A191760 Digital root of the n-th odd square.

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

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(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).

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

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).

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

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).

a(n) = A056992(2n-1). - 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 *)
PadRight[{}, 120, {1, 9, 7, 4, 9, 4, 7, 9, 1}] (* Harvey P. Dale, Jun 26 2021 *)

Crossrefs

Cf. A010888, A016754, A056992.

Sequence in context: A010546 A361603 A232735 * A237841 A109846 A096230

Adjacent sequences: A191757 A191758 A191759 * A191761 A191762 A191763

Keyword

nonn,base,easy

Author

Ant King, Jun 17 2011

Status

approved