A376418 a(n) = n - A276086(A276085(n)), where A276085 and A276086 are primorial base log and exp-functions.

0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 7, 0, 0, 0, 5, 0, 0, 0, 6, 0, 0, 22, 7, 0, 0, 0, 14, 0, 0, 0, 31, 0, 0, 0, 10, 0, 0, 0, 11, 0, 0, 0, 43, 0, 0, 0, 13, 0, 44, 0, 14, 0, 0, 0, 15, 0, 0, 0, 59, 0, 0, 0, 17, 0, 0, 0, 62, 0, 0, 0, 19, 0, 0, 0, 35, 66, 0, 0, 21, 0, 0, 0, 22, 0, 0, 0, 23, 0, 0, 0, 86, 0, 0, 0, 25

Offset: 1

Author

Antti Karttunen, Nov 03 2024

Keywords

• nonn

Comments

All terms are nonnegative because for all n, x = A276086(A276085(n)) <= n, as any factor prime(i)^k || n (with k >= prime(i)) will propagate carries (in the image of fully additive A276085) towards more significant digit positions, which A276086 will convert back to the exponents of larger primes, but for each new instance of such larger prime present in x, enough instances of smaller primes in n have been eliminated (by the carry process) so that the net change of magnitude is negative, unless there are no such factors present at all in n (i.e., when n is a term of A048103), then A276086(A276085(n)) = n, and a(n) = 0.

This implies also that the least k for which A276085(k) = n is k = A276086(n).

There are several conspicuous patterns among the terms. For example, for n = 392, 3992, 39992, 399992, 3999992, ..., a(n) = 98, 998, 9998, 99998, 999998, ..., = n/4, but this holds only if n/8 is not in A100716, as generally, for all terms x that are in the intersection of A051062 and A168183 and x/8 is in A048103, it follows that a(x) = x/4. There are many other similar identities.

Differs from similar A376417 for the first time at n=625, 1250, 1875, 2500, 3125, 3375, 3750, 4375, 4500, 5000, 5625, ...

Examples

			a(4) = 1, as 4 = prime(1)^2, thus A276085(4) = 2 * A002110(1-1) = 2, and A276086(2) = prime(2) = 3, and 4-3 = 1.
a(625) = 0, as 625 = prime(3)^4, thus A276085(625) = 4 * A002110(3-1) = 4*6 = 24, and A276086(24) = prime(3)^4 [because A049345(24) = 400] = 625, and 625-625 = 0.
a(2500) = 625, as 2500 = 2^2 * 5^4 = prime(1)^2 * prime(3)^4, thus A276085(2500) = 2 * A002110(1-1) + 4 * A002110(3-1) = 2*1 + 4*6 = 26, but on the other hand, A276086(26) = prime(2) * prime(3)^4 [because A049345(26) = 410] = 3 * 5^4 = 1875, and 2500 - 1875 = 625.
a(3999999992) = 999999998, as 3999999992 = 2^3 * 691 * 723589 = prime(1)^3 * prime(125) * prime(58312), thus x = A276085(3999999992) = A002110(1-1) + A002110(2-1) + A002110(125-1) + A002110(58312-1), so A276086(x) = prime(1) * prime(2) * prime(125) * prime(58312), therefore a(3999999992) = (8-6)*prime(125)*prime(58312) = 3999999992/4 = 999999998. Note that A049345(8) = "110", as 8 = 6+2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences related to primorial base

Crossrefs

Cf. A049345, A048103 (indices of 0's), A100716 (of terms > 0), A276085, A276086, A376417.

Programs

PARI

A276085(n) = { my(f=factor(n)); sum(k=1, #f~, f[k, 2]*prod(i=1,primepi(f[k, 1]-1),prime(i))); };
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A376418(n) = (n - A276086(A276085(n)));