A239937 Numbers k such that DigitSum(k^2) > DigitSum((k+1)^2).

3, 7, 8, 9, 14, 17, 19, 24, 26, 28, 29, 31, 33, 34, 37, 38, 39, 43, 44, 47, 48, 53, 54, 57, 59, 63, 64, 67, 69, 70, 74, 77, 78, 79, 83, 84, 87, 88, 89, 93, 94, 97, 98, 99, 104, 107, 109, 114, 117, 118, 119, 122, 124, 126, 128, 129, 133, 134, 137, 138, 141, 143

Offset: 1

Author

Oliver Bel, Mar 29 2014

Keywords

• nonn
• base

Examples

			For k=3, we have DigitSum(3^2) = 9 > 7 = DigitSum(4^2).
		

Crossrefs

Cf. A007953 (sum of digits of n), A004159 (sum of digits of n^2).

Programs

Mathematica

lis = Table[Total[IntegerDigits[n^2, 10]], {n, 1, 100}];
Flatten[Position[Greater @@@ Partition[lis, 2, 1], True]]
Position[Partition[Total[IntegerDigits[#]]&/@(Range[150]^2),2,1],?(#[[1]]> #[[2]]&),1,Heads->False]//Flatten (* _Harvey P. Dale, Feb 19 2020 *)

PARI

isok(k) = sumdigits(k^2) > sumdigits((k+1)^2); \\ Michel Marcus, Jul 03 2021

Extensions

More terms from Jon E. Schoenfield, Mar 29 2014