

A165808


Expansion of x*(403+2967*x+1047*x^2x^3)/(1x)^4.


10



403, 4579, 16945, 41917, 83911, 147343, 236629, 356185, 510427, 703771, 940633, 1225429, 1562575, 1956487, 2411581, 2932273, 3522979, 4188115, 4932097, 5759341, 6674263, 7681279, 8784805, 9989257, 11299051, 12718603, 14252329, 15904645
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OFFSET

1,1


COMMENTS

Old name was: As mentioned in short description of A165806, polynomials have the following unique property: let f(x) be a polynomial in x. Then f(x+k*f(x)) is congruent to 0 (mod(f(x)); here k belongs to N. The present case pertains to f(x) = x^3 + 2x + 11 when x is complex (2 + 3i). The quotient f(x+k*f(x))/f(x), for any given k, consists of two parts: a) a rational integer part and b) rational integer coefficient of sqrt(1). This sequence pertains to a.


LINKS

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (4,6,4,1).


FORMULA

From R. J. Mathar, Sep 30 2009: (Start)
a(n) = 113*n321*n^2+736*n^3.
G.f.: x*(403+2967*x+1047*x^2x^3)/(1x)^4. (End)
From G. C. Greubel, Apr 08 2016: (Start)
a(n) = 4*a(n1)  6*a(n2) + 4*a(n3)  a(n4).
E.g.f.: (1 333*x  318*x^2 + x^3)*exp(x). (end)


EXAMPLE

f(x)= x^3 + 2x + 11. When x = 2 + 3i, we get f(x) = 31 + 15i. x + f(x) = 29 + 18i. f(29 + 18i) = 3752 + 39618i. When this value is divided by (31 + 15i) we get 403  1083i; needless to say, PARI takes care of necessary rationalization.


MATHEMATICA

LinearRecurrence[{4, 6, 4, 1}, {403, 4579, 16945, 41917}, 100](* G. C. Greubel, Apr 08 2016 *)


PROG

(PARI) Vec((403+2967*x+1047*x^2x^3)/(1x)^4+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012


CROSSREFS

Cf. A165806, A165809.
Sequence in context: A083815 A250893 A261857 * A283662 A097741 A117836
Adjacent sequences: A165805 A165806 A165807 * A165809 A165810 A165811


KEYWORD

nonn,easy


AUTHOR

A.K. Devaraj, Sep 29 2009


EXTENSIONS

More terms from R. J. Mathar, Sep 30 2009
Edited by Jon E. Schoenfield, Dec 12 2013


STATUS

approved



