A234344 -id:A234344 - cp's OEIS frontend

A234470 Number of ways to write n = k + m with k > 0 and m > 2 such that p(k + phi(m)/2) is prime, where p(.) is the partition function (A000041) and phi(.) is Euler's totient function.

0, 0, 0, 1, 2, 3, 4, 5, 5, 4, 4, 4, 2, 2, 3, 5, 4, 2, 4, 2, 3, 2, 3, 2, 3, 1, 0, 3, 1, 1, 2, 1, 2, 0, 1, 2, 1, 1, 4, 2, 1, 4, 2, 1, 2, 3, 3, 3, 1, 0, 4, 2, 4, 1, 1, 2, 2, 3, 2, 2, 0, 2, 2, 1, 2, 2, 1, 1, 2, 2, 4, 2, 1, 0, 1, 3, 1, 0, 2, 4, 3, 1, 6, 2, 2, 1, 2, 4, 3, 1, 2, 6, 2, 3, 2, 2, 2, 2, 3, 3

Offset: 1

Zhi-Wei Sun, Dec 26 2013

nonn

Conjecture: a(n) > 0 if n > 3 is not among 27, 34, 50, 61, 74, 78, 115, 120, 123, 127.

This implies that there are infinitely many primes in the range of the partition function p(n).

			a(26) = 1 since 26 = 2 + 24 with p(2 + phi(24)/2) = p(6) = 11 prime.
a(54) = 1 since 54 = 27 + 27 with p(27 + phi(27)/2) = p(36) = 17977 prime.
a(73) = 1 since 73 = 1 + 72 with p(1 + phi(72)/2) = p(36) = 17977 prime.
a(110) = 1 since 110 = 65 + 45 with p(65 + phi(45)/2) = p(77) = 10619863 prime.
a(150) = 1 since 150 = 123 + 27 with p(123 + phi(27)/2) = p(132) = 6620830889 prime.
a(170) = 1 since 170 = 167 + 3 with p(167 + phi(3)/2) = p(168) = 228204732751 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..6500
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Cf. A000010, A000040, A000041, A049575, A232504, A233307, A233346, A233918, A234200, A234246, A234309, A234337, A234344, A234347, A234359, A234360, A234451, A234475

f[n_,k_]:=PartitionsP[k+EulerPhi[n-k]/2]
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-3}]
Table[a[n],{n,1,100}]

A234309 a(n) = |{2 < k <= n/2: 2^{phi(k)} + 2^{phi(n-k)} - 1 is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Dec 23 2013

nonn

Conjecture: (i) a(n) > 0 for all n > 5.
(ii) For any integer n > 1, 2^k +2^{phi(n-k)} - 1 is prime for some 0 < k < n, and 2^{sigma(j)} + 2^{phi(n-j)} - 1 is prime for some 0 < j < n, where sigma(j) is the sum of all positive divisors of j.
As phi(k) is even for any k > 2, part (i) of the conjecture implies that there are infinitely many primes of the form 4^a + 4^b - 1 with a and b positive integers (cf. A234310). Note that any Mersenne prime greater than 3 has the form 2^{2a+1} - 1 = 4^a + 4^a - 1.

			a(6) = 1 since 2^{phi(3)} + 2^{phi(3)} - 1 = 2^2 + 2^2 - 1 = 7 is prime.
a(7) = 1 since 2^{phi(3)} + 2^{phi(4)} - 1 = 2^2 + 2^2 - 1 = 7 is prime.
a(8) = 2 since 2^{phi(3)} + 2^{phi(5)} - 1 = 2^2 + 2^4 - 1 = 19 and 2^{phi(4)} + 2^{phi(4)} - 1 = 2^2 + 2^2 - 1 = 7 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..8000

Cf. A000010, A000040, A000079, A000668, A233307, A233390, A233542, A233544, A233547, A233566, A233867, A233918, A234200, A234246, A234308, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361

a[n_]:=Sum[If[PrimeQ[2^(EulerPhi[k])+2^(EulerPhi[n-k])-1],1,0],{k,3,n/2}]
Table[a[n],{n,1,100}]

A234337 a(n) = |{0 < k < n - 2: 4^k + 2^{phi(n-k)} - 1 is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Dec 23 2013

nonn

Conjecture: Let a be 2 or 3 or 4. If n > 3, then a^k + a^{phi(n-k)/2} - 1 is prime for some 0 < k < n - 2.

