OFFSET
0,2
COMMENTS
If one Pisano period (of 48 members) is partitioned sequentially as 8 groups of 6, then we observe that in each group, the 6th member is 0.
There are 8 instances of this residue 0.
The set of residues {3,7,11,15,19,23,27,31} is seen in the 2nd members of each group of 6, and similarly seen in the 4th, and in the 5th members. This set appears in a total of 3 instances.
The set of residues {6,14,22,30} has 2 instances, seen when considering the 3rd members of each group of 6.
The set of residues {1,5,9,13,17,21,25,29} appears as 1 instance when considering the 1st members of each group of 6.
For the 8 instances case, each residue is congruent with 8 (mod 4).
For the 3 instances case, each residue is congruent with 3 (mod 4).
For the 2 instances case, each residue is congruent with 2 (mod 4).
For the 1 instances case, each residue is congruent with 1 (mod 4).
8,3,2,1 are Fibonacci numbers.
REFERENCES
C. N. Menhinick, The Fibonacci Resonance and other new Golden Ratio discoveries, Onperson, (2015), pages 419-420.
LINKS
E. T. Jacobson, Distribution of the Fibonacci numbers mod 2^k, Fibonacci Quarterly, 30:3, (1992), pages 211-215.
Index entries for linear recurrences with constant coefficients, signature (0,1,0,-1,0,1,0,-1, 0,1,0,-1,0,1,0,-1,0,1, 0,-1,0,1,0,-1, 0,1,0,-1,0,1,0,-1,0,1,0,-1, 0,1,0,-1,0,1,0,-1,0,1).
FORMULA
a(n) = (F(n+4)-2) mod 32. (Based on the F(n) partial sums formula: F(n+2)-1, while here omitting F(1)=1 and F(0)=0.)
MATHEMATICA
Table[Mod[Fibonacci[n + 4] - 2, 32], {n, 0, 64}] (* Michael De Vlieger, Mar 31 2016 *)
Mod[Accumulate[Fibonacci[Range[2, 50]]], 32] (* Harvey P. Dale, Jul 19 2018 *)
PROG
(PARI) a(n)=(fibonacci(n%48+4)-2)%32 \\ Charles R Greathouse IV, Mar 31 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clive N. Menhinick, Mar 30 2016
STATUS
approved