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Least integer coefficients of A(x), where 1<=a(n)<=3, such that A(x)^(1/3) consists entirely of integer coefficients.
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%I #13 Jul 19 2015 01:37:59

%S 1,3,3,1,3,3,3,3,3,3,3,3,1,3,3,2,3,3,2,3,3,1,3,3,2,3,3,3,3,3,2,3,3,3,

%T 3,3,3,3,3,2,3,3,2,3,3,3,3,3,2,3,3,1,3,3,2,3,3,1,3,3,1,3,3,1,3,3,1,3,

%U 3,1,3,3,3,3,3,1,3,3,1,3,3,3,3,3,1,3,3,3,3,3,1,3,3,2,3,3,1,3,3,3,3,3,1,3,3

%N Least integer coefficients of A(x), where 1<=a(n)<=3, such that A(x)^(1/3) consists entirely of integer coefficients.

%C More generally, "least integer coefficients of A(x), where 1<=a(n)<=m, such that A(x)^(1/m) consists entirely of integer coefficients", appears to have a unique solution for all m. Is this sequence periodic?

%H Robert G. Wilson v, <a href="/A083953/b083953.txt">Table of n, a(n) for n = 0..5000</a>.

%H N. Heninger, E. M. Rains and N. J. A. Sloane, <a href="http://arXiv.org/abs/math.NT/0509316">On the Integrality of n-th Roots of Generating Functions</a>, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.

%F a(k)=1 at k=0, 3, 12, 21, 51, 57, 60, 63, 66, ...; a(k)=2 at k=15, 18, 24, 30, 39, 42, 48, 54, ...

%t a[0]=1; a[n_] :=a[n] = Block[{k=1, s=Sum[a[i]*x^i, {i, 0, n-1}]}, While[ Union[ IntegerQ /@ CoefficientList[ Series[(s+k*x^n)^(1/3), {x, 0, n}], x]] != {True}, k++ ]; k]; Table[ a[n], {n, 0, 104}] (* _Robert G. Wilson v_, Jul 25 2005 *)

%Y Cf. A083952, A083954, A083945, A083946.

%K nonn

%O 0,2

%A _Paul D. Hanna_, May 09 2003

%E More terms from _Robert G. Wilson v_, Jul 25 2005