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%I #16 Jul 02 2023 15:45:03
%S 9901009901,990099010099009901,99009900990100990099009901,
%T 9900990099009901009900990099009901,
%U 990099009900990099010099009900990099009901
%N a(n) = A*10^(4*n+1)+B with A=99000*(10^(4*n)-1)/9999+10 and B=9900*(10^(4*n)-1)/9999+1.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (100010001, -1000100010000, 1000000000000).
%F a(n) = ( 100^(2*n+1) + 1 )^2 / 101. [_Bruno Berselli_, Jul 23 2013]
%F G.f.: 101*x*(98029801-1000000010000*x+1000000000000*x^2)/((1-x)*(1-10000*x)*(1-100000000*x)). [_Bruno Berselli_, Jul 23 2013]
%e a(2) = (99000*(10^8-1)/9999+10)*10^9+9900*(10^8-1)/9999+1 = 990099010099009901.
%e Note that 990099010099009901 = 990099010^2+099009901^2.
%t Table[(100^(2 n + 1) + 1)^2/101, {n, 5}] (* _Bruno Berselli_, Jul 23 2013 *)
%o (PARI) a(n) = (99000*(10^(4*n)-1)/9999+10)*10^(4*n+1)+9900*(10^(4*n)-1)/9999+1 \\ _Michel Marcus_, Jul 23 2013
%o (Magma) [(100^(2*n+1)+1)^2/101: n in [1..5]]; // _Bruno Berselli_, Jul 23 2013
%Y Subsequence of A055616.
%K nonn,easy
%O 1,1
%A Ulrich Schimke (ulrschimke(AT)aol.com)