OFFSET
1,3
COMMENTS
From a problem proposed by Paul Erdős (see link).
The smallest of these integers m is A396378(n).
Are there integers n with a(n) = 4, 5, ... or a(n) always <= 3 ?
a(n) = 3 for 7 consecutive values of n : from 3030 up to 3036.
The least integer with a(n) = 4 is n = 407049; a(n) < 5 for all n <= 5*10^8. - David Radcliffe, Jun 10 2026
LINKS
Paul Erdős, Quelques problèmes de Théorie des Nombres, Monographies de l'Enseignement Mathématique, No. 6, pp. 81-135, L'Enseignement Mathématique, Université, Geneva, 1963, Problem 15, p. 90.
EXAMPLE
a(6) = 1 because A396377(6) = 3 and in {6,7,8,9}, only 6 | 7*8*9.
a(8) = 2 because A396377(8) = 4 and in {8,9,10,11,12}, we get 8 | 9*10*11*12 and also 12 | 8*9*10*11.
a(10) = 3 because A396377(10) = 5 and in {10,11,12,13,14,15}, we get 10 | 11*12*13*14*15, then 12 | 10*11*13*14*15 and also 15 | 10*11*12*13*14.
MAPLE
c := proc(n)
local k, L, i, j, t, g, C;
for k from 1 do
L := [seq(n+r, r=0..k)];
C := 0;
for i from 1 to k+1 do
t := L[i];
for j from 1 to k+1 do
if j <> i then
g := gcd(t, L[j]);
t := iquo(t, g);
end if;
end do;
if t = 1 then
C := C+1;
end if;
end do;
if C > 0 then
return C;
end if;
end do;
end proc:
seq(c(n), n=1..50);
PROG
(Python)
def A396380(n: int) -> int:
k = 1
p = n * (n + 1)
while (count := sum(p % (j*j) == 0 for j in range(n, n+k+1))) == 0:
k += 1
p *= n + k
return count # David Radcliffe, Jun 10 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Jun 04 2026
STATUS
approved
