Yifan Xie - cp's OEIS frontend

A386376 a(n) is the smallest integer k such that b(n,k) is squarefree, where b(n,1) = n and b(n, k+1) = b(n,k) + rad(b(n, k)) for k >= 1.

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

Offset: 1

Yifan Xie, Aug 20 2025

rad(k) = A007947(k) is the largest squarefree integer dividing n.
Proof of a(n)'s existence: (Start)
Let c(n,k) = b(n,k) / rad(b(n,k)). Suppose that c(n,k) > 1 for all positive integers k. Pick k such that c(n,k) = m is the smallest. Let p be the smallest prime not dividing b(n,k). There exists an integer 1 <= i' <= p-1 such that p divides m + i'. So there exists a smallest integer 1 <= i <= i' such that m + i has a prime divisor not dividing b(n,k). From the minimality of i, for 1 <= j <= i-1, c(n,k+j) = m + j does not produce prime divisors not dividing b(n,k).
From the choice of i, there exists a prime q >= p such that q does not divide b(n,k) but q divides m + i. Then c(n,k+i) = (rad(b(n,k+i-1) * (c(n,k+i-1) + 1)) / rad(rad(b(n,k+i-1) * (c(n,k+i-1) + 1)) <= (c(n,k+i-1) + 1) / q <= (m + i) / p <= (m + p - 1) / p < m, because m > 1, a contradiction from the minimality of m. (End)

			For n = 9, b(9,1) = 9, b(9,2) = 9 + rad(9) = 12, b(9,3) = 12 + rad(12) = 18, b(9,4) = 18 + rad(18) = 24, b(9,5) = 24 + rad(24) = 30 is squarefree, so a(9) = 5.

Cf. A005117, A007947.

rad[m_]:=Times@@(First@# & /@ FactorInteger@ m);a[n_]:=Module[{k=1,b=n},While[!SquareFreeQ[b],k++;b=b+rad[b]];k];Array[a,87] (* James C. McMahon, Aug 25 2025 *)

A007947(n) = factorback(factorint(n)[, 1]);
a(n) = {my(cnt = 1, x = n); while(x != A007947(x), x += A007947(x); cnt++); cnt};

A385675 For positive integers n, a multiset A of positive integers is called "n-good" if for any number 1 <= i <= n, we can find a submultiset B of A such that the sum of B is equal to i. a(n) is the number of minimal n-good multisets.

Original entry on oeis.org

1, 2, 3, 7, 9, 18, 27, 47, 62, 101, 140, 226, 301, 437, 579, 838, 1077, 1525, 1985, 2721, 3470, 4674, 5899, 7843, 9773, 12703, 15803, 20431, 25129, 32167, 39519, 49982, 60928, 76373, 92537, 115313, 138969, 171372, 205847, 252604, 301444, 367890, 438145, 531202, 630209

Offset: 1

Yifan Xie, Aug 05 2025

nonn

An n-good multiset A is minimal if it is impossible to get another n-good multiset by deleting one element from A.

An n-good multiset must contain 1 and have a sum of elements >= n. - Michael S. Branicky, Aug 13 2025

			For n = 6, {1, 2, 3} and {1, 1, 3, 6} are minimal n-good multisets.

Cf. A126796.

# for illustrative purposes
from functools import cache
from itertools import chain, combinations, combinations_with_replacement as cwr
def f(t): return tuple(e for e in t if e != 0)
def powerset(s): return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
@cache
def good(n, A): return 1 in A and sum(A) >= n and set(range(1, n+1))-set(sum(B) for B in powerset(A)) == set()
def a(n): return sum(good(n, A) and all(not good(n, A[:i]+A[i+1:]) for i in range(len(A))) for A in map(f, cwr(range(n+1), n)))
print([a(n) for n in range(1, 11)]) # Michael S. Branicky, Aug 13 2025

A386820 a(n) is the size of largest subset of {1, 1/2, ..., 1/n} that can be partitioned into two parts, the sum of elements of which are equal.

Original entry on oeis.org

0, 0, 0, 0, 0, 4, 4, 4, 4, 4, 4, 6, 6, 6, 9, 9, 9, 11, 11, 11, 14, 14, 14

Offset: 1

Yifan Xie, Aug 03 2025

			For a(6) = 4, the set {1, 1/2, 1/3, 1/6} is chosen because 1 = 1/2 + 1/3 + 1/6.
The two parts for a(18) = 11 are 1 + 1/5 + 1/6 + 1/15 = 1/2 + 1/3 + 1/4 + 1/9 + 1/10 + 1/12 + 1/18.
The two parts for a(20) = 11 are 1 + 1/9 + 1/10 + 1/15 + 1/18 = 1/2 + 1/3 + 1/5 + 1/6 + 1/12 + 1/20.

Art of Problem Solving, 2020 GQMO Problem 4, which shows that a(n) >= 2*n/5 for sufficiently large n.
Vincent Jugé, Proof that a(n) > c*n for all real numbers c < 1 and sufficiently large n

Cf. A001008, A098181.

a(n) = a(n-1) if n/k = p is a prime and p > A001008(k).

