A191759 Least significant decimal digit of (2n-1)^2.

1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9

Offset

1,2

Comments

Bisection of A008959.

This sequence is periodic with repeating part <1,9,5,9,1> of length five. Hence, as the members of each cycle sum to 25, the terms satisfy the fifth-order homogeneous recurrence a(n) = a(n-5) and the fourth-order inhomogeneous recurrence a(n) = 25 - a(n-1) - a(n-2) - a(n-3) - a(n-4).

Links

Table of n, a(n) for n=1..87.

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

Formula

a(n) = (2n-1)^2 mod 10.

G.f.: x*(1+9*x+5*x^2+9*x^3+x^4)/(1-x^5) (note that the coefficients of x in the numerator are precisely the terms that constitute the periodic cycle of the sequence).

a(n) = 5 + 4*A080891(n+2). - R. J. Mathar, Jun 16 2011

Continued fraction of (97+17*sqrt(3077))/938. - R. J. Mathar, Jun 25 2011

a(n) = (-n^2 + n + 1) mod 10. - Arkadiusz Wesolowski, Jul 03 2012

Example

The fifth odd square number is 81 which has least significant digit 1. Hence a(5)=1.

Mathematica

Mod[(2#-1)^2, 10]&/@Range[50]
LinearRecurrence[{0, 0, 0, 0, 1}, {1, 9, 5, 9, 1}, 87] (* Ray Chandler, Aug 25 2015 *)
PadRight[{}, 120, {1, 9, 5, 9, 1}] (* Harvey P. Dale, Aug 04 2019 *)

Prog

(PARI) a(n)=[1, 9, 5, 9, 1][n%5+1] \\ Charles R Greathouse IV, Jun 15 2011

Crossrefs

Cf. A016754 (odd squares).

Sequence in context: A090463 A272795 A246765 * A289505 A371528 A334427

Adjacent sequences: A191756 A191757 A191758 * A191760 A191761 A191762

Keyword

nonn,easy,base

Author

Ant King, Jun 15 2011

Extensions

More terms from Arkadiusz Wesolowski, Jul 03 2012

Status

approved