A396378 Smallest m belonging to {n, n+1, n+2, ..., n+A396377(n)} that divides the product of the other integers.

1, 2, 3, 4, 6, 6, 8, 8, 12, 10, 12, 12, 15, 15, 15, 16, 18, 18, 24, 24, 21, 24, 24, 24, 30, 32, 28, 28, 30, 30, 36, 36, 36, 40, 40, 36, 40, 40, 40, 40, 48, 48, 48, 48, 45, 48, 48, 48, 56, 50, 54, 60, 60, 60, 60, 56, 60, 60, 60, 60, 63, 63, 63, 64, 72, 72, 72, 72, 72, 70, 72, 72

Offset

1,2

Comments

After problem proposed by Paul Erdős (see link).

Links

Table of n, a(n) for n=1..72.

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.

Formula

a(p) = a(p+1) for p >= 5.

Fixed points are in A396379.

Example

A396377(3) = 3 because in the set (3, 4, 5, 6), we have 3 | 4*5*6, then 4 does not divide 3*5*6, 5 does not divide 3*4*6 and 6 | 3*4*5. As 3 < 6, a(3) = 3.
A396377(9) = 5 because in the set (9, 10, 11, 12, 13, 14), we have 12 | 9*10*11*13*14, and 12 is the only integer belonging to (9, 10, 11, 12, 13, 14) that divides the product of the other integers, so a(9) = 12.

Maple

a := proc(n)
local k, L, i, j, t, g;
for k from 1 do
    L := [seq(n+r, r=0..k)];
    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
            return L[i];
        end if;
     end do;
end do;
end proc;
seq(a(n), n=1..100);

Prog

(PARI) isok(k, n) = my(v=vector(k+1, i, n+i-1), P=vecprod(v)); for (i=1, k+1, if (denominator(P/v[i]^2) == 1, return(i)));
a(n) = my(k=1, j); while (!(j=isok(k, n)), k++); n+j-1; \\ Michel Marcus, May 28 2026

Crossrefs

Cf. A396377, A396379, A396380, A396950.

Sequence in context: A142727 A347388 A112275 * A335876 A306974 A278400

Adjacent sequences: A396375 A396376 A396377 * A396379 A396380 A396381

Keyword

nonn

Author

Bernard Schott, May 24 2026

Status

approved