.
T(n,7) = 840*n^3 - 840*n^2 + 392*n + 1. (End)

A282252 Exponential Riordan array [Bessel_I(0,2*x)^2, x].

Original entry on oeis.org

1, 0, 1, 4, 0, 1, 0, 12, 0, 1, 36, 0, 24, 0, 1, 0, 180, 0, 40, 0, 1, 400, 0, 540, 0, 60, 0, 1, 0, 2800, 0, 1260, 0, 84, 0, 1, 4900, 0, 11200, 0, 2520, 0, 112, 0, 1, 0, 44100, 0, 33600, 0, 4536, 0, 144, 0, 1, 63504, 0, 220500, 0, 84000, 0, 7560, 0, 180, 0, 1

Offset: 0

Peter Bala, Feb 12 2017

Bessel_I(0,2*x) = Sum_{n >= 0} binomial(2*n,n)*x^(2*n)/(2*n)! is a modified Bessel function of the first kind.
Consider the infinite 2-dimensional square lattice Z x Z with an oriented self-loop at each vertex. Then the triangle entry T(n,k) equals the number of walks of length n from the origin to itself having k loops. An example is given below.
See A069466 for walks an infinite 2-dimensional square lattice without self-loops.
This is the square of triangle A109187, whose entries give the number of walks of length n from a vertex to itself having k loops on a 1-dimensional integer lattice with an oriented self-loop at each vertex.
A109187 is the exponential Riordan array [Bessel_I(0,2*x), x]. Note that Bessel_I(0,2*x)^2 = (Sum_{n >= 0} binomial(2*n,n)* x^(2*n)/(2*n)!)^2 = Sum_{n >= 0} binomial(2*n,n)^2*x^(2*n) /(2*n)!.

			The triangle begins
    1;
    0,   1;
    4,   0,   1;
    0,  12,   0,   1;
   36,   0,  24,   0,   1;
    0, 180,   0,  40,   0,   1;
  400,   0, 540,   0,  60,   0,   1;
  ...
T(3,1) = 12: on the square lattice, let L, R, U, D denote a left step, right step, up step and down step respectively. The 12 walks of length 3 containing a single loop are
    loop L R, loop R L, loop U D, loop D U,
    L loop R, R loop L, U loop D, D loop U,
    L R loop, R L loop, U D loop, D U loop.
The infinitesimal generator of this array has integer entries and begins
      0;
      0,    0;
      4,    0,    0;
      0,   12,    0,    0;
    -12,    0,   24,    0,    0;
      0,  -60,    0,   40,    0,    0;
    160,    0, -180,    0,   60,    0,    0;
      0, 1120,    0, -420,    0,   84,    0,    0;
  -4620,    0, 4480,    0, -840,    0,  112,    0,    0;
  ...
It is the generalized exponential Riordan array [ 2*log(Bessel_I(0,2*x)), x ].

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

A201805 gives row sums. Cf. A069466, A109187.

T := (n, k) -> (1/2)*binomial(n, k)*binomial(n-k, floor((1/2)*n-(1/2)*k))^2*(1+(-1)^(n-k)):
seq(seq(T(n, k), k = 0..n), n = 0..9);

Table[Binomial[n, k] Binomial[n - k, Floor[(n - k)/2]]^2*(1 + (-1)^(n - k))/2, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 12 2017 *)

for(n=0,10, for(k=0,n, print1(binomial(n,k)*binomial(n-k,floor((n-k)/2))^2*(1 + (-1)^(n-k))/2, ", "))) \\ G. C. Greubel, Aug 16 2017

