Base 5
Patterns that can be factored: \[[1^*01^*_5]_n={5\cdot5^{2n}-4\cdot5^{n}-1\over4}=\left(5^{n}-1\right)\left(5\cdot5^{n}+1\right)/4=\left(4\cdot[1^*_5]_n\right)\left(2\cdot[2^*3_5]_n\right)/4\] \[[1^*21^*_5]_n={5\cdot5^{2n}+4\cdot5^{n}-1\over4}=\left(5^{n}+1\right)\left(5\cdot5^{n}-1\right)/4=\left(2\cdot[2^*3_5]_{n-1}\right)\left(4\cdot[1^*_5]_{n+1}\right)/4\] \[[2^*02^*_5]_n={5\cdot5^{2n}-4\cdot5^{n}-1\over2}=\left(5^{n}-1\right)\left(5\cdot5^{n}+1\right)/2=\left(4\cdot[1^*_5]_n\right)\left(2\cdot[2^*3_5]_n\right)/2\] \[[2^*42^*_5]_n={5\cdot5^{2n}+4\cdot5^{n}-1\over2}=\left(5^{n}+1\right)\left(5\cdot5^{n}-1\right)/2=\left(2\cdot[2^*3_5]_{n-1}\right)\left(4\cdot[1^*_5]_{n+1}\right)/2\] \[[3^*03^*_5]_n=3\cdot{5\cdot5^{2n}-4\cdot5^{n}-1\over4}=3\cdot\left(5^{n}-1\right)\left(5\cdot5^{n}+1\right)/4=3\cdot\left(4\cdot[1^*_5]_n\right)\left(2\cdot[2^*3_5]_n\right)/4\] \[[3^*23^*_5]_n={15\cdot5^{2n}-4\cdot5^{n}-3\over4}=\left(3\cdot5^{n}+1\right)\left(5\cdot5^{n}-3\right)/4=\left(4\cdot[3^*4_5]_{n-1}\right)\left(2\cdot[2^*1_5]_n\right)/4\] \[[3^*43^*_5]_n={15\cdot5^{2n}+4\cdot5^{n}-3\over4}=\left(3\cdot5^{n}-1\right)\left(5\cdot5^{n}+3\right)/4=\left(2\cdot[12^*_5]_n\right)\left(4\cdot[1^*2_5]_n\right)/4\] \[[4^*04^*_5]_n=5\cdot5^{2n}-4\cdot5^{n}-1=\left(5^{n}-1\right)\left(5\cdot5^{n}+1\right)=\left(4\cdot[1^*_5]_n\right)\left(2\cdot[2^*3_5]_n\right)\]