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A130893 Truncanacci (1,3) numbers : numbers calculated with the rule of Fibonnacci starting from the two first terms 1 and 3 and truncated mod 10. +0
2
1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2 (list; graph; listen)
OFFSET

0,2

COMMENT

This sequence is a loop of 12 steps repeated indefinitely from the 13th term.

The sequence starting with any pair of successive terms of the sequence runs also in the same loop.

Note that the parity of the terms of the loop has a periodicity of 3 :odd,odd,even,odd,odd,even . . . then one at least of the digits of any pair of successive terms of the loop id odd;*

The same rule applied from the fisrt terms 0 an 1 gives A003893 (=truncanacci(0,1)) which is a loop of 60 steps. That is true of course also for any of the 59 other pairs (A003893(i),A003893(1+i))

Another loop of 3 steps is given for the 3 last pairs of digits, one of them at least being odd : Truncanacci(0,5) =0,5,5,0,5,5

Among the pairs of two even digits, 20 of them generate the same loop of 20 steps :Truncanacci(0,2)=0,2,2,4,6,0,6,6,2,8,0,8,8,6,4,0,4,4,8,2,0,2,2,4,6,0,6,6,2,8,0,8,8,6,4,0,4,4,8,2

4 more pairs generate a loop of this size : Truncanacci(4,6)= 2,6,8,4,2,6,8,4

The last pair of even digits (0,0) gives the constant term 0 or a loop of one step.

As we have studied 60+12+3=75 starting pairs with at least an odd term and 20+4+1=25 pairs of even digits, all possible pairs from (0,0) to (9,9) are exhausted

FORMULA

t(0)=1 t(1)=1 t(n)= (t(n-2)+t(n-1))mod 10

EXAMPLE

1+3 = 4 = 4 mod 10, then a(3) = 4

3+4 = 7 = 7 mod 10, then a(4) = 7

4+7 = 11 = 2 mod 10, then a(5) = 1

CROSSREFS

Cf. A003983.

Sequence in context: A095877 A024476 A093087 this_sequence A072079 A116073 A021292

Adjacent sequences: A130890 A130891 A130892 this_sequence A130894 A130895 A130896

KEYWORD

easy,nonn

AUTHOR

Philippe Lallouet (philip;lallouet(AT)wanadoo;fr), Aug 22 2007

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Last modified August 19 23:53 EDT 2008. Contains 142930 sequences.


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