A280642 Numbers k such that k^3 has an odd number of digits and the middle digit is 2.

5, 9, 103, 113, 133, 146, 151, 154, 165, 180, 198, 202, 470, 473, 493, 496, 504, 507, 521, 531, 538, 542, 566, 569, 581, 591, 593, 599, 612, 618, 620, 650, 654, 673, 681, 686, 703, 711, 715, 728, 729, 732, 740, 779, 801, 829, 841, 850, 855, 856, 857, 858, 874

Offset: 1

Author

Lars Blomberg, Jan 07 2017

Keywords

• nonn
• base
• easy

Comments

The sequence of cubes starts: 125, 729, 1092727, 1442897, 2352637, 3112136, 3442951, 3652264, ...

Examples

			5^3 = 1(2)5, 180^3 = 583(2)000, 618^3 = 2360(2)9032.
		

Links

• Lars Blomberg, Table of n, a(n) for n = 1..10000
• Jeremy Gardiner, Middle digit in cube numbers, Seqfan Mailing list, Dec 12 2016.

Crossrefs

Cf. A280640-A280641, A280643-A280649, A181354.

See A279420-A279429 for a k^2 version.

See A279430-A279431 for a k^2 version in base 2.

Programs

Mathematica

a[n_]:=Part[IntegerDigits[n],(Length[IntegerDigits[n]]+1)/2];
Select[Range[0,874],OddQ[Length[IntegerDigits[#^3]]] && a[#^3]==2 &] (* Indranil Ghosh, Mar 06 2017 *)

PARI

isok(k) = my(d=digits(k^3)); (#d%2 == 1) && (d[#d\2 +1] == 2);
for(k=0, 874, if(isok(k)==1, print1(k, ", "))); \\ Indranil Ghosh, Mar 06 2017

Python

i=0
j=1
while i<=874:
    n=str(i**3)
    l=len(n)
    if l%2 and n[(l-1)//2]=="2":
        print(str(i), end=', ')
        j+=1
    i+=1 # Indranil Ghosh, Mar 06 2017