Please try the new web page for sending in a new sequence!

N. J. A. Sloane njas at
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)


[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): 


... and we sum the terms: 

   9+46+131+483+695+200285 = 201649 

Find the integer which, properly chunked, will 
give the smallest possible sum. 


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