|
%I #15 Sep 08 2022 08:45:51
%S 1,4,5,8,10,13,14,17,19,22,23,26,28,31,32,35,37,40,41,44,46,49,50,53,
%T 55,58,59,62,64,67,68,71,73,76,77,80,82,85,86,89,91,94,95,98,100,103,
%U 104,107,109,112,113,116,118,121,122,125,127,130,131,134,136
%N Numbers congruent to {1,4,5,8} mod 9.
%H G. C. Greubel, <a href="/A174396/b174396.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1,-1).
%F a(n) = 3*(n-1-floor((n-1)/4)) + (-1)^floor((n-1)/2).
%F From _Wesley Ivan Hurt_, Oct 17 2015: (Start)
%F G.f.: x*(1+3*x+x^2+3*x^3+x^4)/((x-1)^2*(1+x+x^2+x^3)).
%F a(n) = a(n-1)+a(n-4)-a(n-5) for n>5.
%F a(n) = (18*n-9+3*(-1)^n-2*(-1)^((2*n+1-(-1)^n)/4))/8. (End)
%F E.g.f.: (1/8)*(2*sin(x) - 2*cos(x) + 18*x*exp(x) + 3*exp(-x) - 9*exp(x) + 8). - _G. C. Greubel_, Oct 18 2015
%p seq(3*(n - floor(n/4)) + (-1)^floor(n/2), n=0..100);
%t CoefficientList[Series[(1 + 3 x + x^2 + 3 x^3 + x^4)/((x - 1)^2*(1 + x + x^2 + x^3)), {x, 0, 100}], x] (* _Wesley Ivan Hurt_, Oct 17 2015 *)
%t RecurrenceTable[{a[1] == 1, a[2] == 4, a[3] == 5, a[4] == 8, a[5] == 10 , a[n+5] == a[n+4] + a[n+1] - a[n] }, a, {n, 1, 100}] (* _G. C. Greubel_, Oct 18 2015 *)
%o (Magma) [(18*n-9+3*(-1)^n-2*(-1)^((2*n+1-(-1)^n) div 4))/8 : n in [1..100]]; // _Wesley Ivan Hurt_, Oct 17 2015
%K nonn,easy
%O 1,2
%A _Gary Detlefs_, Mar 18 2010
%E Formula corrected by _Gary Detlefs_, Mar 19 2010
|
