sum of consecutive triangular numbers is a square
Paulo Sachs
sachs6 at yahoo.de
Sat Feb 24 14:53:50 CET 2007
Max Alekseyev,
your 2 sequences following:
> The numbers k such that there exist k consecutive
> triangular numbers
> whose sum is a square:
>
> 1, 2, 3, 4, 11, 13, 22, 23, 25, 27, 32, 37, 39, 46,
> 47, 48, 49, 50,
> 52, 59, 66, 71
>
> The numbers k such that there exist unique k
> consecutive triangular
> numbers whose sum is a square:
>
> 32, 50
may be merged in a more complex one:
Ordered set of the numbers of consecutive triangular
numbers terms in perfect square sums.
So each number appears as many times in the sequence
as how many ways k consecutive triangular numbers sum
a perfect square.
If the density of such numbers is 1, it would be even
better to have just a sequence with the numbers of
ways in which k consecutive triangular numbers sum a
perfect square. This sequence would be 0 in the
5th-10th, 12th, 14th-21st, 24th, 26th... terms. Would
be 1 in the 32nd, 50th,... And so on. But there is
also the possibility that there exist numbers k, such
that there are a infinity number of ways to have k
consecutive triangular number summing a perfect
square.
By the way, Alekseyev, how did you compute those first
terms? How can you be sure that no 5 consecutive very
large triangular numbers add to a perfect square?
Sory for my bad English,
Paulo Sachs
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