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Denominators of u(n) where u(n) = (u(n-1) + u(n-2)) / u(n-3), with u(1) = u(2) = u(3) = 1.
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5
1, 1, 1, 1, 1, 1, 1, 1, 5, 10, 6, 21, 77, 55, 145, 899, 868, 1988, 38411, 90347, 95357, 637807, 3506263, 2382501, 19519203, 649945741, 911672929, 3857971277, 130630182325, 366719420575, 764101349503, 12533062448579, 136235802233249
FORMULA
u(4 - n) = u(n) for all n in Z.
0 = u(n) * u(n+3) - u(n+1) - u(n+2) for all n in Z. - Michael Somos, Nov 01 2014
EXAMPLE
u(1), ... = 1, 1, 1, 2, 3, 5, 4, 3, 7/5, 11/10, 5/6, 29/21, 155/77, 224/55, 639/145, ...
MATHEMATICA
Denominator[RecurrenceTable[{u[1] == u[2] == u[3] == 1, u[n] == (u[n - 1] + u[n - 2])/u[n - 3]}, u, {n, 50}]] (* G. C. Greubel, Jun 27 2017 *)
PROG
(PARI) {u(n) = local(v = [1, 1, 1]); if( n<1, n = 4-n); if( n<4, 1, for( k=4, n, v = [v[2], v[3], (v[2] + v[3]) / v[1]]); denominator( v[3] ))};
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.
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4
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
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).
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
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 *)
PROG
(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]};
a(n) = round((a(n-1) + a(n-2))/a(n-3)) starting with a(1)=a(2)=a(3)=1.
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3
1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1
COMMENTS
While this sequence has period 8, the unrounded version b(n) = (b(n-1) + b(n-2))/b(n-3) seems to have a quasi-period of about 8.7 for this particular starting point.
EXAMPLE
a(7) = round((a(6) + a(5))/a(4)) = round((5+3)/2) = 4.
a(n) = floor(b(n)), where b(1) = b(2) = b(3) = 1 and b(n) = (b(n-1) + b(n-2))/b(n-3) for n > 3.
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0
1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 0, 1, 2, 4, 4, 4, 2, 1, 0, 1, 1, 2, 3, 5, 3, 2, 1, 1, 0, 1, 2, 4, 4, 3, 1, 1, 0, 1, 1, 3, 4, 4, 2, 1, 0, 1, 1, 2, 3, 5, 3, 2, 1, 1, 0, 1, 2, 4, 4, 3, 1, 1, 0, 1, 1, 3, 4, 4, 2, 1, 0, 1, 1, 2, 3, 5, 3, 2, 1, 1, 0, 1, 2, 4, 4, 3, 1, 1, 0, 1, 1, 3, 4, 4, 2, 1, 0, 1, 1, 2, 3, 5
COMMENTS
This sequence seems quasiperiodic with the same quasiperiod of A185332(n)/ A185341(n) (see related comments of Michael Somos in A068508). Conjecture: 0 <= a(n) <= 5.
MATHEMATICA
c[1] = 1; c[2] = 1; c[3] = 1;
c[n_] := c[n] = (c[n - 2] + c[n - 1])/c[n - 3];
Table[Floor@c[j], {j, 1, 2^6}]
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