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.

Showing 1-10 of 10 results.

A271432 Number of n-step excursions on the 4-dimensional f.c.c. lattice.

Original entry on oeis.org

1, 0, 24, 192, 3384, 51840, 911040, 16369920, 307009080, 5902176000, 116083727424, 2323941903360, 47232891389376, 972252599205888, 20233078205573376, 425067670281526272, 9004456318854367800, 192148701659269774848
Offset: 0

Views

Author

Christoph Koutschan, Apr 07 2016

Keywords

Comments

a(n) = number of walks in Z^4 starting and ending at the origin, using only the steps (a,b,0,0), (a,0,b,0), ..., (0,0,a,b), where a,b can be +1 or -1.

Examples

			There is one walk with no steps.
No walk with a single steps returns to the origin.
The number of returning walks with two steps is exactly the number of allowed steps (called the coordination number of the lattice): a(2) = 4*binomial(4,2).
		

Crossrefs

Cf. A002899 (d = 3, i.e., excursions on the 3-dimensional f.c.c. lattice), this sequence (d = 4), A271650 (d = 5), A271651 (d = 6), A271670 (d = 7), A271671 (d = 8), A271672 (d = 9), A271673 (d = 10), A271674 (d = 11).

Programs

  • Maple
    nmax := 50: tt := [seq([seq(add(binomial(2*p,p)*binomial(2*j,2*p-n)*binomial(2*n+2*j-2*p,n+j-p), p = floor((n+1)/2)..floor((n+2*j)/2)), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: for d1 from 3 to 4 do tt := [seq([seq(add(binomial(n,p)*add(binomial(2*j,2*q-p)*binomial(2*j+2*p-2*q,j+p-q)*tt[n-p+1,q+1], q = floor((p+1)/2)..floor((p+2*j)/2)), p = 0..n), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: od: [seq(tt[n+1,1], n = 0..nmax)];
  • Mathematica
    a[0] = 1; a[1] = 0; a[2] = 24; a[3] = 192; a[4] = 3384; a[n_] := a[n] = (27648*(-4 + n)*(-3 + n)^2*(-2 + n)*(-8 + 35*n^2)*a[-5 + n] + 6912*(-3 + n)*(-2 + n)*(-132 + 62*n + 676*n^2 - 525*n^3 + 105*n^4)*a[-4 + n] + 144*(-2 + n)*(1440 - 352*n - 11430*n^2 + 15435*n^3 - 7350*n^4 + 1225*n^5)*a[-3 + n] + 8*(72 - 3738*n + 17065*n^2 - 29745*n^3 + 25150*n^4 - 10500*n^5 + 1750*n^6)*a[-2 + n] - (-1 + n)*(144 - 540*n + 487*n^2 + 151*n^3 - 315*n^4 + 105*n^5)*a[-1 + n])/(n^4*(27 - 70*n + 35*n^2)); Array[a, 30, 0]
    nmax = 50; T = Table[Sum[Binomial[2 p, p]*Binomial[2 j, 2 p - n]*Binomial[2 n + 2 j - 2 p, n + j - p], {p, Floor[(n + 1)/2], Floor[(n + 2 j)/2]}], {n, 0, nmax}, {j, 0, Floor[(nmax - n)/2]}]; Do[T = Table[Sum[Binomial[n, p]*Sum[Binomial[2 j, 2 q - p]*Binomial[2 j + 2 p - 2 q, j + p - q]*T[[n - p + 1, q + 1]], {q, Floor[(p + 1)/2], Floor[(p + 2 j)/2]}], {p, 0, n}], {n, 0, nmax}, {j, 0, If[d1 < 4, Floor[(nmax - n)/2], 0]}], {d1, 3, 4}]; First /@ T

