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A239679 Least number k > 0 such that k*2^n+1 is a cube. 0
7, 13, 31, 91, 307, 1123, 4291, 16771, 66307, 263683, 1051651, 4200451, 16789507, 67133443, 268484611, 1073840131, 4295163907, 17180262403, 68720263171, 274879479811, 1099514773507, 4398052802563, 17592198627331, 70368769343491, 281475027042307, 1125900007505923 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

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

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

This is a particular case of general formula

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

     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

LINKS

Table of n, a(n) for n=0..25.

Index entries for linear recurrences with constant coefficients, signature (7,-14,8).

FORMULA

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

G.f.: -(38*x^2-36*x+7) / ((x-1)*(2*x-1)*(4*x-1)). - Colin Barker, Mar 24 2014

EXAMPLE

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

PROG

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

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

(PARI) a(n) = (3+3*2^n+4^n) \\ Colin Barker, Mar 24 2014

(PARI) Vec(-(38*x^2-36*x+7)/((x-1)*(2*x-1)*(4*x-1)) + O(x^100)) \\ Colin Barker, Mar 24 2014

CROSSREFS

Cf. A239678.

Sequence in context: A215932 A091432 A102002 * A195353 A342184 A247178

Adjacent sequences:  A239676 A239677 A239678 * A239680 A239681 A239682

KEYWORD

nonn,easy

AUTHOR

Derek Orr, Mar 23 2014

STATUS

approved

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Last modified October 2 23:51 EDT 2022. Contains 357230 sequences. (Running on oeis4.)