OFFSET
1,1
COMMENTS
Assuming digits are random, the expected value for the size of the partitions is 3.66021568 = Sum_{k=1..10} k^2*9!/(10^k*(10-k)!).
LINKS
Sean A. Irvine, Java program (github)
EXAMPLE
Pi=3.1415926535897932384626433... => ignore lead 3 and partition as such: 0.|14|15926|53|5897|932|38462|643|3... => 2,5,2,4,3,5,3,...
PROG
(PARI)
F(v)={my(L=List(), S=Set()); for(i=1, #v, if(setsearch(S, v[i]), listput(L, #S); S=Set()); S=setunion(S, [v[i]])); Vec(L)}
{ localprec(10^3); my(t=Pi-3); F(digits(floor(t*10^precision(t)))) } \\ Andrew Howroyd, Aug 10 2020
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Ryan Brooks, May 10 2020
STATUS
approved