# Matching first n signs of differences between consecutive terms

Leroy Quet q1qq2qqq3qqqq at yahoo.com
Sun Feb 24 21:57:43 CET 2008

```I just submitted this sequence.

%I A137947
%S A137947 3,13,13,13,13,13,13,13
%N A137947 a(n) = k: k is smallest integer > 1
such that sign(d(1)-d(2)) = sign(d(k)-d(k+1)),
sign(d(2)-d(3)) =
sign(d(k+1)-d(k+2)),...,sign(d(n-1)-d(n)) =
sign(d(k+n-1)-d(k+n)), where sign is (-,0,+) and
d(m) = the number of positive divisors of m.
%Y A137947 A000005
%O A137947 1
%K A137947 ,more,nonn,
%A A137947 Leroy Quet (qq-quet at mindspring.com),
Feb 24 2008

First, could someone please extend the sequence?

Second, is it certain that the sequence is
infinite? (Maybe the same pattern of signs, for
some n, never repeats again -- But that would
seem unlikely.)

Third, I post this as a suggestion to others that
might want to apply the same process to other
sequences besides A000005.
(Suggestion: Check the signs of differences
between consecutive terms of sequence A000010.)

Fourth, might there be something interesting if
we compared the signs, not with the FIRST n
differences of sequence A000005, but with other
stretches of terms?
For example, compare (d(n),d(n+1),...d(2n)) to
(d(k),d(k+1),d(k+n), k > n.

I hope there is something interesting here. I
don't know for sure that there is.

Thanks,
Leroy Quet

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