A335742 Pseudoperfect (or semiperfect) numbers k having more than one set of contiguous proper divisors whose sum equals k.

12978, 13338, 34920, 41382, 76626, 176946, 253422, 455202, 1336734, 2410254, 3187782, 3214458, 3277800, 3347838, 3387240, 3427866, 3507894, 3587922, 3614598, 3694626, 3747978, 3774654, 3908034, 4094766, 4148118, 4174794, 4228146, 4414878, 4494906, 4628286

Offset

1,1

Comments

Observation of some pseudoperfect numbers with an attribute similar to multiperfect numbers.

A total of 84 of the 96 terms (representing all terms less than 10^7) are equal to 0 (mod 13338).

Many of the terms greater than (13338*239)-1 are in the form of 13338*p where p>=239. Prime(52)*1338 through Prime(50188)*1338 were tested and are all terms in this sequence.

There are numbers greater than (13338*239)-1 in this sequence that do not have 13338 as a divisor, for example; 3277800, 3387240, 5007222 and 9233154.

(Uni-)Perfect numbers cannot be in this sequence.

Links

Table of n, a(n) for n=1..30.

Example

The proper divisors of 12978 are (1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 103, 126, 206, 309, 618, 721, 927, 1442, 1854, 2163, 4326, 6489).
The contiguous divisor lists of (3+6+7+9+14+18+21+42+63+103+126+206+309+618+721+927+1442+1854+2163+4326) and (2163+4326+6489) equals 12978.

Mathematica

pspQ[n_] := Module[{d = Divisors[n]}, c = Accumulate[d]; Length @ Intersection[c, c + n] > 2]; Select[Range[10^6], pspQ] (* Amiram Eldar, Jul 02 2020 *)

Prog

(Python)
# Pseudoperfect (or semiperfect) numbers having more than one set of contiguous proper divisors whose sum equals n.
import sympy
A335742_list = []
for n in range(1, (10**7)+1):
    # create an ascending list of divisors of n.
    n_divs = list(sympy.divisors(n))
    # pop last divisor, which equals n, so only proper divisors are examined.
    n_divs.pop()
    # reset iterator for sets of contiguous proper divisors whose sum equals n.
    itr = 0
    # run the outer loop for each proper divisor of n.
    for i in range(len(n_divs)+1):
        # run the inner loop for each divisor >= i.
        for j in range(i, len(n_divs)+1):
            # if sum of divisors i:j is greater than n; continue to next n.
            if sum(n_divs[i:j]) > n:
                continue
            # elif sum of divisors i:j equals n; increment itr; if itr > 1; append n to sequence.
            elif sum(n_divs[i:j]) == n:
                itr += 1
                if itr > 1:
                    A335742_list.append(n)

Crossrefs

Subsequence of A005835 and A236359.

Sequence in context: A145333 A190468 A028385 * A238142 A190947 A203910

Adjacent sequences: A335739 A335740 A335741 * A335743 A335744 A335745

Keyword

nonn

Author

Matthew Schuster, Jul 02 2020

Status

approved