login
Partial sums of round(5^n/7).
1

%I #41 Sep 08 2022 08:45:54

%S 0,1,5,23,112,558,2790,13951,69755,348773,1743862,8719308,43596540,

%T 217982701,1089913505,5449567523,27247837612,136239188058,

%U 681195940290,3405979701451,17029898507255,85149492536273

%N Partial sums of round(5^n/7).

%H Vincenzo Librandi, <a href="/A178873/b178873.txt">Table of n, a(n) for n = 0..160</a>

%H Mircea Merca, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL14/Merca/merca3.html">Inequalities and Identities Involving Sums of Integer Functions</a> J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (7,-12,11,-5).

%F a(n) = round((5*5^n + 7)/28).

%F a(n) = floor((5*5^n + 19)/28).

%F a(n) = ceiling((5*5^n - 5)/28).

%F a(n) = a(n-6) + 558*5^(n-5), n>5.

%F a(n) = 5*a(n-1) + a(n-6) - 5*a(n-7), n>6.

%F a(n) = 7*a(n-1) - 12*a(n-2) + 11*a(n-3) - 5*a(n-4), n>3.

%F G.f.: -(2*x^2-x)/((x-1)*(5*x-1)*(x^2-x+1)).

%F a(n) = 5^(n+1)/28 + 1/4 + A117373(n+2)/7 = (5*5^n+7)/28 - ((9-i*sqrt(3))*(1-i*sqrt(3))^n + (9+i*sqrt(3))*(1+i*sqrt(3))^n) / (42*2^n) where i is the imaginary unit. - _Bruno Berselli_, Jan 12 2011

%e a(6) = 0 + 1 + 4 + 18 + 89 + 446 + 2232 = 2790.

%p A178873 := proc(n) add( round(5^i/7),i=0..n) ; end proc:

%t Accumulate[Round[5^Range[0,25]/7]] (* _Harvey P. Dale_, Feb 01 2011 *)

%o (Magma) [Floor((5*5^n+19)/28): n in [0..40]]; // _Vincenzo Librandi_, Apr 28 2011

%K nonn,less

%O 0,3

%A _Mircea Merca_, Dec 28 2010