A266236 Least m > 0 such that m*n^3 + 1 is a cube.
1, 7, 91, 37, 4291, 16003, 1801, 17, 263683, 19927, 1003003, 1775557, 111169, 506115, 17145, 423001, 16789507, 24152311, 1261657, 3266062, 64024003, 5080, 113411851, 148072393, 7082497, 244187503, 1922636, 14355469, 3132736, 594896491, 27009001, 8341522, 1073840131
Offset: 0
Keywords
Examples
17*7^3+1 = 18^3, and 17 is the smallest positive m such that m*7^3+1 is a cube, so a(7)=17.
Links
- Robert G. Wilson v, Table of n, a(n) for n = 0..1000
Programs
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Mathematica
f[n_] := Block[{x = 2, n3 = n^3}, While[ Mod[x^3 - 1, n3] != 0, x++]; (x^3 - 1)/n3]; f[0] = 1; Array[f, 34, 0] (* Robert G. Wilson v, Mar 24 2016 *)
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PARI
a(n) = {my(m = 1, cn = n^3); while (!ispower(m*cn + 1, 3), m++); m;} \\ Michel Marcus, Feb 09 2016
Formula
a(n) = A076947(n^3). - Robert Israel, Dec 25 2015
Comments