Formula

a(n) satisfies the fifth-order linear recurrence equation (35*n^2-70*n+27)*n^4*a(n) +(n-1)*(105*n^5-315*n^4+151*n^3 +487*n^2 -540*n+144)*a(n-1) -8*(1750*n^6-10500*n^5+25150*n^4 -29745*n^3 +17065*n^2-3738*n+72)*a(n-2) -144*(n-2) *(1225*n^5-7350*n^4 +15435*n^3 -11430*n^2-352*n+1440)*a(n-3)-6912*(n-3)*(n-2)*(105*n^4 -525*n^3+676*n^2 +62*n-132)*a(n-4)-27648*(n-4)*(n-3)^2*(n-2)*(35*n^2-8) *a(n-5) = 0.
The generating function P(z) = Sum_{n>=0} a(n)*(z/24)^n is given by the 4-fold integral (1/Pi)^4 Int_{0..Pi} ... Int_{0..Pi} 1/(1-z*lambda_4) dk_1 ... dk_4, where the structure function is defined as lambda_4 = (1/binomial(4,2)) Sum_{i=1..4} Sum_{j=(i+1)..4} cos(k_i)*cos(k_j). The function P(z) satisfies the fourth-order linear ODE 12*z*(256+632*z+702*z^2+382*z^3+98*z^4+9*z^5)*P(z)+12*(-384+224*z+3716*z^2+7633*z^3 +6734*z^4+2939*z^5+604*z^6+45*z^7) *P'(z)+6*z*(-5376-5248*z+11080*z^2 +25286*z^3 +19898*z^4+7432*z^5 +1286*z^6+81*z^7) *P''(z)+2*z^2*(4+3*z)*(-3456-2304*z+3676*z^2+4920 *z^3+2079*z^4+356*z^5 +21*z^6)*P'''(z)+(-1+z)*z^3*(2+z)*(3+z)*(6+z)*(8+z)*(4+3*z)^2*P''''(z) = 0.
a(n) ~ 2^(3*n+1) * 3^n / (Pi^2 * n^2). - Vaclav Kotesovec, Apr 08 2016

A271650 Number of n-step excursions on the 5-dimensional f.c.c. lattice.

Original entry on oeis.org

1, 0, 40, 480, 11880, 281280, 7506400, 210268800, 6166993000, 187069411200, 5833030976640, 186014056166400, 6044435339896800, 199561060892793600, 6679216425794140800, 226213441773789550080, 7741313040820500484200
Offset: 0

Views

Author

Christoph Koutschan, Apr 11 2016

Keywords

Comments

a(n) = number of walks in the integer lattice Z^5 starting and ending at the origin, using only the steps of the form (s_1, ..., s_5) with s_1^2 + ... + s_5^2 = 2, i.e., each possible step has precisely two nonzero entries which can be +1 or -1.

Examples

			There is one walk with no steps.
No walk with a single steps returns to the origin.
The number of returning walks with two steps is exactly the number of allowed steps (called the coordination number of the lattice): a(2) = 4*binomial(5,2).
		

Crossrefs

Cf. A002899 (d = 3, i.e., excursions on the 3-dimensional f.c.c. lattice), A271432 (d = 4), this sequence (d = 5), A271651 (d = 6), A271670 (d = 7), A271671 (d = 8), A271672 (d = 9), A271673 (d = 10), A271674 (d = 11).

Programs

  • Maple
    nmax := 50: tt := [seq([seq(add(binomial(2*p,p)*binomial(2*j,2*p-n)*binomial(2*n+2*j-2*p,n+j-p), p = floor((n+1)/2)..floor((n+2*j)/2)), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: for d1 from 3 to 5 do tt := [seq([seq(add(binomial(n,p)*add(binomial(2*j,2*q-p)*binomial(2*j+2*p-2*q,j+p-q)*tt[n-p+1,q+1], q = floor((p+1)/2)..floor((p+2*j)/2)), p = 0..n), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: od: [seq(tt[n+1,1], n = 0..nmax)];
  • Mathematica
    nmax = 50; T = Table[Sum[Binomial[2 p, p]*Binomial[2 j, 2 p - n]*Binomial[2 n + 2 j - 2 p, n + j - p], {p, Floor[(n + 1)/2], Floor[(n + 2 j)/2]}], {n, 0, nmax}, {j, 0, Floor[(nmax - n)/2]}]; Do[T = Table[Sum[Binomial[n, p]*Sum[Binomial[2 j, 2 q - p]*Binomial[2 j + 2 p - 2 q, j + p - q]*T[[n - p + 1, q + 1]], {q, Floor[(p + 1)/2], Floor[(p + 2 j)/2]}], {p, 0, n}], {n, 0, nmax}, {j, 0, If[d1 < 5, Floor[(nmax - n)/2], 0]}], {d1, 3, 5}]; First /@ T

