cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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.

Original entry on oeis.org

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

Views

Author

Lechoslaw Ratajczak, Aug 20 2024

Keywords

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

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)))));

A378275 Numbers m which satisfy the equation: (m - floor((m - k)/k)) mod k = 1 (1 <= k <= m) only for k = 2 and m - 1.

Original entry on oeis.org

3, 4, 7, 11, 19, 23, 59, 83, 167, 227, 491, 659, 839, 983, 1019, 1091, 1319, 1459, 1523, 1847, 2179, 2503, 2963, 3719, 3767, 4519, 4871, 4919, 5059, 6563, 9239, 9419, 10883, 12107, 12539, 14891, 15383, 20071, 20747, 23819, 25219, 26759, 33851, 35591, 37379, 45191
Offset: 1

Views

Author

Lechoslaw Ratajczak, Nov 21 2024

Keywords

Comments

Every term greater than 4 has the form 4*t + 3.
Let b(z) be the number of elements of this sequence <= z:
-------------
z | b(z)
-------------
10^2 | 8
10^3 | 14
10^4 | 32
10^5 | 55
10^6 | 125
10^7 | 347
10^8 | 950
-------------
Every term greater than 4 is prime.

Examples

			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 ...
-----+------------------
   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
 ...
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).
7 is in this sequence because the only k's satisfying the equation (7 - floor((7 - k)/k)) mod k = 1 are 2 and (7-1).
		

Crossrefs

Programs

  • Maxima
    (f(i, j):=mod((i-floor((i-j)/j)), j),
    (n:3, for t:7 thru 100000 step 4 do
    (for k:3 while f(t, k)#1 and k
    				
  • PARI
    is(m) = if(m%4==3, for(k=3, m\2, if((m-m\k)%k==0, return(0))); 1, m==4); \\ Jinyuan Wang, Jan 14 2025

A374870 Let e(m) be the sum of all values of k satisfying the equation: (m mod k = floor((m - k)/k) mod k), minus 2*m (1 <= k <= m); then a(n) is the smallest m for which e(m) = n, or 0 if no e(m) has value n.

Original entry on oeis.org

39, 23, 5847, 735, 65, 29, 35, 77, 111, 173, 415, 185, 79, 47, 113, 137, 317, 867, 307, 543, 4843, 2153, 1203, 161, 59, 159, 351, 531, 1577, 475, 617, 89, 5321, 95, 11405, 1371, 107, 83, 219, 197, 199, 1855, 365, 6521, 3667, 8597, 131
Offset: 0

Views

Author

Lechoslaw Ratajczak, Sep 16 2024

Keywords

Comments

The three smallest values of n (n_1, n_2, n_3) for which a(n) is unknown after computing consecutive e(t) for 1 <= t <= z:
z | n_1 | n_2 | n_3 |
----------------------------------------
10^5 | 309 | 343 | 352 |
2*10^5 | 394 | 556 | 558 |
3*10^5 | 647 | 706 | 755 |
4*10^5 | 941 | 951 | 962 |
5*10^5 | 951 | 964 | 1069 |
Are there any values of n for which a(n) = 0?

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.
a(0) = 39 because 39 is the smallest m for which e(m) = 0 (only k's satisfying the equation: 39 mod k = floor((39 - k)/k) mod k are: 1, 3, 7, 9, 19, 39, hence: 1+3+7+9+19+39-2*39 = 0 = e(39)).
a(2) = 5847 because 5847 is the smallest m for which e(m) = 2 (only k's satisfying the equation: 5847 mod k = floor((5847 - k)/k) mod k are: 1, 85, 135, 171, 343, 730, 1461, 2923, 5847, hence: 1+85+135+171+343+730+1461+2923+5847-2*5847 = 2 = e(5847)).
		

Crossrefs

Programs

  • VBA
    Sub calcul()
    For m = 1 To 500000
    s = 0
          For k = 1 To WorksheetFunction.Floor(m / 2, 1)
                If (m - WorksheetFunction.Floor((m - k) / k, 1)) Mod k = 0 Then
                s = s + k
                End If
          Next k
                       If s > m Then
                       e = s - m
                       v = WorksheetFunction.Ceiling(e / 1000000, 1)
                            If IsEmpty(Cells(e - (v - 1) * 1000000, v)) = False Then
                            Else
                            Cells(e - (v - 1) * 1000000, v).Value = m
                            End If
                       End If
    Next m
    End Sub
Showing 1-3 of 3 results.