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 1-7 of 7 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).

A271651 Number of n-step excursions on the 6-dimensional f.c.c. lattice.

Original entry on oeis.org

1, 0, 60, 960, 30780, 996480, 36560400, 1430553600, 59089923900, 2543035488000, 113129280527760, 5170796720812800, 241741903350301200, 11520044551208793600, 558061378022616811200, 27421336248833005839360
Offset: 0

Views

Author

Christoph Koutschan, Apr 11 2016

Keywords

Comments

a(n) = number of walks in the integer lattice Z^6 starting and ending at the origin, using only the steps of the form (s_1, ..., s_6) with s_1^2 + ... + s_6^2 = 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(6,2).
		

Crossrefs

Cf. A002899 (d = 3, i.e., excursions on the 3-dimensional f.c.c. lattice), A271432 (d = 4), A271650 (d = 5), this sequence (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 6 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 < 6, Floor[(nmax - n)/2], 0]}], {d1, 3, 6}]; First /@ T

Formula

a(n) satisfies a twelfth-order linear recurrence equation with polynomial coefficients of degree 33 (see link above).
The probability generating function P(z) = Sum_{n>=0} a(n)*(z/60)^n is given by the 6-fold integral (1/Pi)^6 Int_{0..Pi} ... Int_{0..Pi} 1/(1-z*lambda_6) dk_1 ... dk_6, where the structure function is defined as lambda_6 = (1/binomial(6,2)) Sum_{i=1..6} Sum_{j=(i+1)..6} cos(k_i)*cos(k_j). The function P(z) satisfies an eighth-order linear ODE with polynomial coefficients of degree 43 (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.

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.
Showing 1-7 of 7 results.