Formula

a(n) satisfies a seventh-order linear recurrence equation with polynomial coefficients of degree 12 (see link above).
The probability generating function P(z) = Sum_{n>=0} a(n)*(z/40)^n is given by the 5-fold integral (1/Pi)^5 Int_{0..Pi} ... Int_{0..Pi} 1/(1-z*lambda_5) dk_1 ... dk_5, where the structure function is defined as lambda_5 = (1/binomial(5,2)) Sum_{i=1..5} Sum_{j=(i+1)..5} cos(k_i)*cos(k_j). The function P(z) satisfies a sixth-order linear ODE with polynomial coefficients of degree 17 (see link above).

A271670 Number of n-step excursions on the 7-dimensional f.c.c. lattice.

Original entry on oeis.org

1, 0, 84, 1680, 66276, 2731680, 128704800, 6555265920, 355588928100, 20247799145280, 1198746727590384, 73266532153214400, 4598338364703822816, 295145004688715301120, 19311431876483926443264
Offset: 0

Views

Author

Christoph Koutschan, Apr 12 2016

Keywords

Comments

a(n) = number of walks in the integer lattice Z^7 starting and ending at the origin, using only the steps of the form (s_1, ..., s_7) with s_1^2 + ... + s_7^2 = 2, i.e., each possible step has precisely two nonzero entries which can be +1 or -1.

Examples

			There is one walk with no steps.
No walk with a single steps returns to the origin.
The number of returning walks with two steps is exactly the number of allowed steps (called the coordination number of the lattice): a(2) = 4*binomial(7,2).
		

Crossrefs

Cf. A002899 (d = 3, i.e., excursions on the 3-dimensional f.c.c. lattice), A271432 (d = 4), A271650 (d = 5), A271651 (d = 6), this sequence (d = 7), A271671 (d = 8), A271672 (d = 9), A271673 (d = 10), A271674 (d = 11).

Programs

  • Maple
    nmax := 50: tt := [seq([seq(add(binomial(2*p,p)*binomial(2*j,2*p-n)*binomial(2*n+2*j-2*p,n+j-p), p = floor((n+1)/2)..floor((n+2*j)/2)), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: for d1 from 3 to 7 do tt := [seq([seq(add(binomial(n,p)*add(binomial(2*j,2*q-p)*binomial(2*j+2*p-2*q,j+p-q)*tt[n-p+1,q+1], q = floor((p+1)/2)..floor((p+2*j)/2)), p = 0..n), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: od: [seq(tt[n+1,1], n = 0..nmax)];
  • Mathematica
    nmax = 50; T = Table[Sum[Binomial[2 p, p]*Binomial[2 j, 2 p - n]*Binomial[2 n + 2 j - 2 p, n + j - p], {p, Floor[(n + 1)/2], Floor[(n + 2 j)/2]}], {n, 0, nmax}, {j, 0, Floor[(nmax - n)/2]}]; Do[T = Table[Sum[Binomial[n, p]*Sum[Binomial[2 j, 2 q - p]*Binomial[2 j + 2 p - 2 q, j + p - q]*T[[n - p + 1, q + 1]], {q, Floor[(p + 1)/2], Floor[(p + 2 j)/2]}], {p, 0, n}], {n, 0, nmax}, {j, 0, If[d1 < 7, Floor[(nmax - n)/2], 0]}], {d1, 3, 7}]; First /@ T

Formula

a(n) conjecturally satisfies a linear recurrence equation of order 15 with polynomial coefficients of degree 56 (see link above).
The probability generating function P(z) = Sum_{n>=0} a(n)*(z/84)^n is given by the 7-fold integral (1/Pi)^7 Int_{0..Pi} ... Int_{0..Pi} 1/(1-z*lambda_7) dk_1 ... dk_7, where the structure function is defined as lambda_7 = (1/binomial(7,2)) Sum_{i=1..7} Sum_{j=(i+1)..7} cos(k_i)*cos(k_j). The function P(z) conjecturally satisfies an eleventh-order linear ODE with polynomial coefficients of degree 68 (see link above).

