%I #48 Dec 14 2023 05:22:02
%S 1,4,9,7,7,9,4,1,9,1,4,9,7,7,9,4,1,9,1,4,9,7,7,9,4,1,9,1,4,9,7,7,9,4,
%T 1,9,1,4,9,7,7,9,4,1,9,1,4,9,7,7,9,4,1,9,1,4,9,7,7,9,4,1,9,1,4,9,7,7,
%U 9,4,1,9,1,4,9,7,7,9,4,1,9,1,4,9,7,7,9
%N Digital roots of square numbers A000290.
%C Cyclic with a period of nine. Note that (7, 9, 4, 1, 9, 1, 4, 9, 7) is palindromic.
%C a(n) is also the decimal expansion of 499264730/333333333. - _Enrique Pérez Herrero_, Jul 28 2009
%C a(n) is also the digital root of the Wonderful Demlo number A002477(n). - _Enrique Pérez Herrero_, Dec 20 2009
%C First comment above by _Enrique Pérez Herrero_ and his formula below together give the following identity: 1+Sum_{n>=2}(1+9*((n^2-1)/9-floor((n^2-1)/9)))/10^(n-1) = 499264730/333333333 = 1.49779419149779419149779419... - _Alexander R. Povolotsky_, Jun 14 201
%H G. C. Greubel, <a href="/A056992/b056992.txt">Table of n, a(n) for n = 1..5000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SquareNumber.html">Square Number.</a>
%F a(n) = 1+9*{(n^2-1)/9} , where the symbol {} means fractional part. - _Enrique Pérez Herrero_, Dec 20 2009
%F a(n) = 3(1 + cos(2n*Pi/3) + cos(4n*Pi/3)) + mod(3n^4+3n^6+4n^8,9). - _Ant King_, Oct 07 2009
%F G.f.: x (1+4x+9x^2+7x^3+7x^4+9x^5+4x^6+x^7+9x^8)/((1-x)(1+x+x^2)(1+x^3+x^6)). - _Ant King_, Oct 20 2009
%F Also a(n) = A010888(A057147(n)). - _Reinhard Zumkeller_, Mar 19 2014
%t DigitalRoot[n_Integer?NonNegative] := 1 + 9*FractionalPart[(n - 1)/9] A056992[n_]:=DigitalRoot[n^2] (* _Enrique Pérez Herrero_, Dec 20 2009 *)
%t Table[FixedPoint[Total[IntegerDigits[#]]&,n^2],{n,90}] (* _Zak Seidov_, Jun 13 2015 *)
%t PadRight[{},120,{1,4,9,7,7,9,4,1,9}] (* _Harvey P. Dale_, Apr 16 2022 *)
%o (Haskell)
%o a056992 = a010888 . a000290 -- _Reinhard Zumkeller_, Mar 19 2014
%Y Cf. A000290, A002477, A010888, A057147, A056991.
%K nonn,base,easy
%O 1,2
%A _Eric W. Weisstein_