A353516 The largest proper divisor of the primorial base exp-function, reduced modulo 4.

1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 3, 3, 1, 1, 3, 3, 3, 3, 1, 1, 3, 3, 3, 3, 1, 1, 3, 3, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 3, 3, 1, 1, 3, 3, 3, 3, 1

Offset: 0

Author

Antti Karttunen, Apr 24 2022

Keywords

• nonn

Comments

All terms are odd (1 or 3) because there can be at most one instance of prime factor 2 in terms of A276086, as it is a permutation of A048103.

Sequence A353487 interleaved with its right shifted version A353517 (= A353487(n-1)). Follows from the lemma below.

Lemma: for all n >= 0, a((2*n)+1) = a((2*n)+2). Because A276086(2n+1) = 2 * A276086(2n) and a(n) = A032742(A276086(n)) mod 4, it follows that a(2n+1) = A276086(2n) mod 4 = A353487(n). However, in general it does not hold as such that A324895((2*n)+2) = A276086(2*n), although in modulo 4 this equation holds, which can be seen by considering how the possible carry will propagate to the left in the primorial base system A049345. Consider first any even x which is a multiple of 6 or of the form 6k+2, with A049345(x) ending with either two zeros or as "...10". Adding 2 to x means that A049345(2+x) now ends either as "...10" or as "...20", and then A276086(2+x) = 3*A276086(x), with 3 the smallest prime factor present, so when it is eliminated by A324895, we are left with A276086(x) again. Then consider even numbers whose primorial expansion ends as "...20", but the third digit from the right is less than 4. E.g., 10 (= "120") or 16 (= "220") or 34 (= "1020"). Adding 2 to such numbers will change the 2 at the penultimate position to zero, and increment the digit at the third rightmost position by one. When mapped with A276086 this means that 9 (= 3^2) is eliminated, and one extra instance of 5 is multiplied to the prime factorization of the product form. However, this extra 5 will be now the smallest prime factor, so again it will be eliminated when we take the largest proper divisor as A324895 does. So in these cases A324895(x+2) = A276086(x)/9. If the third digit from the right is also "full", but the fourth digit is < 6 (i.e., the primorial expansion ends as "..420" but not as "..6420", e.g., numbers like 28, 58, 88, 118, 148, etc.), then A324895(x+2) = A276086(x)/5625, where 5625 = (3^2)*(5^4). Apart from the least significant digit at right, all the full digits in primorial base system that are changed to zeros by the propagating carry are even (because all primes are odd after 2), so when mapped by A276086 they all correspond to odd square factors that will be eliminated. But all odd squares are of the form 4m+1, so their elimination does not affect the result when the product is reduced modulo 4, from which the stated lemma follows. - Minor corrections from Antti Karttunen, Dec 04 2022

Links

• Antti Karttunen, Table of n, a(n) for n = 0..65537
• Index entries for sequences related to primorial base

Crossrefs

Cf. A010873, A048103, A276086, A324895, A353486, A353490, A353526.

Cf. A353487, A353517 (bisections).

Programs

PARI

A032742(n) = if(1==n,n,n/vecmin(factor(n)[,1]));
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A324895(n) = A032742(A276086(n));
A353516(n) = (A324895(n)%4);

Formula

a(n) = A010873(A324895(n)).

a(n) = A353490(A276086(n)).

For all n >= 0, a(2*n) = a((2*n)-1) for all n >= 1. [See the lemma in comments]

For all n >= 0, a(2n+1) = A353487(n) and a(2n) = A353517(n) [= A353487(n-1) for n > 0].