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A328933 For any negative number, add the digits (assigning the negative sign just to the first digit), square the result and add it to the original number. This sequence shows negative numbers which give a positive answer.

%I A328933 #32 Nov 29 2019 21:09:21
%S A328933 -2,-3,-4,-5,-6,-7,-8,-9,-15,-16,-17,-18,-19,-28,-29,-159,-168,-169,
%T A328933 -178,-179,-187,-188,-189,-197,-198,-199
%N A328933 For any negative number, add the digits (assigning the negative sign just to the first digit), square the result and add it to the original number. This sequence shows negative numbers which give a positive answer.
%C A328933 These numbers are the Zombie Numbers.
%C A328933 Start with any negative (dead) number, add the digits (attaching the negative to the first digit), square the result and add it to the original number. If your answer is positive then you have a 'zombie number' which has 'risen from the dead'.
%C A328933 The list is finite with 26 terms.
%C A328933 Negative integer k such that (digitsum(-k) - 2*(1st digit of -k))^2 > -k. - _Stefano Spezia_, Nov 01 2019
%H A328933 Ed Southall, <a href="https://twitter.com/solvemymaths/status/1189824239408422912">Twitter post about Halloween maths</a>, SolveMyMaths, Oct 31 2019.
%e A328933 -27 is not a zombie number because -2 + 7 = 5 and -27 + (5)^2 = -2.
%e A328933 -28 is a zombie number because -2 + 8 = 6 and -28 + (6)^2 = 8.
%t A328933 -Select[Range[200],(Total[IntegerDigits[#]]-2*First[IntegerDigits[#]])^2-#>0&] (* _Stefano Spezia_, Nov 01 2019 *)
%o A328933 (PARI) f(n) = my(d=digits(n), s = sumdigits(n) - 2*d[1]); s^2 + n;
%o A328933 isok(n) = f(n) > 0;
%o A328933 forstep(n=-1, -10000, -1, if (isok(n), print1(n, ", "))) \\ _Michel Marcus_, Oct 31 2019
%K A328933 sign,base,dumb,fini,full
%O A328933 1,1
%A A328933 _Andrew Toothill_, Oct 31 2019