T(n,k) = binomial(n,k)*binomial(n-k,floor((n-k)/2))^2*(1 + (-1)^(n-k))/2.
T(n,n-2*k) = n/(n - 2*k)*T(n-1,n-2*k-1).
T(n,k) = the coefficient of t^k in the expansion of (t + X + 1/X + Y + 1/Y)^n.
T(n,k) = 1/Pi^2 * Integral_{y = 0..Pi} Integral_{x = 0..Pi} ( t + 2*cos(x) + 2*cos(y) )^n dx dy.
E.g.f.: exp(x*t)*Bessel_I(0,2*x)^2 = 1 + t*x + (4 + t^2)*x^2/2! + (12*t + t^3)*x^3/3! + (36 + 24*t^2 + t^4)*x^4/4! + ....
The n-th row polynomial R(n,t) = Sum_{k = 0..floor(n/2)} binomial(n,2*k)*binomial(2*k,k)^2 * t^(n-2*k).
Recurrence: n^2*R(n,t) = t*(3*n^2 - 3*n + 1)*R(n-1,t) + (16 - 3*t^2)*(n - 1)^2*R(n-2,t) + t*(t^2 - 16)*(n - 1)*(n - 2)*R(n-3,t) with R(0,t) = 1, R(1,t) = t and R(2,t) = 4 + t^2.
d/dt(R(n,t)) = n*R(n-1,t).
The zeros of the row polynomials appear to lie on the imaginary axis in the complex plane. Also, the zeros of R(n,t) and R(n+1,t) appear to interlace on the imaginary axis.

A328713 Constant term in the expansion of (1 + x + y + z + 1/x + 1/y + 1/z)^n.

Original entry on oeis.org

1, 1, 7, 19, 127, 511, 3301, 16297, 103279, 570367, 3595177, 21167917, 133789789, 818625133, 5207248879, 32649752779, 209258291599, 1333828204303, 8612806088761, 55546469634733, 361143420408337, 2349709451702737, 15370341546766939, 100695951740818903, 662213750028892429

Offset: 0

Seiichi Manyama, Oct 26 2019

nonn

a(n) is the number of n-step closed walks (from origin to origin) in cubic lattice where each step changes at most one component by -1 or by +1. - Alois P. Heinz, Oct 26 2019

			(1+x+y+z+1/x+1/y+1/z)^2 = x^2 + 1/x^2 + y^2 + 1/y^2 + z^2 + 1/z^2 + 2 * (xy + 1/(xy) + yz + 1/(yz) + zx + 1/(zx) + x/y + y/x + y/z + z/y + z/x + x/z + x + 1/x + y + 1/y + z + 1/z) + 7. So a(2) = 7.

Alois P. Heinz, Table of n, a(n) for n = 0..1189

Row 3 of A328718.

Cf. A002426, A002896, A201805.

{a(n) = polcoef(polcoef(polcoef((1+x+y+z+1/x+1/y+1/z)^n, 0), 0), 0)}

From Vaclav Kotesovec, Oct 26 2019: (Start)
Recurrence: n^3*a(n) = (2*n - 1)*(2*n^2 - 2*n + 1)*a(n-1) + (n-1)*(34*n^2 - 68*n + 41)*a(n-2) - 38*(n-2)*(n-1)*(2*n - 3)*a(n-3) - 105*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ 7^(n + 3/2) / (8 * Pi^(3/2) * n^(3/2)). (End)
E.g.f.: exp(x) * BesselI(0,2*x)^3. - Ilya Gutkovskiy, Oct 26 2019

A361677 Constant term in the expansion of (1 + x + y + z + 1/(xy) + 1/(yz) + 1/(z*x))^n.

Original entry on oeis.org

1, 1, 1, 19, 73, 181, 1711, 10081, 38809, 256033, 1696861, 8388271, 49449511, 326195299, 1847392093, 10789655059, 69202030969, 418647580489, 2498113460881, 15735859252147, 97919649290053, 598317173139313, 3748943081117323

Offset: 0

Seiichi Manyama, Mar 20 2023

nonn

Cf. A201805, A361678.

Cf. A275047, A344560.

Table[Sum[(3*k)!/k!^3 * Binomial[3*k,k] * Binomial[n,3*k], {k,0,n/3}], {n,0,25}] (* Vaclav Kotesovec, Mar 22 2023 *)

a(n) = sum(k=0, n\3, (3*k)!/k!^3*binomial(3*k, k)*binomial(n, 3*k));

a(n) = Sum_{k=0..floor(n/3)} (3*k)!/k!^3 * binomial(3*k,k) * binomial(n,3*k).
From Vaclav Kotesovec, Mar 22 2023: (Start)
Recurrence: 2*n^3*(2*n - 3)*a(n) = 2*(10*n^4 - 32*n^3 + 38*n^2 - 22*n + 5)*a(n-1) - 2*(n-1)*(2*n - 3)*(10*n^2 - 24*n + 17)*a(n-2) + (n-2)*(n-1)*(769*n^2 - 2331*n + 1594)*a(n-3) - 2*(n-3)*(n-2)*(n-1)*(739*n - 1481)*a(n-4) + 733*(n-4)*(n-3)*(n-2)*(n-1)*a(n-5).
a(n) ~ sqrt(733/108 + 1/2^(2/3) + 9/2^(4/3)) * (1 + 9/2^(2/3))^n / (2 * Pi^(3/2) * n^(3/2)). (End)

