OFFSET
0,1
COMMENTS
With the Fibonacci sequence A000045 without 0 i.e. here F(0)=F(1)=1, F(2)=2, F(3)=3, ..., this sequence satisfies: a(F(2*n)-2)=(43*9^(n-1)-3)/8; a(F(2*n+1)-2)=(129*9^(n-1)-1)/8; a(F(2*n)+F(2*n-2)-2)=(51*9^(n-2)-3)/8; a(F(2*n+1)+F(2*n-1)-2)=(17*9^(n-1)-1)/8; and the recurrence rules: with A the finite sequence between a(F(2*n-2)-2) and a(F(2*n-1)-2), and B the finite sequence between a(F(2*n-1)-2) and a(F(2*n)-2), the next step of the sequence between a(F(2*n)-2) and a(F(2*n+1)-2) is given by: "B - a(F(2*n)+F(2*n-2)-2) - A". In the same vein, with A' the finite sequence between a(F(2*n-1)-2) and a(F(2*n)-2) and B' the finite sequence between a(F(2*n)-2) and a(F(2*n+1)-2), the next step of the sequence between a(F(2*n+1)-2) and a(F(2*n+2)-2) is given by: "B' - a(F(2*n+1)+F(2*n-1)-2) - A'". It seems that this sequence gives the numbers of 2 in the successive sets of 2 in the sequence A174664.
EXAMPLE
a(F(4)-2)=a(3)=(9*43-3)/8=48. a(F(5)-2)=a(6)= (129*9-1)/8=145. a(F(6)-2)=a(11)=(43*9^2-3)/8=435. Between a(F(5)-2)=a(7)=145 and a(F(6)-2)=a(11)=435, there is: "2, 6, 19, 2". Between a(F(6)-2)=435 and a(F(7)-2)=a(19)=1306 there is "2, 6, 19, 2, 57, 2, 6". Then between a(19)=1306 and a(F(8)-2)=a(32)=3918, there is "2, 6, 19, 2, 57, 2, 6, a(F(7)+F(5)-2)=a(27)=172, 2, 6, 19, 2".
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Richard Choulet, Mar 26 2010
STATUS
approved