Lechoslaw Ratajczak - cp's OEIS frontend

A386213 Integers t having at least one nonempty subset of the set of its proper divisors for which the equation sigma(t) + r = m*t (m is any integer > 1, r is the sum of elements of such subset) is true.

2, 4, 6, 8, 10, 12, 15, 16, 18, 20, 21, 24, 28, 30, 32, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56, 60, 63, 64, 66, 70, 72, 75, 78, 80, 84, 88, 90, 96, 99, 100, 102, 104, 105, 108, 112, 114, 117, 120, 126, 128, 130, 132, 135, 136, 138, 140, 144, 150, 152, 153, 154, 156, 160, 162, 165

Offset: 1

Lechoslaw Ratajczak, Aug 12 2025

The following table lists sequences which give k-deficient-m-perfect numbers:
------------------------------------------------------------
 k/m   |     any m     |       2       |         3         |
------------------------------------------------------------
 any k | this sequence | A331627 \ {1} |         -         |
------------------------------------------------------------
 1     | A385462       | A271816 \ {1} | A364977 \ A000396 |
------------------------------------------------------------
 2     |       -       | A331628       |         -         |
------------------------------------------------------------
 3     |       -       | A331629       |         -         |
------------------------------------------------------------
This sequence contains all, and only, (any k)-deficient-m-perfect numbers (m = 2,3,4,...), equivalently it contains all, and only, k-deficient-(any m)-perfect numbers (k = 1,2,3,...).

			24 is a term because for 24 the set of proper divisors is {1, 2, 3, 4, 6, 8, 12} and it has exactly 6 subsets which sum up to r satisfying the equation sigma(24) + r = k*24:
  (1) sigma(24) + d_7(24) = 60 + 12 = 72 and 72 = 3*24,
  (2) sigma(24) + (d_4(24) + d_6(24)) = 60 + (4 + 8) = 72 and 72 = 3*24,
  (3) sigma(24) + (d_2(24) + d_4(24) + d_5(24)) = 60 + (2 + 4 + 6) = 72 and 72 = 3*24,
  (4) sigma(24) + (d_1(24) + d_3(24) + d_6(24)) = 60 + (1 + 3 + 8) = 72 and 72 = 3*24,
  (5) sigma(24) + (d_1(24) + d_2(24) + d_3(24) + d_5(24)) = 60 + (1 + 2 + 3 + 6) = 72 and 72 = 3*24,
  (6) sigma(24) + (d_1(24) + d_2(24) + d_3(24) + d_4(24) + d_5(24) + d_6(24) + d_7(24)) = 60 + (1 + 2 + 3 + 4 + 6 + 8 + 12) = 96 and 96 = 4*24.
So 24 is (1, 2, 3 (in 2 variants), 4)-deficient-3-perfect and 7-deficient-4-perfect number.

Cf. A000005, A000203, A007691, A331627, A385462.

n = 1;l={};Do[x = 1;s=DivisorSigma[1,t];A=Most[Divisors[t]];B=Subsets[A];  Do[r=Total[B[[i]]];If[Mod[s+r,t]==0,x=x+1],{i,2,2^Length[A]}];  If[x>1,AppendTo[l,t];n=n+1],{t,1,165}];l (* James C. McMahon, Aug 25 2025 *)

(n:1, for t:1 thru 300 do (x:1, s:divsum(t), A:delete(t, divisors(t)), B:args(powerset(A)),
              for i:2 thru 2^(length(args(A))) do (r:apply("+", args(B[i])),
                      if mod(s+r, t)=0 then (x:x+1)),
                                       if x>1 then (print(n, "", t), n:n+1)));

A386317 Integers t which satisfy 3/2 <= abundancy(t) < 2 but which are not k-deficient-perfect numbers A331627.

Original entry on oeis.org

14, 22, 26, 34, 38, 46, 58, 62, 68, 74, 76, 82, 86, 92, 94, 98, 106, 110, 116, 118, 122, 124, 134, 142, 146, 147, 148, 158, 164, 166, 171, 172, 178, 188, 194, 202, 206, 212, 214, 218, 225, 226, 236, 242, 244, 248, 254, 255, 262, 268, 274, 278, 284, 285, 286, 292, 296, 298, 302, 314

Offset: 1

Lechoslaw Ratajczak, Jul 18 2025

nonn

A necessary (but not sufficient) condition for an integer t to be a k-deficient-m-perfect number: (m + 1)/2 <= abundancy(t) < m:
- for m = 2: 3/2 <= abundancy(t) < 2,
- for m = 3: 2 <= abundancy(t) < 3,
- for m = 4: 5/2 <= abundancy(t) < 4.

			13 is not in this sequence because abundancy(13) = 14/13 (14/13 < 3/2).
14 is in this sequence because abundancy(14) = 12/7 (3/2 <= 12/7 < 2) but 14 is not a k-deficient-perfect number (therefore is not included in A331627).
15 is not in this sequence because abundancy(15) = 8/5 (3/2 <= 8/5 < 2) but 15 is a k-deficient-perfect number (therefore is included in A331627).

Cf. A000203, A271816, A331627.

