|
%I #35 Nov 09 2016 02:35:26
%S 1,2,-1,1,3,-1,2,6,-8,3,2,10,10,-10,3,4,20,10,-50,46,-15,17,119,245,
%T 35,-217,161,-45,34,238,406,-350,-644,1372,-1056,315,62,558,1722,1638,
%U -1092,-1008,1828,-1188,315,124,1116,3138,1134,-5838,1134,9452,-14724,10134,-2835
%N Triangle of coefficients of polynomials concerning Newman-like phenomenon of multiples of b+1 in even base b in interval [0,b^n) (see comment).
%C In 1969, D. J. Newman (see the reference) proved that difference between numbers of multiples of 3 with even and odd binary digit sums in interval [0,x] is always positive. This fact now is known as Newman phenomenon.
%C Consider difference between numbers of multiples of b+1 with even and odd digit sums in even base b in interval [0, b^n). It is a polynomial in b P_n(b) of degree n-1 and multiple of b, if n is even, and n-2, if n is odd, such that all polynomials Q_n(b):=A156769(n/2)*P_n(b)/b, if n is even, and Q_n(b):=A156769((n-1)/2)*P_n(b), if n is odd, presumably have integer coefficients and are of degree n-2. The sequence is triangle of coefficients of polynomials Q_n(b).
%C The r-th row contains r-1 entries.
%C Since, evidently, P_n(1)=1, then the row sums form sequence A156769 repeated.
%H D. J. Newman, <a href="http://dx.doi.org/10.1090/S0002-9939-1969-0244149-8">On the number of binary digits in a multiple of three</a>, Proc. Amer. Math. Soc. 21 (1969) 719-721.
%H Vladimir Shevelev, <a href="http://arxiv.org/abs/0710.3177">On monotonic strengthening of Newman-like phenomenon on (2m+1)-multiples in base 2m</a>, arXiv:0710.3177v2 [math.NT], 2007.
%F If n>=2 is even, then P_(n+1)(b) = (-1)^((n-2)/2)*(C(b+1,n)-C(b-1,n))-sum{i=2,4,...,n-2}(-1)^((n+i)/2)*C(b+1, n-i)*P_(i+1)(b), where P_n(b)=b*Q_n(b)/A156769(n/2);
%F if n>=3 is odd, then P_(n+1)(b) = (-1)^((n-1)/2)*(C(b,n)-b*C(b+1,n-1))+sum{i=3,5,...,n-2}(-1)^((n+i)/2)*C(b+1, n-i)*P_(i+1)(b), where
%F P_n(b) = Q_n(b)/A156769((n-1)/2).
%F P_n(b) = 2/(b+1)*Sum_{j=1..b/2}(tan(j*Pi/(b+1)))^n, if n is even, and
%F P_n(b) = 2/(b+1)*Sum_{j=1..b/2}(tan(j*Pi/(b+1)))^n*sin(j*Pi/(b+1)), if n is odd.
%e Triangle begins (r is the number of row or the number of polynomial; coefficients of b^k, k=r-2-i, i=0,1,..., r-2)
%e r/i.|..0......1......2.....3.....4......5......6.....7
%e ======================================================
%e .2..|..1
%e .3..|..2.....-1
%e .4..|..1......3.....-1
%e .5..|..2......6.....-8.....3
%e .6..|..2.....10.....10...-10.....3
%e .7..|..4.....20.....10...-50....46....-15
%e .8..|.17....119....245....35..-217....161....-45
%e .9..|.34....238....406..-350..-644...1372..-1056....315
%e For example, if r=4, the polynomial
%e P_4(b)=b*(b^2+3*b-1)/A156769(4/2)=b/3*(b^2+3*b-1) (b==0 mod 2)
%e gives difference between multiples of b+1 with even and odd digit sums in base b in interval [0, b^4). Note also that P_2(b)=b. Therefore, setting in the formula n=r=3, again for P_4(b) we have P_4(b)=b*C(b+1,2)-C(b,3)=b/3*(b^2+3*b-1).
%t A156769[n_] := Denominator[(2^(2*n-2)/Factorial[2*n-1])]; poly[1, b_] := 1; poly[2, b_] := b; poly[n_, b_] := poly[n, b] = If[OddQ[n], (-1)^((n - 1)/2) (FunctionExpand[Binomial[b - 1, n - 1]] - Sum[(-1)^(k/2) FunctionExpand[Binomial[b + 1, n - k - 1]] poly[k + 1, b], {k, 0, n - 2, 2}]), (-1)^((n - 2)/2) (FunctionExpand[Binomial[b, n - 1]] - Sum[(-1)^((k - 1)/2) FunctionExpand[Binomial[b + 1, n - k - 1]] poly[k + 1, b], {k, 1, n - 2, 2}])]; Table[If[EvenQ[z], Most[Reverse[CoefficientList[poly[z, b] A156769[z/2], b]]], Reverse[CoefficientList[poly[z, b] A156769[(z - 1)/2], b]]], {z, 2, 12}]
%Y Cf. A156769, A038754, A084990, A091042, A212500, A212592, A212593, A212594, A212668, A212669, A212670, A212705, A212706.
%K sign,base,tabl
%O 2,2
%A _Vladimir Shevelev_ and _Peter J. C. Moses_, May 28 2012
|
