A182456 a(0)=1; for n>0, a(n) = ( a(n-1) mod (n+3) )*(n+3).

1, 4, 20, 12, 35, 24, 54, 40, 77, 60, 104, 84, 135, 112, 170, 144, 209, 180, 252, 220, 299, 264, 350, 312, 405, 364, 464, 420, 527, 480, 594, 544, 665, 612, 740, 684, 819, 760, 902, 840, 989, 924, 1080, 1012, 1175, 1104, 1274, 1200, 1377, 1300, 1484

Offset

0,2

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Formula

For k>0, a(2*k) = A182455(2*k+3)-1, a(2k+1) = A182455(2k).

From Alexander R. Povolotsky, May 01 2012: (Start)

for the same sequence with index starting from 1 instead of 0, i.e. k=1,2,...

a(k+1) = (k+3)^2 -((k+3)*a(k))/(k+2).

G.f.: (-1-3*x-14*x^2+14*x^3+8*x^4-8*x^5)/((x-1)^3*(1+x)^2). (End)

Example

a(6) = (a(5) mod 9) * 9 = (24 mod 9) * 9 = 6*9 = 54.

Mathematica

CoefficientList[Series[(-1 - 3*x - 14*x^2 + 14*x^3 + 8*x^4 - 8*x^5)/((x - 1)^3*(1 + x)^2), {x, 0, 50}], x] (* G. C. Greubel, Feb 25 2017 *)
RecurrenceTable[{a[0]==1, a[n]==Mod[a[n-1], n+3](n+3)}, a, {n, 50}] (* Harvey P. Dale, Oct 21 2018 *)

Prog

(Python)
a=1
for n in range(1, 55):
    print(a, end=", ")
    a = (a%(n+3)) * (n+3)
(PARI) x='x+O('x^50); Vec((-1-3*x-14*x^2+14*x^3+8*x^4-8*x^5)/((x-1)^3*(1+x)^2)) \\ G. C. Greubel, Feb 25 2017

Crossrefs

Cf. A182455.

Sequence in context: A263964 A180855 A213822 * A196380 A227997 A130316

Adjacent sequences: A182453 A182454 A182455 * A182457 A182458 A182459

Keyword

nonn,easy

Author

Alex Ratushnyak, Apr 30 2012

Status

approved