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

1,4

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.

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

Index entries for sequences generated by sieves

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).

Example

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].

Prog

(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));

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).

Sequence in context: A291328 A280497 A280495 * A280498 A280496 A291327

Adjacent sequences: A302039 A302040 A302041 * A302043 A302044 A302045

Keyword

nonn

Author

Antti Karttunen, Mar 31 2018

Status

approved