

A095265


A sequence generated from a 4th degree Pascal's Triangle polynomial.


0



1, 22, 103, 284, 605, 1106, 1827, 2808, 4089, 5710, 7711, 10132, 13013, 16394, 20315, 24816, 29937, 35718, 42199, 49420, 57421, 66242, 75923, 86504, 98025, 110526, 124047, 138628, 154309, 171130, 189131, 208352, 228833, 250614, 273735, 298236
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OFFSET

1,2


COMMENTS

The characteristic polynomial of M = x^4  4x^3 + 6x^2  4x + 1. (the recursive multipliers are seen in the polynomial with changed signs: (4), (6), (4), (1).


LINKS

Table of n, a(n) for n=1..36.
Index entries for linear recurrences with constant coefficients, signature (4,6,4,1).


FORMULA

a(n+4) = 4*a(n+3)  6*a(n+2) + 4*a(n+1)  a(n), (multipliers which are present with changed signs in the characteristic polynomial, x^4  4x^3 + 6x^2  4x + 1. Given the 4 X 4 matrix derived from an A056939 triangle (fill in with zeros): M = [1 0 0 0 / 1 1 0 0 / 1 4 1 0 / 1 10 10 1], then M^n * [1 0 0 0] = [1 n A000384(n) a(n)] where A000384 is the hexagonal series 1, 6, 15, 28... 3. a(n) = (20/3)n^3  10n^2 + (13/3)n.
G.f.: x*(21*x^2+18*x+1) / (x1)^4.  Colin Barker, May 25 2013


EXAMPLE

a(13) = 13013 = 4*a(12)  6*a(11) + 4*a(10)  a(9) = 4*10132  6*7711 + 4*5710  4089.
a(6) = 1106 since M^6 * [1 0 0 0] = [ 1 6 66 1106].
a(6) = 1106 = f(n) = (20/3)(6)^3 10*(6^2) +(13/3)*6 = 1440  360 + 26.


MAPLE

a:= n> (20*n^230*n+13)*n/3:
seq(a(n), n=1..50); # Alois P. Heinz, May 25 2013


MATHEMATICA

a[n_] := (MatrixPower[{{1, 0, 0, 0}, {1, 1, 0, 0}, {1, 4, 1, 0}, {1, 10, 10, 1}}, n].{{1}, {0}, {0}, {0}})[[4, 1]]; Table[ a[n], {n, 36}] (* Robert G. Wilson v, Jun 05 2004 *)


CROSSREFS

Cf. A056939, A000384.
Sequence in context: A044273 A044654 A156795 * A066450 A231225 A124950
Adjacent sequences: A095262 A095263 A095264 * A095266 A095267 A095268


KEYWORD

nonn,easy


AUTHOR

Gary W. Adamson, May 31 2004


EXTENSIONS

Edited and corrected by Robert G. Wilson v, Jun 05 2004


STATUS

approved