A328494 Constant term in the expansion of (1+x+y+1/x+1/y)^n without assuming commutativity.

Original entry on oeis.org

1, 1, 5, 13, 53, 181, 713, 2689, 10661, 41989, 168785, 680329, 2770409, 11331529, 46639157, 192762013, 800228069, 3333843685, 13936599857, 58432259977, 245665962113, 1035412181761, 4373982501245, 18516210906853, 78536526586553, 333712398776281, 1420364536094093

Offset: 0

Robin Hankin, Oct 16 2019

nonn

Related to A035610 which is the constant term of (x+y+1/x+1/y)^(2n).

If commutativity is assumed then the constant term of (1+x+y+1/x+1/y)^n is given by A201805(n). - Andrew Howroyd, Oct 25 2019

Andrew Howroyd, Table of n, a(n) for n = 0..500
Mark Haiman, Non-commutative rational power series and algebraic generating functions, European Journal of Combinatorics, 14(4):335-9 (1993).
Robin Hankin, Discussion of this and similar sequences
Pakawut Jiradilok and Supanat Kamtue, Transportation Distance between Probability Measures on the Infinite Regular Tree, arXiv:2107.09876 [math.CO], 2021.

Cf. A035610, A201805.

h := n -> GAMMA(n+1/2)/GAMMA(n+2)*hypergeom([2, 1-n], [n+2], -3):
a := n -> 3-(-3)^n-5^n+(1/sqrt(Pi))*add(12^(k+1)*binomial(n, 2*k)*h(k), k=1..n/2):
seq(simplify(a(n)), n=0..26); # Peter Luschny, Oct 25 2019

a(n)={my(p=3/(1+2*sqrt(1-12*x+O(x*x^(n\2))))); sum(k=0, n\2, binomial(n, 2*k)*polcoef(p,k))} \\ Andrew Howroyd, Oct 25 2019

library("freealg")
g <- function(p,string){constant(as.freealg(string)^p)} sapply(0:7,g,"1+x+y+X+Y")

a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*A035610(k). - Andrew Howroyd, Oct 25 2019

Terms a(8) and beyond from Andrew Howroyd, Oct 25 2019

A361678 Constant term in the expansion of (1 + w + x + y + z + 1/(xyz) + 1/(wyz) + 1/(wxz) + 1/(wxy))^n.

Original entry on oeis.org

1, 1, 1, 1, 97, 481, 1441, 3361, 77281, 647137, 3195361, 11674081, 116286721, 1147935361, 7611379777, 37451144641, 263670781921, 2456043418081, 19073086806241, 115319128034017, 748239468100417, 6179458007222977, 50636218964639617, 350400618132423937

Offset: 0

Seiichi Manyama, Mar 20 2023

nonn

Cf. A201805, A361677.

Cf. A361637.

Table[Sum[(4*k)!/k!^4 * Binomial[4*k,k] * Binomial[n,4*k], {k,0,n/4}], {n,0,25}] (* Vaclav Kotesovec, Mar 22 2023 *)

a(n) = sum(k=0, n\4, (4*k)!/k!^4*binomial(4*k, k)*binomial(n, 4*k));

