A236998 -id:A236998 - cp's OEIS frontend

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.

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, 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, 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

Offset: 1

Zhi-Wei Sun, Feb 02 2014

nonn

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

			a(9) = 1 since 9 = 1 + 3 + 5 with phi(1)*phi(3)*phi(5) = 1*2*4 = 2^3.
a(21) = 1 since 21 = 5 + 8 + 8 with phi(5)*phi(8)*phi(8) = 4*4*4 = 4^3.

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

Cf. A000010, A000578, A236998.

CQ[n_]:=IntegerQ[n^(1/3)]
p[i_,j_,k_]:=CQ[EulerPhi[i]*EulerPhi[j]*EulerPhi[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}]

A237016 a(n) = |{0 < k < n: phi(k)*sigma(n-k) is a square}|, where phi(.) is Euler's totient function and sigma(j) is the sum of all positive divisors of j.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 3, 1, 4, 2, 2, 1, 0, 1, 2, 2, 2, 2, 6, 4, 2, 2, 4, 2, 2, 4, 1, 6, 5, 6, 3, 3, 8, 3, 2, 4, 6, 1, 2, 4, 3, 3, 3, 5, 6, 5, 5, 3, 2, 5, 4, 4, 3, 6, 5, 7, 10, 7, 4, 2, 1, 4, 6, 7, 9, 6, 12, 3, 3, 4, 12, 6, 6, 5, 6, 4, 5, 8, 6, 5, 10, 7, 7, 2, 5, 8, 4, 2, 4, 3, 8, 4, 4, 11, 6, 6

Offset: 1

Zhi-Wei Sun, Feb 02 2014

nonn

Conjecture: (i) a(n) > 0 except for n = 1, 7, 17.
(ii) If n > 5, then phi(k)*sigma(n-k) + 1 is a square for some 0 < k < n.
(iii) If n > 309, then there is a positive integer k < n/2 such that sigma(k)*sigma(n-k) is a square.
See also A236998 for a similar conjecture.

			a(9) = 1 since phi(8)*sigma(1) = 4*1 = 2^2.
a(16) = 1 since phi(6)*sigma(10) = 2*18 = 6^2.
a(31) = 1 since phi(24)*sigma(7) = 8*8 = 8^2.
a(65) = 1 since phi(19)*sigma(46) = 18*72 = 36^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000010, A000203, A000290, A234246, A236567, A236998.

sigma[n_]:=DivisorSigma[1,n]
SQ[n_]:=IntegerQ[Sqrt[n]]
p[n_,k_]:=SQ[EulerPhi[k]*sigma[n-k]]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

A237523 a(n) = |{0 < k < n/2: phi(k*(n-k)) + 1 is a square}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 2, 1, 2, 1, 2, 5, 4, 4, 1, 3, 3, 3, 2, 4, 4, 4, 2, 4, 5, 6, 5, 3, 3, 3, 6, 5, 4, 4, 6, 6, 2, 6, 6, 6, 2, 6, 5, 5, 2, 4, 4, 7, 7, 4, 3, 5, 5, 9, 5, 5, 3, 5, 2, 3, 10, 10, 9, 7, 5, 8, 5, 9, 8, 6, 4, 5, 6, 11, 5

Offset: 1

Zhi-Wei Sun, Feb 08 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 7, and a(n) = 1 only for n = 8, 9, 13, 15, 20, 132.
(ii) If n > 5 is not among 10, 15, 20, 60, 105, then phi(k*(n-k)) is a square for some 0 < k < n/2.
See also A237524 for a similar conjecture involving higher powers.

			a(8) = 1 since phi(3*5) + 1 = 8 + 1 = 3^2.
a(9) = 1 since phi(4*5) + 1 = 8 + 1 = 3^2.
a(13) = 1 since phi(3*10) + 1 = 8 + 1 = 3^2.
a(15) = 1 since phi(7*8) + 1 = 24 + 1 = 5^2.
a(20) = 1 since phi(6*14) + 1 = 24 + 1 = 5^2.
a(132) = 1 since phi(46*(132-46)) + 1 = 1848 + 1 = 43^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000010, A000290, A234246, A236998, A237497, A237524.

SQ[n_]:=IntegerQ[Sqrt[n]]
s[n_]:=SQ[EulerPhi[n]+1]
a[n_]:=Sum[If[s[k(n-k)],1,0],{k,1,(n-1)/2}]
Table[a[n],{n,1,80}]

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.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Feb 02 2014

nonn

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.

			 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.

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

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

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}]

A237050 Number of ways to write n = i_1 + i_2 + i_3 + i_4 + i_5 (0 < i_1 <= i_2 <= i_3 <= i_4 <= i_5) with i_1, i_2, ..., i_5 not all equal such that the product i_1i_2i_3i_4i_5 is a fifth power.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 5, 4, 3, 3, 3, 5, 4, 5, 7, 3, 5, 3, 4, 3, 3, 4, 6, 4, 4, 4, 4, 2, 4, 3, 5, 5, 3, 5, 4, 8, 7, 7, 9, 10, 9, 12, 7, 6, 9, 10, 9, 9, 8, 8, 7, 10, 7, 10, 10, 10, 10, 5, 8, 13, 10, 9, 8, 12, 15, 10, 12, 9, 8

Offset: 1

Zhi-Wei Sun, Feb 02 2014

nonn

Conjecture: For each k = 3, 4, ... there is a positive integer M(k) such that any integer n > M(k) can be written as i_1 + i_2 + ... + i_k with i_1, i_2, ..., i_k positive and not all equal such that the product i_1*i_2*...*i_k is a k-th power. In particular, we may take M(3) = 486, M(4) = 23, M(5) = 26, M(6) = 36 and M(7) = 31.

