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%I #36 Aug 27 2023 16:23:28
%S 0,1,10,32,100,1000,10000,31623,100000,316228,1000000,3162278,
%T 10000000,31622777,100000000,1000000000,10000000000,31622776602,
%U 100000000000,316227766017,1000000000000,10000000000000,31622776601684,100000000000000,316227766016838
%N Numbers k such that k^3 starts with k itself (in base 10).
%C Replace the first term with 4, then add 1 to all the others to find numbers k where k^3 starts with k+2. Similar processes can be used for any k+2m. (conjectured) - _Dhilan Lahoti_, Aug 30 2015
%C 10^k is in the sequence for all k. For odd k, m = ceil(10^(k/2)) is in the sequence if and only if m^3/(m+1) < 10^k. A necessary condition for this is that m - 1/2 > 10^(k/2), i.e. the first digit after the decimal point in 10^(k/2) is at least 5. Is this sufficient as well as necessary? - _Robert Israel_, Aug 31 2015
%H Andrew Howroyd, <a href="/A052210/b052210.txt">Table of n, a(n) for n = 1..400</a>
%e 32^3=32768, which starts with 32.
%t Join[{0}, Sort[ Table[ 10^i,{i,0,22} ]~Join~Select[ Table[ Ceiling[ Sqrt[ 10 ]*10^i ],{i,0,22} ], Take[ IntegerDigits[ #^3 ],Length[ IntegerDigits[ # ] ] ]==IntegerDigits[ # ]& ] ] ]
%o (PARI) r=29; print1(1, ", "); e=3; for(n=2, r, p=round((10^(1/(e-1)))^n); f=p^e; b=10^(#Str(f)-#Str(p)); if((f-lift(Mod(f, b)))/b==p, print1(p, ", "))); \\ _Arkadiusz Wesolowski_, Dec 10 2013
%Y Cf. A052211, A233451, A233452, A233453, A233454, A233455, A233456, A261751.
%K nonn,base,easy
%O 1,3
%A _Erich Friedman_, Jan 29 2000
%E 0 inserted by _Juhani Heino_, Aug 31 2015