Duplicate hunting
Ralf Stephan
ralf at ark.in-berlin.de
Wed Apr 11 09:52:39 CEST 2007
> A061433 and A069659
> A033553 and A050991
I'm glad you didn't find more below 90k as that was
when I did the last duplicate purge.
As to the first, both A069659 and A061433 have the conjecture
of equality stated but nothing proved.
For A033553 and A050991, they are identical IMHO and should be
merged. But please someone double check it.
ralf
I'm busy with a conference this week, so have not
been able to do anything yet about these duplicates.
But let me say something about this pair:
> Finally, there are a large number (well over half of the sequences I've
> reviewed) which are similar to A116133 and A116188, all by the same
> author. I wonder if there's some way to relate them.
that Andrew found:
%S A116133 64,1876,3214,8776,17227120144584448354,19391087990479091186,
%T A116133 21557280540171744040,23970390649400564404,26051259424903756364,
%U A116133 28697643366269876506,31320461807899711990,34215988012600009664
%N A116133 n concatenated with n-4 gives the product of two numbers which differ by 9.
%e A116133 8776//8772 = 9364 * 9373, where // denotes concatenation.
and
%S A116188 64,1876,3214,8776,17227120144584448354,19391087990479091186,
%T A116188 21557280540171744040,23970390649400564404,26051259424903756364,
%U A116188 28697643366269876506,31320461807899711990,34215988012600009664
%N A116188 n concatenated with n+4 gives the product of two numbers which differ by 7.
%e A116188 8776//8780 = 9365 * 9372, where // denotes concatenation.
Since
n(10^k+1) + 4 = m(m+7) iff n(10^k+1) - 4 = m(m+7) - 4 = (m-1)(m+8) = M(M+9)
these are trivially the same.
Neil
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