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%I #34 Sep 08 2022 08:44:51
%S 1,5,26,130,651,3255,16276,81380,406901,2034505,10172526,50862630,
%T 254313151,1271565755,6357828776,31789143880,158945719401,
%U 794728597005,3973642985026,19868214925130,99341074625651,496705373128255
%N Base-5 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.
%C Partial sums of A015531. - _Mircea Merca_, Dec 28 2010
%H Vincenzo Librandi, <a href="/A033115/b033115.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (5,1,-5).
%F a(n) = 5*a(n-1) + a(n-2) - 5*a(n-3). - _Joerg Arndt_, Jan 08 2011
%F From _Paul Barry_, Nov 12 2003: (Start)
%F a(n) = floor(5^(n+2)/24);
%F a(n) = Sum_{k=0..floor(n/2)} 5^(n-2*k);
%F a(n) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*5^j.
%F Partial sums of A083425.
%F G.f.: 1/((1-x)*(1+x)*(1-5*x));
%F a(n) = 4*a(n-1) + 5*a(n-2) + 1. (End)
%F From _Mircea Merca_, Dec 28 2010: (Start)
%F a(n) = (1/3)*floor(5^(n+1)/8) = floor((5*5^n - 1)/24) = round((5*5^n - 3)/24) = round((5*5^n - 5)/24) = ceiling((5*5^n - 5)/24);
%F a(n) = a(n-2) + 5^(n-1), n > 1. (End)
%p seq(1/3*floor(5^(n+1)/8),n=1..32); # _Mircea Merca_, Dec 26 2010
%t Table[FromDigits[PadRight[{},n,{1,0}],5],{n,30}] (* or *) LinearRecurrence[ {5,1,-5},{1,5,26},30] (* _Harvey P. Dale_, Jan 28 2017 *)
%o (Magma) [Round((5*5^n-3)/24): n in [1..30]]; // _Vincenzo Librandi_, Jun 25 2011
%Y Cf. A015531.
%K nonn,base,easy
%O 1,2
%A _Clark Kimberling_
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