A233386 -id:A233386 - cp's OEIS frontend

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.

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}]

A237524 Number of ordered ways to write n = i + j + k with 0 < i <= j <= k such that phi(ijk) is a cube, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 1, 1, 0, 0, 0, 1, 4, 3, 2, 2, 1, 1, 1, 1, 2, 5, 2, 3, 2, 3, 6, 5, 4, 4, 4, 5, 4, 5, 4, 6, 6, 5, 5, 9, 6, 10, 8, 7, 7, 5, 5, 4, 11, 10, 8, 10, 5, 8, 8, 10, 10, 8, 11, 16, 11, 13, 14, 16, 18, 19, 18, 16, 24, 19, 21, 18, 15, 21, 9, 15, 14, 13, 15, 18, 19, 20, 15, 19

Offset: 1

Zhi-Wei Sun, Feb 09 2014

nonn

Conjecture: For each k = 3, 4, ..., any integer n > 2*k + 1 can be written as a sum of k positive integers n_1, n_2, ..., n_k such that phi(n_1*n_2*...*n_k) is a k-th power.
Note that 2*k + 2 = (k-1)*2 + 4 with phi(2^(k-1)*4) = 2^k.
See also A237523 for a similar conjecture with k = 2.

			a(4) = 1 since 4 = 1 + 1 + 2 with phi(1*1*2) = 1^3.
a(13) = 1 since 13 = 1 + 2 + 10 with phi(1*2*10) = 2^3.
a(16) = 1 since 16 = 4 + 4 + 8 with phi(4*4*8) = phi(2^7) = 4^3.

Zhi-Wei Sun, Table of n, a(n) for n = 1..1500

Cf. A000010, A000578, A233386, A237050, A237123, A237523.

CQ[n_]:=IntegerQ[n^(1/3)]
q[n_]:=CQ[EulerPhi[n]]
a[n_]:=Sum[If[q[i*j(n-i-j)],1,0],{i,1,n/3},{j,i,(n-i)/2}]
Table[a[n],{n,1,80}]

A290399 Number of solutions to Diophantine equation x + y + z = prime(n) with xyz = k^3 (0 < x <= y <= z).

Original entry on oeis.org

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

Offset: 1

XU Pingya, Jul 29 2017

nonn

			a(11) = 2 because the equation x + y + z = 31 (prime(11)) has exactly 2 solutions with x*y*z = k^3: (x, y, z) = (1, 5, 25) and (1, 12, 18), which satisfy 1*5*25 = 5^3 and 1*12*18 = 6^3.

Tianxin Cai and Deyi Chen, A new variant of the Hilbert-Waring problem, Math. Comp. 82 (2013), 2333-2341.

Cf. A000040, A000578, A233386.

a[n_] := Length@ Select[ IntegerPartitions[ Prime[n], {3}], IntegerQ[ (Times @@ #)^(1/3)] &]; Array[a, 50] (* Giovanni Resta, Aug 07 2017 *)

A290400 Primes p such that Diophantine equation x + y + z = p with xyz = k^3 (0 < x <= y <= z) has a unique solution.

Original entry on oeis.org

3, 7, 13, 17, 19, 23, 29, 37, 53, 71, 101, 149, 157, 317, 347

Offset: 1

XU Pingya, Jul 29 2017

			7 is in the sequence since, of the triples whose sum is 7, i.e., (1, 1, 5), (1, 2, 4), (1, 3, 3), and (2, 2, 3), only one (i.e., (1, 2, 4)), yields a cube as its product: 1 * 2 * 4 = 8 = 2^3.
31 is not here, since the corresponding equation has two solutions: (1, 5, 25) and (1, 12, 18).

Tianxin Cai and Deyi Chen, A new variant of the Hilbert-Waring problem, Math. Comp. 82 (2013), 2333-2341.

Cf. A000040, A000578, A233386.

cp's OEIS Frontend

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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A237524 Number of ordered ways to write n = i + j + k with 0 < i <= j <= k such that phi(i*j*k) is a cube, where phi(.) is Euler's totient function.

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A290399 Number of solutions to Diophantine equation x + y + z = prime(n) with x*y*z = k^3 (0 < x <= y <= z).

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A290400 Primes p such that Diophantine equation x + y + z = p with x*y*z = k^3 (0 < x <= y <= z) has a unique solution.

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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.

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.

A237524 Number of ordered ways to write n = i + j + k with 0 < i <= j <= k such that phi(ijk) is a cube, where phi(.) is Euler's totient function.

A290399 Number of solutions to Diophantine equation x + y + z = prime(n) with xyz = k^3 (0 < x <= y <= z).

A290400 Primes p such that Diophantine equation x + y + z = p with xyz = k^3 (0 < x <= y <= z) has a unique solution.