A070089 -id:A070089 - cp's OEIS frontend

A100387 a(n) is the largest number x such that for m=n to n+x-1, A006530(m) decreases.

1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 4, 3, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 4, 3, 2, 1, 2, 1, 3, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 4, 3, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 3, 2, 1, 2, 1, 1, 3, 2, 1, 1, 2, 1, 1, 3, 2, 1, 2, 1, 1, 1, 1, 3, 2, 1, 2, 1, 2, 1, 2, 1, 3, 2, 1

Offset: 2

Labos Elemer, Dec 10 2004

nonn

A006530(m) is the largest prime factor of m.

			a(13)=4 because the largest prime factors of 13,14,15,16 are 13,7,5,2; but A006530(17)=17.

Cf. A006530, A070087, A070089, A071870, A079748, A100376, A100385, A100386.

Mathematica
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From Pontus von Brömssen, Nov 09 2022: (Start)
a(n) = 1 if and only if n >= 2 and n is a term of A070089.
If a(n) > 1 then a(n) = a(n+1)+1.
(End)

Edited by Don Reble, Jun 13 2007

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.

Original entry on oeis.org

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

Tamas Sandor Nagy, Jan 19 2023

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

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

Cf. A000040, A006530, A070089, A087429.

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

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

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

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

A334073 Decimal expansion of Sum_{k >= 1} e(k)/2^k, where e(k) = 1 if gpf(k+1) > gpf(k) and 0 otherwise, and gpf(k) is the greatest prime dividing k (A006530).

Original entry on oeis.org

8, 3, 5, 2, 2, 5, 9, 2, 2, 4, 2, 0, 5, 2, 4, 5, 9, 4, 3, 4, 8, 7, 8, 2, 9, 8, 0, 5, 7, 5, 1, 7, 6, 2, 4, 1, 1, 9, 4, 0, 4, 3, 3, 1, 7, 1, 0, 5, 3, 2, 5, 3, 6, 6, 9, 4, 3, 8, 9, 1, 5, 7, 5, 3, 1, 5, 9, 3, 0, 3, 1, 8, 5, 7, 9, 4, 0, 5, 1, 0, 5, 3, 3, 8, 3, 3, 5

Offset: 0

Amiram Eldar, Apr 13 2020

This constant is irrational (Erdős and Pomerance, 1978).

It is assumed that gpf(1) = A006530(1) = 1.

			0.83522592242052459434878298057517624119404331710532...

Paul Erdős and Carl Pomerance, On the largest prime factors of n and n + 1, Aequationes mathematicae, Vol. 17, No. 1 (1978), pp. 115-115, alternative link.

Cf. A006530, A070089.

gpf[n_] := FactorInteger[n][[-1, 1]]; e[n_] := Boole[gpf[n+1] > gpf[n]]; RealDigits[Sum[e[n]/2^n, {n, 1, 500}], 10, 100][[1]]

cp's OEIS Frontend

A100387 a(n) is the largest number x such that for m=n to n+x-1, A006530(m) decreases.

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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.

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A334073 Decimal expansion of Sum_{k >= 1} e(k)/2^k, where e(k) = 1 if gpf(k+1) > gpf(k) and 0 otherwise, and gpf(k) is the greatest prime dividing k (A006530).

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