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

Author

Erich Friedman, Jan 29 2000

Keywords

• nonn
• base
• easy

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

Examples

			32^3=32768, which starts with 32.
		

Links

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

Crossrefs

Cf. A052211, A233451, A233452, A233453, A233454, A233455, A233456, A261751.

Programs

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[ # ]& ] ] ]

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

Extensions

0 inserted by Juhani Heino, Aug 31 2015