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Numbers k such that DigitSum(k^2) > DigitSum((k+1)^2).
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%I #21 Jul 03 2021 04:20:34

%S 3,7,8,9,14,17,19,24,26,28,29,31,33,34,37,38,39,43,44,47,48,53,54,57,

%T 59,63,64,67,69,70,74,77,78,79,83,84,87,88,89,93,94,97,98,99,104,107,

%U 109,114,117,118,119,122,124,126,128,129,133,134,137,138,141,143

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

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

%t lis = Table[Total[IntegerDigits[n^2, 10]], {n, 1, 100}];

%t Flatten[Position[Greater @@@ Partition[lis, 2, 1], True]]

%t Position[Partition[Total[IntegerDigits[#]]&/@(Range[150]^2),2,1],_?(#[[1]]> #[[2]]&),1,Heads->False]//Flatten (* _Harvey P. Dale_, Feb 19 2020 *)

%o (PARI) isok(k) = sumdigits(k^2) > sumdigits((k+1)^2); \\ _Michel Marcus_, Jul 03 2021

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

%K nonn,base

%O 1,1

%A _Oliver Bel_, Mar 29 2014

%E More terms from _Jon E. Schoenfield_, Mar 29 2014