A126255 Number of distinct terms i^j for 2 <= i,j <= n.

1, 4, 8, 15, 23, 34, 44, 54, 69, 88, 106, 129, 152, 177, 195, 226, 256, 291, 324, 361, 399, 442, 483, 519, 564, 600, 648, 703, 755, 814, 856, 915, 976, 1039, 1085, 1156, 1224, 1295, 1365, 1444, 1519, 1602, 1681, 1762, 1846, 1937, 2023, 2095, 2184, 2279

Offset

2,2

Comments

An easy upper bound is (n-1)^2 = A000290(n-1).

Links

Eric M. Schmidt, Table of n, a(n) for n = 2..10000

Project Euler, Problem 29: Distinct powers.

Example

a(4) = 8 as there are 8 distinct terms in 2^2=4, 2^3=8, 2^4=16, 3^2=9, 3^3=27, 3^4=81, 4^2=16, 4^3=64, 4^4=256.

Mathematica

SetAttributes[a, {Listable, NumericFunction}]
a[n_ /; n < 2] := "error"
a[2] := 1
a[n_Integer?IntegerQ /; n > 2] :=
 Length[DeleteDuplicates[
   Distribute[f[Range[2, n], Range[2, n]], List,
     f] /. {f ->
      Power}]](*By using Distribute instead of Outer I avoid having to use Flatten on Outer*)
a[Range[2, 100]]
(* Peter Cullen Burbery, Aug 15 2023 *)

Prog

(PARI) lim=51; z=listcreate((lim-1)^2); for(m=2, lim, for(i=2, m, x=factor(i); x[, 2]*=m; s=Str(x); f=setsearch(z, s, 1); if(f, listinsert(z, s, f))); t=factor(m); for(j=2, m, x=t; x[, 2]=j*t[, 2]; s=Str(x); f=setsearch(z, s, 1); if(f, listinsert(z, s, f))); print1(#z, ", "))
(Python)
def A126255(n): return len({i**j for i in range(2, n+1) for j in range(2, n+1)}) # Chai Wah Wu, Oct 17 2023

Crossrefs

Cf. A027424, A061786, A126254, A126256, A126257.

Sequence in context: A001182 A264599 A122247 * A267682 A194804 A169953

Adjacent sequences: A126252 A126253 A126254 * A126256 A126257 A126258

Keyword

easy,nonn

Author

Nick Hobson, Dec 24 2006

Status

approved