Base-10 Skolem-Langford numbers

Dean Hickerson dean at
Tue Aug 28 04:48:02 CEST 2007

Hans Havermann wrote:

> Conjecture: Half the sum of the digits of these numbers is even.
> The conjecture is trivially true in the sense that I can verify it  
> for all 20120 terms. However, I wonder if someone has a simple proof  
> of it.

A108116 is only finite because of the artificial constraint that the
digits be <= 9.  In fact Hans's conjecture is true more generally, not
just for those cases.

Theorem:  Suppose {a_1, ..., a_(2n)} is a sequence of nonnegative integers
such that (A) any number which occurs in the sequence occurs exactly twice
and (B) if a_i and a_j (i<j) are the two occurrences of the number d, then
j-i=d+1.  Then  a_1 + a_2 + ... + a_(2n)  is a multiple of 4.

Proof:  Let D be the set of numbers that occur in the sequence; D has
exactly n elements.  For each d in D, let l(d) and r(d) be the smallest
and largest indices i such that  a_i = d;  we have  r(d)-l(d) = d+1.  Let
s  be the sum of the elements of D.

(For example, for the sequence 15120025, we have  n=4,  D={0,1,2,5},  s=8,
and l and r are given by this table:

    d  l(d)  r(d)
    0   5     6
    1   1     3
    2   4     7
    5   2     8 )

Note that

     SUM   (r(d) - l(d))  =    SUM   (d+1)  =  s + n
   d in D                    d in D

Also, every number from 1 to 2n occurs exactly once as either l(d) or r(d), so

     SUM   (r(d) + l(d))  =  1 + 2 + ... + 2n  =  2 n^2 + n
   d in D

The difference between these two sums is twice the sum of l(d), so it's
even; hence  s  is even.  The sum  a_1 + a_2 + ... + a_(2n)  equals  2s,
so it's divisible by 4.

Dean Hickerson
dean at

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