cp's OEIS Frontend

For all i, j:

A305800(i) = A305800(j) => a(i) = a(j),

a(i) = a(j) => A001222(i) = A001222(j),

a(i) = a(j) => A322826(i) = A322826(j).

The initial term a(1)=1 is purely conventional: The unit 1 is not a prime number, although it has been considered so in the past. 1 is the empty product of prime numbers, thus 1 has no largest prime factor. - Daniel Forgues, Jul 05 2011

Greatest noncomposite number dividing n, (cf. A008578). - Omar E. Pol, Aug 31 2013

Conjecture: Let a, b be nonzero integers and f(n) denote the maximum prime factor of a*n + b if a*n + b <> 0 and f(n)=0 if a*n + b=0 for any integer n. Then the set {n, f(n), f(f(n)), ...} is finite of bounded size. - M. Farrokhi D. G., Jan 10 2021

From Antti Karttunen, May 12 2014: (Start)

a(A003961(n)) = n for all n. [This is a left inverse function for the injection A003961.]

Bisections are A064216 (the terms at odd indices) and A064989 itself (the terms at even indices), i.e., a(2n) = a(n) for all n.

(End)

From Antti Karttunen, Dec 18-21 2014: (Start)

When n represents an unordered integer partition via the indices of primes present in its prime factorization (for n >= 2, n corresponds to the partition given as the n-th row of A112798) this operation subtracts one from each part. If n is of the form 2^k (a partition having just k 1's as its parts) the result is an empty partition (which is encoded by 1, having an "empty" factorization).

For all odd numbers n >= 3, a(n) tells which number is located immediately above n in square array A246278. Cf. also A246277.

(End)

Alternatively, if numbers are represented as the multiset of indices of prime factors with multiplicity, this operation subtracts 1 from each element and discards the 0's. - M. F. Hasler, Dec 29 2014

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.

From Antti Karttunen, Jun 27 2014: (Start)

The odd bisection (containing even terms) halved gives A244153.

The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.

(End)

Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

Factor n; replace each prime(i) with i, take the conjugate partition, replace parts i with prime(i) and multiply out.

From Antti Karttunen, May 12-19 2014: (Start)

For all n >= 1, A001222(a(n)) = A061395(n), and vice versa, A061395(a(n)) = A001222(n).

Because the partition conjugation doesn't change the partition's total sum, this permutation preserves A056239, i.e., A056239(a(n)) = A056239(n) for all n.

(Similarly, for all n, A001221(a(n)) = A001221(n), because the number of steps in the Ferrers/Young-diagram stays invariant under the conjugation. - Note added Apr 29 2022).

Because this permutation commutes with A241909, in other words, as a(A241909(n)) = A241909(a(n)) for all n, from which follows, because both permutations are self-inverse, that a(n) = A241909(a(A241909(n))), it means that this is also induced when partitions are conjugated in the partition enumeration system A241918. (Not only in A112798.)

(End)

From Antti Karttunen, Jul 31 2014: (Start)

Rows in arrays A243060 and A243070 converge towards this sequence, and also, assuming no surprises at the rate of that convergence, this sequence occurs also as the central diagonal of both.

Each even number is mapped to a unique term of A102750 and vice versa.

Conversely, each odd number (larger than 1) is mapped to a unique term of A070003, and vice versa. The permutation pair A243287-A243288 has the same property. This is also used to induce the permutations A244981-A244984.

Taking the odd bisection and dividing out the largest prime factor results in the permutation A243505.

Shares with A245613 the property that each term of A028260 is mapped to a unique term of A244990 and each term of A026424 is mapped to a unique term of A244991.

Conversely, with A245614 (the inverse of above), shares the property that each term of A244990 is mapped to a unique term of A028260 and each term of A244991 is mapped to a unique term of A026424.

(End)

The Maple program follows the steps described in the first comment. The subprogram C yields the conjugate partition of a given partition. - Emeric Deutsch, May 09 2015

The Heinz number of the partition that is conjugate to the partition with Heinz number n. The Heinz number of a partition p = [p_1, p_2, ..., p_r] is defined as Product(p_j-th prime, j=1...r). Example: a(3) = 4. Indeed, the partition with Heinz number 3 is [2]; its conjugate is [1,1] having Heinz number 4. - Emeric Deutsch, May 19 2015

It seems that a(n) = Max_{j=n+1..2n-1} gcd(n,j). - Labos Elemer, May 22 2002

This is correct: No integer in the range [n+1, 2n-1] has n as its divisor, but certainly at least one multiple of the largest proper divisor of n will occur there (e.g., if it is n/2, then at n + (n/2)). - Antti Karttunen, Dec 18 2014

The slopes of the visible lines made by the points in the scatter plot are 1/2, 1/3, 1/5, 1/7, ... (reciprocals of primes). - Moosa Nasir, Jun 19 2022

a(n) tells how many iterations of A252463 are needed before 1 is reached, i.e., the distance of n from 1 in binary trees like A005940 and A163511.

Similarly for A253553 in trees A253563 and A253565. - Antti Karttunen, Apr 14 2019