A385366 a(n) = Sum_{permutations p of [n]} des(p^2), where des(p) is the number of descents of p.

Original entry on oeis.org

0, 0, 2, 24, 192, 1560, 13680, 131040, 1370880, 15603840, 192326400, 2554675200, 36404121600, 554204851200, 8979363993600, 154305575424000, 2803653844992000, 53708801642496000, 1082001156268032000, 22869278876860416000, 506043617700741120000, 11699825757321461760000

Offset: 1

Yifan Xie, Jun 26 2025

			For the permutation p = (2, 3, 4, 1), p^2 = (3, 4, 1, 2), and des(p) = des(p^2) = 1 (because 4 > 1).

Paolo Xausa, Table of n, a(n) for n = 1..445

Cf. A001286.

A385366[n_] := If[n <= 2, 0, (n - 1)!*(n^2 - n - 4)/2];
Array[A385366, 25] (* Paolo Xausa, Jul 14 2025 *)

PARI
```
a(n)=if(n>2,(n-1)!*(n^2-n-4)/2, 0);
```

a(n) = 0 if n <= 2; a(n) = (n-1)!*(n^2-n-4)/2 if n >= 3.

A384424 The maximal possible number of 'good' steps in a Hamiltonian cycle on the n X n king's graph, as is specified in the comments.

Original entry on oeis.org

0, 0, 5, 8, 16, 24, 36, 44

Offset: 1

Yifan Xie, May 28 2025

The cycle is drawn on an n X n square grid. Denote the geometric center of the grid by O. An edge a -> b on the cycle is 'good' if the distance from b to O is strictly less than the distance from a to O.

The n = 8 case is Grade 9, Problem 4 of 2025 All-Russian Olympiad.

			Proof that a(6) <= 24: Mark the cells on a 6 X 6 grid with the following symbols:
  ABXXBA
  BXXXXB
  XXOOXX
  XXOOXX
  BXXXXB
  ABXXBA
The 4 steps to A and the 4 steps from O must be non-'good'. For each A, the 2 steps to the neighboring B's cannot both be 'good', or they must both be A -> B. So there are at least 4 + 4 + 4 = 12 non-'good' steps, hence a(6) <= 24.

Yifan Xie, Illustration of a(n), n = 2..8, 'good' steps are green, the others are blue and the red circles are the centers of the grids (the construction for n = 6 from Sean A. Irvine).

Cf. A308129.

A383622 First differences of A383621.

Original entry on oeis.org

1, 4, 5, 7, 11, 13, 14, 17, 19, 20, 23, 25, 28, 29, 31, 35, 37, 41, 43, 44, 46, 47, 49, 52, 53, 55, 59, 61, 65, 67, 68, 70, 71, 73, 76, 77, 79, 83, 85, 89, 91, 92, 95, 97, 98, 100, 101, 103, 107, 109, 113, 115, 116, 119, 121, 124, 125, 127, 131, 133, 137, 139, 140, 143

Offset: 1

Yifan Xie, May 10 2025

The multiplication table T(i,j) = A027649(i) * A007310(j) for i >= 0 and j >= 1, sorted in ascending order. There are repeating terms: 322 appears twice.

Yifan Xie, Table of n, a(n) for n = 1..10000
Yifan Xie, On a solvable minimal sum antichain problem

Cf. A001047, A007310, A027649, A383621.

A027649(n) = 2*3^n-2^n;
upto(nn) = {v=[]; for(n=0, logint(nn,3), d = A027649(n); m = floor(nn/d); for(i=0, floor(m/6), if(6*i+1 <= m, v=concat(v, d*(6*i+1))); if(6*i+5 <= m, v=concat(v, d*(6*i+5))))); v=vecsort(v); v};

a(n) ~ c*n, where c = 3*(Sum_{k=0..oo} 1/A027649(k))^(-1) = 2.216821...

A383621 a(n) is the minimum possible value of x_1 + x_2 + ... + x_n where x_1, x_2, ..., x_n are positive integers such that x_i does not divide x_j for any i != j.

Original entry on oeis.org

1, 5, 10, 17, 28, 41, 55, 72, 91, 111, 134, 159, 187, 216, 247, 282, 319, 360, 403, 447, 493, 540, 589, 641, 694, 749, 808, 869, 934, 1001, 1069, 1139, 1210, 1283, 1359, 1436, 1515, 1598, 1683, 1772, 1863, 1955, 2050, 2147, 2245, 2345, 2446, 2549, 2656, 2765, 2878

Offset: 1

Yifan Xie, May 10 2025

The sequence is the solution to Problem 9 of 2022 Chinese Team Selection Test. - Yifan Xie, Jun 27 2025

			For n <= 6, the construction is given by the n smallest primes.
For n = 7, the numbers 4, 5, 6, 7, 9, 11, 13 are mutually indivisible and their sum is a(7) = 55.