This is motivated by the conjectures in A233386 and A237049.

			a(25) = 1 since 25 = 1 + 4 + 4 + 8 + 8 with 1*4*4*8*8 = 4^5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..300
Tianxin Cai and Deyi Chen, A new variant of the Hilbert-Waring problem, Math. Comp. 82 (2013), 2333-2341.

Cf. A000584, A233386, A236998, A237016, A237049.

QQ[n_]:=IntegerQ[n^(1/5)]
a[n_]:=Sum[If[QQ[i*j*h*k*(n-i-j-h-k)],1,0],{i,1,(n-1)/5},{j,i,(n-i)/4},{h,j,(n-i-j)/3},{k,h,(n-i-j-h)/2}]
Table[a[n],{n,1,100}]

A237123 Number of ways to write n = i + j + k with 0 < i < j < k such that phi(i), phi(j) and phi(k) are all cubes, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 0, 0, 1, 2, 1, 0, 1, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1

Offset: 1

Zhi-Wei Sun, Feb 03 2014

nonn

Conjecture: For each k = 2, 3, ... there is a positive integer s(k) such that any integer n >= s(k) can be written as i_1 + i_2 + ... + i_k with 0 < i_1 < i_2 < ... < i_k such that all those phi(i_1), phi(i_2), ..., phi(i_k) are k-th powers. In particular, we may take s(2) = 70640, s(3) = 935 and s(4) = 3273.

			a(18) = 1 since 18 = 1 + 2 + 15 with phi(1) = 1^3, phi(2) = 1^3 and phi(15) = 2^3.
a(101) = 1 since 101 = 1 + 15 + 85 with phi(1) = 1^3, phi(15) = 2^3 and phi(85) = 4^3.
a(1613) = 1 since 1613 = 192 + 333 + 1088 with phi(192) = 4^3, phi(333) = 6^3 and phi(1088) = 8^3.

Zhi-Wei Sun, Table of n, a(n) for n = 1..7000

Cf. A000010, A000578, A039771, A233386, A236998.

CQ[n_]:=IntegerQ[EulerPhi[n]^(1/3)]
a[n_]:=Sum[If[CQ[i]&&CQ[j]&&CQ[n-i-j],1,0],{i,1,n/3-1},{j,i+1,(n-1-i)/2}]
Table[a[n],{n,1,70}]

A237203 Least positive integer k < n/2 with phi(k)*phi(n-k) a square, or 0 if such a number k does not exist.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 2, 0, 1, 2, 1, 2, 1, 2, 5, 7, 5, 1, 2, 4, 6, 3, 3, 4, 6, 6, 3, 4, 12, 3, 4, 14, 1, 2, 1, 2, 5, 1, 2, 8, 1, 2, 16, 6, 5, 7, 10, 8, 1, 2, 17, 7, 5, 3, 4, 8, 3, 1, 2, 6, 1, 2, 7, 1, 2, 11, 3, 4, 12, 6

Offset: 1

Zhi-Wei Sun, Feb 05 2014

nonn

Conjecture: a(n) < sqrt(n)*log(2*n) for all n > 0.
We have verified this for n up to 2*10^5. Note that a(211) = 85 > sqrt(211)*log(211) and a(373) = 117 > sqrt(373)*log(373).
According to the conjecture in A236998, a(n) should be positive for all n > 8.

			 a(7) = 2 since phi(2)*phi(7-2) = 1*4 = 2^2 but phi(1)*phi(7-1) = 2 is not a square.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000010, A000290, A236998.

SQ[k_,m_]:=IntegerQ[Sqrt[EulerPhi[k]*EulerPhi[m]]]
Do[Do[If[SQ[k,n-k],Print[n," ",k];Goto[aa]],{k,1,(n-1)/2}];
Print[n," ",0];Label[aa];Continue,{n,1,70}]

cp's OEIS Frontend

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.

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A237016 a(n) = |{0 < k < n: phi(k)*sigma(n-k) is a square}|, where phi(.) is Euler's totient function and sigma(j) is the sum of all positive divisors of j.

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A237523 a(n) = |{0 < k < n/2: phi(k*(n-k)) + 1 is a square}|, where phi(.) is Euler's totient function.

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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.

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A237050 Number of ways to write n = i_1 + i_2 + i_3 + i_4 + i_5 (0 < i_1 <= i_2 <= i_3 <= i_4 <= i_5) with i_1, i_2, ..., i_5 not all equal such that the product i_1*i_2*i_3*i_4*i_5 is a fifth power.

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A237123 Number of ways to write n = i + j + k with 0 < i < j < k such that phi(i), phi(j) and phi(k) are all cubes, where phi(.) is Euler's totient function.

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A237203 Least positive integer k < n/2 with phi(k)*phi(n-k) a square, or 0 if such a number k does not exist.

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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.

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.

A237050 Number of ways to write n = i_1 + i_2 + i_3 + i_4 + i_5 (0 < i_1 <= i_2 <= i_3 <= i_4 <= i_5) with i_1, i_2, ..., i_5 not all equal such that the product i_1i_2i_3i_4i_5 is a fifth power.