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A264613 Numbers n such that the Shevelev polynomial {m, n} has a root at m = -1. 3

%I #38 Sep 06 2018 11:10:14

%S 2,5,8,11,23,32,47,95,128,191,383,512,767,1535,2048,3071,6143,8192,

%T 12287,24575,32768,49151,98303,131072,196607,393215,524288,786431,

%U 1572863,2097152,3145727,6291455,8388608

%N Numbers n such that the Shevelev polynomial {m, n} has a root at m = -1.

%C From _Peter J. C. Moses_, Dec 12 2015: (Start)

%C This appears to split into 3 sequences:

%C b(n) = 3*4^(n-1)-1, n>=1: 2,11,47,191,767,3071,12287,49151,...,

%C c(n) = 3*2^(2*n-1)-1, n>=1: 5,23,95,383,1535,6143,24575,98303,...,

%C d(n) = 2^(2*n+1), n>=1: 8,32,128,512,2048,8192,32768,...;

%C If this is true, then the next few terms of the sequence are 12582911, 25165823, 33554432, 50331647, 100663295, ...

%C (End)

%H Vladimir Shevelev, <a href="http://arxiv.org/abs/0801.0072">On the Basis Polynomials in the Theory of Permutations with Prescribed Up-Down Structure</a>, arXiv:0801.0072 [math.CO], 2007-2010. See Appendix.

%H Vladimir Shevelev, <a href="http://www.emis.de/journals/INTEGERS/papers/m1/m1.Abstract.html">The number of permutations with prescribed up-down structure as a function of two variables</a>, INTEGERS, 12 (2012), #A1. (See Section 11, Problem 3.)

%H V. Shevelev and J. Spilker, <a href="http://dx.doi.org/10.4171/EM/229">Up-down coefficients for permutations</a>, Elemente der Mathematik, Vol. 68 (2013), no.3, 115-127.

%F Conjectured g.f.: (2 + x*(5 + x*(8 + x*(1 + (-2 - 8*x)* x)))) / (1 + x^3*(-5 + 4*x^3)). - _Peter J. C. Moses_, Dec 12 2015

%t upDown[n_, k_] := upDown[n, k] = Module[{t, m}, t = Flatten[ Reverse[ Position[ Reverse[ IntegerDigits[k, 2]], 1]]]; m = Length[t]; (-1)^m + Sum[upDown[t[[j]], k - 2^(t[[j]] - 1)]*Binomial[n, t[[j]]], {j, 1, m}]];

%t Reap[For[k = 2, k <= 2^15, k++, If[(upDown[n, k] /. n -> -1) == 0, Print[k]; Sow[k]]]][[2, 1]] (* _Jean-François Alcover_, Sep 06 2018 *)

%Y Cf. A133457 (positive integer roots of {m,n}), A263848.

%K nonn

%O 1,1

%A _N. J. A. Sloane_, Nov 28 2015

%E More terms (starting at a(6)) from _Peter J. C. Moses_, Dec 12 2015

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