Many thanks Martin again. Really Excellent.

What Weil Aerts showed you is the "proper way" to go about looking for such a formula (graphing data and using one's experience about recognizing functions). However, an alternative that I just recently saw at Mathematica StackExchange is a "lazy person's way" (or "efficient person's way") which consists of going to https://oeis.org/ and typing in the sequence of numbers.

Typing you your sequence results in a(n) = 4n^2 - 6n + 3.

Many thanks Jim. Really Excellent.

Many thanks Martin. Really Excellent.

Hello Lea,

I analyzed your data in the following way: first make a graph (ListPlot): this looks like a second degree. Apply NonLinearModelFit with a quadratic equation. Inspect the residues. So the function is known. Use this function in a definition (series). Reproduce your series. And just another number (sort of infinity).

Good Luck, Wiel

data = {1, 7, 21, 43, 73, 111, 157, 211, 273, 343, 421, 507, 601, 703,
    813};

model = NonlinearModelFit[data, {a x^2 + b x + c}, {a, b, c}, x];
model["BestFitParameters"]
{a -> 3.999999999999997`, b -> -5.999999999999966`, 
 c -> 2.9999999999998295`}

model["FitResiduals"]

{1.3988810110276972`*^-13, 1.1546319456101628`*^-13, \
9.947598300641403`*^-14, 8.526512829121202`*^-14, \
7.105427357601002`*^-14, 8.526512829121202`*^-14, \
8.526512829121202`*^-14, 1.1368683772161603`*^-13, \
1.1368683772161603`*^-13, 1.1368683772161603`*^-13, \
2.2737367544323206`*^-13, 2.2737367544323206`*^-13, \
3.410605131648481`*^-13, 2.2737367544323206`*^-13, \
3.410605131648481`*^-13}

series[n_] := 3 - 6 n + 4 n^2

Table[series[k], {k, 1, 15}]
{1, 7, 21, 43, 73, 111, 157, 211, 273, 343, 421, 507, 601, 703, 813}

series[1000]
3994003

Of course it would be accurate,fits is perfect. My solution is exact, will work for any integer, billion, trillion, google... . Moreover I gave you all the constant components, all the linear components and quadratic et cetera. What do you want more?

It can be done like this:

seq = {1, 7, 21, 43, 73, 111, 157, 211, 273, 343, 421, 507, 601, 703, 813};
seqlen = Length[seq];
prefactors = CoefficientList[FindSequenceFunction[seq][x], x];
powers = Range[Length[prefactors]] - 1;
values = MapThread[#2 Range[seqlen]^#1 &, {powers, prefactors}];
values
Total[values]

{
    {3,3,3,3,3,3,3,3,3,3,3,3,3,3,3},
    {-6,-12,-18,-24,-30,-36,-42,-48,-54,-60,-66,-72,-78,-84,-90},
    {4,16,36,64,100,144,196,256,324,400,484,576,676,784,900}
}

{1,7,21,43,73,111,157,211,273,343,421,507,601,703,813}

Given that we have now found the function for the series

data = {1, 7, 21, 43, 73, 111, 157, 211, 273, 343, 421, 507, 601, 703, 813}
series[n_]:=3-6 n+4 n^2

If I create 2 different series as multiplications of this series - for example:-

series1[n_]:= (3-6 n+4 n^2) * 30
series2[n_]:= (3-6 n+4 n^2) * 47

I tried using

Intersection[ series1 , series2 ] 

with a million elements & found nothing.

Question - Is there one or more common values between series1 & series2 within the Integer domain from 1 to infinity?

Please could someone check & confirm whether any values are shared or not.

Many thanks for your help & attention. Lea...

Usually I start by graphing the available data.

data = {1, 7, 21, 43, 73, 111, 157, 211, 273, 343, 421, 507, 601, 703, 813};
ListPlot[data]

Hummm. Looks roughly like a parabola, but that is only a first guess.

Hint: Look at the documentation for Fit and see if that might let you find something interesting and useful.

That might be enough if you work quickly before someone blurts out the answer and spoils the learning opportunity.