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User: Matthew Schuster

Matthew Schuster has authored 2 sequences.

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

Matthew Schuster, Jul 02 2020

nonn

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.

			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.

Subsequence of A005835 and A236359.

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 *)

# 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)

A236359 Pseudoperfect (or semiperfect) numbers in which a sum of contiguous proper divisors of n equals n.

Original entry on oeis.org

6, 18, 24, 28, 36, 42, 54, 66, 78, 102, 108, 114, 126, 132, 138, 162, 174, 186, 196, 198, 222, 234, 246, 258, 282, 288, 294, 306, 318, 324, 342, 354, 360, 366, 378, 402, 414, 426, 432, 438, 462, 474, 486, 496, 498, 504, 522, 534, 540, 546, 558, 582, 594, 600, 606, 618, 642, 654, 666, 678, 684, 690, 696, 702, 714, 726

Offset: 1

Matthew Schuster, Jan 23 2014

nonn

Also includes perfect numbers.

Are there numbers that contain multiple contiguous divisor sums?

			The proper divisors of 132 are [1,2,3,4,6,11,12,22,33,44,66]; the contiguous divisor set 4,6,11,12,22,33,44 sums to 132.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 1000 terms from Matthew Schuster)
Matthew Schuster, A236359.cpp; source file

Subsequence of A005835.

aQ[n_] := Catch@Block[{d = Most@Divisors@n, s, i=1}, s = Accumulate@d; While[s != {}, If[MemberQ[s, n], Throw@True, s = Rest[s - d[[i++]]]]]; False]; Select[ Range@ 726, aQ] (* Giovanni Resta, Jan 23 2014 *)
Select[Range[800],MemberQ[Flatten[Table[Total/@Partition[Most[Divisors[ #]],n,1],{n,DivisorSigma[0,#]-1}]],#]&] (* Harvey P. Dale, Apr 25 2015 *)

is(n)=my(d=divisors(n),i=1,j=1,s=1); while(i<#d, s+=d[i++]; while(s>n, s-=d[j]; j++); if(s==n, return(i<#d))); 0 \\ Charles R Greathouse IV, Jan 23 2014

from sympy import divisors
A236359_list = []
for n in range(1,10**3):
    d = divisors(n)
    d.pop()
    ld = len(d)
    if sum(d) >= n:
        s, j = d[0], 1
        for i in range(ld-1):
            while s < n and j < ld:
                s += d[j]
                j += 1
            if s == n:
                A236359_list.append(n)
                break
            j -= 1
            s -= d[i]+d[j] # Chai Wah Wu, Sep 16 2014

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User: Matthew Schuster

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

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