A302042 A032742 analog for a nonstandard factorization process based on the sieve of Eratosthenes (A083221).

1, 1, 1, 2, 1, 3, 1, 4, 3, 5, 1, 6, 1, 7, 5, 8, 1, 9, 1, 10, 9, 11, 1, 12, 5, 13, 7, 14, 1, 15, 1, 16, 15, 17, 7, 18, 1, 19, 11, 20, 1, 21, 1, 22, 21, 23, 1, 24, 7, 25, 25, 26, 1, 27, 25, 28, 27, 29, 1, 30, 1, 31, 13, 32, 11, 33, 1, 34, 33, 35, 1, 36, 1, 37, 17, 38, 11, 39, 1, 40, 39, 41, 1, 42, 35, 43, 35, 44, 1, 45, 49, 46, 45, 47, 13, 48

Offset: 1

Author

Antti Karttunen, Mar 31 2018

Keywords

• nonn

Comments

Like [A020639(n), A032742(n)] also ordered pair [A020639(n), a(n)] is unique for each n. 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 multiset of primes in ascending order, unique for each natural number n >= 1. Permutation pair A250245/A250246 maps between this non-standard prime factorization of n and the ordinary prime factorization of n.

Examples

			For n = 66, A020639(66) [its smallest prime factor] is 2. Because A055396(66) = A000720(2) = 1, a(66) is just A078898(66) = 66/2 = 33.
For n = 33, A020639(33) = 3 and A055396(33) = 2, so a(33) = A250469(A078898(33)) = A250469(6) = 15. [15 is under 6 in array A083221].
For n = 15, A020639(15) = 3 and A055396(15) = 2, so a(15) = A250469(A078898(15)) = A250469(3) = 5. [5 is under 3 is array A083221].
For n = 5, A020639(5) = 5 and A055396(5) = 3, so a(5) = A250469(A250469(A078898(5))) = A250469(A250469(1)) = 1.
Collecting the primes given by A020639 we get a multiset of factors: [2, 3, 3, 5]. Note that 2*3*3*5 = 90 = A250246(66).
If we start from n = 66, iterating the map n -> A302044(n) [instead of n -> A302042(n)] and apply A020639 to each term obtained we get just a single instance of each prime: [2, 3, 5]. Then by applying A302045 to the same terms we get the corresponding exponents (multiplicities) of those primes: [1, 2, 1].
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences generated by sieves

Crossrefs

Cf. A020639, A032742, A055396, A078898, A083221, A250245, A250246, A250469, A268674, A276151, A280496, A302032.

Cf. also following analogs: A302041 (omega), A253557 (bigomega), A302043, A302044, A302045 (exponent of the least prime present), A302046 (prime signature filter), A302050 (Moebius mu), A302051 (tau), A302052 (char.fun of squares), A302039, A302055 (Arith. derivative).

Programs

PARI

\\ Assuming A250469 and its inverse A268674 have been precomputed, then the following is fast enough:
A302042(n) = if(1==n,n,my(k=0); while((n%2), n = A268674(n); k++); n = n/2; while(k>0, n = A250469(n); k--); (n));

PARI

A020639(n) = if(n>1, if(n>n=factor(n, 0)[1, 1], n, factor(n)[1, 1]), 1); \\ From A020639
A078898(n) = if(n<=1,n, my(spf=A020639(n),k=1,m=n/spf); while(m>1,if(A020639(m)>=spf,k++); m--); (k));
\\ Faster if we precompute A078898 as an ordinal transform of A020639:
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; };
v078898 = ordinal_transform(vector(up_to,n,A020639(n)));
A078898(n) = v078898[n];
A302042(n) = if((1==n)||isprime(n),1,my(c = A078898(n), p = prime(-1+primepi(A020639(n))+primepi(A020639(c))), d = A078898(c), k=0); while(d, k++; if((1==k)||(A020639(k)>=p),d -= 1)); (k*p));

Formula

For n > 1, a(n) = A250469^(r)(A078898(n)), where r = A055396(n)-1 and A250469^(r)(n) stands for applying r times the map x -> A250469(x), starting from x = n.

a(n) = A250245(A032742(A250246(n))) = A250245(A280496(n)) = A250245(A250246(n) / A020639(n)).

a(n) = n - A302043(n).