Gordon Hamilton - cp's OEIS frontend

A387019 Cubefree numbers organized based on their prime factorization represented in base three.

1, 2, 4, 3, 6, 12, 9, 18, 36, 5, 10, 20, 15, 30, 60, 45, 90, 180, 25, 50, 100, 75, 150, 300, 225, 450, 900, 7, 14, 28, 21, 42, 84, 63, 126, 252, 35, 70, 140, 105, 210, 420, 315, 630, 1260, 175, 350, 700, 525, 1050, 2100, 1575, 3150, 6300, 49, 98, 196, 147, 294, 588, 441, 882, 1764

Offset: 0

Gordon Hamilton, Aug 13 2025

These are cubefree numbers organized by the highest factor. By converting to a different base, we avoid the row-by-row triangular entry used in the analogous squarefree A339195.

			a(11) = a(102_3) and so we get prime(3)^1 * prime(2)^0 * prime(1)^2 = 5^1 * 3^0 * 2^2 = 5 * 1 * 4 = 20.
a(33) = a(1020_3) and so we get prime(4)^1 * prime(3)^0 * prime(2)^2 * prime(1)^0 = 7^1 * 5^0 * 3^2 * 2^0 = 7 * 1 * 9 * 1 = 63.

Paolo Xausa, Table of n, a(n) for n = 0..10000

Cf. A004709, A339195.

A387019[n_] := Times @@ (Prime[Range[Length[#], 1, -1]]^#) & [IntegerDigits[n, 3]];
Array[A387019, 100, 0] (* Paolo Xausa, Sep 02 2025 *)

a(n) = {my(d = digits(n, 3), pr = primes(#d)); prod(i = 1, #d, pr[#d + 1 - i]^d[i])} \\ David A. Corneth, Aug 13 2025

a(n) = Product_{k=1..m} prime(k)^d(k) where d(m)d(m-1)...d(2)d(1) is the ternary representation of n. - David A. Corneth, Aug 19 2025

a(45) corrected and more terms from David A. Corneth, Aug 13 2025

A380455 Maximum number of prime polyomino factors of an n-polyomino.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 4, 1, 5, 1, 5, 1

Offset: 2

Gordon Hamilton, Jun 22 2025

Grade one students are capable of understanding "prime" and "composite" and "is a factor of" when presented in terms of polyomino tilings. Exploring these ideas is a 10/10 classroom activity, even if limited to only pentomino factors of larger polyominoes.

Conjectured terms: a(14)..a(17) = 10?, 3, 12?, 1.

			a(6) = 3 because the 2x3 rectangular hexomino can be tiled by three prime polyominoes:
The domino:
  XOY
  XOY
The bent tromino:
  XXO
  XOO
The straight tromino:
  XXX
  OOO
a(9) = 1 because no 9-polyomino can be tiled by both all straight and all bent trominoes.
It is conjectured that a(14) = 10 because this 14-polyomino can be tiled with 9 prime heptominoes and by the domino:
    X
   XXXX
   XXXX
   XXXX
     X
It is also conjectured that a(16) = 12 because this 16-polyomino can be tiled with twelve prime 8-polyominoes:
     XXX
    XXXXX
   XXXXX
    XXX

Cf. A342430 (number of prime polyominoes with n cells).

a(p) = 1 if p is prime. - Pontus von Brömssen, Jun 24 2025

A386243 a(n) is the smallest possible g(k) in a set of increasing numbers g(1) < g(2) < ... < g(k) having Frobenius number n.

Original entry on oeis.org

3, 5, 5, 7, 4, 7, 5, 9, 7, 9, 5, 9, 8, 11, 10, 11, 7, 13, 6, 9, 11, 10, 7, 13, 11, 11, 8, 13, 7, 13, 9, 13, 14, 13, 11, 13, 12, 13, 11, 17, 8, 17, 12, 17, 14, 16, 9, 19, 11, 17, 14, 17, 10, 13, 9, 17, 15, 17, 11, 18, 15, 19, 16, 15, 12, 16, 16, 18, 11, 17, 10, 19, 17, 17, 18, 18, 15

Offset: 1

Gordon Hamilton, Jul 16 2025

			a(15) = 10 because the set {6,7,10} has the Frobenius number of 15. No set of the form {..., 9} or {..., 8}, etc. has a Frobenius number of 15.

David A. Corneth, Table of n, a(n) for n = 1..415
David A. Corneth, All solutions for n = 1..415 where no number is a divisor of another number
Shunichi Matsubara, The Computational Complexity of the Frobenius Problem, arXiv:1602.05657, 2016. [Background information]
Wikipedia, Coin problem

Cf. A028387, A079326, A124506, A138984, A138985.