(n:1, abundancy(x):=divsum(x)/x,
     for t:1 thru 500 do
        (if abundancy(t)>=3/2 and abundancy(t)<2 then
        (A:append(args(powerset(delete(t,divisors(t)))),[{0}]), b:length(A),
            for i:1 unless (divsum(t)+apply("+" , args(A[i])))/t=2 or i>=b do j:i,
               if j>=b-1 then (print(n , "" , t), n:n+1))));

isok(m) = my(d=divisors(m), ss=vecsum(d), ab=sigma(m)/m); if ((ab>=3/2) && (ab<2), d = Vec(d, #d-1); forsubset(#d, s, if (#s && (sum(i=1, #s, d[s[i]]) == 2*m - ss), return(0))); return(1)); \\ Michel Marcus, Jul 19 2025

A385462 Numbers t which have a proper divisor d_i(t) such that (d_i(t) + sigma(t))/t is an integer k.

Original entry on oeis.org

2, 4, 8, 10, 16, 24, 32, 44, 60, 64, 84, 128, 136, 152, 168, 184, 252, 256, 270, 336, 512, 630, 752, 756, 792, 864, 884, 924, 936, 1024, 1140, 1170, 1488, 1638, 2048, 2144, 2268, 2272, 2528, 2808, 2970, 3672, 4096, 4320, 4464, 4680, 5148, 5472, 6804, 7308, 7644, 8192, 8384

Offset: 1

Lechoslaw Ratajczak, Jun 29 2025

nonn

Consecutive elements of this sequence for which k = 2 are consecutive deficient-perfect numbers (A271816) > 1.
Consecutive elements of this sequence for which k = 3 are consecutive non-perfect elements of A364977.
Let b_k(m) be the number of elements of this sequence with the same k and <= m.
--------------------------------------------
   m   | b_2(m) | b_3(m) | b_4(m) | b_5(m) |
--------------------------------------------
 10^3  |   16   |   13   |    -   |    -   |
 10^4  |   24   |   31   |    2   |    -   |
 10^5  |   37   |   62   |    5   |    -   |
 10^6  |   54   |  107   |   19   |    -   |
 10^7  |   73   |  175   |   43   |    1   |
 10^8  |   98   |  254   |   80   |    3   |
 10^9  |  128   |  357   |  141   |   13   |
--------------------------------------------
Are there any odd terms in this sequence for which k > 2? If they exist, they are > 10^9.
Contains 2^k * (2^(k+1) + 2^j - 1) if 0 <= j <= k and 2^(k+1) + 2^j - 1 is prime. - Robert Israel, Jun 30 2025

			4 is in this sequence because sigma(4) + d_1(4) = 7 + 1 = 8 and 8/4 = 2.
24 is in this sequence because sigma(24) + d_7(24) = 60 + 12 = 72 and 72/24 = 3.
4320 is in this sequence because sigma(4320) + d_47(4320) = 15120 + 2160 = 17280 and 17280/4320 = 4.

Cf. A000005, A000203, A007691, A054027, A271816, A364977.

filter:= proc(n) local s;
  s:= - numtheory:-sigma(n) mod n;
  ormap(d -> d mod n = s, numtheory:-divisors(n) minus {n})
end proc:
select(filter, [$1..10^4]); # Robert Israel, Jun 30 2025

Select[Range[8384],AnyTrue[(Drop[Divisors[#],-1]+DivisorSigma[1,#])/#,IntegerQ]&] (* James C. McMahon, Jul 05 2025 *)

(n:1, for t:1 thru 10000 do (s:divsum(t), (A:args(divisors(t)),
              for i:1 thru length(A)-1 do (y:s+A[i],
                      if mod(y,t)=0 then (print(n,"",t), n:n+1)))));

isok(t) = my(s=sigma(t)); fordiv(t, d, if ((dMichel Marcus, Jun 30 2025

A381060 Numbers t which are the sum of some subset of the values of k satisfying the equation (t - floor((t - k)/k)) mod k = 0 (t > 1, 1 <= k < t).

Original entry on oeis.org

23, 29, 39, 41, 53, 59, 65, 71, 77, 79, 83, 89, 99, 101, 107, 111, 113, 119, 125, 137, 143, 149, 155, 161, 167, 173, 179, 185, 191, 197, 199, 209, 221, 227, 233, 239, 245, 251, 257, 263, 269, 279, 281, 287, 293, 299, 305, 311, 317, 323, 329, 335, 339, 341, 349, 353, 359, 365, 371

Offset: 1

Lechoslaw Ratajczak, Feb 12 2025

nonn

The sequence is based on the triangle in A380305, which is a variant of the triangle in A048158. Thus, the elements of this sequence are counterparts of pseudoperfect numbers (A005835), such as the elements of A375595 are counterparts of abundant numbers (A005101).
The sequence includes all elements of A380153.
The first even element of this sequence is a(768) = 4094.

