A392684 - OEIS

A392684

Number of ways to write n as x + y + z with x >= y >= z >= 1 such that lcm(x,y,z) is a product of three consecutive integers.

0, 0, 0, 0, 0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 4, 2, 2, 4, 3, 5, 4, 5, 5, 8, 6, 6, 6, 10, 8, 10, 9, 11, 10, 11, 9, 12, 8, 12, 11, 16, 7, 12, 9, 16, 13, 11, 14, 11, 12, 14, 12, 14, 9, 21, 11, 16, 12, 19, 10, 13, 13, 18, 17, 16, 18, 23, 18, 18, 13, 24, 15, 27, 18, 20, 22, 26, 21, 26

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OFFSET

1,8

COMMENTS

Conjecture: (i) For any integer k > 2, each integer n >= 2^k - 2 can be written as x_1 + ... + x_k with x_1,...,x_k positive integers such that lcm(x_1,...,x_k) is a product of k consecutive integers.

(ii) For each integer k > 2, let N(k) be the least positive integer N such that any integer n >= N can be written as x_1 + ... + x_k with x_1,...,x_k positive integers such that lcm(x_1,...,x_k) is a product of k consecutive integers. Then N(3) = 6, N(4) = 13, N(5) = 18, N(6) = 33, N(7) = 40, N(8) = 153 and N(9) = 226.

It seems that for every integer k > 2 any positive integer n < N(k) cannot be written as x_1 + ... + x_k with x_1,...,x_k positive integers such that lcm(x_1,...,x_k) is a product of k consecutive integers.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..1600

EXAMPLE

a(7) = 1 since 7 = 3 + 2 + 2 with lcm(3,2,2) = 6 = 1*2*3.

a(9) = 1 since 9 = 6 + 2 + 1 with lcm(6,2,1) = 6 = 1*2*3.

a(11) = 1 since 11 = 6 + 3 + 2 with lcm(6,3,2) = 6 = 1*2*3.

a(16) = 2 since 16 = 8 + 6 + 2 with lcm(8,6,2) = 24 = 2*3*4, and 16 = 8 + 5 + 3 with lcm(8,5,3) = 120 = 4*5*6.

a(17) = 2 since 17 = 10 + 4 + 3 with lcm(10,4,3) = 60 = 3*4*5, and 17 = 8 + 6 + 3 with lcm(8,6,3) = 24 = 2*3*4.

MATHEMATICA

P[n_]:=P[n]=n(n+1)(n+2);

f[n_]:=f[n]=Sum[If[P[k]<=n&&P[k+1]>n, k, 0], {k, 1, n^(1/3)}];

YQ[n_]:=YQ[n]=P[f[n]]==n;