incorrect A082109 comment about A000217

[Russ Cox]

Hi all,

Takao Komatsu (cc'ed) wrote to me pointing out their paper https://arxiv.org/abs/2409.14788 that cites A082109. However, there is an incorrect comment they found in A082109 that I cannot tell how to fix either.

The comment reads:

Define b(n) = A000217(n), the triangular numbers. Using six consecutive terms to create the vertices of a triangle at points (b(n-2), b(n-1)), (b(n), b(n+1)), and (b(n+2), b(n+3)), one fourth the area of these triangles = a(n). - J. M. Bergot, Jul 30 2013

Takao Komatsu points out that any such triangle created this way seems to have an area of 4, no matter what value of n is used. A constant four sequence divided by four is a constant one sequence, which is not the same as A082109. So something is not right about this comment. Can anyone determine what the comment intended to say?

As a potential hint, A000217 contains this comment:

Plot the three points (0,0), (a(n), a(n+1)), (a(n+1), a(n+2)) to form a triangle. The area will be a(n+1)/2. - J. M. Bergot, May 04 2012

So perhaps something more along those lines was intended? 

Any ideas? Thanks!

Best,

Russ

[Neil Sloane]

I seem to recall that Robert Israel was a good friend of J. M. Bergot.  I will copy this to Robert.

Robert, can you help with this?

Best regards

Neil 

Neil J. A. Sloane, Chairman, OEIS Foundation.

Also Visiting Scientist, Math. Dept., Rutgers University, 

Email: njas...@gmail.com

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[Robert Israel]

Unfortunately my friendship with Bergot did not give me any special insight regarding these sequences. 

I did find that A082109(n) is 1/4 the area of a triangle with vertices [0,0], [A053755(n), A053755(n+1)] and [A053755(n+1), A053755(n+2)].

I have no idea if this has anything to do with what Bergot thought he had found.

Cheers,

Robert

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[Daniel Mondot]

This might be irrelevant, but...
As Russ Cox and Takao Komatsu mentioned above, the surface of the triangle at points (b(n-2), b(n-1)), (b(n), b(n+1)), and (b(n+2), b(n+3)), is always 4 when b(n) is  A000217(n) (triangular numbers).
If b(n) = A000290 (square numbers), then, the surface of the triangle at points (b(n-2), b(n-1)), (b(n), b(n+1)), and (b(n+2), b(n+3)), is always 16.

for b(n) = A000326 (pentagonal numbers), the surface is always 36

for b(n) = A000384 (hexagonal numbers), the surface is always 64

for b(n) = heptagonal numbers, the surface is always 100

for b(n) = octagonal numbers, the surface is always 144

for b(n) = nonagonal numbers, the surface is always 196
for b(n) = decagonal numbers, the surface is always 256

generally, for b(n) = g-gonal numbers, the surface is always (2g-4)^2

And if b(n) = A082109(n), then, the surface of the triangle at points (b(n-2), b(n-1)), (b(n), b(n+1)), and (b(n+2), b(n+3)), is always 256.

That might be one way that A082109 is related to polygonal numbers, they have similar equations after all.

Daniel.

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