This conjecture for a = 4 implies that there are infinitely many terms of the sequence A234310. The conjecture for a = 3 implies that there are infinitely many primes of the form 3^k + 3^m - 1 (cf. A234346), where k and m are positive integers.

			a(4) = 1 since 4^1 + 2^{phi(3)} - 1 = 7 is prime.
a(5) = 2 since 4^1 + 2^{phi(4)} - 1 = 7 and 4^2 + 2^{phi(3)} - 1 = 19 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..4000

Cf. A000010, A000040, A000244, A000302, A234309, A234310, A234344, A234346, A234347, A234359, A234360, A234361

f[n_,k_]:=4^k+2^(EulerPhi[n-k])-1
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-3}]
Table[a[n],{n,1,100}]

A234347 a(n) = |{0 < k < n: 3^k + 3^{phi(n-k)/2} - 1 is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Dec 24 2013

nonn

Conjecture: a(n) > 0 for all n > 3.

A234475 Number of ways to write n = k + m with 2 < k <= m such that q(phi(k)*phi(m)/4) + 1 is prime, where phi(.) is Euler's totient function and q(.) is the strict partition function (A000009).

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Dec 26 2013

nonn

Conjecture: a(n) > 0 for all n > 5.

This implies that there are infinitely many primes p with p - 1 a term of A000009.

			a(6) = 1 since 6 = 3 + 3 with q(phi(3)*phi(3)/4) + 1 = q(1) + 1 = 2 prime.
a(76) = 1 since 76 = 18 + 58 with q(phi(18)*phi(58)/4) + 1 = q(42) + 1 = 1427 prime.
a(197) = 1 since 197 = 4 + 193 with q(phi(4)*phi(193)/4) + 1 = q(96) + 1 = 317789.
a(356) = 1 since 356 = 88 + 268 with q(phi(88)*phi(268)/4) + 1 = q(1320) + 1 = 35940172290335689735986241 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..2525
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Cf. A000009, A000010, A000040, A232504, A233307, A233346, A233547, A233390, A233393, A234309, A234310, A234337, A234344, A234347, A234359, A234360, A234361, A234451, A234470, A234514, A234530, A234567, A234569

f[n_,k_]:=PartitionsQ[EulerPhi[k]*EulerPhi[n-k]/4]+1
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,3,n/2}]
Table[a[n],{n,1,100}]

A234346 Primes of the form 3^k + 3^m - 1, where k and m are positive integers.

Original entry on oeis.org

5, 11, 17, 29, 53, 83, 89, 107, 251, 269, 809, 971, 2213, 2267, 4373, 6563, 6569, 6803, 8747, 13121, 19709, 19763, 20411, 59051, 65609, 177173, 183707, 531521, 538001, 590489, 1062881, 1594331, 1594403, 1595051, 1596509, 4782971, 4782977, 4783697, 14348909

Offset: 1

Zhi-Wei Sun, Dec 23 2013

nonn

Clearly, all terms are congruent to 5 modulo 6.
By a conjecture in A234337 or A234347, this sequence should have infinitely many terms.
Conjecture: For any integer a > 1, there are infinitely many primes of the form a^k + a^m - 1, where k and m are positive integers.

			a(1) = 5 since 3^1 + 3^1 - 1 = 5 is prime.
a(2) = 11 since 3^2 + 3^1 - 1 = 11 is prime.