More terms from David A. Corneth, Jul 16 2025

A385829 Numbers k that are the largest k such that k cannot be partitioned into parts that are a set of at least two consecutive primes.

Original entry on oeis.org

1, 4, 7, 9, 13, 16, 23, 27, 30, 31, 35, 41, 42, 49, 53, 54, 59, 63, 64, 65, 66, 67, 79, 80, 83, 85, 95, 101, 102, 105, 107, 110, 113, 114, 116, 117, 119, 121, 125, 131, 135, 136, 138, 143, 145, 150, 160, 162, 163, 169, 174, 175, 178, 187, 191, 194, 197, 199, 200, 203

Offset: 1

Gordon Hamilton, Jul 09 2025

nonn

If we consider partitions into one distinct prime then no such largest number k exists.

			1 is a term as it is the largest positive integer that cannot be partitioned into parts 2 and 3. We have 2 = 2, 3 = 3 and so any positive integer at least two can be partitioned into parts 2 and 3.
30 is a term as 30 is the largest number that cannot be partitions into parts 7, 11 and 13. Proof:
30 cannot be written as a partition of 7, 11, 13 and we have 31 = 7 + 11 + 13, 32 = 3*7 + 11, 33 = 3*11, 34 = 3*7 + 13, 35 = 5*7, 36 = 2*7 + 2*11, 37 = 11 + 2*13 which proves that the next 7 positive integers after 30 can be partitioned into parts 7, 11, 13. Any larger number than that can have more sevens added.

David A. Corneth, Terms with their corresponding list of consecutive primes

Frobenius numbers for k successive primes: A037165 (k=2), A138989 (k=3), A138990 (k=4), A138991 (k=5), A138992 (k=6), A138993 (k=7), A138994 (k=8).

More terms from David A. Corneth, Jul 09 2025

A385811 Numbers k such that there exists a partition of the sum of prime factors of k (cf. A001414) into bigomega(k) (cf. A001222) prime parts where the product of parts is more than k.

Original entry on oeis.org

20, 21, 28, 33, 39, 40, 42, 44, 51, 52, 56, 57, 60, 63, 65, 66, 68, 69, 76, 78, 80, 84, 85, 87, 88, 92, 93, 95, 99, 100, 102, 104, 105, 111, 112, 114, 115, 116, 117, 119, 120, 123, 124, 126, 129, 130, 132, 133, 136, 138, 140, 141, 145, 147, 148, 152, 153, 155

Offset: 1

Gordon Hamilton, Jul 09 2025

nonn

As a temporary name, let's call these numbers "inefficient".
A number n is inefficient if there is a larger number N which has the same number of prime factors (counted with multiplicity) and the sum of the prime factors of n and N are the same.
The density of terms in the positive integers is 1.
This is a good sequence for students exploring prime factorization for the first time. When teaching elementary school students, refrain from divulging the rules upfront. Instead, seek high engagement by delivering a series of epic communal fails. I ask the students to give me an inefficient number, knowing they know nothing about what that means. When they supply "15" I'll show that 3*5 = 15 and 3+5 = 8 and complain that they've failed. When they do stumble upon a number like "28" then I use the opportunity to explain more of the rules.
Some numbers are noticeably absent from this list:
 - primes,
 - powers of primes,
 - numbers whose prime factors include only two consecutive prime numbers,
 - double any of the above three.

			60 is inefficient because its prime factors are 2,2,3,5. These factors add to 12. 81 is larger than 60 and also has its four prime factors adding to 12. 60 is therefore inefficient.
63 is inefficient because its prime factors are 3,3,7. These factors add to 13. 75 is larger than 63 and also has its three prime factors (3,5,5) adding to 13. 63 is therefore inefficient.

Cf. A001222, A001414, A385755, A385756.

More terms from Alois P. Heinz, Jul 09 2025

A384428 a(n) is the minimal area of a polyomino without holes having a product of edge lengths equal to n, or 0 if no solution is possible.

Original entry on oeis.org

1, 0, 4, 2, 7, 0, 10, 5, 3, 0, 16, 4, 19, 0, 6, 4, 25, 0, 28, 6, 9, 0, 34, 5, 5, 0, 6, 9, 43, 0, 46, 6, 15, 0, 8, 5, 55, 0, 18, 6, 61, 0, 64, 15, 8, 0, 70, 6, 7, 0

Offset: 1

Gordon Hamilton, May 28 2025

Good sequence for elementary school students learning multiplication.
If p is the largest prime factor dividing n, then a(n) >= p because there needs to be at least one edge of length k*p for some k>=1.
a(51) > 21. - Sean A. Irvine, Jun 13 2025

			a(36)=5 because the V pentomino is the smallest polyomino whose edges multiply together to give 36. The edges of the V pentomino are: 3,3,2,2,1,1.
   XXX
   X
   X
a(45)=8 because of the following polyomino with edges 5,3,3,1,1,1,1,1.
   XXXXX
   XX
   X

Sean A. Irvine, Java program (github)

Cf. A000104, A027709.

a(4*n+2) = 0.
a(p) = p + (p-1)/2 for any odd prime p.
a(p^2) = p for any prime p.

a(33)-a(50) from Sean A. Irvine, Jun 13 2025

A384327 Minimal Trips Around The Collatz Galaxy: a(n) is the minimal cycle length containing n. Each step in the cycle must be either to the next larger integer or follow a Collatz trajectory: k -> 3k+1 if k is odd or k -> k/2 if k is even.

