On Fri, 6 Jun 2025 15:25:51 -0000 (UTC), B. Pym wrote:

B. Pym wrote:

Given the sequence 0, 1, 9, 36, 100, 225, 441... it is possible to
calculate the polynomial expression for the sum of the entries in a multiplication table of n rows and n columns. 2025 is the 9th entry in
this sequence as it is the sum for the entries in a 9 x 9 multiplication
table.

Can you calculate that function?

It'd be interesting to see how the various functions perform with respect
to that polynomial for large n. They may well implement it.

If you don't know how to proceed, there is a pointer below to the method I
used (any other methods would be of interest):

V hfrq gur zrgubq bs svavgr qvssreraprf.



--
David Entwistle

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On Mon, 9 Jun 2025 11:25:49 -0000 (UTC), David Entwistle wrote:

It'd be interesting to see how the various functions perform with
respect to that polynomial for large n. They may well implement it.

Unless I've made a mistake, the polynomial indicates the sum of all the
entries in a 46341 by 46341 multiplication table is 1152866771266955745.
This is the largest solution that fits in a long variable. The answer is reported in less than 1 ms.
--
David Entwistle

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In article <1026gbt$gsd0$[email protected]>,
David Entwistle <[email protected]> wrote:

Given the sequence 0, 1, 9, 36, 100, 225, 441... it is possible to
calculate the polynomial expression for the sum of the entries in a >multiplication table of n rows and n columns. 2025 is the 9th entry in
this sequence as it is the sum for the entries in a 9 x 9 multiplication >table.

Can you calculate that function?

sum(x=1..n) sum(y=1..n) [xy]
= sum(x=1..n) [x sum(y=1..n) y]
= [sum(x=1..n) x] [sum(y=1..n) y]
= [sum(x=1..n)]^2

sum(x=1..n) is well known and easily seen geometrically.

-- Richard

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On Mon, 9 Jun 2025 16:32:37 -0000 (UTC), Richard Tobin wrote:

sum(x=1..n) sum(y=1..n) [xy]
= sum(x=1..n) [x sum(y=1..n) y]
= [sum(x=1..n) x] [sum(y=1..n) y]
= [sum(x=1..n)]^2

sum(x=1..n) is well known and easily seen geometrically.

Well blow me down with a feather.

SPOILER
POILER
OILER
ILER
LER
ER
R

To calculate the sum of the numbers 1 through n:

1, 2, 3 ... n-2, n-1, n

We can write the series in reverse:

n, n-1, n-2 ... 3, 2, 1

Adding the two rows together each element provides the same sum, (n+1).
so:

2[sum(x=1..n)] = n(n+1)

and

[sum(x=1..n)] = n(n+1)/2

sum(x=1..n) sum(y=1..n) [xy]
= sum(x=1..n) [x sum(y=1..n) y]
= [sum(x=1..n) x] [sum(y=1..n) y]
= [sum(x=1..n)]^2

The value we are looking for, the sum of the elements in a n x n
multiplication table, is given by

[sum(x=1..n)]^2 = [n(n+1)/2]^2

= (n^4 + 2n^3 + n^2)/4

Thanks, that's brill.
--
David Entwistle

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