			23 is in this sequence because the only k's < 23 satisfying the equation (23 - floor((23 - k)/k)) mod k = 0 are: 1, 5, 7, 11, hence: 5+7+11 = 23.
29 is in this sequence because the only k's < 29 satisfying the equation (29 - floor((29 - k)/k)) mod k = 0 are: 1, 2, 3, 5, 9, 14, hence: 1+2+3+9+14 = 29 and 1+5+9+14 = 29.
47 is not in this sequence because the only k's < 47 satisfying the equation (47 - floor((47 - k)/k)) mod k = 0  are: 1, 3, 7, 11, 15, 23 and no subset of these numbers adds to 47.

Cf. A005835, A048158, A375595, A380153, A380305.

(kill(all), s(y):=(f(i,j):=mod(i-floor((i-j)/j),j),s:0,x:1,
      for k:1 thru floor(y/2) do
              (if f(y,k)=0 then
              (s:s+k, B[x]:k, x:x+1)),
B:setify(makelist(B[r],r,1,x-1)), s),
n:1, for t:2 thru 1000 do
              (if s(t)>=t  then
                     (for b:2 while b<=x-1 and e#t do
                     (C:args(powerset(B,b)),
                            for h:1 while h<=length(C) and e#t do
                            (e:apply("+" , args(C[h])),
                                    if e=t then
                                    (print(n , " " , t), n:n+1))))));

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.

Original entry on oeis.org

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

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

Lechoslaw Ratajczak, Nov 21 2024

nonn

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.

			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).

Jinyuan Wang, Table of n, a(n) for n = 1..3000

Cf. A000040, A048158, A051731, A375007, A375595.

(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

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

A377417 Numbers t such that t + s(t) = s(s(t)), where s(t) is the sum of aliquot divisors of t (A001065(t)).

Original entry on oeis.org

1272, 4632, 31632, 266712, 805152, 4897392, 94177392, 2928675264

Offset: 1

Lechoslaw Ratajczak, Oct 27 2024

Equivalently, numbers t which satisfy the equation t = sigma(sigma(t) - t) - 2*(sigma(t) - t), where sigma(t) is the sum of divisors of t (A000203(t)).
The sum of aliquot divisors is also called the sum of aliquot parts, the aliquot sum or the sum of proper divisors (A001065).
a(9) > 10^10 (if it exists).
Let s^[m](t) denote m-fold iteration of s(t) (i.e., s^[0](t) = t and s^[m](t) = s(s^[m-1](t))).
The following table shows consecutive solutions t (t <= z) of the equation s^[m](t) = Sum_{i=0..m-1} s^[i](t) for m = 3,4,5,...,30 (solutions for m = 1 are perfect numbers, solutions for m = 2 are in the data section):
---------------------------------------------------------------------------------------------
     m   |   z  |    t    |     (m+1)-tuple (s^[0](t), s^[1](t),...,s^[m](t))
---------------------------------------------------------------------------------------------
3        | 10^9 | 8880    | 8880, 19392, 32424, 60696 = 8880+19392+32424
         |      | 1468584 | 1468584, 2933016, 4399584, 8801184 = 1468584+2933016+4399584
4        | 10^9 | 285816  | 285816, 428784, 679032, 1160208, 2553840 = 285816+428784+
         |      |         | +679032+1160208
5        | 10^9 | 3280592 | 3280592, 4415344, 4196952, 7343928, 12546072, 31782888 =
         |      |         | = 3280592+4415344+4196952+7343928+12546072
6,...,30 | 10^8 |    -    |
---------------------------------------------------------------------------------------------

			a(1) = 1272 because s(1272) = 1968, s(1968) = 3240 = 1272 + 1968.
a(2) = 4632 because s(4632) = 7008, s(7008) = 11640 = 4632 + 7008.
a(3) = 31632 because s(31632) = 50208, s(50208) = 81840 = 31632 + 50208.

Cf. A000203, A000396, A001065, A003416, A051027, A063990.

(n:1, for t:2 thru 100000000 do
(x:divsum(t)-t, y:divsum(x)-x,
if t+x=y then
(print(n, "" , t ), n:n+1)));

isok(t) = t == sigma(sigma(t) - t) - 2*(sigma(t) - t); \\ Michel Marcus, Oct 29 2024

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

Lechoslaw Ratajczak, Sep 16 2024

nonn

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?

			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)).

Cf. A005101, A033880, A051731, A294347, A375007, A375595.

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

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

Lechoslaw Ratajczak, Aug 20 2024

nonn

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

			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.

Cf. A005101, A051731, A375007.

(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)))));

cp's OEIS Frontend

User: Lechoslaw Ratajczak

A386213 Integers t having at least one nonempty subset of the set of its proper divisors for which the equation sigma(t) + r = m*t (m is any integer > 1, r is the sum of elements of such subset) is true.

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A386317 Integers t which satisfy 3/2 <= abundancy(t) < 2 but which are not k-deficient-perfect numbers A331627.

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A385462 Numbers t which have a proper divisor d_i(t) such that (d_i(t) + sigma(t))/t is an integer k.

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Maple

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A381060 Numbers t which are the sum of some subset of the values of k satisfying the equation (t - floor((t - k)/k)) mod k = 0 (t > 1, 1 <= k < t).

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

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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.

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A377417 Numbers t such that t + s(t) = s(s(t)), where s(t) is the sum of aliquot divisors of t (A001065(t)).

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