A234309 -id:A234309 - 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}]

A234310 Primes of the form 4^k + 4^m - 1, where k and m are positive integers.

Original entry on oeis.org

7, 19, 31, 67, 79, 127, 271, 1039, 1087, 1279, 4099, 4111, 4159, 5119, 8191, 16447, 20479, 65539, 65551, 65599, 81919, 131071, 262147, 262399, 263167, 266239, 524287, 1049599, 1114111, 1310719, 4194319, 4194559, 4195327, 16842751, 17825791, 67108879

Offset: 1

Zhi-Wei Sun, Dec 23 2013

nonn

Clearly each term is congruent to 1 modulo 6.
By the conjecture in A234309, this sequence should have infinitely many terms.
Note that any Mersenne prime greater than 3 has the form 2^{2*k+1} - 1 = 4^k + 4^k - 1, where k is a positive integer.

			a(1) = 7 since 7 = 4^1 + 4^1 - 1 is prime.
a(2) = 19 since 19 = 4^1 + 4^2 - 1 is prime.
a(3) = 31 since 31 = 4^2 + 4^2 - 1 is prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..800

Cf. A000040, A000302, A000668, A233346, A233393, A234309.

n=0;Do[If[PrimeQ[4^k+4^m-1],n=n+1;Print[n," ",4^m+4^k-1]],{m,1,250},{k,1,m}]

for(k=1,30,for(m=1,k,if(ispseudoprime(t=4^k+4^m-1),print1(t", ")))) \\ Charles R Greathouse IV, Dec 23 2013

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}]

A234344 a(n) = |{0 < k < n: 2^{phi(k)/2} + 3^{phi(n-k)/2} is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 6, 8, 7, 9, 12, 12, 10, 10, 10, 10, 16, 7, 11, 9, 6, 14, 11, 17, 12, 15, 15, 17, 16, 15, 19, 18, 12, 13, 9, 20, 11, 8, 17, 19, 19, 12, 17, 14, 16, 9, 21, 16, 13, 12, 16, 19, 17, 11, 21, 15, 16, 15, 17, 19, 16, 23, 11, 20, 15

Offset: 1

Zhi-Wei Sun, Dec 23 2013

nonn

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

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

			a(6) = 1 since 2^{phi(3)/2} + 3^{phi(3)/2} = 5 is prime.
a(8) = 3 since 2^{phi(3)/2} + 3^{phi(5)/2} = 11, 2^{phi(4)/2} + 3^{phi(4)/2} = 5, and 2^{phi(5)/2} + 3^{phi(3)/2} = 7 are all prime.

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

Cf. A000010, A000040, A000079, A000244, A004051, A234309, A234310, A234337, A234346, A234347

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

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.

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A234310 Primes of the form 4^k + 4^m - 1, where k and m are positive integers.

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

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A234344 a(n) = |{0 < k < n: 2^{phi(k)/2} + 3^{phi(n-k)/2} is prime}|, where phi(.) is Euler's totient function.

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

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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).

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

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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.

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