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A237049 Number of ordered ways to write n = i + j + k (0 < i <= j <= k) with i,j,k not all equal such that sigma(i)*sigma(j)*sigma(k) is a cube, where sigma(m) denotes the sum of all positive divisors of m. 2
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 2, 1, 0, 0, 2, 2, 1, 1, 1, 3, 4, 2, 3, 3, 1, 2, 3, 2, 3, 2, 2, 3, 2, 1, 0, 5, 3, 4, 3, 1, 2, 4, 1, 2, 3, 5, 7, 5, 6, 3, 4, 6, 7, 6, 7, 3, 8, 2, 7, 6, 4, 3, 8, 7, 6, 6, 2, 7, 5, 7, 2, 8, 4, 8, 6, 5, 7, 7, 9, 10, 5, 9, 7, 11, 3, 6, 7, 8, 8, 7, 5, 6, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,15

COMMENTS

Conjecture: For every k = 2, 3, ... there is a positive integer N(k) such that any integer n > N(k) can be written as n_1 + n_2 + ... + n_k with n_1, n_2, ..., n_k positive and distinct such that the product sigma(n_1)*sigma(n_2)*...*sigma(n_k) is a k-th power. In particular, we may take N(2) = 309, N(3) = 42, N(4) = 25, N(5) = 24, N(6) = 27 and N(7) = 32.

This is similar to the conjecture in A233386.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..1000

EXAMPLE

a(9) = 1 since 9 = 1 + 1 + 7 with sigma(1)*sigma(1)*sigma(7) = 1*1*8 = 2^3.

a(41) = 1 since 41 = 2 + 6 + 33 with sigma(2)*sigma(6)*sigma(33) = 3*12*48 = 12^3.

a(50) = 1 since 50 = 2 + 17 + 31 with sigma(2)*sigma(17)*sigma(31) = 3*18*32 = 12^3.

MATHEMATICA

sigma[n_]:=DivisorSigma[1, n]

CQ[n_]:=IntegerQ[n^(1/3)]

p[i_, j_, k_]:=CQ[sigma[i]*sigma[j]*sigma[k]]

a[n_]:=Sum[If[p[i, j, n-i-j], 1, 0], {i, 1, (n-1)/3}, {j, i, (n-i)/2}]

Table[a[n], {n, 1, 100}]

CROSSREFS

Cf. A000203, A000578, A233386, A236998, A237016.

Sequence in context: A136487 A178108 A021501 * A101662 A091064 A275760

Adjacent sequences:  A237046 A237047 A237048 * A237050 A237051 A237052

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Feb 02 2014

STATUS

approved

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Last modified February 22 16:26 EST 2019. Contains 320399 sequences. (Running on oeis4.)