A271671 Number of n-step excursions on the 8-dimensional f.c.c. lattice.

Original entry on oeis.org

1, 0, 112, 2688, 126000, 6316800, 364887040, 23038364160, 1562288430640, 112014905049600, 8399872737107712, 653454438359331840, 52412319029000899584, 4313870772211888183296, 362994066330649023029760
Offset: 0

Views

Author

Christoph Koutschan, Apr 12 2016

Keywords

Comments

a(n) = number of walks in the integer lattice Z^8 starting and ending at the origin, using only the steps of the form (s_1, ..., s_8) with s_1^2 + ... + s_8^2 = 2, i.e., each possible step has precisely two nonzero entries which can be +1 or -1.

Examples

			There is one walk with no steps.
No walk with a single steps returns to the origin.
The number of returning walks with two steps is exactly the number of allowed steps (called the coordination number of the lattice): a(2) = 4*binomial(8,2).
		

Crossrefs

Cf. A002899 (d = 3, i.e., excursions on the 3-dimensional f.c.c. lattice), A271432 (d = 4), A271650 (d = 5), A271651 (d = 6), A271670 (d = 7), this sequence (d = 8), A271672 (d = 9), A271673 (d = 10), A271674 (d = 11).

Programs

  • Maple
    nmax := 50: tt := [seq([seq(add(binomial(2*p,p)*binomial(2*j,2*p-n)*binomial(2*n+2*j-2*p,n+j-p), p = floor((n+1)/2)..floor((n+2*j)/2)), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: for d1 from 3 to 8 do tt := [seq([seq(add(binomial(n,p)*add(binomial(2*j,2*q-p)*binomial(2*j+2*p-2*q,j+p-q)*tt[n-p+1,q+1], q = floor((p+1)/2)..floor((p+2*j)/2)), p = 0..n), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: od: [seq(tt[n+1,1], n = 0..nmax)];
  • Mathematica
    nmax = 50; T = Table[Sum[Binomial[2 p, p]*Binomial[2 j, 2 p - n]*Binomial[2 n + 2 j - 2 p, n + j - p], {p, Floor[(n + 1)/2], Floor[(n + 2 j)/2]}], {n, 0, nmax}, {j, 0, Floor[(nmax - n)/2]}]; Do[T = Table[Sum[Binomial[n, p]*Sum[Binomial[2 j, 2 q - p]*Binomial[2 j + 2 p - 2 q, j + p - q]*T[[n - p + 1, q + 1]], {q, Floor[(p + 1)/2], Floor[(p + 2 j)/2]}], {p, 0, n}], {n, 0, nmax}, {j, 0, If[d1 < 8, Floor[(nmax - n)/2], 0]}], {d1, 3, 8}]; First /@ T

Formula

a(n) conjecturally satisfies a linear recurrence equation of order 20 with polynomial coefficients of degree 109 (see link above).
The probability generating function P(z) = Sum_{n>=0} a(n)*(z/112)^n is given by the 8-fold integral (1/Pi)^8 Int_{0..Pi} ... Int_{0..Pi} 1/(1-z*lambda_8) dk_1 ... dk_8, where the structure function is defined as lambda_8 = (1/binomial(8,2)) Sum_{i=1..8} Sum_{j=(i+1)..8} cos(k_i)*cos(k_j). The function P(z) conjecturally satisfies a linear ODE of order 14 with polynomial coefficients of degree 126 (see link above).

A271672 Number of n-step excursions on the 9-dimensional f.c.c. lattice.

Original entry on oeis.org

1, 0, 144, 4032, 219024, 12942720, 887135040, 67057079040, 5484251057040, 477369708721920, 43704143706754944, 4170816570389736960, 412062922497680790336, 41920366214226928716288, 4372905161028532447478016
Offset: 0

Views

Author

Christoph Koutschan, Apr 12 2016

Keywords

Comments

a(n) = number of walks in the integer lattice Z^9 starting and ending at the origin, using only the steps of the form (s_1, ..., s_9) with s_1^2 + ... + s_9^2 = 2, i.e., each possible step has precisely two nonzero entries which can be +1 or -1.

