cp's OEIS Frontend

a(n) <= A265125(n), a(n) < A265125(n) for n > 10. - Reinhard Zumkeller, Dec 02 2015

T(n,1) = A020639(n); T(n,A001222(n)) = n.

T(n,1) = A006530(n); T(n,A001222(n)) = n.

This is an enumeration of all partitions.

Technically this is an enumeration of all multisets (finite weakly increasing sequences of positive integers) rather than integer partitions. - Gus Wiseman, Dec 12 2016

A000040(a(n)) is a prime factor of A082288(n). - Reinhard Zumkeller, Feb 03 2008

Row n is the partition with Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product(p_j-th prime, j=1..r) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. For a given n, the 2nd Maple program yields row n; for example, we obtain at once B(2436) = [1,1,2,4,10]. - Emeric Deutsch, Jun 04 2015

From Emeric Deutsch, May 05 2015: (Start)

Number of entries in row n is bigomega(n) (i.e., the number of prime factors of n, multiplicities included).

Product of entries in row n = A003963(n).

Row n contains the Matula numbers of the rooted trees obtained from the rooted tree with Matula number n by deleting the edges emanating from the root. Example: row 8 is 1,1,1; indeed the rooted tree with Matula number 8 is \|/ and deleting the edges emanating from the root we obtain three one-vertex trees, having Matula numbers 1, 1, 1. Example: row 7 is 4; indeed, the rooted tree with Matula number 7 is Y and deleting the edges emanating from the root we obtain the rooted tree V, having Matula number 4.

The Matula (or Matula-Goebel) number of a rooted tree can be defined in the following recursive manner: to the one-vertex tree there corresponds the number 1; to a tree T with root degree 1 there corresponds the t-th prime number, where t is the Matula-Goebel number of the tree obtained from T by deleting the edge emanating from the root; to a tree T with root degree m >= 2 there corresponds the product of the Matula-Goebel numbers of the m branches of T. (End)

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

Also, the largest number of distinct integers such that all their pairwise differences are coprime to n. - Max Alekseyev, Mar 17 2006

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 least prime factor. - Daniel Forgues, Jul 05 2011

a(n) = least m > 0 for which n! + m and n - m are not relatively prime. - Clark Kimberling, Jul 21 2012

For n > 1, a(n) = the smallest k > 1 that divides n. - Antti Karttunen, Feb 01 2014

For n > 1, records are at prime indices. - Zak Seidov, Apr 29 2015

The initials "lpf" might be mistaken for "largest prime factor" (A009190), using "spf" for "smallest prime factor" would avoid this. - M. F. Hasler, Jul 29 2015

n = 89 is the first index > 1 for which a(n) differs from the smallest k > 1 such that (2^k + n - 2)/k is an integer. - M. F. Hasler, Aug 11 2015

From Stanislav Sykora, Jul 29 2017: (Start)

For n > 1, a(n) is also the smallest k, 1 < k <= n, for which the binomial(n,k) is not divisible by n.

Proof: (A) When k and n are relatively prime then binomial(n,k) is divisible by n because k*binomial(n,k) = n*binomial(n-1,k-1). (B) When gcd(n,k) > 1, one of its prime factors is the smallest; let us denote it p, p <= k, and consider the binomial(n,p) = (1/p!)*Product_{i=0..p-1} (n-i). Since p is a divisor of n, it cannot be a divisor of any of the remaining numerator factors. It follows that, denoting as e the largest e > 0 such that p^e|n, the numerator is divisible by p^e but not by p^(e+1). Hence, the binomial is divisible by p^(e-1) but not by p^e and therefore not divisible by n. Applying (A), (B) to all considered values of k completes the proof. (End)

From Bob Selcoe, Oct 11 2017, edited by M. F. Hasler, Nov 06 2017: (Start)

a(n) = prime(j) when n == J (mod A002110(j)), n, j >= 1, where J is the set of numbers <= A002110(j) with smallest prime factor = prime(j). The number of terms in J is A005867(j-1). So:

a(n) = 2 when n == 0 (mod 2);

a(n) = 3 when n == 3 (mod 6);

a(n) = 5 when n == 5 or 25 (mod 30);

a(n) = 7 when n == 7, 49, 77, 91, 119, 133, 161 or 203 (mod 210);

etc. (End)

For n > 1, a(n) is the leftmost term, other than 0 or 1, in the n-th row of A127093. - Davis Smith, Mar 05 2019

Meyers (see Guy reference) conjectures that for all r >= 1, the least odd number not in the set {a(i): i < prime(r)} is prime(r+1). - N. J. A. Sloane, Jan 08 2021

Meyers' conjecture would be refuted if and only if for some r there were such a large gap between prime(r) and prime(r+1) that there existed a composite c for which prime(r) < c < a(c) < prime(r+1), in which case (by Bertrand's postulate) c would necessarily be a term of A246281. - Antti Karttunen, Mar 29 2021

a(n) is odd for all n and for each odd m there exists a k with a(k) = m (see A064216). a(n) > n for n > 1: bijection between the odd and all numbers. - Reinhard Zumkeller, Sep 26 2001

a(n) and n have the same number of distinct primes with (A001222) and without multiplicity (A001221). - Michel Marcus, Jun 13 2014

From Antti Karttunen, Nov 01 2019: (Start)

More generally, a(n) has the same prime signature as n, A046523(a(n)) = A046523(n). Also A246277(a(n)) = A246277(n) and A287170(a(n)) = A287170(n).

Many permutations and other sequences that employ prime factorization of n to encode either polynomials, partitions (via Heinz numbers) or multisets in general can be easily defined by using this sequence as one of their constituent functions. See the last line in the Crossrefs section for examples.

(End)

MacMahon calls this the potency of n.

Downgrades the operators in a prime decomposition. E.g., 40 factors as 2^3 * 5 and sopfr(40) = 2 * 3 + 5 = 11.

Consider all ways of writing n as a product of zero, one, or more factors; sequence gives smallest sum of terms. - Amarnath Murthy, Jul 07 2001

a(n) <= n for all n, and a(n) = n iff n is 4 or a prime.

Look at the graph of this sequence. At the lower edge of the logarithmic scatterplot there is a set of fuzzy but unmistakable diagonal fringes, sloping toward the southeast. Their spacing gradually increases, and their slopes gradually decrease; they are more distinct toward the lower edge of the range. Is any explanation known? - Allan C. Wechsler, Oct 11 2015

For n >= 2, the glb and lub are: 3 * log(n) / log(3) <= a(n) <= n, where the lub occurs when n = 3^k, k >= 1. (Jakimczuk 2012) - Daniel Forgues, Oct 12 2015

Except for the initial term, row sums of A027746. - M. F. Hasler, Feb 08 2016

Atanassov proves that a(n) <= A065387(n) - n. - Charles R Greathouse IV, Dec 06 2016

From Robert G. Wilson v, Aug 15 2022: (Start)

Differs from A337310 beginning with n at 64, 192, 256, 320, 448, 512, ..., .

The number of terms which equal k is A000607(k).

The first occurrence of k>1 is A056240(k).

The last occurrence of k>1 is A000792(k).

The Amarnath Murthy comment of Jul 07 2001 is a result of the fundamental theorem of arithmetic.

(End)