A117606 a(n) = ones digit minus tens digit of the square of a(n-1), with a(0) = 2.

2, 4, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7, 5, 3, 9, -7

Offset

0,1

Comments

From the 2006 Collaborative Problem Solving Contest, written by Tom Clymer. It repeats forever, obviously. If the terms are thought of as the absolute value of the difference of the digits, then the -7's should all be 7 instead, but the remaining terms are unchanged. (The original puzzle sequence had only 2,4,5,3,9 given).

Links

Colin Barker, Table of n, a(n) for n = 0..1000

National Assessment and Testing, the sponsor of the contest from which this problem comes.

The 2006 Collaborative Problem Solving Contest, 2006 is the source of this puzzle.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Formula

From Colin Barker, Oct 21 2017: (Start)

G.f.: (2 + 4*x + 5*x^2 + 3*x^3 + 7*x^4 - 11*x^5) / ((1 - x)*(1 + x)*(1 + x^2)).

a(n) = (5 + 9*(-1)^n + (2 - 5*i)*(-i)^n + (2+5*i)*i^n) / 2 for n>1.

a(n) = a(n-4) for n>6.

(End)

Example

Since a(0) = 2, a(1) = 2^2 = 4 and since a(1) = 4, a(2) = the difference of the digits of 16 = 5.

Prog

(PARI) Vec((2 + 4*x + 5*x^2 + 3*x^3 + 7*x^4 - 11*x^5) / ((1 - x)*(1 + x)*(1 + x^2)) + O(x^100)) \\ Colin Barker, Oct 21 2017

Crossrefs

Sequence in context: A026198 A026206 A353708 * A245814 A060736 A097292

Adjacent sequences: A117603 A117604 A117605 * A117607 A117608 A117609

Keyword

base,sign,easy

Author

Joshua Zucker, Apr 06 2006

Status

approved