A061100 Squares with digital root 4.

4, 49, 121, 256, 400, 625, 841, 1156, 1444, 1849, 2209, 2704, 3136, 3721, 4225, 4900, 5476, 6241, 6889, 7744, 8464, 9409, 10201, 11236, 12100, 13225, 14161, 15376, 16384, 17689, 18769, 20164, 21316, 22801, 24025, 25600, 26896, 28561, 29929

Offset: 1

Author

Amarnath Murthy, Apr 19 2001

Keywords

• nonn
• base

Examples

			256 = 16^2, 2 + 5 + 6 = 13, 1 + 3 = 4;
1849 = 43^2, 1 + 8 + 4 + 9 = 22, 2 + 2 = 4.
		

Links

• Harry J. Smith, Table of n, a(n) for n = 1..1000
• Amarnath Murthy & Charles Ashbacher, Fabricating a perfect square with a given valid digit sum, in Generalized Partitions and New Ideas On Number Theory and Smarandache Sequences, pp. 154-156.

Crossrefs

Cf. A056991.

Programs

Maple

seq(seq((a+9*k)^2,a=[2,7]),k=0..20); # Robert Israel, Jun 13 2018

Mathematica

fdsQ[n_]:=NestWhile[Total[IntegerDigits[#]]&,n,#>9&]==4; Select[Range[ 200]^2,fdsQ] (* Harvey P. Dale, Dec 15 2011 *)

PARI

a(n)=(n\2*9-2*(-1)^n)^2 \\ Charles R Greathouse IV, Sep 21 2012

Formula

From Colin Barker, Feb 18 2013: (Start)

Conjecture:

a(n) = (16-72*n+81*n^2)/4 for n even;

a(n)=(25-90*n+81*n^2)/4 for n odd;

g.f.: -x*(4*x^4+45*x^3+64*x^2+45*x+4) / ((x-1)^3*(x+1)^2). (End)

Conjecture is true since x^2 == 4 (mod 9) if and only if x == 2 or 7 (mod 9). The odd-numbered terms are (2+9*k)^2 and the even-numbered terms are (7+9*k)^2. - Robert Israel, Jun 13 2018

Extensions

More terms from Harry J. Smith, Jul 18 2009