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A330135 a(n) = ((10^(n+1))^4 - 1)/9999 for n >= 0.

%I A330135 #34 May 06 2022 13:13:51
%S A330135 1,10001,100010001,1000100010001,10001000100010001,
%T A330135 100010001000100010001,1000100010001000100010001,
%U A330135 10001000100010001000100010001,100010001000100010001000100010001
%N A330135 a(n) = ((10^(n+1))^4 - 1)/9999 for n >= 0.
%C A330135 This sequence was the subject of the 6th problem of the 15th British Mathematical Olympiad in 1979 (see the link BMO).
%C A330135 There are no prime numbers in this infinite sequence. Why?
%C A330135 a(0) = 1 and a(1) = 10001 = 73 * 137;
%C A330135 if n even = 2*k, k >= 1, then A094028(n) divides a(n);
%C A330135 if n odd = 2*k+1, k >= 1, then a(k) divides a(n).
%D A330135 A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Pb 6 pp. 68 and 201 (1979).
%H A330135 Colin Barker, <a href="/A330135/b330135.txt">Table of n, a(n) for n = 0..200</a>
%H A330135 British Mathematical Olympiad, <a href="https://bmos.ukmt.org.uk/home/bmo-1979.pdf">1979 - Problem 6</a>.
%H A330135 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10001,-10000).
%F A330135 a(n) = (10^(4*n+4) - 1)/9999 for n >= 0.
%F A330135 G.f.: 1 / ((1 - x)*(1 - 10000*x)). - _Colin Barker_, Dec 05 2019
%e A330135 a(2) = ((10^3)^4 - 1)/9999 = 100010001 = 10101 * 9901 where 10101 = A094028(2).
%e A330135 a(3) = ((10^4)^4 - 1)/9999 = 1000100010001 = 10001 * 100000001 where 10001 = a(1).
%e A330135 From _Omar E. Pol_, Dec 04 2019: (Start)
%e A330135 Illustration of initial terms:
%e A330135                   1;
%e A330135                 10001;
%e A330135               100010001;
%e A330135             1000100010001;
%e A330135           10001000100010001;
%e A330135         100010001000100010001;
%e A330135       1000100010001000100010001;
%e A330135     10001000100010001000100010001;
%e A330135   100010001000100010001000100010001;
%e A330135 ...
%e A330135 (End)
%p A330135 A: = seq((10^(4*n+4)-1)/9999, n=1..4);
%t A330135 Table[((10^(n+1))^4 - 1)/9999, {n, 0, 8}] (* _Amiram Eldar_, Dec 04 2019 *)
%o A330135 (PARI) Vec(1 / ((1 - x)*(1 - 10000*x)) + O(x^11)) \\ _Colin Barker_, Dec 05 2019
%Y A330135 Cf. A000533 (1000...0001), A094028 (10101...101), A261544 (1001001...1001).
%Y A330135 Cf. A131865 (similar, with 2^(n+1)).
%K A330135 nonn,easy
%O A330135 0,2
%A A330135 _Bernard Schott_, Dec 02 2019