A095265 A sequence generated from a 4th degree Pascal's Triangle polynomial.

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

Offset: 1

Author

Gary W. Adamson, May 31 2004

Keywords

• nonn
• easy

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).

Examples

			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.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A056939, A000384.

Programs

Maple

a:= n-> (20*n^2-30*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 *)

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) / (x-1)^4. - Colin Barker, May 25 2013

Extensions

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