^{ }– 1)

^{3 }= 0.

And this number sequence is intimately related to the corresponding number sequence associated with the Zeta 1 (Riemann) function where s = – 2, i.e. ζ

_{1}( – 2), i.e. 1, 4, 9, 16, 25, ...

Then when we look at the numbers in this later sequence, we can see that they are related to the corresponding numbers of the former sequence in the following manner

1 = 0 + 1; 4 = 1 + 3; 9 = 3 + 6; 16 = 6 + 10; 25 = 10 + 15 and so on.

So in more general terms, t

_{k – 1}+ t

_{k }(with respect to the former infinite sequence) = t

_{k }(with respect to the latter).

With respect to the former sequence,

1 – 0
= 1 and 1 +
0 = 1 (i.e. 1

^{2});
3 – 1 = 2 and 3 +
1 = 4 (i.e. 2

^{2});
6 – 3 = 3 and 6 + 3 =
9 (i.e. 3

^{2});
10 – 6 = 4 and 10 + 6
= 16 (i.e. 4

^{2});
So in general terms the t

^{th}– (t – 1)^{th}term = t; and t^{th}+ (t – 1)^{th}term = t^{2}Again with respect to the former sequence,

0 + 1 = 1 and 0

^{2 }+ 1

^{2 }= 1 (i.e. the 1

^{st}term);

1 + 3

^{ }= 4 and 1^{2 }+ 3^{2 }= 10 (i.e. the 4^{th}term);
3 + 6 = 9 and 3

^{2 }+ 6^{2 }= 45 (i.e. the 9^{th}term);
6 + 10 = 16 and 6

^{2 }+ 10^{2 }= 136 (i.e. the 16^{th}term);
Thus in general terms (t

_{k – 1})^{2}+ (t_{k}^{ 2})_{ }= the (t_{k – 1 }+ t_{k})^{th }term of the infinite sequence.
In like manner, with respect to the same sequence,

1 – 0 = 1 and 1

^{2 }– 0^{2 }= 1 (i.e. 1^{3})_{ }
3 – 1 = 2 and 3

^{2 }– 1^{2 }= 8 (i.e. 2^{3})_{ }
6 – 3 = 3 and 6

^{2 }– 3^{2 }= 27 (i.e. 3^{3})_{ }
10 – 6 = 4 and 10

^{2 }– 4^{2 }= 64 (i.e. 4^{3})_{ }
So once more in general terms,

(t

With respect to the sequence 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ..., the 1st term of each successive group of 3 leaves a remainder of 1 when divided by 9

So, 1 divided by 9 = 0 (+ remainder of 1);

10 divided by 9 = 1(+ remainder of 1);

28 divided by 9 = 3 (+ remainder of 1);

55 divided by 9 = 6 (+ remainder of 1);

Thus continuing on, when we ignore the reminder of 1, we obtain the same series 0, 1, 3, 6,...

We have already noted the remarkable fact that with respect to the Zeta 1 (Riemann) function,

_{k})^{2}– (t_{k – 1})^{2 }= (t_{k}– t_{k – 1})^{3 }for the infinite sequence.With respect to the sequence 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ..., the 1st term of each successive group of 3 leaves a remainder of 1 when divided by 9

So, 1 divided by 9 = 0 (+ remainder of 1);

10 divided by 9 = 1(+ remainder of 1);

28 divided by 9 = 3 (+ remainder of 1);

55 divided by 9 = 6 (+ remainder of 1);

Thus continuing on, when we ignore the reminder of 1, we obtain the same series 0, 1, 3, 6,...

We have already noted the remarkable fact that with respect to the Zeta 1 (Riemann) function,

_{ ∞}

∑{ζ

s = 2

= 1/2 + 1/6 + 1/12 + 1/20 + … = 1/2( 1 + 1/3 + 1/6 + 1/10 + …) = 1.

_{1}(s) – 1} = .64493... + .20205... + .08232.. + .03692... + ...s = 2

= 1/2 + 1/6 + 1/12 + 1/20 + … = 1/2( 1 + 1/3 + 1/6 + 1/10 + …) = 1.

And once again 1 + 1/3 + 1/6 + 1/10 + ... represents the sum of reciprocals of the unique number sequence associated with (x

Then 1/2 (1 + 1/3 + 1/6 + 1/10 + ...) +

1/6 (1 + 1/4 + 1/10 + 1/20 + ...) +

1/12 (1 + 1/5 + 1/15 + 1/35 + ... ) +

1/20 (1 + 1/6 + 1/21 + 1/42 + ...) +

...

= 1/2(2/1) + 1/6(3/2) + 1/12(4/3) + 1/20(5/4) + ... ...

= 1 + 1/4 + 1/9 + 1/16 + ...

In other words by combining each successive term of 1/2 + 1/6 + 1/12 + 1/20 + ...

with the sum of reciprocals of the unique number sequences associated with (x

^{ }– 1)^{3 }= 0.Then 1/2 (1 + 1/3 + 1/6 + 1/10 + ...) +

1/6 (1 + 1/4 + 1/10 + 1/20 + ...) +

1/12 (1 + 1/5 + 1/15 + 1/35 + ... ) +

1/20 (1 + 1/6 + 1/21 + 1/42 + ...) +

...

= 1/2(2/1) + 1/6(3/2) + 1/12(4/3) + 1/20(5/4) + ... ...

= 1 + 1/4 + 1/9 + 1/16 + ...

In other words by combining each successive term of 1/2 + 1/6 + 1/12 + 1/20 + ...

with the sum of reciprocals of the unique number sequences associated with (x

^{ }– 1)^{n }= 0, where n = 3, 4, 5, 6, ..., we obtain the Zeta 1(Riemann) function ζ_{1}(s), where s = – 2.