A126254 Number of distinct terms i^j for 1 <= i,j <= n.
1, 3, 7, 11, 19, 28, 40, 50, 60, 76, 96, 115, 139, 163, 189, 207, 239, 270, 306, 340, 378, 417, 461, 503, 539, 585, 621, 670, 726, 779, 839, 881, 941, 1003, 1067, 1113, 1185, 1254, 1326, 1397, 1477, 1553, 1637, 1717, 1799, 1884, 1976, 2063, 2135, 2225
Offset: 1
Examples
a(4) = 11, as there are 11 distinct terms in 1^1=1, 1^2=1, 1^3=1, 1^4=1, 2^1=2, 2^2=4, 2^3=8, 2^4=16, 3^1=3, 3^2=9, 3^3=27, 3^4=81, 4^1=4, 4^2=16, 4^3=64, 4^4=256.
Links
- Nick Hobson and John Silberholz, Table of n, a(n) for n = 1..10000 [Terms 1 through 1000 were computed by N. Hobson; terms 1001 through 10000 by John Silberholz, Feb 23 2015]
- John Silberholz, Combinatorial approach to calculate sequence
Programs
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Maple
seq(nops({seq(seq(i^j, i=1..n),j=1..n)}),n=1..100); # Robert Israel, Feb 23 2015 -
PARI
lim=50; z=listcreate(lim*(lim-1)+1); for(m=1, lim, for(i=1, 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=1, m, x=t; x[, 2]=j*t[, 2]; s=Str(x); f=setsearch(z, s, 1); if(f, listinsert(z, s, f))); print1(#z, ", "))
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Python
def A126254(n): return len({i**j for i in range(1,n+1) for j in range(1,n+1)}) # Chai Wah Wu, Oct 17 2023
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R
A126254 <- function(limit) { if (limit == 1) { return(1) } ; num.powers <- c(1, rep(0, limit-1)) ; handled <- c(T, rep(F, limit-1)) ; for (base in 2:ceiling(sqrt(limit))) { if (!handled[base]) { num.handle <- floor(log(limit, base)) ; handled[base^(1:num.handle)] <- T ; num.powers[base] <- length(unique(as.vector(outer(1:num.handle, 1:limit)))) }} ; num.powers[!handled] <- limit ; sum(num.powers) } ; A126254(50) # John Silberholz, Feb 23 2015
Comments