The numbers 0, 3, 20 and 119 are the only numbers n under 500 such that for the
series S(0) = -n-1; S(1) = n; S(i) = 6*S(i-1) - S(i-2) + 2 the triangular
numbers T(S(i)) = S(i) *(S(i) +1)/2 can be factored into R*(R + K) where K is dependent upon n. For
n = 0, K is 1: T(-1) = T(0) = 0*1 and T(3)=2*3 and T(20) = 15*15 ... . For n = 3, K is 5: T(-4) = T(3) = 1*6 & T(24) = 15*20 & T(119) = 99*104 ... . Find the K values for n = 20 and n = 119.
Sorry but I meant to say T(20) = 14*15 for K = 1.
series S(0) = -n-1; S(1) = n; S(i) = 6*S(i-1) - S(i-2) + 2 the triangular
numbers T(S(i)) = S(i) *(S(i) +1)/2 can be factored into R*(R + K) where K is dependent upon n. For
n = 0, K is 1: T(-1) = T(0) = 0*1 and T(3)=2*3 and T(20) = 15*15 ... . For n = 3, K is 5: T(-4) = T(3) = 1*6 & T(24) = 15*20 & T(119) = 99*104 ... . Find the K values for n = 20 and n = 119.
Sorry but I meant to say T(20) = 14*15 for K = 1.