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
Examples
The fifth odd square number is 81 which has least significant digit 1. Hence a(5)=1.
Links
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1).
Crossrefs
Cf. A016754 (odd squares).
Programs
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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 *)
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PARI
a(n)=[1,9,5,9,1][n%5+1] \\ Charles R Greathouse IV, Jun 15 2011
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
Extensions
More terms from Arkadiusz Wesolowski, Jul 03 2012
Comments