A205303 a(n) where a(n) * a(n-5) * a(n-10) = a(n-1) * a(n-6) * a(n-8) + a(n-2) * a(n-4) * a(n-9), with a(1) = ... = a(10) = 1.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 5, 8, 18, 21, 44, 60, 174, 372, 1344, 2556, 12984, 24048, 82224, 160848, 904032, 1967328, 14812992, 43671744, 374004864, 1108847232, 8442489600, 18677267712, 211090572288, 612702392832, 6883734979584

Offset: 1

Author

Michael Somos, Jan 28 2012

Keywords

• nonn

Comments

The recursion has the Laurent property. If a(1), ..., a(10) are variables, then a(n) is a Laurent polynomial -- a rational function with a monomial denominator.

Similar to the Somos-5 sequence, the sequence a(n) can be expressed in terms of the Jacobi theta_3(u, q) function as a(n) = c1 * c2^(n - c6)^2 * theta_3(c4*n - c5, c3) where both c1 and c5 depend on the residue class of n modulo 12, c5 linearly with slope 0.2347354... with c5 = 0.4030547... if n=12*k+6, c6 = 5.5 + (-1)^n * 0.1844232..., c2 = 1.0303784..., c4 = 0.6231417..., q = c3 = 0.116755251... = exp(Pi i tau) and 3 * (72961 / 432)^3 / 1367 = 10572.4060... the corresponding invariant j(tau).

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..282
• S. Fomin and A. Zelevinsky, The Laurent phenomenon, arXiv:math/0104241 [math.CO], 2001.
• B. Grammaticos, A. Ramani and T. Tamizhmani, Investigating the integrability of the Lyness mappings, J. Phys. A: Math. Theor. 42 454009.
• Eric Weisstein's World of Mathematics, Laurent Polynomial

Crossrefs

Cf. A185332, A185341.

Programs

Mathematica

nxt[{a10_,a9_,a8_,a7_,a6_,a5_,a4_,a3_,a2_,a1_}]:={a9,a8,a7,a6,a5,a4,a3,a2,a1,(a1*a6*a8+a2*a4*a9)/(a5*a10)}; Transpose[ NestList[ nxt,Table[1, {10}],40]][[1]] (* Harvey P. Dale, Mar 27 2015 *)
a[ n_] := Which[ n < 6, a[11 - n], n < 11, 1, True, (a[n - 1] a[n - 6] a[n - 8] + a[n - 2] a[n - 4] a[n - 9]) / (a[n - 5] a[n - 10])]; (* Michael Somos, Oct 21 2018 *)
a[ n_] := Which[ n < 6, a[11 - n], n < 11, 1, n < 13, n - 9, True, (-a[n - 1] a[n - 12] + 13 a[n - 4] a[n - 9]) / a[n - 13]]; (* Michael Somos, Oct 21 2018 *)

PARI

{a(n) = my(v); if( n<1, n = 11-n); v = vector( n, k, 1); for( k=11, n, v[k] = (v[k-1] * v[k-6] * v[k-8] + v[k-2] * v[k-4] * v[k-9]) / (v[k-5] * v[k-10])); v[n]};

PARI

{a(n) = my(v); if( n<1, n = 11-n); v = vector( n, k, 1); for( k=11, n, v[k] = ( -v[k-3] * v[k-4] + v[k-1] * v[k-6] * [2, 2, 2, 3] [k%4 + 1]) / v[k-7]); v[n]};

PARI

{a(n) = my(v); if( n<1, n = 11-n); v = vector( n, k, 1); for( k=11, n, v[k] = ( v[k-1] * v[k-4] * [3, 3, 4] [k%3 + 1] - v[k-2] * v[k-3] * [3, 2, 2, 2] [k%4 + 1]) / v[k-5]); v[n]};

Formula

Let u(n) := (a(n) * a(n+7)) / (a(n+3) * a(n+4)) = A185332(n) / A185341(n), then u(n) = (u(n-1) + u(n-2)) / u(n-3), u(1) = u(2) = u(3) = 1.

a(n) = a(11-n) for all n in Z.

a(n+7) * a(n-6) = -a(n+6) * a(n-5) + 13 * a(n+3) * a(n-2) for all n in Z. [see Grammaticos et al., Equation (3.2) for the general form of this equation.]

a(n+7) * a(n-7) = a(n+5) * a(n-5) + 13 * a(n+1) * a(n-1) for all n in Z.

a(n+11) * a(n-11) = 156*a(n+5) * a(n-5) + 612 * a(n+1) * a(n-1) for all n in Z. - Michael Somos, Oct 19 2023

a(n+3) * a(n-2) = (3 + [3|(n+1)]) * a(n+2) * a(n-1) - (2 + [4|n]) * a(n+1) * a(n) for all n in Z where [] is the Iverson bracket. - Michael Somos, Oct 19 2023