Joshua Zelinsky - cp's OEIS frontend

User: Joshua Zelinsky

Joshua Zelinsky has authored 3 sequences.

A383148 k-facile numbers: Numbers m such that the sum of the divisors of m is equal to 2*m+s where s is a product of distinct divisors of m.

12, 18, 20, 24, 30, 40, 42, 54, 56, 60, 66, 78, 84, 88, 90, 102, 104, 114, 120, 132, 138, 140, 168, 174, 186, 196, 204, 222, 224, 234, 246, 252, 258, 264, 270, 280, 282, 308, 312, 318, 348, 354, 360, 364, 366, 368, 380, 402, 414, 420, 426, 438, 440, 456, 464, 468, 474, 476

Offset: 1

Joshua Zelinsky, Apr 17 2025

nonn

Subsequence of A005101 but seem to be much rarer.

			The sum of the divisors of 60 is 168, and 168 = 2*60 + 48, and 48 = 4*12 and 4 and 12 are divisors of 60, so 60 is in the sequence.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
S. Flora Jeba, Anirban Roy, and Manjil P. Saikia, On k-Facile Perfect Numbers, Algebra and Its Applications (ICAA-2023), Springer Proc. Math. Stat., Vol. 474 (2025), 111-121.

Subsequence of A005101.

Cf. A000203, A000396, A181595.

q[m_] := Module[{d = Divisors[m], ab}, ab = Total[d] - 2*m; ab > 0 && AnyTrue[Subsets[d], Times @@ # == ab &]]; Select[Range[500], q] (* Amiram Eldar, Apr 18 2025 *)

prodDistinctDiv(n, k, f=factor(n))=my(D=divisors([n,f])); helper(D[2..#D], k)
helper(v,k)=if(k==1, return(1)); v=select(d->k%d==0, v); if(#v<3, if(#v==2, return(v[2]==k || vecprod(v)==k)); return(#v && v[1]==k)); my(u=v[1..#v-1]); helper(u,k) || helper(u,k/v[#v])
is(n,f=factor(n))=my(t=sigma([n,f])-2*n); t>1 && prodDistinctDiv(n, t, f) \\ Charles R Greathouse IV, Apr 24 2025

def facile_candidates(n):
    from itertools import combinations
    divs = divisors(n)
    sigma_n = sigma(n, 1)
    candidates = set()
    # Generate all products of distinct combinations of divisors
    for r in range(2, len(divs)+1):  # start from 2-element products to avoid m=n
        for combo in combinations(divs, r):
            product = prod(combo)
            if product < sigma_n:
                candidates.add(product)
    return sorted(candidates)
def find_facile_perfects(x):
    result = []
    for n in range(1, x+1):
        sig = sigma(n, 1)
        if sig < 2*n:
            continue
        candidates = facile_candidates(n)
        for m in candidates:
            if sig == 2*n + m:
                print(n,m)
                result.append(n)
                break
    return result

A358566 Number of distinct spans of length n with no 3-term arithmetic progression, containing zero, and with maximum element smallest possible.

Original entry on oeis.org

1, 1, 2, 1, 4, 7, 6, 1, 2, 2, 2, 1, 2, 2, 20, 1, 14, 2, 2, 2, 4

Offset: 1

Joshua Zelinsky, Nov 22 2022

Offset by 1 from the same function in the referenced paper, which does not include zero in their count of how long spans are for the "well-spaced rows".

			a(5)=4 since we have as possible spans 0,1,3,7,8; 0,1,5,6,8; 0,1,5,7,8; 0,2,3,7,8. (It is easy to check that if the maximum term is 7 or lower then there must be fewer than 5 elements in the sequence.)

Noam Benson-Tilsen, Samuel Brock, Brandon Faunce, Monish Kumar, Noah Dokko Stein, and Joshua Zelinsky, Total Difference Labeling of Regular Infinite Graphs, arXiv:2107.11706 [math.CO], 2021-2022.

Cf. A005047.

A334876 Numbers m with the property that sigma(2^m+1)/(2^m+1) > sigma(2^k+1)/(2^k+1) for all k < m, where sigma is the sum of divisors function, A000203.

Original entry on oeis.org

1, 3, 5, 9, 15, 45, 135, 315, 945

Offset: 1

Joshua Zelinsky, May 13 2020

Set h(m) = sigma(m)/m. Then the sequence lists the numbers m at which record values of h(2^m+1) occur. This sequence is essentially defined similarly to A004394 but restricted to looking just at numbers which are one more than a power of 2.
The sequence is infinite. This can be proved by seeing that we can make h(2^m+1) arbitrarily large. Note that if p is a prime which is 3 (mod 8), then p|2^m+1 for any odd m such that (p-1)/2|m. By the strong version of Dirichlet's theorem the sum of the reciprocals of the primes which are 3 (mod 8) diverges. So we can make h(2^m+1) arbitrarily large by taking m as the largest odd divisor of k! for large k.
It appears that every term in the sequence is odd. This seems likely to be true since if m is even then 2^m+1 is not divisible by 3, which makes it much harder to make h(2^m+1) large.

Joshua Zelinsky, On the small prime factors of a non-deficient number, arXiv:2005.12118 [math.NT], 2020-2022.

Cf. A000203, A004394, A069061.

r[n_] := DivisorSigma[1, 2^n+1]/(2^n + 1); seq = {}; rm = 1; Do[r1 = r[n]; If[r1 > rm, rm = r1; AppendTo[seq, n]], {n, 1, 50}]; seq (* Amiram Eldar, May 15 2020 *)

def h(n):
    return (sigma(n,1))/n
def hchecker(k):
    s=0
    for i in range(1,k):
        j=2^i+1
        a=h(j)
        if a> s:
            print(i)
            s=a

More terms from Amiram Eldar, May 13 2020, using A069061.

cp's OEIS Frontend

User: Joshua Zelinsky

A383148 k-facile numbers: Numbers m such that the sum of the divisors of m is equal to 2*m+s where s is a product of distinct divisors of m.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Sage

A358566 Number of distinct spans of length n with no 3-term arithmetic progression, containing zero, and with maximum element smallest possible.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A334876 Numbers m with the property that sigma(2^m+1)/(2^m+1) > sigma(2^k+1)/(2^k+1) for all k < m, where sigma is the sum of divisors function, A000203.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Sage

Extensions