Examples

			There is one walk with no steps.
No walk with a single steps returns to the origin.
The number of returning walks with two steps is exactly the number of allowed steps (called the coordination number of the lattice): a(2) = 4*binomial(9,2).
		

Crossrefs

Cf. A002899 (d = 3, i.e., excursions on the 3-dimensional f.c.c. lattice), A271432 (d = 4), A271650 (d = 5), A271651 (d = 6), A271670 (d = 7), A271671 (d = 8), this sequence (d = 9), A271673 (d = 10), A271674 (d = 11).

Programs

  • Maple
    nmax := 50: tt := [seq([seq(add(binomial(2*p,p)*binomial(2*j,2*p-n)*binomial(2*n+2*j-2*p,n+j-p), p = floor((n+1)/2)..floor((n+2*j)/2)), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: for d1 from 3 to 9 do tt := [seq([seq(add(binomial(n,p)*add(binomial(2*j,2*q-p)*binomial(2*j+2*p-2*q,j+p-q)*tt[n-p+1,q+1], q = floor((p+1)/2)..floor((p+2*j)/2)), p = 0..n), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: od: [seq(tt[n+1,1], n = 0..nmax)];
  • Mathematica
    nmax = 50; T = Table[Sum[Binomial[2 p, p]*Binomial[2 j, 2 p - n]*Binomial[2 n + 2 j - 2 p, n + j - p], {p, Floor[(n + 1)/2], Floor[(n + 2 j)/2]}], {n, 0, nmax}, {j, 0, Floor[(nmax - n)/2]}]; Do[T = Table[Sum[Binomial[n, p]*Sum[Binomial[2 j, 2 q - p]*Binomial[2 j + 2 p - 2 q, j + p - q]*T[[n - p + 1, q + 1]], {q, Floor[(p + 1)/2], Floor[(p + 2 j)/2]}], {p, 0, n}], {n, 0, nmax}, {j, 0, If[d1 < 9, Floor[(nmax - n)/2], 0]}], {d1, 3, 9}]; First /@ T

Formula

a(n) conjecturally satisfies a linear recurrence equation of order 22 with polynomial coefficients of degree 151 (see link above).
The probability generating function P(z) = Sum_{n>=0} a(n)*(z/144)^n is given by the 9-fold integral (1/Pi)^9 Int_{0..Pi} ... Int_{0..Pi} 1/(1-z*lambda_9) dk_1 ... dk_9, where the structure function is defined as lambda_9 = (1/binomial(9,2)) Sum_{i=1..9} Sum_{j=(i+1)..9} cos(k_i)*cos(k_j). The function P(z) conjecturally satisfies a linear ODE of order 18 with polynomial coefficients of degree 169 (see link above).
Hence a(n) conjecturally satisfies a linear recurrence equation with polynomial coefficients.

A271673 Number of n-step excursions on the 10-dimensional f.c.c. lattice.

Original entry on oeis.org

1, 0, 180, 5760, 355860, 24226560, 1923670800, 169658496000, 16291413249300, 1674631754611200, 181989927592033680, 20709782925396364800, 2449425950787336166800, 299337868552812779289600, 37621311095831818078152000
Offset: 0

Views

Author

Christoph Koutschan, Apr 12 2016

Keywords

Comments

a(n) = number of walks in the integer lattice Z^10 starting and ending at the origin, using only the steps of the form (s_1, ..., s_10) with s_1^2 + ... + s_10^2 = 2, i.e., each possible step has precisely two nonzero entries which can be +1 or -1.

Examples

			There is one walk with no steps.
No walk with a single steps returns to the origin.
The number of returning walks with two steps is exactly the number of allowed steps (called the coordination number of the lattice): a(2) = 4*binomial(10,2).
		

Crossrefs

Cf. A002895, A002899 (d = 3, i.e., excursions on the 3-dimensional f.c.c. lattice), A271432 (d = 4), A271650 (d = 5), A271651 (d = 6), A271670 (d = 7), A271671 (d = 8), A271672 (d = 9), this sequence (d = 10), A271674 (d = 11).

