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A178873 Partial sums of round(5^n/7).

%I A178873 #41 Sep 08 2022 08:45:54
%S A178873 0,1,5,23,112,558,2790,13951,69755,348773,1743862,8719308,43596540,
%T A178873 217982701,1089913505,5449567523,27247837612,136239188058,
%U A178873 681195940290,3405979701451,17029898507255,85149492536273
%N A178873 Partial sums of round(5^n/7).
%H A178873 Vincenzo Librandi, <a href="/A178873/b178873.txt">Table of n, a(n) for n = 0..160</a>
%H A178873 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 A178873 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (7,-12,11,-5).
%F A178873 a(n) = round((5*5^n + 7)/28).
%F A178873 a(n) = floor((5*5^n + 19)/28).
%F A178873 a(n) = ceiling((5*5^n - 5)/28).
%F A178873 a(n) = a(n-6) + 558*5^(n-5), n>5.
%F A178873 a(n) = 5*a(n-1) + a(n-6) - 5*a(n-7), n>6.
%F A178873 a(n) = 7*a(n-1) - 12*a(n-2) + 11*a(n-3) - 5*a(n-4), n>3.
%F A178873 G.f.: -(2*x^2-x)/((x-1)*(5*x-1)*(x^2-x+1)).
%F A178873 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 A178873 a(6) = 0 + 1 + 4 + 18 + 89 + 446 + 2232 = 2790.
%p A178873 A178873 := proc(n) add( round(5^i/7),i=0..n) ; end proc:
%t A178873 Accumulate[Round[5^Range[0,25]/7]] (* _Harvey P. Dale_, Feb 01 2011 *)
%o A178873 (Magma) [Floor((5*5^n+19)/28): n in [0..40]]; // _Vincenzo Librandi_, Apr 28 2011
%K A178873 nonn,less
%O A178873 0,3
%A A178873 _Mircea Merca_, Dec 28 2010