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.

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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).

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

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Author

Lechoslaw Ratajczak, Feb 12 2025

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Maxima
    (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))))));
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