A378275 -id:A378275 - cp's OEIS frontend

A380153 Numbers m for which the sum of all values of k satisfying the equation: (m - floor((m - k)/k)) mod k = 0 (1 <= k <= m) equals 2*m.

39, 4395, 29055, 57931, 81115, 152571, 164955, 410731, 664747, 877435, 2080875, 2521087, 2539515

Offset: 1

Lechoslaw Ratajczak, Jan 13 2025

a(14) > 3*10^6 (if it exists). Is there any even term?

			Let T(i,j) be the triangle read by rows: T(i,j) = (i - floor((i - j)/j)) mod j for 1 <= j <= i. The triangle begins:
 i\j | 1 2 3 4 5 6 7 8 9 10 11 ...
-----+------------------------
   1 | 0
   2 | 0 0
   3 | 0 1 0
   4 | 0 1 1 0
   5 | 0 0 2 1 0
   6 | 0 0 2 2 1 0
   7 | 0 1 0 3 2 1 0
   8 | 0 1 1 3 3 2 1 0
   9 | 0 0 1 0 4 3 2 1 0
  10 | 0 0 2 1 4 4 3 2 1  0
  11 | 0 1 0 2 0 5 4 3 2  1  0
  ...
The j-th column has period j^2, r-th element of this period has the form (r - 1 - floor((r - 1)/j)) mod j (1 <= r <= j^2). The period of j-th column consists of the sequence (0,1,2,...,j-1) and its consecutive j-1 right rotations (moving rightmost element to the left end).
39 is in this sequence because the only k's <= 39 satisfying the equation (39 - floor((39 - k)/k)) mod k = 0 are: 1, 3, 7, 9, 19, 39, hence: 1+3+7+9+19+39 = 78 = 2*39.

Cf. A048158, A375595, A378275.

(f(i, j):=mod(i-floor((i-j)/j), j),
(n:0, for m:2 thru 5000 do
(s:0, for k:1 thru floor(m/2) do
(if f(m, k)=0 then
(s:s+k)), if s=m then
(n:n+1, print(n , "" , m)))));

a(9)-a(13) from Jinyuan Wang, Jan 14 2025

A380305 Triangle read by rows: T(n,k) = (n - floor((n - k)/k)) mod k, for 0 < k <= n.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 2, 1, 0, 0, 0, 2, 2, 1, 0, 0, 1, 0, 3, 2, 1, 0, 0, 1, 1, 3, 3, 2, 1, 0, 0, 0, 1, 0, 4, 3, 2, 1, 0, 0, 0, 2, 1, 4, 4, 3, 2, 1, 0, 0, 1, 0, 2, 0, 5, 4, 3, 2, 1, 0, 0, 1, 0, 2, 1, 5, 5, 4, 3, 2, 1, 0, 0, 0, 1, 3, 2, 0, 6, 5, 4, 3, 2, 1, 0

Offset: 1

Lechoslaw Ratajczak, Jan 19 2025

The triangle is a variant of the triangle in A048158. The period of the k-th column consists of the period of the k-th column in the triangle in A048158 (0,1,2,...,k-1) and its consecutive k-1 right rotations (moving the rightmost element to the left end). Thus the k-th column has period k^2 and the r-th element of this period has the form (r - 1 - floor((r - 1)/k)) mod k (1 <= r <= k^2).

Such as the triangle in A048158 may be the basis for definitions of different kinds of numbers (abundant numbers, perfect numbers, etc.), this triangle may be the basis for definitions of counterparts of these numbers (elements of A375595 as counterparts of abundant numbers, elements of A380153 as counterparts of perfect numbers).

			Triangle begins:
n\k| 1 2 3 4 5 6 7 8 9 10 11 ...
----------------------------
  1| 0
  2| 0 0
  3| 0 1 0
  4| 0 1 1 0
  5| 0 0 2 1 0
  6| 0 0 2 2 1 0
  7| 0 1 0 3 2 1 0
  8| 0 1 1 3 3 2 1 0
  9| 0 0 1 0 4 3 2 1 0
 10| 0 0 2 1 4 4 3 2 1  0
 11| 0 1 0 2 0 5 4 3 2  1  0
 ...

Cf. A048158, A375595, A378275, A380153.

T[n_,k_]:=Mod[n - Floor[(n - k)/k], k]; Table[T[n,k], {n,13},{k,n}]//Flatten (* Stefano Spezia, Jan 20 2025 *)

(for n:1 thru 25 do
(for k:1 thru n do
(T[n,k]:mod(n-floor((n-k)/k),k)),
print(makelist(T[n,i], i, 1, n))));

row(n) = vector(n, k, (n - floor((n - k)/k)) % k); \\ Michel Marcus, Jan 20 2025

cp's OEIS Frontend

A380153 Numbers m for which the sum of all values of k satisfying the equation: (m - floor((m - k)/k)) mod k = 0 (1 <= k <= m) equals 2*m.

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