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

Cf. A000040, A000079, A000244, A234309, A234310, A234337, A234344, A234347

n=0;Do[If[PrimeQ[3^k+3^m-1],n=n+1;Print[n," ",3^k+3^m-1]],{m,1,310},{k,1,m}]

A234359 a(n) = |{2 < k < n-2: 5^{phi(k)} + 5^{phi(n-k)/2} - 1 is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 1, 2, 4, 2, 4, 4, 3, 4, 3, 6, 5, 4, 6, 7, 8, 6, 7, 11, 7, 10, 9, 9, 7, 10, 11, 8, 7, 11, 10, 9, 6, 11, 15, 4, 14, 5, 14, 11, 13, 9, 13, 6, 12, 10, 12, 11, 10, 10, 13, 9, 7, 11, 7, 11, 4, 11, 9, 10, 6, 11, 8, 4, 10, 12, 13, 9, 7, 9, 6, 12, 10

Offset: 1

Zhi-Wei Sun, Dec 24 2013

nonn

Conjecture: For any integer a > 1, there is a positive integer N(a) such that if n > N(a) then a^{phi(k)} + a^{phi(n-k)/2} - 1 is prime for some 2 < k < n-2. Moreover, we may take N(2) = N(3) = ... = N(6) = N(8) = 5 and N(7) = 17.

Clearly, this conjecture implies that for each a = 2, 3, ... there are infinitely many primes of the form a^{2*k} + a^m - 1, where k and m are positive integers.

			a(6) = 1 since 5^{phi(3)} + 5^{phi(3)/2} - 1 = 29 is prime.
a(11) = 2 since 5^{phi(4)} + 5^{phi(7)/2} - 1 = 149 and 5^{phi(7)} + 5^{phi(4)/2} - 1 = 15629 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..2500

Cf. A000010, A000040, A000351, A234309, A234310, A234337, A234344, A234346, A234347

f[n_,k_]:=5^(EulerPhi[k])+5^(EulerPhi[n-k]/2)-1
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,3,n-3}]
Table[a[n],{n,1,100}]

A234360 a(n) = |{0 < k < n: (k+1)^{phi(n-k)} + k is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Dec 24 2013

nonn

Conjecture: (i) a(n) > 0 for all n > 1. Also, for any n > 5 there is a positive integer k < n with (k+1)^{phi(n-k)/2} - k prime.

(ii) If n > 1, then k*(k+1)^{phi(n-k)} + 1 is prime for some 0 < k < n. If n > 3, then k*(k+1)^{phi(n-k)/2} - 1 is prime for some 0 < k < n.

			a(74) = 2 since (2+1)^{phi(72)} + 2 = 3^{24} + 2 =
282429536483 and (14+1)^{phi(60)} + 14 = 15^{16} + 14 = 6568408355712890639 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..2500

Cf. A000010, A000040, A234309, A234310, A234337, A234344, A234346, A234347, A234359

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

A234361 a(n) = |{0 < k < n: 2^{phi(k)/2}*3^{phi(n-k)/4} + 1 is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 1, 3, 2, 3, 4, 2, 6, 3, 6, 6, 5, 7, 4, 6, 4, 5, 7, 9, 4, 6, 4, 10, 7, 2, 11, 9, 12, 6, 9, 10, 9, 12, 11, 10, 6, 12, 13, 8, 11, 9, 10, 7, 8, 7, 11, 8, 9, 6, 14, 4, 15, 5, 14, 7, 15, 5, 12, 11, 9, 10, 9, 10, 8, 10, 7, 12, 11, 15, 10

Offset: 1

Zhi-Wei Sun, Dec 24 2013

nonn

Conjecture: a(n) > 0 for all n > 7.

This implies that there are infinitely many primes of the form 2^k*3^m + 1, where k and m are positive integers.

			a(10) = 1 since 2^{phi(5)/2}*3^{phi(5)/4} + 1 = 13 is prime.
a(12) = 1 since 2^{phi(4)/2}*3^{phi(8)/4} + 1 = 13 is prime.
a(35) = 2 since 2^{phi(3)/2}*3^{phi(32)/4} + 1 = 2*3^4 + 1 = 163 and 2^{phi(5)/2}*3^{phi(30)/4} + 1 = 2^2*3^2 + 1 = 37 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..5000

Cf. A000010, A000040, A000079, A000244, A234309, A234310, A234337, A234344, A234346, A234347, A234359, A234360.

