Please try the new web page for sending in a new sequence!
N. J. A. Sloane
njas at research.att.com
Thu Aug 28 04:52:30 CEST 2008
Hello Math-Fun & SeqFans,
Dan Hoey found this (finite) sequence -- do you think
it is worth adding it to the OEIS ? I would say yes !
S = 4, 5, 6, 7, 8, 41, 51, 63, 72, 83, 200
(there are 4 digits between the two 4's
5 digits between the two 5's
6 digits between the two 6's
7 digits between the two 7's
8 digits between the two 8's
1 digit between the two 1's
...
0 digit between the two 0's)
Best,
É.
---
[Dan Hoey]:
> The best is 456784151637283200, which can be chunked into
> 4+5+6+7+8+41+51+63+72+83+200 = 540.
>
> (...) There are 10216 such 18-digit strings, or 9060 such
> integers if we prohibit leading zeroes.
---
[Eric A., a week ago]:
We are looking for an 18-digit integer (like
946131483695200285) where we have "d" digits
between two d's (here: one digit between two
1's, zero digit between the two 0's, nine
digits between the two 9's, etc.)
Those 18 digits form thus 9 pairs of digits --
the 9 pairs being different one from another.
Now cut this integer into chunks so to make
a finite monotonically increasing sequence
(like this one, for instance, among others):
9,46,131,483,695,200285
... and we sum the terms:
9+46+131+483+695+200285 = 201649
Question:
Find the integer which, properly chunked, will
give the smallest possible sum.
(...)
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