A233544 -id:A233544 - cp's OEIS frontend

A233793 Least odd prime p such that 2*n - p = sigma(k) for some k > 0, or 0 if such an odd prime p does not exist, where sigma(k) is the sum of all (positive) divisors of k.

0, 3, 3, 5, 3, 5, 7, 3, 3, 5, 7, 11, 11, 13, 17, 17, 3, 5, 7, 37, 3, 5, 7, 17, 11, 13, 23, 17, 19, 3, 5, 7, 3, 5, 7, 41, 11, 13, 47, 17, 19, 53, 23, 31, 59, 29, 3, 3, 5, 7, 11, 11, 13, 17, 17, 19, 23, 23, 61, 29, 29, 3, 5, 7, 3, 5, 7, 3, 5, 7, 79, 11, 13, 109, 17, 19, 61, 23, 31, 67, 29, 31, 73, 41, 37, 79, 3, 5, 7, 47, 11, 13, 3, 5, 7, 59, 11, 13, 3, 5

Offset: 1

Zhi-Wei Sun, Dec 15 2013

nonn

Conjecture: a(n) > 0 for all n > 1. Moreover, if n > 180 is not among 284, 293, 371, 542, 788, 1274, then 2*n can be written as p + sigma(m^2), where p is an odd prime and m is a positive integer.
See also part (i) of the conjecture in A233654.
Note that if sigma(k) is odd, then the order of k at each odd prime must be even, and hence k has the form m^2 or 2*m^2, where m is a positive integer.
We have verified part (i) of the conjecture for n up to 10^9.

			a(2) = 3 since 2*2 = 3 + sigma(1), but 2*2 = 2 + sigma(k) for no k > 0.

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

Cf. A000040, A000203, A232270, A233544, A233654.

sigma[n_]:=Sum[If[Mod[n,d]==0,d,0],{d,1,n}]
S[n_]:=Union[Table[sigma[j^2],{j,1,Sqrt[n]}],Table[sigma[2*j^2],{j,1,Sqrt[n/2]}]]
Do[Do[If[MemberQ[S[2n],2n-Prime[k]],Print[n," ",Prime[k]];Goto[aa]],{k,2,PrimePi[2n]}];
Print[n," ",0];Label[aa];Continue,{n,1,100}]

A233864 a(n) = |{0 < m < 2n: m = sigma(k) for some k > 0, and 2n - 1 - m and 2*n - 1 + m are both prime}|, where sigma(k) is the sum of all (positive) divisors of k.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Dec 16 2013

nonn

Conjecture: (i) a(n) > 0 for all n > 3.
(ii) For any even number 2*n > 0, 2*n + sigma(k) is prime for some 0 < k < 2*n.
See also A233793 for a related conjecture.
Clearly part (i) of the conjecture implies Goldbach's conjecture for even numbers 2*(2*n - 1) with n > 3; we have verified part (i) for n up to 10^8. Concerning part (ii), we remark that 1024 is the unique positive integer k < 1134 with 1134 + sigma(k) prime, and that sigma(1024) = 2047 > 1134.

			a(7) = 1 since sigma(5) = 6, and 2*7 - 1 - 6 = 7 and 2*7 - 1 + 6 = 19 are both prime.
a(10) = 1 since sigma(6) = sigma(11) = 12, and 2*10 - 1 - 12 = 7 and 2*10 - 1 + 12 = 31 are both prime.
a(11) = 1 since sigma(7) = 8, and 2*11 - 1 - 8 = 13 and 2*11 - 1 + 8 = 29 are both prime.

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

Cf. A000040, A000203, A002372, A002375, A232270, A233544, A233654, A233793.

f[n_]:=Sum[If[Mod[n,d]==0,d,0],{d,1,n}]
S[n_]:=Union[Table[f[j],{j,1,n}]]
PQ[n_]:=n>0&&PrimeQ[n]
a[n_]:=Sum[If[PQ[2n-1-Part[S[2n-1],i]]&&PQ[2n-1+Part[S[2n-1],i]],1,0],{i,1,Length[S[2n-1]]}]
Table[a[n],{n,1,100}]

A234963 Number of ways to write n = k + m with k > 0 and m > 2 such that C(2*sigma(k) + phi(m), sigma(k) + phi(m)/2) - 1 is prime, where sigma(k) is the sum of all positive divisors of k and phi(.) is Euler's totient function.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Jan 01 2014

nonn

Conjecture: a(n) > 0 for all n >= 180.
Clearly, this implies that there are infinitely many primes of the form C(2*n,n) - 1. We have verified the conjecture for n up to 10000.
Note that every n = 400, ..., 9123 can be written as k + m with k > 0 and m > 0 such that f(k, m) = sigma(k) + phi(m) is even and C(f(k, m) + 2, f(k, m)/2 + 1) + 1 is prime, but this fails for n = 9124.

			a(5) = 1 since 5 = 1 + 4 with C(2*sigma(1) + phi(4), sigma(1) + phi(4)/2) - 1 = C(4, 2) - 1 = 5 prime.
a(28) = 1 since 28 = 2 + 26 with C(2*sigma(2) + phi(26), sigma(2) + phi(26)/2) - 1 = C(18, 9) - 1 = 48619 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..2000

Cf. A000010, A000040, A000203, A000984, A066726, A066699, A232270, A233544, A234451, A234470, A234503, A234514, A234567, A234615.

