OK, I did'nt understood you made same seqs with other beginnings than ours... So I agree, there are a lot of loops around there... :-) Thanx Alexandre De : Graeme McRae <g_m@mcraefamily.com> Date : Thu, 8 Dec 2005 10:28:11 -0800 À : Alexandre Wajnberg <alexandre.wajnberg@skynet.be> Objet : Re: [math-fun] Sum of last ten digits Just taking the first one, if the sequence begins 0, 6, 1, 8, 7, 8, 6, 6, 3, 0, then the sequence through the repetition (loop length 8) is 0, 6, 1, 8, 7, 8, 6, 6, 3, 0, 45, 48, 51, 42, 36, 42, 39, 39, 45, 48, 51, 51, 45, 42, 39, 39, 45, 48, 51, 51, 45. --Graeme

----- Original Message -----

De : Graeme McRae <g_m@mcraefamily.com> Date : Thu, 8 Dec 2005 07:47:23 -0800

Searching randomly, I found that: A loop of length 8 is possible, starting at 0, 6, 1, 8, 7, 8, 6, 6, 3, 0. A loop of length 24 is possible, starting at 0, 4, 2, 1, 9, 7, 1, 7, 7, 4. A loop of length 26 is possible, starting at 2, 3, 2, 7, 0, 9, 8, 7, 8, 4. A loop of length 78 is possible, starting at 2, 2, 5, 2, 6, 0, 3, 8, 5, 7. --Graeme

...

The nine sequences (beginning with ...0 1; ...2; 9) involve digits in their construction and our loops are made of terms (numbers).

I tried your digit-propositions without success, no "digit loops" hidden inside. I did this way: -pasting the nine complete sequences (the beginning + the first loop + the ten first terms of the second loop) on a new page of my "word" program; -compacting them (removing all the "blanks" and the ",") -compacting the same way your ten digits propositions; -searching, using the "find" function of my "word" program. I found only my control. So I don't understand. May be I miss a point? Are you shure of your finding? Or of your sequences? I send you separately our complete sequences. Thank you for your interest.

Alexandre