OFFSET
1,6
COMMENTS
Iterating n, a(n), a(a(n)), a(a(a(n))), ..., until 1 is reached, and taking the smallest prime factor (A020639) of each term gives a sequence of distinct primes in ascending order, while applying A302045 to the same terms gives the corresponding exponents (multiplicities) of those primes. Permutation pair A250245/A250246 maps between this non-standard prime factorization and the ordinary factorization of n. See also comments and examples in A302042.
LINKS
FORMULA
PROG
(PARI)
\\ Assuming A250469 and its inverse A268674 have been precomputed, then the following is fast enough:
A302044(n) = if(1==n, n, my(k=0); while((n%2), n = A268674(n); k++); n = (n/2^valuation(n, 2)); while(k>0, n = A250469(n); k--); (n));
(PARI)
up_to = 65537;
ordinal_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), pt); for(i=1, length(invec), if(mapisdefined(om, invec[i]), pt = mapget(om, invec[i]), pt = 0); outvec[i] = (1+pt); mapput(om, invec[i], (1+pt))); outvec; };
A000265(n) = (n/2^valuation(n, 2));
v078898 = ordinal_transform(vector(up_to, n, A020639(n)));
A078898(n) = v078898[n];
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Mar 31 2018
STATUS
approved