

A160862


a(n) = length of period of transformation sumdigitsquare: given a number m(k) define m(k+1) to be the sum of the squares of the decimal digits of m(k) and iterate until m(s) is found such that m(s)=m(k) with k<s. Numbers repeat periodically with a period of length a(n).


3



1, 8, 12, 8, 11, 16, 5, 12, 11, 1, 9, 12, 2, 13, 10, 8, 12, 12, 4, 8, 11, 13, 3, 8, 10, 12, 13, 3, 9, 13, 2, 3, 12, 11, 12, 15, 8, 9, 13, 8, 13, 8, 11, 4, 14, 11, 11, 13, 4, 11, 10, 10, 12, 14, 12
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OFFSET

1,2


REFERENCES

C. Suriano, Miniature Matematiche, unpublished, 2009, section 30


LINKS



EXAMPLE

a(1)=1 since 1^2=1; a(2)=8 since the period contains 8 numbers: 4,16,37,58,89,145,42,20 the following being 4 again.
a(5)=11 since the sequence runs as follows:
25,29, 85, 89, 145, 42, 20, 4, 16, 37, 58 the next term being 89 that is already there.


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



