cp's OEIS Frontend

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

%I A191759 #31 Aug 03 2023 01:41:56
%S A191759 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,
%T A191759 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,
%U A191759 9,1,1,9,5,9,1,1,9,5,9,1,1,9,5,9,1,1,9
%N A191759 Least significant decimal digit of (2n-1)^2.
%C A191759 Bisection of A008959.
%C A191759 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).
%H A191759 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,1).
%F A191759 a(n) = (2n-1)^2 mod 10.
%F A191759 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).
%F A191759 a(n) = 5 + 4*A080891(n+2). - _R. J. Mathar_, Jun 16 2011
%F A191759 Continued fraction of (97+17*sqrt(3077))/938. - _R. J. Mathar_, Jun 25 2011
%F A191759 a(n) = (-n^2 + n + 1) mod 10. - _Arkadiusz Wesolowski_, Jul 03 2012
%e A191759 The fifth odd square number is 81 which has least significant digit 1. Hence a(5)=1.
%t A191759 Mod[(2#-1)^2,10]&/@Range[50]
%t A191759 LinearRecurrence[{0, 0, 0, 0, 1},{1, 9, 5, 9, 1},87] (* _Ray Chandler_, Aug 25 2015 *)
%t A191759 PadRight[{},120,{1,9,5,9,1}] (* _Harvey P. Dale_, Aug 04 2019 *)
%o A191759 (PARI) a(n)=[1,9,5,9,1][n%5+1] \\ _Charles R Greathouse IV_, Jun 15 2011
%Y A191759 Cf. A016754 (odd squares).
%K A191759 nonn,easy,base
%O A191759 1,2
%A A191759 _Ant King_, Jun 15 2011
%E A191759 More terms from _Arkadiusz Wesolowski_, Jul 03 2012