# 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.

(...)

```