A233567 -id:A233567 - cp's OEIS frontend

A234246 a(n) = |{0 < k < n: k*phi(n-k) + 1 is a square}|, where phi(.) is Euler's totient function.

0, 0, 0, 1, 1, 0, 2, 1, 1, 3, 2, 1, 1, 2, 3, 4, 5, 4, 2, 2, 2, 5, 4, 1, 5, 4, 4, 3, 2, 8, 5, 2, 1, 3, 9, 5, 9, 4, 4, 6, 2, 4, 9, 5, 5, 7, 9, 3, 1, 10, 6, 8, 3, 6, 4, 5, 7, 8, 3, 5, 5, 4, 6, 6, 10, 14, 8, 3, 3, 6, 9, 5, 7, 7, 9, 2, 8, 8, 9, 5, 6, 6, 6, 8, 9, 7, 9, 4, 5, 9, 10, 8, 8, 7, 14, 9, 5, 7, 6, 10

Offset: 1

Zhi-Wei Sun, Dec 21 2013

nonn

Conjecture: (i) a(n) > 0 if n is not a divisor of 6. The only values of n with a(n) = 1 are 4, 5, 8, 9, 12, 13, 24, 33, 49.
(ii) If n >= 60, then k + phi(n-k) is a square for some 0 < k < n. If n > 60, then sigma(k) + phi(n-k) is a square for some 0 < k < n, where sigma(k) is the sum of all positive divisors of k.
(iii) If n > 7 is not equal to 10 or 20, then phi(k)*phi(n-k) + 1 is a square for some 0 < k < n.
(iv) If n > 7 is not equal to 10 or 19, then (phi(k) + phi(n-k))/2 is a triangular number for some 0 < k < n.
Note that (n - 1)*phi(1) + 1 = n. So a(n) > 0 if n is a square.

			a(4) = 1 since 3*phi(1) + 1 = 2^2.
a(5) = 1 since 3*phi(2) + 1 = 2^2.
a(8) = 1 since 4*phi(4) + 1 = 3^2.
a(9) = 1 since 8*phi(1) + 1 = 3^2.
a(12) = 1 since 2*phi(10) + 1 = 3^2.
a(13) = 1 since 4*phi(9) + 1 = 5^2.
a(14) = 2 since 2*phi(12) + 1 = 3^2 and 6*phi(8) + 1 = 5^2.
a(24) = 1 since 12*phi(12) + 1 = 7^2.
a(33) = 1 since 3*phi(30) + 1 = 5^2.
a(49) = 1 since 48*phi(1) + 1 = 7^2.

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

Cf. A000010, A000290, A233542, A233544, A233547, A233566, A233567, A233867, A233918, A234200

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

A233918 a(n) = |{0 < k <= n/2: (phi(k) + phi(n-k))/2 is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 2, 2, 1, 2, 3, 1, 3, 2, 4, 3, 2, 7, 1, 3, 3, 4, 7, 2, 4, 5, 5, 5, 5, 6, 6, 4, 7, 5, 6, 4, 4, 11, 5, 5, 5, 11, 4, 3, 5, 7, 12, 4, 6, 11, 3, 6, 7, 8, 6, 7, 8, 11, 10, 5, 9, 7, 9, 5, 4, 14, 8, 9, 6, 10, 7, 6, 10, 9, 10, 7, 10, 11, 7, 7, 13, 11, 13, 5, 8, 11, 9, 9, 3, 12, 4, 11, 13, 11, 19, 8, 12, 11, 7

Offset: 1

Zhi-Wei Sun, Dec 21 2013

nonn

Conjecture: (i) a(n) > 0 for all n > 5.
(ii) If n > 5 is not equal to 19, then phi(k) + phi(n-k) - 1 and phi(k) + phi(n-k) + 1 are both prime for some 0 < k < n.
(iii) If n > 5, then (phi(k)/2)^2 + (phi(n-k)/2)^2 is prime for some 0 < k < n.
(iv) If n > 8, then (sigma(k) + phi(n-k))/2 is prime for some 0 < k < n, where sigma(k) is the sum of all positive divisors of k.

