A383614 The unique sequence such that Sum_{d|n} d*a(d)^(n/d) = sigma(n)^2 for every n.

1, 4, 5, 4, 7, -10, 9, -44, -23, -197, 13, -845, 15, -2340, -701, -9164, 19, -31578, 21, -124979, -11355, -381326, 25, -1778580, -3323, -5162265, -212899, -21915630, 31, -70256029, 33, -311369996, -4439583, -1010580635, -129393, -4135827284, 39, -14467258386

Offset: 1

Author

Yifan Xie, May 02 2025

Keywords

• sign
• easy

Comments

Replace the sequence A072861 on the right-hand side of the equation with any integer sequence. It can be proved that the resulting sequence {a(n)} contain only integer terms if and only if for any prime p and positive integer n such that val(n, p) = k, p^k divides s(n) - s(n/p). The simplest sequence satisfying this property is A000203, and the resulting sequence {a(n)} is the constant sequence of 1's.

In 2025 China Team Selection Test, Test 4, Day 1, Problem 1, this sequence gives {z(n)} when {x(n)} and {y(n)} are constant sequences of 1's.

Examples

			For n = 1, the equation gives a(1) = sigma(1)^2 = 1;
For n = 6, the equation gives 1*1^6 + 2*4^3 + 3*5^2 + 6*a(6) = sigma(6)^2 = 144, so a(6) = -10.
		

Links

• Art of Problem Solving, 2025 China Team Selection Test, Test 4, Day 1, Problem 1

Crossrefs

Cf. A000203, A072861.

Programs

PARI

lista(nn) = {my(v=vector(nn)); v[1] = 1; for(n=2, nn, s = sigma(n)^2; fordiv(n, d, s -= d*v[d]^(n/d)); v[n]=s/n); v}

Formula

For prime p, a(p) = p + 2.