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 1-4 of 4 results.

A233542 Number of ways to write n = k^2 + m with k > 0 and m > 0 such that phi(k^2)*phi(m) - 1 is prime, where phi(.) is Euler's totient function (A000010).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 12 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 5.
(ii) Any integer n > 7 can be written as k^2 + m with k > 0 and m > 0 such that phi(k)^2*phi(m) - 1 is prime.
(iii) If n > 1 is not equal to 36, then n can be written as k^2 + m with k > 0 and m > 0 such that sigma(k)^2*phi(m) + 1 is prime, where sigma(k) is the sum of all (positive) divisors of k.
We have verified part (i) of the conjecture for n up to 2*10^7.

Examples

			a(6) = 1 since 6 = 1^2 + 5 with phi(1^2)*phi(5) - 1 = 1*4 - 1 = 3 prime.
a(7) = 1 since 7 = 2^2 + 3 with phi(2^2)*phi(3) - 1 = 2*2 - 1 = 3 prime.
a(23) = 1 since 23 = 4^2 + 7 with phi(4^2)*phi(7) - 1 = 8*6 - 1 = 47 prime.
a(42) = 1 since 42 = 6^2 + 6 with phi(6^2)*phi(6) - 1 = 12*2 - 1 = 23 prime.
a(49) = 1 since 49 = 2^2 + 45 with phi(2^2)*phi(45) - 1 = 2*24 - 1 = 47 prime.
a(83) = 1 since 83 = 9^2 + 2 with phi(9^2)*phi(2) - 1 = 54*1 - 1 = 53 prime.
a(188) = 1 since 188 = 6^2 + 152 with phi(6^2)*phi(152) - 1 = 12*72 - 1 = 863 prime.
a(327) = 1 since 327 = 5^2 + 302 with phi(5^2)*phi(302) - 1 = 20*150 - 1 = 2999 prime.
a(557) = 1 since 557 = 12^2 + 413 with phi(12^2)*phi(413) - 1 = 48*348 - 1 = 16703 prime.

Links

Crossrefs

Cf. A000010, A000040, A000203, A233544, A233547, A233549.

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[EulerPhi[k^2]*EulerPhi[n-k^2]-1],1,0],{k,1,Sqrt[n-1]}]
Table[a[n],{n,1,100}]

A233566 a(n) = |{0 < p < n: p and p*phi(n-p) - 1 are both prime}|, where phi(.) is Euler's totient function (A000010).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 13 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3. Also, for any n > 2 there is a prime p < n with p^2*phi(n-p) - 1 prime.

Examples

			a(4) = 1 since 3 and 3*phi(4-3) - 1 = 2 are both prime.
a(5) = 2 since 2 and 2*phi(5-2) - 1 = 3 are both prime, and also 3 and 3*phi(5-3) - 1 = = 2 are both prime.

Links

Crossrefs

Cf. A000010, A000040, A233542, A233547, A233549.

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[Prime[k]*EulerPhi[n-Prime[k]]-1],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,100}]

A233654 |{prime p < n: n - p = sigma(k) for some k > 0}|, where sigma(k) is the sum of all (positive) divisors of k.

Original entry on oeis.org

0, 0, 1, 1, 1, 3, 1, 3, 3, 3, 3, 2, 2, 4, 4, 3, 5, 4, 4, 6, 4, 3, 5, 3, 6, 5, 5, 1, 4, 4, 7, 5, 6, 4, 8, 3, 8, 5, 5, 2, 9, 5, 10, 8, 8, 4, 10, 3, 11, 6, 10, 2, 8, 4, 11, 5, 8, 3, 12, 5, 13, 7, 7, 3, 13, 3, 14, 7, 7, 5, 12, 3, 14, 9, 11, 6, 12, 2, 16, 7, 11, 5, 12, 3, 18, 8, 12, 2, 11, 3, 19, 6, 11, 4, 13, 4, 17, 8, 10, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 14 2013

Keywords

Comments

Conjecture: (i) Let n > 1 be an integer. Then we have a(2*n) > 0. Also, 2*n + 1 can be written as p + sigma(k), where p is a Sophie Germain prime and k is a positive integer.
(ii) Each odd number greater than one can be written as sigma(k^2) + phi(m), where k and m are positive integers, and phi(.) is Euler's totient function.
That a(2*n+1) > 0 for n > 1 is a consequence of Goldbach's conjecture, for, if 2*n = p + q with p and q both prime, then 2*n + 1 = p + sigma(q) = q + sigma(p).

Examples

			a(3) = 1 since 3 = 2 + 1 = 2 + sigma(1) with 2 prime.
a(7) = 1 since 7 = 3 + 4 = 3 + sigma(3) with 3 prime.
a(10) = 3 since 10 = 2 + 8 = 2 + sigma(7) with 2 prime, 10 = 3 + 7 = 3 + sigma(4) with 3 prime, and 10 = 7 + 3 = 7 + sigma(2) with 7 prime.
a(13) = 2 since 13 = 5 + 8 = 5 + sigma(7) with 5 prime, and 13 = 7 + 6 = 7 + sigma(5) with 7 prime.
a(28) = 1 since 28 = 13 + 15 = 13 + sigma(8) with 13 prime.
a(36) = 3 since 36 = 5 + 31 = 5 + sigma(16) = 5 + sigma(25) with 5 prime, 36 = 23 + 13 = 23 + sigma(9) with 23 prime, and 36 = 29 + 7 = 29 + sigma(4) with 29 prime.
a(148) = 1 since 148 = 109 + 39 = 109 + sigma(18) with 109 prime.

Links

Crossrefs

Cf. A000010, A000040, A000203, A002372, A002375, A005384, A232270, A233542, A233544, A233547, A233549.

Programs

  • Mathematica
    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[n-Part[S[n],i]],1,0],{i,1,Length[S[n]]}]
Table[a[n],{n,1,100}]

A233567 Number of ways to write n = p + q (q > 0) with p and p^4 + phi(q)^4 both prime, where phi(.) is Euler's totient function (A000010).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 13 2013

Keywords

Comments

Conjecture: If n > 2 is not equal to 5, then we have a(n) > 0, also there is a prime p < n with p^2 + phi(n-p)^2 prime.
We have verified this for n up to 10^7. The first assertion in the conjecture implies that there are infinitely many primes of the form p^4 + q^4, where p is a prime and q is a positive integer.

Examples

			a(7) = 1 since 7 = 3 + 4 with 3 and 3^4 + phi(4)^4 = 81 + 16 = 97 both prime.
a(12) = 1 since 12 = 7 + 5 with 7 and 7^4 + phi(5)^4 = 7^4 + 4^4 = 2657 both prime.
a(31) = 1 since 31 = 23 + 8 with 23 and 23^4 + phi(8)^4 = 23^4 + 4^4 = 280097 both prime.
a(36) = 1 since 36 = 3 + 33 with 3 and 3^4 + phi(33)^4 = 3^4 + 20^4 = 160081 both prime.
a(90) = 1 since 90 = 79 + 11 with 79 and 79^4 + phi(11)^4 = 79^4 + 10^4 = 38960081 both prime.

Links

Crossrefs

Cf. A000010, A000040, A002645, A233542, A233544, A233547, A233549, A233566.

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[Prime[k]^4+EulerPhi[n-Prime[k]]^4],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,100}]
Showing 1-4 of 4 results.

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