OFFSET
1,12
COMMENTS
For a given positive integer n consider the set D(n) of the sums d(1), d(1)+d(2), d(1)+d(2)+d(3), ..., d(1)+...+d(k_n), where 1<d(1)<d(2)<...<d(k_n) are all divisors of n greater than 1. Then a(n) is the number of elements of D(n) that do not belong to any D(m) for m<n.
Erdős and Graham asked about these numbers in their 1980 problem book on p. 93. A related problem on partial sums of divisors including 1 was posed by Erdős and it was mentioned towards the end of Section B2 of Guy's book from 1981.
Indices attaining record values of the sequence are: 2, 12, 36, 144, 336, 1320, 1980, 5040, 8400, 25200, 75600, etc.
The average value of the first 100000 terms of the sequence equals 0.8349.
The smallest m such that D(m) contains a given positive integer n is sequence A167485 with indices shifted by 1.
REFERENCES
P. Erdős and R. L. Graham, Old and new problems and results in combinatorial number theory, Monographies de L'Enseignement Mathématique, vol. 28, Université de Genève, 1980, p. 93.
R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981, Section B2.
LINKS
Vjekoslav Kovac, Table of n, a(n) for n = 1..10000
Thomas Bloom, Erdős problems, Erdős problem #468
EXAMPLE
For example, D(1)={}, D(2)={2}, D(3)={3}, D(4)={2,6}, so the only debut sum in D(4) is 6 and thus a(4)=1. Similarly, D(5)={5}, D(6)={2,5,11}, D(7)={7}, D(8)={2,6,14}, D(9)={3,12}, D(10)={2,7,17}, D(11)={11}, D(12)={2,5,9,15,27}, but the only debut sums in D(12) are 9, 15, and 27, so a(12)=3.
MATHEMATICA
For[n=1, n<=1000, n++, dd[n]=Complement[Table[Plus@@(Divisors[n][[2;; Length[Divisors[n]]]])[[1;; j]], {j, 1, Length[Divisors[n]]-1}], Flatten[Table[dd[k], {k, 1, n-1}]]]]; Table[Length[dd[n]], {n, 1, 1000}]
PROG
(PARI) lista(n)={my(a=vector(n), S=List([1])); for(n=1, n, my(s=0, r=0); fordiv(n, d, s+=d; while(#S<s, listput(S, 0)); if(!S[s], S[s]=n; r++)); a[n]=r); a} \\ Andrew Howroyd, Sep 01 2025
(Python)
from sympy import divisors
from itertools import accumulate, count, islice
def D(m): return set(accumulate(divisors(m)[1:]))
def agen(): # generator of terms
U = set()
for m in count(1):
debut = D(m) - U
yield len(debut)
U |= debut
print(list(islice(agen(), 100))) # Michael S. Branicky, Sep 06 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Vjekoslav Kovac, Aug 31 2025
STATUS
approved