A359953 a(1) = 0, a(2) = 1. For n >= 3, if the greatest prime dividing n is greater than the greatest prime dividing n-1, then a(n) = a(n-1) + 1. Otherwise a(n) = a(n-1) - 1.

0, 1, 2, 1, 2, 1, 2, 1, 2, 3, 4, 3, 4, 3, 2, 1, 2, 1, 2, 1, 2, 3, 4, 3, 4, 5, 4, 5, 6, 5, 6, 5, 6, 7, 6, 5, 6, 5, 4, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 3, 4, 3, 4, 3, 4, 5, 6, 5, 6, 5, 4, 3, 4, 3, 4, 3, 4, 3, 4, 3, 4, 3, 2, 3, 2, 3, 4, 3, 2, 3, 4, 3

Offset: 1

Author

Tamas Sandor Nagy, Jan 19 2023

Keywords

• easy
• sign

Comments

The first negative value is at a(3888). Within the first 1000000 values are the negative record values a(n) = -4 at n = {3913, 3915, 3927, 3933}. - Thomas Scheuerle, Jan 20 2023

Examples

			a(5) = a(4) + 1 = 1 + 1 = 2 because A006530(5) = 5 > A006530(4) = 2.
		

Crossrefs

Cf. A000040, A006530, A070089, A087429.

Programs

MATLAB

function a = A359953(max_n)
    a = [0 cumsum(sign(diff([0 arrayfun(@(x)(max(factor(x))),[2:max_n])])))];
end % Thomas Scheuerle, Jan 20 2023

Mathematica

Join[{0}, Accumulate@ Sign@ Differences@ Table[FactorInteger[n][[-1, 1]], {n, 1, 100}]] (* Amiram Eldar, Jan 20 2023, after the MATLAB code *)

PARI

lista(nn) = my(va = vector(nn)); va[1] = 0; va[2] = 1; for (n=3, nn, if (vecmax(factor(n)[,1]) > vecmax(factor(n-1)[,1]), va[n] = va[n-1] + 1, va[n] = va[n-1] - 1);); va; \\ Michel Marcus, Jan 31 2023

Formula

For n >= 2, if A006530(n) > A006530(n-1), then a(n) = a(n-1) + 1; a(n) = a(n-1) - 1 otherwise.

a(n) = (-1)*Sum_{i=1..n-1} (-1)^A087429(i).

a(1 + A070089(n)) = 1 + a(A070089(n)). - Thomas Scheuerle, Jan 20 2023