cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A236998 a(n) = |{0 < k < n/2: phi(k)*phi(n-k) is a square}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 02 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 8.
(ii) If n > 20, then phi(k)*phi(n-k) + 1 is a square for some 0 < k < n/2.
(iii) If n > 1 is not among 4, 7, 60, 199, 267, then k*phi(n-k) is a square for some 0 < k < n.
We have verified part (i) of the conjecture for n up to 2*10^6.

Examples

			a(17) = 1 since phi(5)*phi(12) = 4*4 = 4^2.
a(24) = 1 since phi(4)*phi(20) = 2*8 = 4^2.
a(56) = 1 since phi(8)*phi(48) = 4*16 = 8^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
    p[n_,k_]:=SQ[EulerPhi[k]*EulerPhi[n-k]]
    a[n_]:=Sum[If[p[n,k],1,0],{k,1,(n-1)/2}]
    Table[a[n],{n,1,100}]

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

Views

Author

Zhi-Wei Sun, Feb 02 2014

Keywords

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.

Examples

			 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.
		

Crossrefs

Programs

  • 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}]

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.

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

Views

Author

Zhi-Wei Sun, Feb 02 2014

Keywords

Comments

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.

Examples

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

Crossrefs

Programs

  • Mathematica
    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}]
Showing 1-3 of 3 results.