Extending A062885

Don Reble djr at nk.ca
Fri Nov 2 04:00:05 CET 2001

Numseq fans, John Layman:

> The interesting part is proving that no such multiple exists, i.e.
> that a(n) is INDEED -1. The proof for n=25 is given in the comment
> above and is also easy for n=40. It might be of some interest to see
> proofs for n=31, 33, 37, 39, 41, etc. Some of these may be easy but,
> who knows, some may be rather challenging.
> John Layman

    Did Someone Say "CHALLENGING"?!?!?
    Read on, or skip to the last paragraph.


For any number N which is not a multiple of 2 nor 5, (1/N) has an
infinite periodic decimal representation which repeats from the start.

Suppose 1/N has M digits in its period. Then 10^M/N has the same
fractional part, and (10^M/N - 1/N) is an integer; that is, N divides

Now, 10^M-1 is just a string of M nines, and it divides the string of
nines with length 5M: 10^M-1 divides 10^(5M)-1. Also, the number
99999 divides 10^(5M)-1. Let Q(N) = ((10^(5M)-1) / 99999). Q(N) looks
like this:
There are M ones, separated by quadruples of zeros.


Let S be 8642, 20864, 42086, 64208, or 86420.

For any N not divisible by 2 nor 5, and for any S, S*Q(N) is a DED
(descending even digits) number. There are M groups of the quadruple.

So, N divides 10^M-1, which divides 10^(5M)-1, which equals
Q(N)*99999. So N divides Q(N)*99999.


The prime factorization of 99999 is 3*3*41*271. If N (already not a
multiple of 2 nor 5) is also not a multiple of 3, 41, nor 271, then N
divides Q(N), and also S*Q(N).

Therefore, if N is not a multiple of 2, 3, 5, 41, nor 271, then N
divides some DED number, and A062885[N] exists.

(Note: S*Q(N) needn't be equal to A062885[N]. But it is an upper bound,
 and establishes existence.)


Now, suppose that N is not a multiple of 2 nor 5. Let P=99999*N.
Then P divides 99999*Q(P) (because P is not a multiple of 2 nor 5).
So (99999*Q(P))/(99999*N) is an integer, and so is Q(P)/N.
And N divides S*Q(P), which is a DED.

Therefore, if N is not a multiple of 2 nor 5, then N divides some DED
number, and A062885[N] exists.


The prime factorizations of S numbers are:
     8642 = 2*29*149
    20864 = 2*2*2*2*2*2*2*163
    42086 = 2*11*1913
    64208 = 2*2*2*2*4013
    86420 = 2*2*5*29*149.

For any N, one factorization of N is (2^A)*(5^B)*C, where C is not a
multiple of 2 nor 5. S*Q(99999*C) is:
    a DED number,
    a multiple of C (because of the previous argument).
If some S is also a multiple of (2^A)*(5^B), then S*Q(99999*C) is:
    a multiple of N (because C is coprime its cofactor).
And therefore N divides S*Q(99999*C).

Therefore if N=(2^A)*(5^B)*C (where C isn't a multiple of 2 nor 5), and
    A<8 and B=0 (because of 20864), or
    A<3 and B<2 (because of 86420),
then N divides some DED (either 20864*Q(99999*C) or 86420*Q(99999*C))
and A062885[N] exists.

So "C" multiples of 128 are ok, but multiples of 256 and 640 might not be.
And "C" multiples of 20 are ok, but multiples of 40 and 100 might not be.


We already know that 25 and 40 do not divide DED numbers; neither do
their multiples. Only multiples of 256=2^8 remain unknown.

A number is divisible by 2^K, if and only if its last K digits are
divisible by 2^K. There are thirty-one possible eight-digit endings of
a DED number:
    2, 20, 208, ..., 4, 42, 420, ..., 86420864.
Of these, only the 420864 ending is a multiple of 256. And the only DED
number with that ending is 420864 itself.

Therefore: N divides a DED (and A062885[N] exists) if and only if
    N divides 420864, or
    N is not divisible by 25, 40, nor 256.

Don Reble       djr at nk.ca

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