[seqfan] Re: a(n) divides the concatenation

[Eric Angelini]

> every a(n) is of the form  2^j * 5^k.

... yes, thank you David, well done!
Best,
É.

Propulsé d'un aPhone



Le 2 avr. 2013 à 01:43, "David Wilson" <davidwwilson at comcast.net<mailto:davidwwilson at comcast.net>> a écrit :

every a(n) is of the form  2^j * 5^k.

Clearly, the base case a(0) = 1 = 2^0 * 5^0 is true.

Now suppose a(n) is of the form 2^j * 5*k. Then a(n+1) | 10^d * a(n) = 2^(j+d) * 5^(k+d), so a(n+1) is also of the form 2^j * 5^k.

I would conjecture