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A192356
Coefficients of x in the reduction of the polynomial p(n,x) = ((x+2)^n + (x-2)^n)/2 by x^2->x+2.
2
0, 1, 1, 15, 29, 211, 561, 3095, 9829, 46971, 164921, 728575, 2707629, 11450531, 43942081, 181348455, 708653429, 2884834891, 11388676041, 46006694735, 182670807229, 734751144051, 2926800830801, 11743814559415, 46865424529029, 187791199242011, 750176293590361
OFFSET
1,4
COMMENTS
For an introduction to reductions of polynomials by substitutions such as x^2->x+2, see A192232.
Direct sums can be obtained for A192355 and A192356 in the following way. The polynomials p_{n}(x) can be given in series form by p_{n}(x) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*4*k*x^(n-2*k). For the reduction x^2 -> x+2 then the general form can be seen as x^n -> J_{n}*x + phi_{n}, where J_{n} = A001045(n) are the Jacobsthal numbers and phi_{n} = A078008. The reduction of p_{n}(x) now takes the form p_{n}(x) = x * Sum_{k=0..floor(n/2)} binomial(n,2*k)*4^k*J_{n-2*k} + Sum_{k=0..floor(n/2)} binomial(n,2*k)*4^k*phi_{n-2*k}. Evaluating the series leads to p_{n}(x) = x * (4^n - (-3)^n - 1 + 2^n*delta(n,0))/6 + (4^n + 2*(-3)^n + 2 + 2^n*delta(n,0))/6, where delta(n,k) is the Kronecker delta. - G. C. Greubel, Oct 29 2018
FORMULA
Empirical g.f.: x^2*(1-x+2*x^2)/((x-1)*(3*x+1)*(4*x-1)). - Colin Barker, Sep 12 2012
From G. C. Greubel, Oct 28 2018: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k) * 4^k * J_{n-2*k}, where J_{n} = A001045(n) are the Jacobsthal numbers.
a(n) = (4^n - (-3)^n - 1 + 2^n*delta(n,0))/6, with delta(n,0) = 1 if n=0, 0 otherwise. (End)
MATHEMATICA
(See A192355.)
Join[{0}, Table[(4^n - (-3)^n - 1)/6, {n, 1, 50}]] (* G. C. Greubel, Oct 20 2018 *)
PROG
(PARI) for(n=0, 50, print1(if(n==0, 0, (4^n - (-3)^n - 1)/6), ", ")) \\ G. C. Greubel, Oct 20 2018
(Magma) [0] cat [(4^n - (-3)^n - 1)/6: n in [1..50]]; // G. C. Greubel, Oct 20 2018
CROSSREFS
Sequence in context: A146427 A202512 A014095 * A350468 A196184 A201136
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jun 29 2011
STATUS
approved