[seqfan] Re: A07421 real world geometry
Brad Klee
bradklee at gmail.com
Thu May 14 08:33:41 CEST 2020
49^2=2401 does not appear in A074216 because sigma(49^2)%3=2,
yet it must occur in the Sangaku circle sequence if 49 does.
For the new sequence, I calculated that:
49, 169, 196, 361, 441, 676, 784, 961, 1225, 1369, 1444, 1521, 1764,
1849, 2401, 2704, 3136, 3249, 3721, 3844, 3969, 4225, 4489, 4900,
5329, 5476, 5776, 5929, 6084, 6241, 7056, 7396, 8281, 8649, 9025,
9409, 9604, 10609, 10816, 11025, 11881, 12321, 12544, 12996, 13689,
14161, 14884, 15376, 15876, 16129, 16641, 16900, 17689, 17956, 19321,
19600, 20449, 21316, 21609, 21904, 22801, 23104, 23716, 24025, 24336,
24649, 24964, 25921, 26569, 28224, 28561 ...
The last entry 28561=169^2 is the first "mistake", which
is not a multiple of 7^2. It is a multiple of 13^2.
So perhaps between the two sequences, the only complement is due
to multiples of a(n) from A074216 such that sigma(k*a(n))%3>0 ?
This interests me, and I don't know why. Nice find!
--Brad
On Wed, May 13, 2020 at 9:09 PM Daniel Palasek <dan.palasek at gmail.com>
wrote:
> A07421 is defined as "Squares satisfying sigma(n)==0 (mod 3)", which is not
> clear how that connects to this:
> It appears to have an interesting property with real world geometry.
> The sequence is perfect squares that are themselves the sums of 3 perfect
> squares of unequal size:
>
>
> *49, 169, 196, 361 ...*
> The 3 unequal squares are the integer radius solutions of Sangaku 3
> touching circles sharing a tangent line
> https://en.wikipedia.org/wiki/Sangak
>
> *49=7^2*
>
> *49=36+9+4 (a+b+c)*
> * or 6^2 + 3^2 + 2^2*
> * 1/sqrt(c) = 1/sqrt(a) + 1/sqrt(b) *
>
> *1/sqrt(4) = 1/sqrt(36) + 1/sqrt(9)*
>
> *i.e. circles of radius 36, 9, 4 will touch when aligned on a tangent.*
>
> *169=13^2*
> *169=144+16+9*
> * or 12^2 + 4^2 + 3^2*
>
> *1/sqrt(9) = 1/sqrt(144) + 1/sqrt(16) *
>
> *196=14^2*
> *196=144+36+16*
> * or 12^2 + 6^2 + 4^2*
>
> *1/sqrt(16) = 1/sqrt(144) + 1/sqrt(36) *
>
> *361=19^2*
> *361=225+100+36 *
> * or 15^2 + 10^2 + 6^2*
> *1/sqrt(36) = 1/sqrt(225) + 1/sqrt(100) *
>
> -Daniel
>
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