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 A233386 Number of ways to write n = i + j + k with 0 < i <= j <= k and i, j, k not all equal such that phi(i)*phi(j)*phi(k) is a cube. 7

%I #16 Feb 02 2014 13:16:35

%S 0,0,0,1,1,0,0,0,1,3,2,3,4,3,4,4,4,5,3,3,1,2,6,5,5,6,7,8,8,7,8,13,7,9,

%T 4,6,8,10,7,11,14,12,8,9,10,14,12,9,9,8,8,11,8,9,19,14,12,9,11,19,12,

%U 19,10,15,13,22,18,27,22,31,20,22,18,25,25,24,18,22,19,21,24,22,30,31,35,25,28,32,23,27,28,29,23,24,30,30,29,30,33,31

%N Number of ways to write n = i + j + k with 0 < i <= j <= k and i, j, k not all equal such that phi(i)*phi(j)*phi(k) is a cube.

%C Conjecture: For each k = 3, 4, ..., any integer n >= 3*k can be written as n_1 + n_2 + ... + n_k with n_1, n_2, ..., n_k positive and not all equal such that the product phi(n_1)*phi(n_2)*...*phi(n_k) is a k-th power.

%C We have verified this conjecture with k = 3 for n up to 10^5 and with k = 4, 5, 6 for n up to 30000.

%C See also A236998 for a similar conjecture with k = 2.

%H Zhi-Wei Sun, <a href="/A233386/b233386.txt">Table of n, a(n) for n = 1..1000</a>

%e a(9) = 1 since 9 = 1 + 3 + 5 with phi(1)*phi(3)*phi(5) = 1*2*4 = 2^3.

%e a(21) = 1 since 21 = 5 + 8 + 8 with phi(5)*phi(8)*phi(8) = 4*4*4 = 4^3.

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

%t p[i_,j_,k_]:=CQ[EulerPhi[i]*EulerPhi[j]*EulerPhi[k]]

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

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

%Y Cf. A000010, A000578, A236998.

%K nonn

%O 1,10

%A _Zhi-Wei Sun_, Feb 02 2014

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Last modified September 19 09:52 EDT 2024. Contains 376008 sequences. (Running on oeis4.)