cp's OEIS Frontend

Version 1 of the "previous prime" function is "largest prime <= n". This produces A007917, the same sequence of numerical values, except the offset (or indexing) starts at 2 instead of 3.

Maple's "prevprime" function uses version 2.

See A007917 for references and further information.

Version 2 of the "next prime" function is "smallest prime > n". This produces A151800.

Maple uses version 2.

According to the "k-tuple" conjecture, a(n) is the initial term of the lexicographically earliest increasing arithmetic progression of n primes; the corresponding common differences are given by A061558. - David W. Wilson, Sep 22 2007

It is easy to show that the initial term of an increasing arithmetic progression of n primes cannot be smaller than a(n). - N. J. A. Sloane, Oct 18 2007

Also, smallest prime bounded by n and 2n inclusively (in accordance with Bertrand's theorem). Smallest prime >n is a(n+1) and is equivalent to smallest prime between n and 2n exclusively. - Lekraj Beedassy, Jan 01 2007

Run lengths of successive equal terms are given by A125266. - Felix Fröhlich, May 29 2022

Conjecture: if n > 1, then a(n) < n^(n^(1/n)). - Thomas Ordowski, Feb 23 2023

a((A003961(n) + 1) / 2) = n and A003961(a(n)) = 2*n - 1 for all n. If the sequence is indexed by odd numbers only, it becomes multiplicative. In this variant sequence, denoted b, even indices don't exist, and we get b(1) = a(1) = 1, b(3) = a(2) = 2, b(5) = 3, b(7) = 5, b(9) = 4 = b(3) * b(3), ... , b(15) = 6 = b(3) * b(5), and so on. This property can also be stated as: a(x) * a(y) = a(((2x - 1) * (2y - 1) + 1) / 2) for x, y > 0. - Reinhard Zumkeller [re-expressed by Peter Munn, May 23 2020]

Not multiplicative in usual sense - but letting m=2n-1=product_j (p_j)^(e_j) then a(n)=a((m+1)/2)=product_j (p_(j-1))^(e_j). - Henry Bottomley, Apr 15 2005

From Antti Karttunen, Jul 25 2016: (Start)

Several permutations that use prime shift operation A064989 in their definition yield a permutation obtained from their odd bisection when composed with this permutation from the right. For example, we have:

A243505(n) = A122111(a(n)).

A243065(n) = A241909(a(n)).

A244153(n) = A156552(a(n)).

A245611(n) = A243071(a(n)).

(End)

Rows equal columns and are periodic. No zero elements are found (conjecture). The recurrence is related to the recurrence for the Mahonian numbers. The main diagonal is the Dirichlet inverse of the Euler totient function A023900 (conjecture). [The 2nd and 3rd formulas state that the conjecture is correct. R. J. Mathar, Sep 16 2017]

The sums from n=1 to infinity of T(n,k)/n converge to the Mangoldt function for column k (conjecture).

If gcd(n,k)=1 then T(n,k)=1 and if T(n,k)=1 then gcd(n,k)=1 (conjecture).

The Dirichlet generating functions for s > 1 for the columns appear to be (see A054535):

Zeta(s)*(1 + (Sum over all possible combinations of products of negative distinct prime factors of k, up to rearrangement, 1/((-1* first distinct prime factor)*(-1*second distinct prime factor)*(-1*third distinct prime factor * ...))^(s-1))).

Examples:

k=1: Zeta(s)

k=2: Zeta(s)*(1 - 1/2^(s-1))

k=3: Zeta(s)*(1 - 1/3^(s-1))

k=4: Zeta(s)*(1 - 1/2^(s-1))

k=5: Zeta(s)*(1 - 1/5^(s-1))

k=6: Zeta(s)*(1 - 1/2^(s-1) - 1/3^(s-1) + 1/6^(s-1))

k=7: Zeta(s)*(1 - 1/7^(s-1))

...

k=30: Zeta(s)*(1 - 1/2^(s-1) - 1/3^(s-1) - 1/5^(s-1) + 1/6^(s-1) + 1/10^(s-1) + 1/15^(s-1) - 1/30^(s-1))

...

(conjecture)

See triangle A142971 for negative distinct prime factors.

This could probably be checked by matrix multiplication.

The signs of the eigenvalues of this matrix are a rearrangement of the Mobius function A008683 (conjecture). The first few eigenvalues are:

{1.0000}

{-1.4142, 1.4142}

{-2.6554, 1.8662, -1.2108}

{-3.4393, 2.1004, -1.6611, 0}

{-4.7711, -3.3867, 2.5910, -1.4332, 0}

{-5.2439, -3.4641, 3.4641, 2.5169, -2.2730, 0}

The relation to Dirichlet characters for the entries in this matrix appears appears to be formulated in terms of the sequence A089026 which is equal to n if n is a prime, otherwise equal to 1. See Mathematica program below. [Mats Granvik, Nov 23 2013]

From Mats Granvik, Jun 19 2016: (Start)

Remark about the Dirichlet generating function for the whole matrix: Subtracting the first column (in the form of zeta(c)) of the matrix gives us the limit: lim_{c->1} zeta(s)*zeta(c)/zeta(c+s-1)-zeta(c) = -zeta'(s)/zeta(s) which is the classical Dirichlet generating function for the von Mangoldt function.

For n >= k, see A231425, this matrix has row sums equal to zero except for the first row:

1=1

1-1=0

1+1-2=0

1-1+1-1=0

1+1+1+1-4=0

...

log(A014963(n)) = Sum_{k>=1} A191898(n,k)/k, for n>1.

log(A014963(k)) = Sum_{n>=1} A191898(n,k)/n, for k>1.

log(A014963(n)) = limit of zeta(s)*(Sum_{d divides n} A008683(d)/d^(s-1)) as s->1, for n>1.

A008683(n) = Sum_{k=1..n} A191898(n,k)*exp(-i*2*Pi*k/n)/n.

A008683(n) = Sum_{n=1..k} A191898(n,k)*exp(-i*2*Pi*n/k)/k.

(End)

From Mats Granvik, Aug 09 2016: (Start)

The Dirichlet generating function for the matrix is zero for any pair c and s = 2 - c and any pair s and c = 2 - s except at the pole c = 1 and s = 1 where it is indeterminate.

In the Mathematica program section, in the expression for the matrix as Dirichlets characters, the variables s and c can apparently be any pair of positive integers.

Limits related to the Dirichlet generating function for the matrix: Let s = ZetaZero(n), then lim_{c->1} zeta(s*c)/zeta(c+s-1) = ZetaZero(n). Let s = ZetaZero(n), then lim_{c->1} zeta(s*c)/zeta(c+s*c-1) = ZetaZero(n)/(1+ZetaZero(n)).

(End)