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 A052210 Numbers n such that n^3 starts with n itself (in base 10). 10
 0, 1, 10, 32, 100, 1000, 10000, 31623, 100000, 316228, 1000000, 3162278, 10000000, 31622777, 100000000, 1000000000, 10000000000, 31622776602, 100000000000, 316227766017, 1000000000000, 10000000000000, 31622776601684, 100000000000000, 316227766016838 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Replace the first term with 4, then add 1 to all the others to find numbers where n^3 starts with n+2.  Similar processes can be used for any n+2k. (conjectured) - Dhilan Lahoti, Aug 30 2015 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 LINKS EXAMPLE 32^3=32768, which starts with 32. MATHEMATICA Sort[ Table[ 10^i, {i, 0, 22} ]~Join~Select[ Table[ Ceiling[ Sqrt[ 10 ]*10^i ], {i, 0, 22} ], Take[ IntegerDigits[ #^3 ], Length[ IntegerDigits[ # ] ] ]==IntegerDigits[ # ]& ] ] PROG (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 CROSSREFS Cf. A052211, A233451, A233452, A233453, A233454, A233455, A233456. Sequence in context: A229720 A024933 A198646 * A084818 A264286 A120969 Adjacent sequences:  A052207 A052208 A052209 * A052211 A052212 A052213 KEYWORD nonn,base,easy AUTHOR Erich Friedman, Jan 29 2000 STATUS approved

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Last modified October 19 03:24 EDT 2019. Contains 328211 sequences. (Running on oeis4.)