OFFSET
1,1
COMMENTS
Conjecture: for every integer n>=1, there always exists at least one integer k such that its proper divisors are palindromic modulo n.
The interest lies in studying the interaction between the prime factorization of k and the cycles of the multiplicative groups Z/n.
a(11) = 14106314, a(12) = 732421875, a(14) = 7130807042, a(15) = 2141286002. - Sean A. Irvine, Apr 02 2026
EXAMPLE
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n a(n) Number of patterns modulo n
proper divisors (palindromes)
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1 6 3 [0,0,0]
2 18 5 [1,0,1,0,1]
3 104 7 [1,2,1,2,1,2,1]
4 162 9 [1,2,3,2,1,2,3,2,1]
5 7502 11 [1,2,1,2,1,2,1,2,1,2,1]
6 46875 13 [1,3,5,3,1,3,5,3,1,3,5,3,1]
7 811246 15 [1,2,1,2,1,2,1,2,1,2,1,2,1,2,1]
8 140450 17 [1,2,5,2,1,2,5,2,1,2,5,2,1,2,5,2,1]
9 19026866 19 [1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2,1]
a(9) = 19^4*37^3, 19==1 (mod 9), 37==1 (mod 9).
MAPLE
with(numtheory):
with(ListTools):
get_an := proc(n)
local k, divs, props, mods, target_divs;
target_divs := 2*n + 2;
for k from 2 do
divs := divisors(k);
if nops(divs) = target_divs then
props := [op(1..nops(divs)-1, divs)];
mods := map(x -> x mod n, props);
# Test of the palindrome
if mods = Reverse(mods) then
# printf("Pour n = %d, a(n) = %d\n", n, k);
# printf("Suite mod %d : %a\n", n, mods);
return k;
fi;
fi;
od:
end proc:
# n from 2 to 9
for i from 2 to 9 do
get_an(i);
od:
PROG
(PARI) isok(k, n) = if (numdiv(k) == (2*n+2), my(v=apply(x->Mod(x, n), setminus(divisors(k), Set(k)))); v == Vecrev(v));
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Mar 05 2026
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Michel Lagneau, Mar 01 2026
EXTENSIONS
a(9) corrected by Michel Marcus, Mar 05 2026
STATUS
approved