regenerating sequences by successive differences

Leroy Quet qq-quet at
Fri Jun 11 20:00:19 CEST 2004

>Amarnath asks:
>Can any one design a sequence which produces itself as k'th 
>successive difference? are these unique?
>Me:  Martin Gardner and I considered this question
>years ago.  Take a look at (i quote from the Index):
>differences = complement: A005228*, A030124

Ah, *these* sequences! About a month and a half ago I had high-hopes that 
they were not in the EIS yet so I myself could submit a post regarding 
them to, and claim them as mine.

But, alas, they were actually well known sequences.


In any case, if A005228 is {a(n)} and A030124 is {b(n)}, both with offset 
1 here *,
then the sequence {a(n)-n} gives the number of elements of
{b(n)-n+1} which are <= n, and
{b(n)-n} gives the number of elements of {a(n)-n+1} which are <= n.

*(In the EIS, A030124 has an offset of 0, A005228 an offset of 1. Again, 
for the purposes of the result above, assume both sequences have an 
offset of 1.)

Now, the result above works for any {a(k)} and {b(k)} which are 
increasing positive integers and are compliments of each other. But I 
thought I might mention it anyway, since we are on the subject.

Leroy Quet

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