

A191760


Digital root of the nth odd square.


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, 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, 9, 1, 1, 9, 7, 4, 9, 4, 7, 9, 1, 1, 9, 7, 4, 9, 4, 7, 9, 1
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OFFSET

1,2


COMMENTS

This sequence is periodic with period <1,9,7,4,9,4,7,9,1> of length nine.
Related to the continued fraction of (153727+sqrt(2207057870693))/1477642 = 1+ 1/(9+1/(7+1/...)).  R. J. Mathar, Jun 27 2011


LINKS

Table of n, a(n) for n=1..81.
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 1).


FORMULA

a(n)= 3*(1+cos(2(n2)pi/3)+cos(4(n2)pi/3)) +mod( (1+n)(1+7n7n^2+7n^3+n^4n^5+3n^6+3n^7), 9).
a(n)=a(n9).
a(n)=51a(n1)a(n2)a(n3)a(n4)a(n5)a(n6)a(n7)a(n8).
a(n) = A010888(A016754(n)).
G.f.: x(1+9x+7x^2+4x^3+9x^4+4x^5+7x^6+9x^7+x^8)/( (1x)*(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).
a(n) = A056992(2n1).  R. J. Mathar, Jun 27 2011


EXAMPLE

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


MATHEMATICA

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


CROSSREFS

Cf. A010888, A016754.
Sequence in context: A155958 A010546 A232735 * A237841 A109846 A096230
Adjacent sequences: A191757 A191758 A191759 * A191761 A191762 A191763


KEYWORD

nonn,base,easy


AUTHOR

Ant King, Jun 17 2011


STATUS

approved



