Correction: .

]]>So, consider families of graphs of the form , where the generating graph consists of wheel graphs with a common central vertex. The number of vertices of is . We want to find the maximum number of vertices for each such graph .

Based on the Tom’s formula, one can calculate the number of vertices for some and . By fixing , one can investigate the obtained sequences of the number of vertices. They are described by polynomials of order .

Here are the coefficients of the first 7 first polynomials, if I did not make a mistake in the calculations:

{1, 3, 3},

{1, 3, 9/2, 3, 3/2},

{1, 18/5, 57/10, 9/2, 3, 9/10, 3/10},

{1, 132/35, 93/14, 63/10, 183/40, 9/5, 3/4, 9/70, 9/280},

{1, 141/35, 2607/350, 54/7, 1713/280, 117/40, 267/200, 9/28, 27/280, 3/280, 3/1400},

{1, 1611/385, 15681/1925, 1593/175, 1527/200, 1143/280, 5643/2800, 837/1400, 549/2800, 9/280, 3/400, 9/15400, 3/30800},

{1, 21738/5005, 1535661/175175, 2844/275, 140331/15400, 531/100, 7767/2800, 2619/2800, 3231/9800, 27/400, 3/175, 9/4400, 3/7700, 9/400400, 9/2802800}

The leading coefficients are of the form . In general, the values of the coefficients fit into the scheme , where the first few functions form the sequence {1 , 1, , , , }. But I still don’t see any pattern here. Maybe I made a mistake somewhere?

]]>For a 31-vertex generating graph on five wheel graphs, we get .

]]>Tom’s formula for a 19-vertex generating graph on three wheel graphs leads to . And it gives the second sequence (out of four above). In particular, it is generated at the angles , , .

For a 25-vertex graph on four wheel graphs, we get .

So far, I have problems with a similar 31-vertex graph on five wheel graphs.

The last sequence is described by the formula starting from the 10th term.

]]>For a sequence with angle , I found a relation that works from the ninth member of the sequence (at least up to the 20th): .

But so far I can’t find a solution for sequences with the angle and some others. Here are some of them:

{19, 163, 865, 3313, 10099, 25765, 55747, 105889, 183127, 295645, 452491, 663787, 940681, 1295347, 1740985, 2291821, 2963107, 3771121, 4733167, 5867575}

{19, 163, 865, 3313, 10099, 26083, 59473, 123121, 236035, 425107, 727057, 1190593, 1878787, 2871667, 4269025, 6193441}

{19, 163, 865, 3313, 10099, 26083, 59473, 123121, 236035, 425107, 727057, 1190593, 1878787, 2871547, 4265545}

{19, 163, 805, 2239, 4015, 6001, 8305, 10957, 13939, 17245, 20893, 24889, 29233, 33925, 38965, 44353, 50089, 56173, 62605, 69385, 76513, 83989, 91813, 99985, 108505, 117373, 126589, 136153, 146065, 156325}

I figured out why {19, 163, 715, … } didn’t work for me. In fact, I worked with the sequence {1, 19, 163, 715, … }, which seemed natural.

For the sequence {31, 451, 3451, 14557, 39973, 87913, …} I found the following relation, which works from the sixth member of the sequence (up to at least the 18th): . The first terms of the sequence according to the formula are equal to {343, 127, 3313, 14467, 39949, 87913, …}

oops, typo!

]]>Thanks, Tom. Very fast and constructive. I’ll figure out why I couldn’t handle {19, 163, 715, … }. But it looks like you wanted to add something after “this”. Or is it a typo?

]]>In the general case, maybe it might be useful to think in terms of “basis” vectors. Wheels a, b, c have basis vectors {a0, a1}, {b0, b1}, {c0, c1}. (a0 and a1 form a 60-degree angle, same for b0, b1 and c0, c1)

Then each vertex in G_m is a 6-dimensional lattice point. P(i0, i1, i2, i3, i4, i5) = a0*i0 + a1*i1 + b0*i2 + b1*i3 + c0*i4 + c1*i5.

There’s a simple way to check if P(i0..i5) is in G_m or not. And my post above counts them. But P(i0..i5) may equal P(j0..j5). This happens when there’s some P(k0..k5) = 0. Then P(i0+k0..i5+k5) equals P(i0..i5).

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