f[n_,k_]:=f[n,k]=2^(EulerPhi[k]/2)*3^(EulerPhi[n-k]/4)+1
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

A234451 Number of ways to write n = k + m with k > 0 and m > 0 such that 2^(phi(k)/2 + phi(m)/6) + 3 is prime, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Dec 26 2013

nonn

Conjecture: (i) a(n) > 0 for all n > 9. Also, any integer n > 13 can be written as k + m with k > 0 and m > 0 such that 2^(phi(k)/2 + phi(m)/6) - 3 is prime.
(ii) Each integer n > 25 can be written as k + m with k > 0 and m > 0 such that 3*2^(phi(k)/2 + phi(m)/8) + 1 (or 3*2^(phi(k)/2 + phi(m)/12) + 1 when n > 38) is prime. Also, any integer n > 14 can be written as k + m with k > 0 and m > 0 such that 3*2^(phi(k)/2 + phi(m)/12) - 1 is prime.
This conjecture implies that there are infinitely many primes in any of the four forms 2^n + 3, 2^n - 3, 3*2^n + 1, 3*2^n - 1.
We have verified the conjecture for n up to 50000.

			a(10) = 1 since 10 = 3 + 7 with 2^(phi(3)/2 + phi(7)/6) + 3 = 7 prime.
a(11) = 1 since 11 = 4 + 7 with 2^(phi(4)/2 + phi(7)/6) + 3 = 7 prime.
a(12) = 2 since 12 = 3 + 9 = 5 + 7 with 2^(phi(3)/2 + phi(9)/6) + 3 = 7 and 2^(phi(5)/2 + phi(7)/6) + 3 = 11 both prime.
a(769) = 1 since 769 = 31 + 738 with 2^(phi(31)/2 + phi(738)/6) + 3 = 2^(55) + 3 prime.
a(787) = 1 since 787 = 112 + 675 with 2^(phi(112)/2 + phi(675)/6) + 3 = 2^(84) + 3 prime.
a(867) = 1 since 867 = 90 + 777 with 2^(phi(90)/2 + phi(777)/6) + 3 = 2^(84) + 3 prime.
a(869) = 1 since 869 = 51 + 818 with 2^(phi(51)/2 + phi(818)/6) + 3 = 2^(84) + 3 prime.
a(913) = 1 since 913 = 409 + 504 with 2^(phi(409)/2 + phi(504)/6) + 3 = 2^(228) + 3 prime.
a(1085) = 1 since 1085 = 515 + 570 with 2^(phi(515)/2 + phi(570)/6) + 3 = 2^(228) + 3 prime.

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

Cf. A000010, A000040, A000079, A050415, A057733, A234309, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361, A234388, A234399, A234470, A236358.

f[n_,k_]:=2^(EulerPhi[k]/2+EulerPhi[n-k]/6)+3
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

cp's OEIS Frontend

A234470 Number of ways to write n = k + m with k > 0 and m > 2 such that p(k + phi(m)/2) is prime, where p(.) is the partition function (A000041) and phi(.) is Euler's totient function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A234309 a(n) = |{2 < k <= n/2: 2^{phi(k)} + 2^{phi(n-k)} - 1 is prime}|, where phi(.) is Euler's totient function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A234337 a(n) = |{0 < k < n - 2: 4^k + 2^{phi(n-k)} - 1 is prime}|, where phi(.) is Euler's totient function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A234347 a(n) = |{0 < k < n: 3^k + 3^{phi(n-k)/2} - 1 is prime}|, where phi(.) is Euler's totient function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A234475 Number of ways to write n = k + m with 2 < k <= m such that q(phi(k)*phi(m)/4) + 1 is prime, where phi(.) is Euler's totient function and q(.) is the strict partition function (A000009).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A234346 Primes of the form 3^k + 3^m - 1, where k and m are positive integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A234359 a(n) = |{2 < k < n-2: 5^{phi(k)} + 5^{phi(n-k)/2} - 1 is prime}|, where phi(.) is Euler's totient function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A234360 a(n) = |{0 < k < n: (k+1)^{phi(n-k)} + k is prime}|, where phi(.) is Euler's totient function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs