A018852 a(n)^3 is smallest cube beginning with n.

0, 1, 3, 7, 16, 8, 4, 9, 2, 21, 10, 48, 5, 11, 52, 25, 55, 12, 57, 27, 59, 6, 61, 62, 29, 63, 64, 3, 66, 31, 67, 68, 32, 15, 7, 33, 154, 72, 73, 34, 16, 161, 35, 76, 164, 77, 36, 78, 169, 17, 37, 8, 174, 81, 38, 82, 178, 83, 18, 39, 182, 85, 184, 86, 4, 87, 188, 189, 19, 191, 89, 193, 9

Offset: 0

Author

David W. Wilson

Keywords

• nonn
• base
• look

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 1..1000 from T. D. Noe)

Crossrefs

Cf. A018797.

Cf. A018851 (k=2), this sequence (k=3), A018853 (k=4), A018872 (k=5), A018874 (k=6), A018876 (k=7), A018878 (k=8), A018880 (k=9), A018882 (k=10).

Programs

Maple

f:= proc(n) local d,m;
  for d from 0 do
    m:= ceil((10^d*n)^(1/3));
    if m^3 < 10^d*(n+1) then return m fi
  od
end proc:
map(f, [$1..100]); # Robert Israel, Jul 26 2015

PARI

a(n)=k=1;while(k,d=digits(k^3);D=digits(n);if(#D<=#d,c=1;for(i=1,#D,if(D[i]!=d[i],c=0;break));if(c,return(k)));k++)
vector(100,n,a(n)) \\ Derek Orr, Jul 26 2015

Python

for n in range(1,10**3):
    for k in range(10**3):
        if str(k**3).startswith(str(n)):
            print(k,end=', ')
            break
    n += 1 # Derek Orr, Aug 03 2014

Formula

a(n) >= n^(1/3), for all n > 0, with equality when n is a cube. - Derek Orr, Jul 26 2015

Extensions

0 prepended by Seiichi Manyama, Jan 30 2017