Basic questions?

[Alexandre Wajnberg]

A108150       C(n+1,1)*C(n+3,1)

0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323,
360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088,
1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024,
2115, 2208, 2303, 2400, 2499, 2600  (list)
OFFSET        0,2
EXAMPLE     If n=0 then C(0+1,1)*C(0+3,1)= C(1,1)*C(3,1)=1*3=3
        If n=5 then C(5+1,1)*C(5+3,1)= C(6,1)*C(8,1)=6*8=48
KEYWORD    easy,nonn
AUTHOR      Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 06 2005
------------------------

Hi,

All terms are <a square -1> so I submitted the formula a(n-1)=(n^2)-1
verified for all terms displayed.

But I wonder:
1) How to be shure it is verified to the infinity?
2) Should we *prove* that a formula is correct for any corresponding
sequence?

Here I would say C(i,1)= always i, so
C(n+1,1)*C(n+3,1)==(n+1)(n+3)=n^2+4n+3=[(n+2)^2]-1,
which is well <a square-1>.

[May be offset should be -1, or the first term should be removed (see
exemples), and the formula would become a(n-2)=(n^2)-1]

I ask these two questions, may be very basic ones, in this case but most of
all *in general*.

Alexandre
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