Original entry on oeis.org

2, 2, 3, 3, 4, 4, 5, 4, 5, 4, 5, 5, 6, 5, 6, 4, 6, 6, 7, 6, 7, 5, 6, 6, 8, 7, 8, 5, 7, 7, 8, 8, 9, 6, 7, 7, 9, 9, 10, 6, 8, 8, 9, 9, 11, 7, 8, 8, 10, 10, 11, 7, 9, 9, 10, 10, 12, 8, 9, 9, 11, 11, 12, 8, 10, 10, 11, 11, 13, 9, 10, 10, 12, 12, 13, 9, 11, 11, 12

Offset: 1

Gordon Hamilton, May 26 2025

nonn

Excellent puzzle to get elementary students engaged in arithmetic. Recommended for use after the Collatz conjecture (3x+1 conjecture) is introduced side-by-side with its 3x-1 cousin.

In a tree of possible choices from n every positive integer occurs at most once which allows for lots of cutting in the search space. - David A. Corneth, May 26 2025

			a(25) is 8 because the shortest cycle containing 25 has 8 elements: 25 => 26 => 13 => 14 => 15 => 46 => 23 => 24 => 25.

David A. Corneth, Table of n, a(n) for n = 1..10000
David A. Corneth, list of n, a(n) and the (or a) shortest path of length a(n)
David A. Corneth, PARI program
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006370.

PARI
```
\\ See Corneth link
```

More terms from Alois P. Heinz, May 26 2025

A383020 G(n) is a graph constructed with nodes labelled with integers n through n+a(n). Edges are drawn between consecutive integers and between integers sharing the same largest prime factor. a(n) is the smallest integer for which G(n) is not planar.

Original entry on oeis.org

8, 7, 10, 10, 12, 11, 12, 12, 11, 14, 13, 12, 11, 13, 12, 11, 10, 14, 13, 15, 17, 16, 15, 17, 16, 15, 16, 19, 18, 18, 17, 16, 16, 15, 14, 17, 16, 15, 15, 14, 13, 13, 12, 15, 19, 18, 17, 16, 18, 19, 20, 22, 21, 22, 21, 23, 23, 22, 21, 24, 23, 22, 24, 23, 24, 23

Offset: 2

Gordon Hamilton, Apr 20 2025

nonn

			a(2) = 8: the graph G(2) has nodes labelled 2-10. Consecutive integers are connected by an edge. Also pairwise connected are: 3, 6, and 9 because they have 3 as the largest prime factor; 2, 4, and 8 because they have 2 as the largest prime factor; 5 and 10 because they have 5 as the largest prime factor. Nodes 2 and 10 are not connected because although 2 is a prime factor of 10, it is not the largest prime factor. This graph is non-planar. a(2) is larger than 7 because the nodes 2-9 make a planar graph. So a(2) = 8.

Pontus von Brömssen, Table of n, a(n) for n = 2..10000

Cf. A006530.

a(n) >= a(n-1) - 1. - Pontus von Brömssen, Apr 22 2025

a(15) corrected and a(41)-a(67) added by Pontus von Brömssen, Apr 21 2025

A382518 Let N = A001481(n), the n-th number that is the sum of two nonnegative squares. a(n) is the index of the first lattice-edge sequence that will accept N so that no sequence contains the edges of a triangle, otherwise if no such sequence exists, a(n) = 0.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 0, 1, 1, 2, 0, 0, 3, 0, 0, 1, 3, 1, 0, 2, 0, 0, 0, 0, 3, 4, 0, 0, 3, 0, 0, 1, 0, 4, 0, 1, 3, 0, 0, 2, 3, 0, 0, 0, 2, 0, 0, 0, 1, 4

Offset: 1

Gordon Hamilton, Mar 29 2025

a(n) is only defined where n is the sum of two nonnegative squares. a(n) = 0 is used in all cases where this is untrue.
Conjecture 1: bin #1 contains the orthogonal and 45-degree diagonal lattice edges.
Conjecture 2: After chessboard coloring the lattice, bin #3 contains only lattice edges that connect black and white points.