Yifan Xie, Table of n, a(n) for n = 1..10000
Yifan Xie, On a solvable minimal sum antichain problem

Cf. A027649.

Partial sums of A383622.

A027649(n) = 2*3^n-2^n;
A383622(nn) = {my(v=[]); for(n=0, logint(nn,3), d = A027649(n); m = floor(nn/d); for(i=0, floor(m/6), if(6*i+1 <= m, v=concat(v, d*(6*i+1))); if(6*i+5 <= m, v=concat(v, d*(6*i+5))))); v=vecsort(v); v};
lista(nn) = {u = A383622(3*nn); my(v=vector(nn)); s=0; for(n=1, nn, s = s + u[n]; v[n] = s); v};

a(n) ~ c*n^2, where c = (3/2)*(Sum_{k=0..oo} 1/A027649(k))^(-1) = 1.108410...

A383614 The unique sequence such that Sum_{d|n} d*a(d)^(n/d) = sigma(n)^2 for every n.

Original entry on oeis.org

1, 4, 5, 4, 7, -10, 9, -44, -23, -197, 13, -845, 15, -2340, -701, -9164, 19, -31578, 21, -124979, -11355, -381326, 25, -1778580, -3323, -5162265, -212899, -21915630, 31, -70256029, 33, -311369996, -4439583, -1010580635, -129393, -4135827284, 39, -14467258386

Offset: 1

Yifan Xie, May 02 2025

Replace the sequence A072861 on the right-hand side of the equation with any integer sequence. It can be proved that the resulting sequence {a(n)} contain only integer terms if and only if for any prime p and positive integer n such that val(n, p) = k, p^k divides s(n) - s(n/p). The simplest sequence satisfying this property is A000203, and the resulting sequence {a(n)} is the constant sequence of 1's.

In 2025 China Team Selection Test, Test 4, Day 1, Problem 1, this sequence gives {z(n)} when {x(n)} and {y(n)} are constant sequences of 1's.

			For n = 1, the equation gives a(1) = sigma(1)^2 = 1;
For n = 6, the equation gives 1*1^6 + 2*4^3 + 3*5^2 + 6*a(6) = sigma(6)^2 = 144, so a(6) = -10.

Art of Problem Solving, 2025 China Team Selection Test, Test 4, Day 1, Problem 1

Cf. A000203, A072861.

lista(nn) = {my(v=vector(nn)); v[1] = 1; for(n=2, nn, s = sigma(n)^2; fordiv(n, d, s -= d*v[d]^(n/d)); v[n]=s/n); v}

For prime p, a(p) = p + 2.

A383194 The least number of times that b(k) = 2*k - 1 for the first n terms of a "mean-central" sequence, as is defined in A383192.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 10, 10, 10, 10, 10, 11, 11

Offset: 1

Yifan Xie, Apr 21 2025

Problem 2 of the 2025 European Girls' Mathematical Olympiad implies that a(n) -> oo as n -> oo.

Art of Problem Solving, European Girls' Mathematical Olympiad 2025 Problem 2
Yifan Xie, Python program

Cf. A383192, A383193.

A383193 The lexicographically earliest "mean-central" sequence, as is defined in A383192.

Original entry on oeis.org

1, 3, 5, 6, 10, 11, 12, 13, 19, 20, 21, 23, 25, 26, 27, 28, 36, 37, 38, 39, 41, 42, 46, 47, 49, 51, 53, 55, 56, 57, 58, 59, 69, 70, 71, 72, 73, 75, 77, 78, 82, 83, 84, 85, 91, 92, 93, 94, 98, 99, 101, 102, 106, 107, 109, 111, 113, 115, 117, 118, 119, 120

Offset: 1

Yifan Xie, Apr 20 2025

nonn

Art of Problem Solving, European Girls' Mathematical Olympiad 2025 Problem 2
Yifan Xie, Python program

Cf. A383192, A383194.

cp's OEIS Frontend

User: Yifan Xie

A386376 a(n) is the smallest integer k such that b(n,k) is squarefree, where b(n,1) = n and b(n, k+1) = b(n,k) + rad(b(n, k)) for k >= 1.

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A385675 For positive integers n, a multiset A of positive integers is called "n-good" if for any number 1 <= i <= n, we can find a submultiset B of A such that the sum of B is equal to i. a(n) is the number of minimal n-good multisets.

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A386820 a(n) is the size of largest subset of {1, 1/2, ..., 1/n} that can be partitioned into two parts, the sum of elements of which are equal.

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A385366 a(n) = Sum_{permutations p of [n]} des(p^2), where des(p) is the number of descents of p.

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A384424 The maximal possible number of 'good' steps in a Hamiltonian cycle on the n X n king's graph, as is specified in the comments.

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A383622 First differences of A383621.

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A383621 a(n) is the minimum possible value of x_1 + x_2 + ... + x_n where x_1, x_2, ..., x_n are positive integers such that x_i does not divide x_j for any i != j.

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A383614 The unique sequence such that Sum_{d|n} d*a(d)^(n/d) = sigma(n)^2 for every n.

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