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A291658 a(n) is the sum of all integers from 5^n to 5^(n+1)-1.

%I A291658 #30 Sep 29 2017 09:07:33
%S A291658 10,290,7450,187250,4686250,117181250,2929656250,73242031250,
%T A291658 1831053906250,45776363281250,1144409160156250,28610229394531250,
%U A291658 715255736816406250,17881393430175781250,447034835803222656250,11175870895324707031250,279396772384338378906250
%N A291658 a(n) is the sum of all integers from 5^n to 5^(n+1)-1.
%C A291658 a(n) is the sum of all (positive) numbers having exactly (n+1) digits when written in base 5. - _Alois P. Heinz_, Sep 25 2017
%H A291658 Colin Barker, <a href="/A291658/b291658.txt">Table of n, a(n) for n = 0..700</a>
%H A291658 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (30,-125).
%F A291658 a(n) = ((5^n)/2)*(5^(n+2) - 5^n - 4), n >= 0.
%F A291658 From _Colin Barker_, Sep 12 2017: (Start)
%F A291658 G.f.: 10*(1 - x) / ((1 - 5*x)*(1 - 25*x)).
%F A291658 a(n) = 30*a(n-1) - 125*a(n-2) for n>1.
%F A291658 (End)
%F A291658 a(n) = A162729(n+1) - A162729(n). - _Alois P. Heinz_, Sep 25 2017
%e A291658 For n=0, the sum is from 1 to 4, so a(0)=10;
%e A291658 for n=1, the sum is from 5 to 24, so a(1)=290, etc.
%p A291658 a:= unapply(sum(i, i=5^n..5^(n+1)-1), n):
%p A291658 seq(a(n), n=0..20);  # _Alois P. Heinz_, Sep 25 2017
%o A291658 (PARI) Vec(10*(1 - x) / ((1 - 5*x)*(1 - 25*x)) + O(x^30)) \\ _Colin Barker_, Sep 12 2017
%Y A291658 Cf. A010036, A226508, A121544, A101291, A162729.
%K A291658 nonn,easy
%O A291658 0,1
%A A291658 _Enrique Navarrete_, Aug 28 2017