A052210 Numbers k such that k^3 starts with k itself (in base 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

Offset

1,3

Comments

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

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

Andrew Howroyd, Table of n, a(n) for n = 1..400

Example

32^3=32768, which starts with 32.

Mathematica

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[ # ]& ] ] ]

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, A261751.

Sequence in context: A024933 A198646 A356277 * A084818 A264286 A120969

Adjacent sequences: A052207 A052208 A052209 * A052211 A052212 A052213

Keyword

nonn,base,easy

Author

Erich Friedman, Jan 29 2000

Extensions

0 inserted by Juhani Heino, Aug 31 2015

Status

approved