A069584 a(n) = n - largest perfect power <= n.

0, 1, 2, 0, 1, 2, 3, 0, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 0, 1, 2, 3, 4, 0, 1, 2, 3, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

Offset: 1

Author

Amarnath Murthy, Mar 24 2002

Keywords

• easy
• nonn

Comments

a(n) = 0 if n = m^p that is if n is a full power (square, cube etc.).

As Catalan's conjecture is now proved, n=8=2^3, n+1=9=3^2 is the only solution for a(n+1) = a(n) = 0.

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Catalan's conjecture

Crossrefs

Cf. A001597, a(n)=n-A081676(n).

Programs

Maple

N:= 100: # for a(1) .. a(N), where N is a perfect power
S:= {seq(seq(i^p,i=1.. floor(N^(1/p))),p=2..ilog2(N))}:
P:= sort(convert(S,list)):
A:= [seq(seq(i-P[j],i=P[j]..P[j+1]-1),j=1..nops(P)-1),0]; # Robert Israel, Apr 04 2025

PARI

a(n) = {m = n; while(!ispower(m), m--; if (m==0, return (n-1))); n-m;} \\ Michel Marcus, Nov 04 2015

Extensions

Edited by Reinhard Zumkeller, Mar 26 2003