Programs

  • Maple
    nmax := 50: tt := [seq([seq(add(binomial(2*p,p)*binomial(2*j,2*p-n)*binomial(2*n+2*j-2*p,n+j-p), p = floor((n+1)/2)..floor((n+2*j)/2)), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: for d1 from 3 to 10 do tt := [seq([seq(add(binomial(n,p)*add(binomial(2*j,2*q-p)*binomial(2*j+2*p-2*q,j+p-q)*tt[n-p+1,q+1], q = floor((p+1)/2)..floor((p+2*j)/2)), p = 0..n), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: od: [seq(tt[n+1,1], n = 0..nmax)];
  • Mathematica
    nmax = 50; T = Table[Sum[Binomial[2 p, p]*Binomial[2 j, 2 p - n]*Binomial[2 n + 2 j - 2 p, n + j - p], {p, Floor[(n + 1)/2], Floor[(n + 2 j)/2]}], {n, 0, nmax}, {j, 0, Floor[(nmax - n)/2]}]; Do[T = Table[Sum[Binomial[n, p]*Sum[Binomial[2 j, 2 q - p]*Binomial[2 j + 2 p - 2 q, j + p - q]*T[[n - p + 1, q + 1]], {q, Floor[(p + 1)/2], Floor[(p + 2 j)/2]}], {p, 0, n}], {n, 0, nmax}, {j, 0, If[d1 < 10, Floor[(nmax - n)/2], 0]}], {d1, 3, 10}]; First /@ T

Formula

a(n) conjecturally satisfies a linear recurrence equation of order 30 with polynomial coefficients of degree 274 (see link above).
The probability generating function P(z) = Sum_{n>=0} a(n)*(z/180)^n is given by the 10-fold integral (1/Pi)^10 Int_{0..Pi} ... Int_{0..Pi} 1/(1-z*lambda_10) dk_1 ... dk_10, where the structure function is defined as lambda_10 = (1/binomial(10,2)) Sum_{i=1..10} Sum_{j=(i+1)..10} cos(k_i)*cos(k_j). The function P(z) conjecturally satisfies a linear ODE of order 22 with polynomial coefficients of degree 300 (see link above).

A271674 Number of n-step excursions on the 11-dimensional f.c.c. lattice.

Original entry on oeis.org

1, 0, 220, 7920, 548460, 42276960, 3818372800, 385303564800, 42556023409900, 5056698223684800, 638162986199119920, 84683717201322993600, 11723112517163129913600, 1682392957299926013542400, 249030549709148521993536000, 37864267170542400351711467520
Offset: 0

Views

Author

Christoph Koutschan, Apr 12 2016

Keywords

Comments

a(n) = number of walks in the integer lattice Z^11 starting and ending at the origin, using only the steps of the form (s_1, ..., s_11) with s_1^2 + ... + s_11^2 = 2, i.e., each possible step has precisely two nonzero entries which can be +1 or -1.

Examples

			There is one walk with no steps.
No walk with a single steps returns to the origin.
The number of returning walks with two steps is exactly the number of allowed steps (called the coordination number of the lattice): a(2) = 4*binomial(11,2).
		

Crossrefs

Cf. A002899 (d = 3, i.e., excursions on the 3-dimensional f.c.c. lattice), A271432 (d = 4), A271650 (d = 5), A271651 (d = 6), A271670 (d = 7), A271671 (d = 8), A271672 (d = 9), A271673 (d = 10), this sequence (d = 11).

Programs

  • Maple
    nmax := 50: tt := [seq([seq(add(binomial(2*p,p)*binomial(2*j,2*p-n)*binomial(2*n+2*j-2*p,n+j-p), p = floor((n+1)/2)..floor((n+2*j)/2)), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: for d1 from 3 to 11 do tt := [seq([seq(add(binomial(n,p)*add(binomial(2*j,2*q-p)*binomial(2*j+2*p-2*q,j+p-q)*tt[n-p+1,q+1], q = floor((p+1)/2)..floor((p+2*j)/2)), p = 0..n), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: od: [seq(tt[n+1,1], n = 0..nmax)];
  • Mathematica
    nmax = 50; T = Table[Sum[Binomial[2 p, p]*Binomial[2 j, 2 p - n]*Binomial[2 n + 2 j - 2 p, n + j - p], {p, Floor[(n + 1)/2], Floor[(n + 2 j)/2]}], {n, 0, nmax}, {j, 0, Floor[(nmax - n)/2]}]; Do[T = Table[Sum[Binomial[n, p]*Sum[Binomial[2 j, 2 q - p]*Binomial[2 j + 2 p - 2 q, j + p - q]*T[[n - p + 1, q + 1]], {q, Floor[(p + 1)/2], Floor[(p + 2 j)/2]}], {p, 0, n}], {n, 0, nmax}, {j, 0, If[d1 < 11, Floor[(nmax - n)/2], 0]}], {d1, 3, 11}]; First /@ T

