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%I #29 Feb 06 2022 18:04:19
%S 5,22,103,504,2505,12506,62507,312508,1562509,7812510,39062511,
%T 195312512,976562513,4882812514,24414062515,122070312516,610351562517,
%U 3051757812518,15258789062519,76293945312520,381469726562521,1907348632812522,9536743164062523
%N a(n) = 4*5^(n-1) + n.
%C The last n decimal digits of 2^a(n) form the number 2^n.
%H Colin Barker, <a href="/A308807/b308807.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-11,5).
%F a(n) = A005054(n) + n.
%F From _Colin Barker_, Jun 26 2019: (Start)
%F G.f.: x*(5 - 13*x + 4*x^2) / ((1 - x)^2*(1 - 5*x)).
%F a(n) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3) for n>3.
%F (End)
%F Conjectures confirmed by _Robert Israel_, Jun 28 2019
%e a(1) = 5, 2^5 = 32, the last digit of 32 is 2, which is 2^1.
%e a(2) = 22, 2^22 = 4194304, the last 2 digits of 4194304 are 04, which is 2^2.
%p seq(4*5^(n-1) + n, n=1..30); # _Robert Israel_, Jun 28 2019
%t Table[4*5^(n-1)+n,{n,30}] (* or *) LinearRecurrence[{7,-11,5},{5,22,103},30] (* _Harvey P. Dale_, Jun 27 2020 *)
%o (PARI) Vec(x*(5 - 13*x + 4*x^2) / ((1 - x)^2*(1 - 5*x)) + O(x^25)) \\ _Colin Barker_, Jun 29 2019
%K nonn,easy
%O 1,1
%A _Clive Tooth_, Jun 25 2019
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