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

1, 10001, 100010001, 1000100010001, 10001000100010001, 100010001000100010001, 1000100010001000100010001, 10001000100010001000100010001, 100010001000100010001000100010001

Offset: 0

Author

Bernard Schott, Dec 02 2019

Keywords

• nonn
• easy

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).

Examples

			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)
		

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).

Crossrefs

Cf. A000533 (1000...0001), A094028 (10101...101), A261544 (1001001...1001).

Cf. A131865 (similar, with 2^(n+1)).

Programs

Maple

A: = seq((10^(4*n+4)-1)/9999, n=1..4);

Mathematica

Table[((10^(n+1))^4 - 1)/9999, {n, 0, 8}] (* Amiram Eldar, Dec 04 2019 *)

PARI

Vec(1 / ((1 - x)*(1 - 10000*x)) + O(x^11)) \\ Colin Barker, Dec 05 2019

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