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Least number k > 0 such that k*2^n+1 is a cube.
0

%I #26 Jan 14 2023 11:18:34

%S 7,13,31,91,307,1123,4291,16771,66307,263683,1051651,4200451,16789507,

%T 67133443,268484611,1073840131,4295163907,17180262403,68720263171,

%U 274879479811,1099514773507,4398052802563,17592198627331,70368769343491,281475027042307,1125900007505923

%N Least number k > 0 such that k*2^n+1 is a cube.

%C Terms are either 1, 3, or 7 mod 10.

%C Using the formula, we see that the cubes produced are (2^n+1)^3 for all n.

%C This is a particular case of general formula

%C (3 + 3 f + f^2) f + 1 = (1 + f)^3 with f = 2^n.

%C For the simplest case f = n, we have (3 + 3 n + n^2) = A002061 (Central polygonal numbers, with corresponding offset). - _Zak Seidov_, Mar 24 2014

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-14,8).

%F a(n) = 3 + 3*2^n + 2^(2*n).

%F G.f.: -(38*x^2-36*x+7) / ((x-1)*(2*x-1)*(4*x-1)). - _Colin Barker_, Mar 24 2014

%e 13*(2^1)+1 = 27 is a cube. Since 13 is the smallest coefficient that returns a cube, then a(1) = 13.

%o (Python) {print(3+3*(2**n)+2**(2*n)) for n in range(100)}

%o (PARI) for(n=0, 100, k=1; while(!ispower(k*2^n+1, 3), k++); print1(k, ", ")) \\ _Colin Barker_, Mar 24 2014

%o (PARI) a(n) = (3+3*2^n+4^n) \\ _Colin Barker_, Mar 24 2014

%o (PARI) Vec(-(38*x^2-36*x+7)/((x-1)*(2*x-1)*(4*x-1)) + O(x^100)) \\ _Colin Barker_, Mar 24 2014

%Y Cf. A239678.

%K nonn,easy

%O 0,1

%A _Derek Orr_, Mar 23 2014