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Partial sums of round(5^n/8).
1

%I #43 Aug 10 2024 02:46:48

%S 0,1,4,20,98,489,2442,12208,61036,305177,1525880,7629396,38146974,

%T 190734865,953674318,4768371584,23841857912,119209289553,596046447756,

%U 2980232238772,14901161193850,74505805969241,372529029846194

%N Partial sums of round(5^n/8).

%H Vincenzo Librandi, <a href="/A178874/b178874.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 (6,-4,-6,5).

%F a(n) = round((5*5^n + 4*n - 1)/32) = round((5*5^n + 4*n - 5)/32).

%F a(n) = floor((5*5^n + 4*n + 3)/32).

%F a(n) = ceiling((5*5^n + 4*n - 5)/32).

%F a(n) = a(n-2) + (3*5^(n-1) + 1)/4 , n > 1.

%F a(n) = 6*a(n-1) - 4*a(n-2) - 6*a(n-3) + 5*a(n-4), n > 3.

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

%F a(n) = (5^(n+1) + 4*n - 4*(-1)^n - 1)/32. - _Bruno Berselli_, Jan 12 2011

%e a(2) = 0 + 1 + 3 = 4.

%p A178874 := proc(n) add( round(5^i/8),i=0..n) ; end proc:

%t Accumulate[Round[5^Range[0,25]/8]] (* _Harvey P. Dale_, Feb 18 2011 *)

%o (Magma) [Floor((5*5^n+4*n+3)/32): n in [0..40]]; // _Vincenzo Librandi_, Apr 28 2011

%K nonn,easy,less

%O 0,3

%A _Mircea Merca_, Dec 28 2010