Formula

The probability generating function P(z) = Sum_{n>=0} a(n)*(z/220)^n is given by the 11-fold integral (1/Pi)^11 Int_{0..Pi} ... Int_{0..Pi} 1/(1-z*lambda_11) dk_1 ... dk_11, where the structure function is defined as lambda_11 = (1/binomial(11,2)) Sum_{i=1..11} Sum_{j=(i+1)..11} cos(k_i)*cos(k_j). The function P(z) conjecturally satisfies a linear ODE of order 27 with polynomial coefficients of degree 409 (see link above).
Hence a(n) conjecturally satisfies a linear recurrence equation with polynomial coefficients.

A292881 Number of n-step closed paths on the E6 lattice.

Original entry on oeis.org

1, 0, 72, 1440, 54216, 2134080, 93993120, 4423628160, 219463602120, 11341793393280
Offset: 0

Views

Author

Samuel Savitz, Sep 26 2017

Keywords

Comments

Calculated by brute force computational enumeration.
The moments of the imaginary part of the suitably normalized E6 lattice Green's function.

Examples

			The 2-step walks consist of hopping to one of the 72 minimal vectors of the E6 lattice and then back to the origin.
		

Crossrefs

Cf. A126869 (Linear A1 lattice), A002898 (Hexagonal A2), A002899 (FCC A3), A271432 (D4), A271650 (D5), A271651 (D6), A292882 (E7), A271670 (D7), A292883 (E8), A271671 (D8).

Formula

Summed combinatorial expressions and recurrence relations for this sequence exist, but have not been determined. These would allow one to write a differential equation or hypergeometric expression for the E6 lattice Green's function.

A292882 Number of n-step closed paths on the E7 lattice.

Original entry on oeis.org

1, 0, 126, 4032, 228690, 14394240, 1020623940, 78353170560, 6393827197170
Offset: 0

Views

Author

Samuel Savitz, Sep 26 2017

Keywords

Comments

Calculated by brute force computational enumeration.
The moments of the imaginary part of the suitably normalized E7 lattice Green's function.

Examples

			The 2-step walks consist of hopping to one of the 126 minimal vectors of the E7 lattice and then back to the origin.
		

Crossrefs

Cf. A126869 (Linear A1 lattice), A002898 (Hexagonal A2), A002899 (FCC A3), A271432 (D4), A271650 (D5), A292881 (E6), A271651 (D6), A271670 (D7), A292883 (E8), A271671 (D8).

Formula

Summed combinatorial expressions and recurrence relations for this sequence exist, but have not been determined. These would allow one to write a differential equation or hypergeometric expression for the E7 lattice Green's function.

A292883 Number of n-step closed paths on the E8 lattice.

Original entry on oeis.org

1, 0, 240, 13440, 1260720, 137813760, 17141798400, 2336327078400, 341350907713200
Offset: 0

Views

Author

Samuel Savitz, Sep 26 2017

Keywords

Comments

Calculated by brute force computational enumeration.
The moments of the imaginary part of the suitably normalized E8 lattice Green's function.

Examples

			The 2-step walks consist of hopping to one of the 240 minimal vectors of the E8 lattice and then back to the origin.
		

Crossrefs

Cf. A126869 (Linear A1 lattice), A002898 (Hexagonal A2), A002899 (FCC A3), A271432 (D4), A271650 (D5), A292881 (E6), A271651 (D6), A292882 (E7), A271670 (D7), A271671 (D8).

Formula

Summed combinatorial expressions and recurrence relations for this sequence exist, but have not been determined. These would allow one to write a differential equation or hypergeometric expression for the E8 lattice Green's function.
Showing 1-10 of 10 results.