A111983 - OEIS

%I #32 Sep 08 2022 08:45:21

%S 1,24,0,320,0,0,3584,0,0,0,36864,0,0,0,0,360448,0,0,0,0,0,3407872,0,0,

%T 0,0,0,0,31457280,0,0,0,0,0,0,0,285212672,0,0,0,0,0,0,0,0,2550136832,

%U 0,0,0,0,0,0,0,0,0,22548578304,0,0,0,0,0,0,0,0,0,0,197568495616,0,0,0,0,0,0

%N G.f.: A(x) = Sum_{n>=0} (2*n+1) * 8^n * x^(n*(n+1)/2).

%C Define F(x,q) = Sum_{n>=0} q^n*(2*n+1)*x^(n*(n+1)/2).

%C (1) F(x,q)^(1/3) is an integer series in x when q == -1, 0, 3 or 6 (mod 9).

%C (2) For q = -1 we have the famous result of Jacobi (Hardy and Wright, Th. 357): F(x,-1)^(1/3) = (1 - 3*x + 5*x^3 - 7*x^6 + 9*x^10 + ...)^(1/3) = 1 + Sum_{n>=1} (-1)^n*[x^(n*(3*n-1)/2)+x^(n*(3*n+1)/2)] = Product_{k>=1} (1-x^k).

%C Concerning (1): For q == 0 (mod 3) we see that F == 1 (mod 9), which by the Heninger-Rains-Sloane paper implies that F^(1/3) has integer coefficients. For q == -1 (mod 9) the same assertion follows from the Jacobi identity mentioned above. For q = 8, F(x,8) = A(x), the current sequence, we see that A == 1 (mod 8), so A^(1/3) == 1 (mod 8) and then, again by our paper, A^(1/12) has integer coefficients. - Eric Rains and _N. J. A. Sloane_, Nov 06 2005

%D G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 285.

%H G. C. Greubel, <a href="/A111983/b111983.txt">Table of n, a(n) for n = 0..5050</a>

%H N. Heninger, E. M. Rains and N. J. A. Sloane, <a href="https://arxiv.org/abs/math/0509316">On the Integrality of n-th Roots of Generating Functions</a>, arXiv:math/0509316 [math.NT], 2005-2006.

%H N. Heninger, E. M. Rains and N. J. A. Sloane, <a href="https://doi.org/10.1016/j.jcta.2006.03.018">On the Integrality of n-th Roots of Generating Functions</a>, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.

%e A(x) = 1 + 3*8*x + 5*8^2*x^3 + 7*8^3*x^6 + 9*8^4*x^10 + ... = 1 + 24*x + 320*x^3 + 3584*x^6 + 36864*x^10 + 360448*x^15 + 3407872*x^21 + 31457280*x^28 + 285212672*x^36 + 2550136832*x^45 + ...

%e Surprisingly, A(x)^(1/3) is an integer series (A111984):

%e A(x)^(1/3) = 1 + 8*x - 64*x^2 + 960*x^3 - 15360*x^4 +- ...

%e In fact (see proof in Comments section), A(x)^(1/12) is also an integer series (A111985):

%e A(x)^(1/12) = 1 + 2*x - 22*x^2 + 364*x^3 - 6490*x^4 +- ...

%p add((2*n+1) * 8^n * x^(n*(n+1)/2), n=0..50);

%t CoefficientList[Sum[(2n+1) 8^n x^(n(n+1)/2), {n, 0, 12}], x] (* _Jean-François Alcover_, Jul 26 2018 *)

%o (PARI) a(n)=polcoeff(sum(k=0,sqrtint(2*n+1),(2*k+1)*8^k*x^(k*(k+1)/2)+x*O(x^n)),n)

%o (Magma) m:=80; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (&+[(2*n+1)*8^n*x^Binomial(n+1,2): n in [0..m]]) )); // _G. C. Greubel_, Oct 24 2019

%o (Sage)

%o def A052961_list(prec):

%o     P.<x> = PowerSeriesRing(ZZ, prec)

%o     return P( sum((2*n+1)*8^n*x^binomial(n+1,2) for n in (0..13))).list()

%o A052961_list(80) # _G. C. Greubel_, Oct 24 2019

%Y Cf. A111984 (g.f. A(x)^(1/3)), A111985 (g.f. A(x)^(1/12)).

%K nonn

%O 0,2

%A _Paul D. Hanna_, Aug 25 2005