A375595 Numbers m for which the sum of all values of k satisfying the equation: m mod k = floor((m - k)/k) mod k (1 <= k <= m) exceeds 2*m.

23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 79, 83, 89, 95, 99, 101, 107, 111, 113, 119, 125, 131, 137, 139, 143, 149, 155, 159, 161, 167, 173, 179, 185, 191, 197, 199, 203, 209, 215, 219, 221, 223, 227, 233, 239, 245, 251, 257, 259, 263, 269

Offset: 1

Author

Lechoslaw Ratajczak, Aug 20 2024

Keywords

• nonn

Comments

The first even element of this sequence is a(817) = 3464.

Examples

			Let T(i,j) be the triangle read by rows: T(i,j) = 1 if i mod j = floor((i - j)/j) mod j, T(i,j) = 0 otherwise, for 1 <= j <= i. The triangle begins:
 i\j| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
-----------------------------------------
   1| 1
   2| 1 1
   3| 1 0 1
   4| 1 0 0 1
   5| 1 1 0 0 1
   6| 1 1 0 0 0 1
   7| 1 0 1 0 0 0 1
   8| 1 0 0 0 0 0 0 1
   9| 1 1 0 1 0 0 0 0 1
  10| 1 1 0 0 0 0 0 0 0  1
  11| 1 0 1 0 1 0 0 0 0  0  1
  12| 1 0 1 0 0 0 0 0 0  0  0  1
  13| 1 1 0 0 0 1 0 0 0  0  0  0  1
  14| 1 1 0 1 0 0 0 0 0  0  0  0  0  1
  15| 1 0 0 0 0 0 1 0 0  0  0  0  0  0  1
 ...
The j-th column has period j^2. Consecutive elements of this period are j X j identity matrix entries, read by rows.
11 is not in this sequence because only k's <= 11 satisfying the equation 11 mod k = floor((11 - k)/k) mod k are: 1, 3, 5, 11, hence 1+3+5+11 = 20 and 20 < 2*11.
23 is in this sequence because only k's <= 23 satisfying the equation 23 mod k = floor((23 - k)/k) mod k are: 1, 5, 7, 11, 23, hence 1+5+7+11+23 = 47 and 47 > 2*23.
		

Crossrefs

Cf. A005101, A051731, A375007.

Programs

Maxima

(f(i,j):=mod(i-floor((i-j)/j),j),
(n:0, for m:2 thru 500 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)))));