			a(6) = 1 since (phi(3) + phi(3))/2 = 2 is prime.
a(7) = 1 since (phi(3) + phi(4))/2 = 2 is prime.
a(10) = 1 since (phi(4) + phi (6))/2 = 2 is prime.
a(13) = 1 since (phi(3) + phi(10))/2 = 3 is prime.
a(20) = 1 since (phi(4) + phi(16))/2 = 5 is prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, New representation problems involving Euler's totient function, a message to Number Theory List, Dec. 18, 2013.

Cf. A000010, A000040, A233542, A233544, A233547, A233566, A233567, A233867, A234200.

a[n_]:=Sum[If[PrimeQ[(EulerPhi[k]+EulerPhi[n-k])/2],1,0],{k,1,n/2}]
Table[a[n],{n,1,100}]

A234200 a(n) = |{0 < k < n/2: kphi(n-k) - 1 and kphi(n-k) + 1 are both prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Dec 21 2013

nonn

Conjecture: (i) a(n) > 0 for all n > 4.
(ii) If n > 3 is different from 9 and 29, then k*sigma(n-k) - 1 and k*sigma(n-k) + 1 are both prime for some 0 < k < n.
Obviously, either of the two parts implies the twin prime conjecture. We have verified part (i) for n up to 10^8.

			a(5) = 1 since 2*phi(3) - 1 = 3 and 2*phi(3) + 1 = 5 are both prime.
a(7) = 1 since 3*phi(4) - 1 = 5 and 3*phi(4) + 1 = 7 are both prime.
a(18) = 1 since 5*phi(13) - 1 = 59 and 5*phi(13) + 1 = 61 are both prime.
a(91) = 1 since 13*phi(78) - 1 = 311 and 13*phi(78) + 1 = 313 are both prime.
a(101) = 1 since 6*phi(95) - 1 = 431 and 6*phi(95) + 1 = 433 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, New representation problems involving Euler's totient function, a message to Number Theory List, Dec. 18, 2013.

Cf. A000010, A000040, A001359, A006512, A014574, A233542, A233544, A233547, A233566, A233567, A233867, A233918.

TQ[n_]:=PrimeQ[n-1]&&PrimeQ[n+1]
a[n_]:=Sum[If[TQ[k*EulerPhi[n-k]],1,0],{k,1,(n-1)/2}]
Table[a[n],{n,1,100}]

cp's OEIS Frontend

A234246 a(n) = |{0 < k < n: k*phi(n-k) + 1 is a square}|, where phi(.) is Euler's totient function.

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A233918 a(n) = |{0 < k <= n/2: (phi(k) + phi(n-k))/2 is prime}|, where phi(.) is Euler's totient function.

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A234200 a(n) = |{0 < k < n/2: kphi(n-k) - 1 and kphi(n-k) + 1 are both prime}|, where phi(.) is Euler's totient function.

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cp's OEIS Frontend

A234246 a(n) = |{0 < k < n: k*phi(n-k) + 1 is a square}|, where phi(.) is Euler's totient function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A233918 a(n) = |{0 < k <= n/2: (phi(k) + phi(n-k))/2 is prime}|, where phi(.) is Euler's totient function.

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Author

Keywords

Comments

Examples

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Crossrefs

Programs

Mathematica

A234200 a(n) = |{0 < k < n/2: k*phi(n-k) - 1 and k*phi(n-k) + 1 are both prime}|, where phi(.) is Euler's totient function.

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Author

Keywords

Comments

Examples

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Programs

Mathematica

A234200 a(n) = |{0 < k < n/2: kphi(n-k) - 1 and kphi(n-k) + 1 are both prime}|, where phi(.) is Euler's totient function.