A056992 Digital roots of square numbers A000290.

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

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• base
• easy

Comments

Cyclic with a period of nine. Note that (7, 9, 4, 1, 9, 1, 4, 9, 7) is palindromic.

a(n) is also the decimal expansion of 499264730/333333333. - Enrique Pérez Herrero, Jul 28 2009

a(n) is also the digital root of A002477(n). - Enrique Pérez Herrero, Dec 20 2009

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 2012

Links

• G. C. Greubel, Table of n, a(n) for n = 1..5000
• Eric Weisstein's World of Mathematics, Square Number.

Crossrefs

Cf. A000290, A002477, A010888, A057147, A056991.

Programs

Haskell

a056992 = a010888 . a000290  -- Reinhard Zumkeller, Mar 19 2014

Mathematica

DigitalRoot[n_Integer?NonNegative] := 1 + 9*FractionalPart[(n - 1)/9] A056992[n_]:=DigitalRoot[n^2] (* Enrique Pérez Herrero, Dec 20 2009 *)
Table[FixedPoint[Total[IntegerDigits[#]]&,n^2],{n,90}] (* Zak Seidov, Jun 13 2015 *)
PadRight[{},120,{1,4,9,7,7,9,4,1,9}] (* Harvey P. Dale, Apr 16 2022 *)

Formula

a(n) = 1+9*{(n^2-1)/9}, where the symbol {} means fractional part. - Enrique Pérez Herrero, Dec 20 2009

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

G.f.: x*(1+4*x+9*x^2+7*x^3+7*x^4+9*x^5+4*x^6+x^7+9*x^8)/((1-x)*(1+x+x^2)*(1+x^3+x^6)). - Ant King, Oct 20 2009

a(n) = A010888(A057147(n)). - Reinhard Zumkeller, Mar 19 2014