a(n) = Sum_{k=0..floor(n/4)} (4*k)!/k!^4 * binomial(4*k,k) * binomial(n,4*k).
From Vaclav Kotesovec, Mar 22 2023: (Start)
Recurrence: 3*n^4*(3*n - 8)*(3*n - 4)*a(n) = 3*(63*n^6 - 405*n^5 + 1015*n^4 - 1355*n^3 + 1049*n^2 - 439*n + 77)*a(n-1) - 3*(n-1)*(189*n^5 - 1485*n^4 + 4685*n^3 - 7575*n^2 + 6313*n - 2163)*a(n-2) + 3*(n-2)*(n-1)*(315*n^4 - 2610*n^3 + 8285*n^2 - 12030*n + 6749)*a(n-3) + (n-3)*(n-2)*(n-1)*(64591*n^3 - 385926*n^2 + 701651*n - 375786)*a(n-4) - 3*(n-4)*(n-3)*(n-2)*(n-1)*(65347*n^2 - 326519*n + 391384)*a(n-5) + 3*(n-5)*(n-4)*(n-3)*(n-2)*(n-1)*(65473*n - 196383)*a(n-6) - 65509*(n-6)*(n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-7).
a(n) ~ sqrt(65563/12288 + 3*sqrt(3)/8 + sqrt(3/8 + 65563/(4096*sqrt(3)))) * (1 + 16/3^(3/4))^n / (Pi^2 * n^2). (End)

A138354 Central moment sequence of tr(A^4) in USp(4).

Original entry on oeis.org

1, 0, 3, 1, 21, 26, 215, 498, 2821, 9040, 43695, 165375, 752785, 3101970, 13881803, 59837183, 267860685, 1184749704, 5337504263, 23996776941, 108964583121, 495544446410, 2267450194443, 10402298479276, 47926692348121

Offset: 0

Andrew V. Sutherland, Mar 16 2008, Mar 31 2008

nonn

Binomial transform of A018224.

If A is a random matrix in the compact group USp(4) (4 X 4 complex are unitary and symplectic), then a(n)=E[(tr(A^4)+1)^n] is the n-th central moment of the trace of A^4, since E[tr(A^4)] = -1 (see A018224).

			a(3) = 1 because E[(tr(A^4)+1)^3] = 1.
a(3) = 1*A018224(0) + 3*A018224(1) + 3*A018224(2) + 1*A018224(1) = 1*1 + 3*(-1) + 3*4 + 1*(-9) = 1.

Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2008-2010.

Cf. A018224.

a18224[n_] := Binomial[n, Floor[n/2]]^2;
a[n_] := Sum[(-1)^i Binomial[n, i] a18224[i], {i, 0, n}];
Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Aug 13 2018 *)

a(n) = (1/2)*Integral_{x=0..Pi,y=0..Pi}(2cos(4x)+2cos(4y)+1)^n*(2cos(x)-2cos(y))^2*(2/Pi*sin^2(x))*(2/Pi*sin^2(y))dxdy.
a(n) = Sum_{i=0..n} (-1)^i binomial(n,i)*A018224(i). [corrected by Jean-François Alcover, Aug 13 2018]
a(n) = (5*A201805(n) - A201805(n+1))/4. - Mark van Hoeij, Nov 29 2024

cp's OEIS Frontend

A150500 Number of walks within N^3 (the first octant of Z^3) starting at (0,0,0) and consisting of n steps taken from {(-1, -1, -1), (-1, -1, 1), (0, 1, 0), (1, 1, -1), (1, 1, 1)}.

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A328718 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(n,k) is the constant term in the expansion of (1 + x_1 + x_2 + ... + x_n + 1/x_1 + 1/x_2 + ... + 1/x_n)^k.

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A282252 Exponential Riordan array [Bessel_I(0,2*x)^2, x].

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A328713 Constant term in the expansion of (1 + x + y + z + 1/x + 1/y + 1/z)^n.

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A361677 Constant term in the expansion of (1 + x + y + z + 1/(x*y) + 1/(y*z) + 1/(z*x))^n.

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A328494 Constant term in the expansion of (1+x+y+1/x+1/y)^n without assuming commutativity.

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A361678 Constant term in the expansion of (1 + w + x + y + z + 1/(x*y*z) + 1/(w*y*z) + 1/(w*x*z) + 1/(w*x*y))^n.

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A138354 Central moment sequence of tr(A^4) in USp(4).

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A361677 Constant term in the expansion of (1 + x + y + z + 1/(xy) + 1/(yz) + 1/(z*x))^n.

A361678 Constant term in the expansion of (1 + w + x + y + z + 1/(xyz) + 1/(wyz) + 1/(wxz) + 1/(wxy))^n.