471, 1372, 50653, 34328125, ?

[brocode]

471, 1372, 50653, 34328125,...

471 = 3*157
1372 = 2*2*7*7*7
50653 = 37*37*37
34328125 = 5*5*5*5*5*5*13*13*13
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d1 = 1372-471 = 901 = 17*53
d2 = 50653-1372 = 49281 = 3*16427
d3 = 34328125-50653 = 34277472 = 2^5 * 3^3 * 97

d2/d1 = r^1
(3*16427)/(17*53) = r ...(1)

d3/d2 = r^2
(2^5 * 3^3 * 97)/(3*16427) = r^2  ...(2)

c/d = r^2
a/b = r

dividing (2) by (1)

no thats not gonna work
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a_n = a_1 * r^n-1

let a_1 = 471

a_2 = 1372
a_2 = 471 * r^1
1372 = 471 * r^1 ...(1)

a_3 =50653
a_3 = 1372 * r^2
50653 = 1372 * r^2 ...(2)

a_4 = 34328125
a_4 = 5063 * r^3
34328125 = 50653 * r^3 ...(3)

dividing (2) by (1)
50653/1372 = (1372 * r^1)/471

r = (50653*471)/(1372^2) = 23857653/1882384
r = (37*37*37*3*157)/(2*2*2*2*7*7*7*7*7*7)

let a_n be some f(a_n-1) on R, such that f(a_n)-f(a_n-1) = r
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fuck man i give up
teach me how to do it

a_n = xyz

then a_{n+1} = y^3 * x^z

in other words middle digits cubed multiplied by first digit to the power of last digit