All the Tribonacci numbers are written successively without commas and spaces resulting in this infinite string:

00112471324448114927450492717053136...........

Reading left to right, what is the 2011th digit in the above string? How about the 20110th digit in the above string?

 10        T1=0:T2=0:T3=1:TriCt=3
 20        CurrLen=3
 30        while EndPos<20111
 40         Temp=T1+T2+T3
 50         T1=T2:T2=T3:T3=Temp:inc TriCt
 60         N$=cutspc(str(T3))
 70         L=len(N$)
 80         StPos=CurrLen+1:EndPos=CurrLen+L
 90         CurrLen=EndPos
100         if 2011>=StPos and 2011<=EndPos then
110           :print TriCt,N$,StPos;EndPos,mid(N$,2011-StPos+1,1)
130         if 20110>=StPos and 20110<=EndPos then
140           :print TriCt,N$,StPos;EndPos,mid(N$,20110-StPos+1,1)
170        wend

finds

 125    119816209721856219780831547518850        1992  2024     8
 392    549435842499922176757513268436445454164615369446056872362025166822986968
4140693338638036540612328930253  20089  20191   5

meaning that the 125th Tribonacci is 119816209721856219780831547518850, occupying positions 1992 - 2024 of the string, so that an 8 is at position 2011; then the 392nd Tribonacci is 5494358424999221767575132684364454541646153694460568723620251668229869684140693338638036540612328930253, occupying positions 20089 - 20191, making a 5 occupy the 20110th position of the string.

Posted by Charlie on 2011-08-30 14:43:28