The well-known Lucas series

1, 3, 4, 7, 11, 18, 29, 47, 76, ……is closely related to the Fibonacci sequence

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,….The Lucas is obtained by continually adding the n and n + 2 terms of the Fibonacci to obtain the corresponding nth term of the Lucas.

Thus 1 - the 1st term of the Lucas - is obtained by adding 0 and 1 (i.e. the 1

^{st}and 3^{rd}terms of the Fibonacci); 3 the 2^{nd}term of the Lucas is obtained by adding 1 and 2 (i.e. the 2^{nd}and 4^{th}terms of the Fibonacci); 4 the 3^{rd}term of the Lucas is obtained by adding 1 and 3 (i.e. the 3^{rd}and 5^{th}terms) etc.In both the Fibonacci and Lucas the ratio of successive terms i.e.

(k+ 1)/kapproximates to the value of Phi (1.61803309887..).Furthermore in the Lucas Series the nth term of the series approximates (phi)

^{n}.Thus

(phi)which is closely approximated by the^{9 =}76.013..9term of the Lucas Series (76).^{th}

Now we can obtain a fascinating Lucas equivalent for the 3-term Fibonacci-like sequence (i.e. where starting with 0, 0, 1 we continually add the last 3terms to obtain the next term).

As we have seen this results in the sequence

0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415,.....

Now a fascinating Lucas type series can be generated for this series using the following formula

(n + 3) + 2(n + 1) + (n - 2)/2 + (n - 6)/2, where n represents thenthterm of the 3-term Fibonacci series = the nth term of the corresponding Lucas series. (When the value of n is negative, it is ignored).Thus when

n = 1, we get 1 + 2(0) =1; The other terms lead to negative values or 0, they are ignored.Therefore the 1

^{st}term of the corresponding Lucas series is 1.

When

n = 2, we get 2 + 2 (1) =4; again the other terms are ignored.Therefore the 2

^{nd}term of the corresponding Lucas series is 4.

When

n = 3, we get 4 + 2(1) =6; again the other terms are ignored.Therefore the 3

^{rd}term of the corresponding Lucas series is 6.

When

n = 4we get 7 + 2 (2) =11.Therefore the 4

^{th}term of the corresponding Lucas series is 11.

When

n = 5, we get 13 + 2(4) + 1/2 =21.5; the final term is ignored. For convenience we can round here to 21 (to express as a whole number).

Therefore the 5

^{th}term of the corresponding Lucas sequence is21.

When

n = 6, we get 24 + 2(7) + 1/2 =38.5; the final term is ignored. For convenience we can round down here to 38 (again to express as a whole number).Therefore the

6term of the corresponding Lucas sequence is^{th}38.

When

n = 7, we get 44 + 2(13) + 2/2 =71; the final term is ignored.Therefore the 7

^{th}term of the corresponding Lucas sequence is 71.

When

n = 8, we get 81 + 2(24) + 4/2 =131.Therefore the 8

^{th}term of the corresponding Lucas sequence is 131.

When

n = 9we get 149 + 2(44) + 7/2 + 1/2 =241.

Continuing on, the

10term we get 274 + 2(81) + 13/2 + 1/2 =^{th}443.

In like manner, the

11term is 504 + 2(149) + 24/2 + 2/2 =^{th}815

The

12term is 927 + 2(274) + 44/2 + 4/2 =^{th}1499.

The

13thterm is 1705 + 2(504) + 81/2 +7/2 =2757

The

14term is 3136 + 2(927) + 149/2 + 13/2 =^{th}5071

The

15term is 5768 + 2(1705) + 274/2 + 24/2 =^{th}9327

Therefore the first 15 terms of the corresponding Lucas series for the 3-term case are

1, 4, 6, 11, 21, 38, 71, 131, 241, 443, 815, 1499, 2757, 5071, 9327, 10609,...

The ratio here (as in the 3-term Fibonacci) approximates tophi_{3}= 1.839286755…For example

t= 9327/5071 = 1.839282.. (which is correct for the first 5 figures after the decimal point)._{15}/ t_{14}

Also

(phi_{3})^{n}_{ }approximates closely the value ofT(in the Lucas series)._{n}

Thus

(phi_{3})^{5}_{ }= (1.839286755..)^{5 }= 21.0497.. This closely approximatesT= 21._{5}

(Phi_{3})^{10}_{ }= (1.839286755..)^{10 }= 443.0925.. This closely approximatesT= 443._{10}

(Phi_{3})^{15}_{ }= (1.839286755..)^{15 }= 9326.9930.. This closely approximatesT= 9327._{15}

T_{20 }= 121415 + 2(35890) + 5768/2 + 504/2 = 196331.

(Phi_{3})^{20}_{ }= (1.839286755..)^{20 }= 196330.99549... This closely approximatesT= 196331._{20}