a number challenge

[MattcAnderson]

Hi Mersenneforum,

For what it's worth, here is the Maple13 code I wrote. The links above to the OEIS entries really answer the original question.

# okay
> # triangular numbers
> sumb := 0; for a to 12 do sumb := sumb+a end do;
0
1
3
6
10
15
21
28
36
> for e to 10000 do f := (1/2)*e*(e+1); if sqrt(f) = floor(sqrt(f)) then print(f, e) end if end do;
1, 1
36, 8
1225, 49
41616, 288
1413721, 1681
48024900, 9800

Regards,
Matt

[science_man_88]

Originally Posted by MattcAnderson

Hi Mersenneforum,

For what it's worth, here is the Maple13 code I wrote. The links above to the OEIS entries really answer the original question.

# okay
> # triangular numbers
> sumb := 0; for a to 12 do sumb := sumb+a end do;
0
1
3
6
10
15
21
28
36
> for e to 10000 do f := (1/2)*e*(e+1); if sqrt(f) = floor(sqrt(f)) then print(f, e) end if end do;
1, 1
36, 8
1225, 49
41616, 288
1413721, 1681
48024900, 9800

Regards,
Matt

with a little rearrangement of the equality you can show n is within the range m to sqrt(2)*m. because in the worst cases n= n^2 and you get it rearranging to 2*sqr(m)=2*sqr(n) and if you ignore the +n that happens you can show that roughly 2*sqr(m) = sqr(n) in the worst case on the other end of things and this leads with square roots to sqrt(2)*m = n. so if an n exist for an m value n is between m and sqrt(2)*m. oh and if they equal you can show that they have a GCD greater than 1 etc.