Shoutbox

quote:

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Originally posted by foaly

quote:

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Originally posted by andrey

quote:

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Originally posted by markee
p = (n+2)(n/2+1.5)

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eh ?



n(n+1)
   2

for n=1,2,3...

That works for the first pattern. The sequence begins with 1, i.e. 1, 3, 6, 10, 15, 21, 28 etc.

(Triangular numbers btw.)

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ehm neither work for 1=n...
p = (1+2)(.5+1.5) = 3*2 = 6
p = (1(1+1))/2 = 1

and it starts with 3...

the correct one is ((n+1)(n+2))/2

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All sequences start at zero, even JA knows that...