OFFSET
0,1
COMMENTS
The multiplying factor 6 (in the recursion formulas below) appears to come from the ratio of b(1)/b(0) of the sequence. Each of the lines of tables (V vs VII) or (VI vs VIII) in oddwheel.com/ImaginaryB.html generates this factor.
k is obtained from the difference of the offsets of two relate sequences. this one, (II), starting at 51 and a second, (I), at 99 (to be submitted separately). Thus, k =[Ic(n)- IIc(n)]*2. When n=0, Ic(0)=99 and IIc(0)=51 giving the value for k of (99-51)*2=96. Furthermore, k is the same constant number for any value of n.
The differences between number in the sequence are identical in both of the related sequences.
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
E. Gutierrez, Table of Tuples for Square of Squares (Part IC)
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).
FORMULA
a(0)= 51, a(1)= 401, a(n+1)= a(n)*6 - a(n-1) + k where k=96.
From Colin Barker, May 18 2016: (Start)
a(n) = (-24+25/2*(3-2*sqrt(2))^(1+n)+25/2*(3+2*sqrt(2))^(1+n)).
a(n) = 7*a(n-1)-7*a(n-2)+a(n-3) for n>2.
G.f.: (51+44*x+x^2) / ((1-x)*(1-6*x+x^2)).
(End)
EXAMPLE
a(2)= 401*6 - (51 - 96)= 2451;
a(3)= 2451*6 - (401 - 96)= 14401;
a(4)= 14401*6 - (2451 - 96)= 84051.
MATHEMATICA
CoefficientList[Series[(51 + 44 x + x^2)/((1 - x) (1 - 6 x + x^2)), {x, 0, 20}], x] (* Michael De Vlieger, May 18 2016 *)
LinearRecurrence[{7, -7, 1}, {51, 401, 2451}, 30] (* Harvey P. Dale, Feb 21 2020 *)
PROG
(PARI) Vec((51+44*x+x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Colin Barker, May 18 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Eddie Gutierrez, May 17 2016
EXTENSIONS
More terms from Colin Barker, May 18 2016
STATUS
approved