A080257 -id:A080257 - cp's OEIS frontend

A386276 Numbers k such that the sequence defined by f(1) = k and f(x+1) = the sum of the three largest proper divisors of f(x), consists entirely of numbers having at least three proper divisors.

6, 18, 42, 54, 66, 72, 78, 102, 114, 126, 138, 162, 174, 186, 198, 216, 222, 234, 246, 258, 282, 294, 306, 318, 342, 354, 366, 378, 402, 414, 426, 438, 462, 474, 486, 498, 504, 522, 534, 546, 558, 582, 594, 606, 618, 642, 648, 654, 666, 678, 702, 714, 726, 738

Offset: 1

Jinyuan Wang, Jul 17 2025

Numbers k of the form 6*i*12^j, where gcd(i, 10) = 1 and j >= 0.

Vincenzo Librandi, Table of n, a(n) for n = 1..3709
Art of Problem Solving, 2025 IMO Problems/Problem 4

Cf. A080257.

is[k_]:=Module[{v},v=IntegerExponent[k,2]; OddQ[v]&&Mod[k,5]!=0&&2*IntegerExponent[k,3]>v] Select[Range[1,1000],is] (* Vincenzo Librandi, Jul 22 2025 *)

is(k) = my(v=valuation(k, 2)); v%2 && k%5 && 2*valuation(k, 3)>v;

a(n) = 55*n/4 + O(log n). - Charles R Greathouse IV, Aug 18 2025

A306427 Least integer m such that there are exactly n quadruples of distinct divisors (d_i, d_j, d_k, d_l) among the divisors of m having the property d_i * d_j - d_k * d_l = 1, for some i, j, k, l.

Original entry on oeis.org

28, 84, 120, 240, 360, 252, 210, 660, 1008, 1848, 630, 1320, 420, 2310, 840, 4830, 1680, 3360, 5880, 11700, 1980, 4200, 1260, 9660, 3960, 3780, 2520, 6930, 4620, 8190, 6300, 7560, 5040, 18900, 19320, 5460, 23760, 7140, 39600, 15120, 27300, 12600, 59220, 45360

Offset: 1

Michel Lagneau, Feb 14 2019

nonn

We observe that a(n) == 0 (mod 6) for n > 1, and a(n) == 0 (mod 30) for n > 10.
Conjecture: for each integer q > 1, there exists a subsequence E(q) of {a(n)} such that q*E(q) is also a subsequence of {a(n)}.
The following table gives the first 10 subsequences E(q).
+----+--------------------------------------------+
|  q |   E(q) such that q*E(q) is a subsequence   |
+----+--------------------------------------------+
|  2 | {120, 210, 420, 630, 660, 840, 1260, ...}  |
|  3 | {28, 84, 120, 210, 420, 660, 840, ...}     |
|  4 | {210, 252, 420, 630, 840, 1260, 3780, ...} |
|  5 | {84, 252, 840, 1008, 1260, 2520, ...}      |
|  6 | {210, 420, 630, 660, 840, 1260, 2520, ...} |
|  7 | {120, 240, 360, 660, 840, ...}             |
|  8 | {210, 420, 630, ...}                       |
|  9 | {28, 420, 840, 1680, 5040, ...}            |
| 10 | {84, 252, 420, 630, 1260, 3960, ...}       |
+----+--------------------------------------------+

			a(7) = 210 because the divisors of 210 are {1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210} with seven following quadruples (1, 7, 2, 3), (1, 15, 2, 7), (1, 21, 2, 10), (2, 3, 1, 5), (3, 5, 1, 14), (3, 5, 2, 7) and (3, 7, 2, 10).

Cf. A000005, A027750, A080257.

with(numtheory):nn:=1000:
for n from 1 to nn do:
ii:=0:it:=0:
for k from 1 to 10^5 while(ii=0) do:
d:=divisors(k):n0:=nops(d):it:=0:
for a from 1 to n0-1 do:
  for b from a+1 to n0 do:
   lst1:={d[a]} union {d[b]}:lst:= d minus lst1:n1:=nops(lst):
     for i from 1 to n1-1 do:
       for j from i+1 to n1 do:
         if d[a]*d[b]-lst[i]*lst[j]=1
         then
          it:=it+1:
          else fi:
        od:
       od:
       od:
      od:
      if it=n then ii:=1:printf (`%d %d \n`,n,k):
      else fi:
      od:
     od:

A363863 Numbers expressible as j^2 - k^2, 1 <= k <= j-2 ("squares with a square hole").

Original entry on oeis.org

8, 12, 15, 16, 20, 21, 24, 27, 28, 32, 33, 35, 36, 39, 40, 44, 45, 48, 51, 52, 55, 56, 57, 60, 63, 64, 65, 68, 69, 72, 75, 76, 77, 80, 81, 84, 85, 87, 88, 91, 92, 93, 95, 96, 99, 100, 104, 105, 108, 111, 112, 115, 116, 117, 119, 120, 123, 124, 125, 128, 129, 132, 133, 135, 136

Offset: 1

Thomas A. Fisher, Jun 25 2023

Inspired by my 4-year-old son, who loves Numberblocks, I decided to work out which numbers appear in the "squares with [square] holes club". These are numbers which, when configured as a square, have a square wholly removed. For example, 8 is 3 X 3 with a 1 X 1 hole in the middle. 24 is both a 5 X 5 with a 1 X 1 hole in the middle and a 7 X 7 with a 5 X 5 hole in the middle. The hole has to be "wholly contained", meaning I can't, for example, have 3^2 - 2^2 = 9 - 4 = 5, as removing a 2 X 2 square from a 3 X 3 square doesn't leave a "hole", as we are working with blocks, i.e., integers.

This sequence contains all natural numbers which factor as (j - k)*(j + k), where j - k >= 2 and k >= 1. That is, all natural numbers which have at least one factor pair of the form u*v such that u and v have the same parity, are distinct, and are both strictly greater than 1. This precisely rules out 1, primes, squares of primes, and the even numbers which are congruent to 2 modulo 4. In other words, this sequence is equal to A080257\A016825.

			8 = 3^2 - 1^2, 12 = 4^2 - 2^2, 15 = 4^2 - 1^2, ...

Subsequence of A024352.

Equals A080257\A016825.

isok(k) = ((omega(k)>1) || (isprimepower(k)>2)) && ((k % 4) != 2); \\ Michel Marcus, Jun 30 2023

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A386276 Numbers k such that the sequence defined by f(1) = k and f(x+1) = the sum of the three largest proper divisors of f(x), consists entirely of numbers having at least three proper divisors.

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A306427 Least integer m such that there are exactly n quadruples of distinct divisors (d_i, d_j, d_k, d_l) among the divisors of m having the property d_i * d_j - d_k * d_l = 1, for some i, j, k, l.

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A363863 Numbers expressible as j^2 - k^2, 1 <= k <= j-2 ("squares with a square hole").

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