OFFSET
0,2
COMMENTS
This sequence was the subject of the 6th problem of the 15th British Mathematical Olympiad in 1979 (see the link BMO).
There are no prime numbers in this infinite sequence. Why?
a(0) = 1 and a(1) = 10001 = 73 * 137;
if n even = 2*k, k >= 1, then A094028(n) divides a(n);
if n odd = 2*k+1, k >= 1, then a(k) divides a(n).
REFERENCES
A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Pb 6 pp. 68 and 201 (1979).
LINKS
Colin Barker, Table of n, a(n) for n = 0..200
British Mathematical Olympiad, 1979 - Problem 6.
Index entries for linear recurrences with constant coefficients, signature (10001,-10000).
FORMULA
a(n) = (10^(4*n+4) - 1)/9999 for n >= 0.
G.f.: 1/((1 - x)*(1 - 10000*x)). - Colin Barker, Dec 05 2019
From Elmo R. Oliveira, Mar 21 2026: (Start)
a(n) = 10001*a(n-1) - 10000*a(n-2).
E.g.f.: exp(x)*(-1 + 10000*exp(9999*x))/9999. (End)
EXAMPLE
a(2) = ((10^3)^4 - 1)/9999 = 100010001 = 10101 * 9901 where 10101 = A094028(2).
a(3) = ((10^4)^4 - 1)/9999 = 1000100010001 = 10001 * 100000001 where 10001 = a(1).
From Omar E. Pol, Dec 04 2019: (Start)
Illustration of initial terms:
1;
10001;
100010001;
1000100010001;
10001000100010001;
100010001000100010001;
1000100010001000100010001;
10001000100010001000100010001;
100010001000100010001000100010001;
...
(End)
MAPLE
seq((10^(4*n+4)-1)/9999, n=0..10);
MATHEMATICA
Table[((10^(n+1))^4 - 1)/9999, {n, 0, 8}] (* Amiram Eldar, Dec 04 2019 *)
PROG
(PARI) Vec(1/((1 - x)*(1 - 10000*x)) + O(x^11)) \\ Colin Barker, Dec 05 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bernard Schott, Dec 02 2019
STATUS
approved