A191760 - OEIS

%I #20 Apr 16 2023 10:42:28

%S 1,9,7,4,9,4,7,9,1,1,9,7,4,9,4,7,9,1,1,9,7,4,9,4,7,9,1,1,9,7,4,9,4,7,

%T 9,1,1,9,7,4,9,4,7,9,1,1,9,7,4,9,4,7,9,1,1,9,7,4,9,4,7,9,1,1,9,7,4,9,

%U 4,7,9,1,1,9,7,4,9,4,7,9,1

%N Digital root of the n-th odd square.

%C This sequence is periodic with period <1,9,7,4,9,4,7,9,1> of length nine.

%C Related to the continued fraction of (153727+sqrt(2207057870693))/1477642 = 1+ 1/(9+1/(7+1/...)). - _R. J. Mathar_, Jun 27 2011

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,1).

%F a(n) = 3*(1+cos(2(n-2)pi/3)+cos(4(n-2)pi/3)) +mod( (1+n)(1+7n-7n^2+7n^3+n^4-n^5+3n^6+3n^7), 9).

%F a(n) = a(n-9).

%F a(n) = 51-a(n-1)-a(n-2)-a(n-3)-a(n-4)-a(n-5)-a(n-6)-a(n-7)-a(n-8).

%F a(n) = A010888(A016754(n)).

%F G.f.: x(1+9x+7x^2+4x^3+9x^4+4x^5+7x^6+9x^7+x^8)/( (1-x)*(1+x+x^2)*(1+x^3+x^6) ) (note that the coefficients of x in the numerator are precisely the terms that constitute the periodic cycle of the sequence).

%F a(n) = A056992(2n-1). - _R. J. Mathar_, Jun 27 2011

%e The fifth, odd square number is 81 which has digital root 9. Hence a(5)=9.

%t DigitalRoot[n_Integer?Positive]:=FixedPoint[Plus@@IntegerDigits[#]&,n];DigitalRoot[#] &/@((2#-1)^2 &/@Range[81])

%t LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 1},{1, 9, 7, 4, 9, 4, 7, 9, 1},81] (* _Ray Chandler_, Aug 25 2015 *)

%t PadRight[{},120,{1,9,7,4,9,4,7,9,1}] (* _Harvey P. Dale_, Jun 26 2021 *)

%Y Cf. A010888, A016754, A056992.

%K nonn,base,easy

%O 1,2

%A _Ant King_, Jun 17 2011