0,1

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.

Colin Barker, Table of n, a(n) for n = 0..1000

E. Gutierrez, Recursion Methods to Generate New Integer Sequences (Part VIF)

E. Gutierrez, Table of Tuples and Use of Magic Ratio for Tuple Conversion (Part IB)

E. Gutierrez, Table of Tuples for Square of Squares (Part IC)

Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

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)

a(2)= 401*6 - (51 - 96)= 2451;

a(3)= 2451*6 - (401 - 96)= 14401;

a(4)= 14401*6 - (2451 - 96)= 84051.

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 *)

(PARI) Vec((51+44*x+x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Colin Barker, May 18 2016

Cf. A178218, A273182, A273187.

Sequence in context: A359026 A152579 A083669 * A222910 A259692 A204215

Adjacent sequences: A273186 A273187 A273188 * A273190 A273191 A273192

nonn,easy

Eddie Gutierrez, May 17 2016

More terms from Colin Barker, May 18 2016

approved