quote:
Originally posted by foaly
quote:
Originally posted by markee
quote:
Originally posted by foaly
quote:
Originally posted by andrey
quote:
Originally posted by markee
p = (n+2)(n/2+1.5)
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.)
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
All sequences start at zero, even JA knows that... 
the ones in the example didn't
They start with n=0, fool
He said, initial value is 3. Implies for n=0, value=3; Then next value is 6 implies for n=1, value=6.
markee, the JA knows *
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