			Let's find a(13). a(13) corresponds to the lattice edge connecting {0,0} to {3,2} because 3^2 = 2^2 = 13. to find a(13) we must know all previous values.
a(1), a(2), a(4), a(8) and a(9) are all in bin#1. a(5) and a(10) are both in bin#2. a(13) cannot be in bin#1 because the lattice edges a(1), a(8) and a(13) make a triangle. a(13) cannot be in bin#2 because a(5), a(10) and a(13) form a triangle. a(13) can go into bin#3. a(13) = 3.
Let's find a(32). It goes into bin#1 because no combination of previous lattice edges added to that bin form a triangle that includes the lattice edge corresponding with a(32). a(32) = 1.

A001481 numbers that are the sum of two nonnegative squares.

A382109 uses the same technique on a cascade of Issai Schur additive sequences.

A382109 a(n) is the index of the first Issai Schur additive sequence that will accept n.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 1, 3, 3, 1, 3, 2, 1, 2, 3, 1, 4, 3, 1, 4, 2, 1, 2, 4, 1, 4, 4, 1, 4, 2, 1, 2, 4, 1, 3, 3, 1, 3, 2, 1, 2, 3, 1, 5, 4, 1, 5, 2, 1, 2, 5, 1, 5, 4, 1, 5, 2, 1, 2, 5, 1, 3, 3, 1, 3, 2, 1, 2, 3, 1, 5, 5, 1, 5, 2, 1, 2, 5, 1, 5, 5, 1, 5, 2, 1, 2, 5, 1, 3, 3

Offset: 1

Gordon Hamilton, Mar 24 2025

nonn

The Issai Schur additive sequences are greedily constructed so that no term in a sequence is the sum of two earlier terms in that sequence. Each positive integer is placed into exactly one additive sequence with priority given to the first sequence that will accept it.
This sequence can be explored by grade 2 students.
Junior high students can be asked what fraction of all numbers have a(n) = 1 (answer 1/3), a(n)=2 (answer 2/9)...
How does this greedy algorithm compare to the best possible results where we are trying to push the first occurrence of a(n) = k to happen for the largest possible n?

			Let's consider where the number 15 goes after the first 14 numbers have been placed:
Issai Schur additive sequence #1 {1,2,4,7,10,13}.
Issai Schur additive sequence #2 {3,5,6,12,14}.
Issai Schur additive sequence #3 {8,9,11}.
We always try to greedily add the next number in the lowest indexed sequence possible. "15" cannot go in #1 because 2+13 = 15. It cannot go in #2 because 3+12 = 15. It goes into #3 because no two distinct numbers in that additive sequence add to 15. So a(15) = 3
Where does 16 go? It can be added to #1, so a(16) = 1
Where does 17 go? It requires that we start a new additive sequence, #4, so a(17) = 4.

Andrew Howroyd, Table of n, a(n) for n = 1..10000

Cf. A033627 is Issai Schur additive sequence #1.

lista(n)={ my(a=vector(n), B=List(), L=List());
  for(n=1, n, my(k=1);
    while(k<=#L && bittest(L[k],n), k++);
    if(k>#L, listput(B,0); listput(L,0));
    a[n] = k; L[k] = bitor(L[k], B[k]<Andrew Howroyd, Mar 25 2025

a(45) onwards from Andrew Howroyd, Mar 25 2025

cp's OEIS Frontend

User: Gordon Hamilton

A387019 Cubefree numbers organized based on their prime factorization represented in base three.

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A380455 Maximum number of prime polyomino factors of an n-polyomino.

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A386243 a(n) is the smallest possible g(k) in a set of increasing numbers g(1) < g(2) < ... < g(k) having Frobenius number n.

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A385829 Numbers k that are the largest k such that k cannot be partitioned into parts that are a set of at least two consecutive primes.

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A385811 Numbers k such that there exists a partition of the sum of prime factors of k (cf. A001414) into bigomega(k) (cf. A001222) prime parts where the product of parts is more than k.

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A384428 a(n) is the minimal area of a polyomino without holes having a product of edge lengths equal to n, or 0 if no solution is possible.

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A384327 Minimal Trips Around The Collatz Galaxy: a(n) is the minimal cycle length containing n. Each step in the cycle must be either to the next larger integer or follow a Collatz trajectory: k -> 3k+1 if k is odd or k -> k/2 if k is even.

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A383020 G(n) is a graph constructed with nodes labelled with integers n through n+a(n). Edges are drawn between consecutive integers and between integers sharing the same largest prime factor. a(n) is the smallest integer for which G(n) is not planar.

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A382518 Let N = A001481(n), the n-th number that is the sum of two nonnegative squares. a(n) is the index of the first lattice-edge sequence that will accept N so that no sequence contains the edges of a triangle, otherwise if no such sequence exists, a(n) = 0.

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