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%I #20 Sep 14 2019 06:37:13
%S 1,6,56,556,5556,55556,555556,5555556,55555556,555555556,5555555556,
%T 55555555556,555555555556,5555555555556,55555555555556,
%U 555555555555556,5555555555555556,55555555555555556,555555555555555556,5555555555555555556,55555555555555555556,555555555555555555556
%N Expansion of (1-5x)/((1-x)(1-10x)).
%C Second binomial transform of 5*A001045(3n)/3+(-1)^n. Partial sums of A093143. A convex combination of 10^n and 1. In general the second binomial transform of k*Jacobsthal(3n)/3+(-1)^n is 1, 1+k, 1+11k, 1+111k, ... This is the case for k=5.
%H Seiichi Manyama, <a href="/A093142/b093142.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-10).
%F a(n) = 5*10^n/9 + 4/9.
%F a(n) = 10*a(n-1) - 4 with a(0)=1. - _Vincenzo Librandi_, Aug 02 2010
%F a(n) = 11*a(n-1) - 10*a(n-2), n > 1. - _Harvey P. Dale_, Aug 23 2014
%t CoefficientList[Series[(1-5x)/((1-x)(1-10x)),{x,0,20}],x] (* or *) LinearRecurrence[{11,-10},{1,6},20] (* _Harvey P. Dale_, Aug 23 2014 *)
%o (PARI) {a(n) = (5*10^n+4)/9} \\ _Seiichi Manyama_, Sep 14 2019
%Y Cf. A001045.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Mar 24 2004
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