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

A232442 a(n) = |{0 < k < n: mprime(m) - 1 and mprime(m) + 1 are both prime with m = sigma(k) + phi(n-k)}|, where sigma(k) is the sum of all positive divisors of k and phi(.) is Euler's totient function.

Original entry on oeis.org

0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 2, 1, 1, 0, 0, 1, 1, 6, 1, 2, 2, 0, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 4, 1, 1, 0, 1, 2, 2, 2, 4, 0, 0, 1, 2, 0, 3, 3, 3, 2, 0, 1, 1, 2, 1, 2, 0, 1, 1, 14, 3, 2, 2, 2, 2, 3, 4, 5, 3, 2, 3, 1, 3, 3, 4, 6, 3, 0, 5, 3, 1, 0, 5, 2, 0, 3, 6, 1

Offset: 1

Zhi-Wei Sun, Jan 14 2014

nonn

Conjecture: a(n) > 0 for all n > 214.
This implies that there are infinitely many twin prime pairs of the special form {m*prime(m) - 1, m*prime(m) + 1}.
We have verified the conjecture for n up to 10^5.

			a(25) = 1 since sigma(6) + phi(19) = 12 + 18 = 30 with {30*prime(30) - 1, 30*prime(30) + 1} = {3389, 3391} a twin prime pair.
a(100) = 1 since sigma(75) + phi(25) = 124 + 20 = 144 with {144*prime(144) - 1, 144*prime(144) + 1} = {119087, 119089} a twin prime pair.

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

Cf. A000010, A000040, A000203, A085637, A086172, A086173, A232270, A232861, A233544, A233539.

sigma[n_]:=DivisorSigma[1,n]
q[n_]:=PrimeQ[n*Prime[n]-1]&&PrimeQ[n*Prime[n]+1]
f[n_,k_]:=sigma[k]+EulerPhi[n-k]
a[n_]:=Sum[If[q[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

A236765 Number of ways to write n = k^2 + m with k > 1 and m > 1 such that sigma(k^2) + prime(m) - 1 is prime, where sigma(j) denotes the sum of all positive divisors of j.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Jan 30 2014

nonn

Conjecture: (i) If n > 6 is not equal to 18, then a(n) > 0.

(ii) Any integer n > 14 can be written as p + q with q > 0 such that p, p + 6 and prime(p) + sigma(q) are all prime.

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

Cf. A000040, A000203, A233544, A233654, A233793, A233864, A236548.

a(10) = 1 since 10 = 2^2 + 6 with sigma(2^2) + prime(6) - 1 = 7 + 13 - 1 = 19 prime.
a(253) = 1 since 253 = 15^2 + 28 with sigma(15^2) + prime(28) - 1 = 403 + 107 - 1 = 509 prime.

p[n_,k_]:=PrimeQ[DivisorSigma[1,k^2]+Prime[n-k^2]-1]
a[n_]:=If[n<6,0,Sum[If[p[n,k],1,0],{k,2,Sqrt[n-2]}]]
Table[a[n],{n,1,100}]

cp's OEIS Frontend

A233793 Least odd prime p such that 2*n - p = sigma(k) for some k > 0, or 0 if such an odd prime p does not exist, where sigma(k) is the sum of all (positive) divisors of k.

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A233864 a(n) = |{0 < m < 2*n: m = sigma(k) for some k > 0, and 2*n - 1 - m and 2*n - 1 + m are both prime}|, where sigma(k) is the sum of all (positive) divisors of k.

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A234963 Number of ways to write n = k + m with k > 0 and m > 2 such that C(2*sigma(k) + phi(m), sigma(k) + phi(m)/2) - 1 is prime, where sigma(k) is the sum of all positive divisors of k and phi(.) is Euler's totient function.

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A232442 a(n) = |{0 < k < n: m*prime(m) - 1 and m*prime(m) + 1 are both prime with m = sigma(k) + phi(n-k)}|, where sigma(k) is the sum of all positive divisors of k and phi(.) is Euler's totient function.

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A236765 Number of ways to write n = k^2 + m with k > 1 and m > 1 such that sigma(k^2) + prime(m) - 1 is prime, where sigma(j) denotes the sum of all positive divisors of j.

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A233864 a(n) = |{0 < m < 2n: m = sigma(k) for some k > 0, and 2n - 1 - m and 2*n - 1 + m are both prime}|, where sigma(k) is the sum of all (positive) divisors of k.

A232442 a(n) = |{0 < k < n: mprime(m) - 1 and mprime(m) + 1 are both prime with m = sigma(k) + phi(n-k)}|, where sigma(k) is the sum of all positive divisors of k and phi(.) is Euler's totient function.