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%I #12 Jun 17 2017 02:59:23
%S 1,22,103,284,605,1106,1827,2808,4089,5710,7711,10132,13013,16394,
%T 20315,24816,29937,35718,42199,49420,57421,66242,75923,86504,98025,
%U 110526,124047,138628,154309,171130,189131,208352,228833,250614,273735,298236
%N A sequence generated from a 4th degree Pascal's Triangle polynomial.
%C 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).
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F 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.
%F G.f.: x*(21*x^2+18*x+1) / (x-1)^4. - _Colin Barker_, May 25 2013
%e a(13) = 13013 = 4*a(12) - 6*a(11) + 4*a(10) - a(9) = 4*10132 - 6*7711 + 4*5710 - 4089.
%e a(6) = 1106 since M^6 * [1 0 0 0] = [ 1 6 66 1106].
%e a(6) = 1106 = f(n) = (20/3)(6)^3 -10*(6^2) +(13/3)*6 = 1440 - 360 + 26.
%p a:= n-> (20*n^2-30*n+13)*n/3:
%p seq(a(n), n=1..50); # _Alois P. Heinz_, May 25 2013
%t 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 *)
%Y Cf. A056939, A000384.
%K nonn,easy
%O 1,2
%A _Gary W. Adamson_, May 31 2004
%E Edited and corrected by _Robert G. Wilson v_, Jun 05 2004
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