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a(n) = 5*2^n + 5.
4

%I #19 Jan 07 2021 20:11:08

%S 10,15,25,45,85,165,325,645,1285,2565,5125,10245,20485,40965,81925,

%T 163845,327685,655365,1310725,2621445,5242885,10485765,20971525,

%U 41943045,83886085,167772165,335544325,671088645,1342177285,2684354565,5368709125,10737418245

%N a(n) = 5*2^n + 5.

%C For n > 2, decimals given by base 2 pattern: 101...zeros...101.

%C The ratio between a(n) and a(n+1) tends to 1/2.

%C Sum(1/a(i), i=0..infinity) = ( psi_2^(0)( 1 - i*Pi/log(2) ) + i*Pi ) / log(32) = 0.252899956069688841838263949451..., where psi_q^(m)(.) is the m-th derivative of the q-digamma function. [_Bruno Berselli_, Nov 15 2013]

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2).

%F a(n) = 5*A000051(n). - _Philippe Deléham_, Nov 15 2013

%F G.f.: 5*(2-3*x)/((1-x)*(1-2*x)). - _Philippe Deléham_, Nov 15 2013

%F a(n) = 3*a(n-1)-2*a(n-2) for n>1, a(0)=10, a(1)=15. - _Philippe Deléham_, Nov 15 2013

%t Table[5*2^n + 5, {n, 0, 50}] (* _T. D. Noe_, Nov 15 2013 *)

%Y Cf. A000079 (powers of 2), A000051.

%K nonn,easy

%O 0,1

%A _Przemyslaw Wcislo_, Nov 12 2013