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%I #26 Mar 16 2023 17:29:29
%S 0,1,0,10,2,100,11,1000,4,3,6,8,19,35,7,16,34,106,13,41,24,17,37,107,
%T 323,43,124,317,67,113,63,114,134,343,83,133,367,1024,167,374,264,314,
%U 386,1043,313,583,1303,3283,707,1183,3316,836,1333,3286,10133
%N a(n) is the least number k such that (sum of digits of k^2) + (number of digits of k^2) = n, or 0 if no such k exists.
%C a(n) < sqrt(10^(n-1)). 0 < a(2m) <= 10^(m-1) with the upper bound reached for 1<=m<=4. - _Chai Wah Wu_, Mar 15 2023
%H Chai Wah Wu, <a href="/A185076/b185076.txt">Table of n, a(n) for n = 1..185</a>
%F n = A004159(a(n)) + A185679(a(n)).
%e a(7)=11 since 7 = sumdigits(121) + numberdigits(121) = 4 + 3.
%t Table[k=1; While[d=IntegerDigits[k^2]; n>Length[d] && n != Total[d] + Length[d], k++]; If[Length[d] >= n, k=0]; k, {n, 50}]
%o (Python)
%o from itertools import count
%o def A185076(n):
%o for k in count(1):
%o if n == (t:=len(s:=str(k**2)))+sum(map(int,s)):
%o return k
%o if t >= n:
%o return 0 # _Chai Wah Wu_, Mar 15 2023
%Y Cf. A004159, A185679.
%K nonn,base
%O 1,4
%A _Carmine Suriano_, Feb 23 2011
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