A101467 - OEIS

%I #29 Apr 15 2021 15:19:54

%S 10,63,841,9279,96601,983583,9919561,99602559,998026681,9990174303,

%T 99950992681,999755323839,9998777694361,99993891685023,

%U 999969468040201,9999847368997119,99999236931275641,999996184915051743,9999980925350886121,99999904629080526399

%N Number of distinct n-term ratios x_1 : x_2 : ... : x_n where each x_i is in the range [1-10].

%C Number of elements of {1,...,10}^n with gcd 1. - _Robert Israel_, Nov 28 2014

%H Colin Barker, <a href="/A101467/b101467.txt">Table of n, a(n) for n = 1..999</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (21,-151,471,-640,300).

%F a(1) = 10; for n>1, a(n) = 10^n - 5^n - 3^n - 2^n + 1.

%F G.f.: x*(2700*x^5-5460*x^4+3579*x^3-1028*x^2+147*x-10) / ((x-1)*(2*x-1)*(3*x-1)*(5*x-1)*(10*x-1)). - _Colin Barker_, Nov 28 2014

%F a(n+4) = -300*a(n)+340*a(n+1)-131*a(n+2)+20*a(n+3)+72 for n >= 2. - _Robert Israel_, Dec 02 2014

%F a(n) = 21*a(n-1) - 151*a(n-2) + 471*a(n-3) - 640*a(n-4) + 300*a(n-5) for n > 6. - _Chai Wah Wu_, Apr 15 2021

%e For n=2: Consider the ratios 1:1, 1:2, ..., 1:10, 2:1, 2:2, ..., 2:10, ..., 10:1, 10:2, ..., 10:10. We get 63 different ratios from the 100 numbers list above after removing duplication. So a(2) = 63, and this is A018805(10).

%p 1, seq(10^n - 5^n - 3^n - 2^n + 1, n=2..20); # _Robert Israel_, Nov 28 2014

%o (PARI) Vec(x*(2700*x^5-5460*x^4+3579*x^3-1028*x^2+147*x-10)/((x-1)*(2*x-1)*(3*x-1)*(5*x-1)*(10*x-1)) + O(x^100)) \\ _Colin Barker_, Nov 28 2014

%Y Cf. A018805 (2 terms), A071778 (3 terms), A082540 (4 terms), A082544 (5 terms).

%K nonn,easy

%O 1,1

%A Su Jianning (sujianning(AT)yahoo.com.cn), Jan 21 2005