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.

A236413 Positive integers m with p(m)^2 + q(m)^2 prime, where p(.) is the partition function (A000041) and q(.) is the strict partition function (A000009).

Original entry on oeis.org

1, 2, 3, 4, 6, 17, 24, 37, 44, 95, 121, 162, 165, 247, 263, 601, 714, 742, 762, 804, 1062, 1144, 1149, 1323, 1508, 1755, 1833, 1877, 2330, 2380, 2599, 3313, 3334, 3368, 3376, 3395, 3504, 3688, 3881, 4294, 4598, 4611, 5604, 5696, 5764, 5988, 6552, 7206, 7540, 7689
Offset: 1

Views

Author

Zhi-Wei Sun, Jan 24 2014

Keywords

Comments

According to the conjecture in A236412, this sequence should have infinitely many terms.
See A236414 for primes of the form p(m)^2 + q(m)^2.
See also A236440 for a similar sequence.

Examples

			a(1) = 1 since p(1)^2 + q(1)^2 = 1^2 + 1^2 = 2 is prime.
a(2) = 2 since p(2)^2 + q(2)^2 = 2^2 + 1^2 = 5 is prime.
a(3) = 3 since p(3)^2 + q(3)^2 = 3^2 + 2^2 = 13 is prime.
		

Crossrefs

Programs

  • Mathematica
    pq[n_]:=PrimeQ[PartitionsP[n]^2+PartitionsQ[n]^2]
    n=0;Do[If[pq[m],n=n+1;Print[n," ",m]],{m,1,10000}]

A236419 a(n) = |{0 < k < n: r = phi(k) + phi(n-k)/6 + 1 and p(r) + q(r) are both prime}|, where phi(.) is Euler's totient function, p(.) is the partition function (A000041) and q(.) is the strict partition function (A000009).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 2, 1, 0, 1, 0, 2, 0, 1, 1, 0, 0, 4, 1, 2, 1, 0, 2, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Jan 25 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 127.
We have verified this for n up to 30000.
The conjecture implies that there are infinitely many primes r with p(r) + q(r) also prime.

Examples

			a(15) = 1 since phi(1) + phi(14)/6 + 1 = 3 with p(3) + q(3) = 3 + 2 = 5 prime.
a(54) = 1 since phi(41) + phi(13)/6 + 1 = 43 with p(43) + q(43) = 63261 + 1610 = 64871 prime.
		

Crossrefs

Programs

  • Mathematica
    pq[n_]:=PrimeQ[n]&&PrimeQ[PartitionsP[n]+PartitionsQ[n]]
    f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/6+1
    a[n_]:=Sum[If[pq[f[n,k]],1,0],{k,1,n-1}]
    Table[a[n],{n,1,100}]

A236418 Primes p with A047967(p) also prime.

Original entry on oeis.org

13, 23, 43, 53, 71, 83, 107, 257, 269, 313, 1093, 2659, 2851, 3527, 8243, 20173, 20717, 24329, 26161, 26237, 31583, 53611, 60719, 74717, 83401, 118259, 118369, 130817, 133811, 145109, 152381, 169111, 178613, 183397, 205963
Offset: 1

Views

Author

Zhi-Wei Sun, Jan 25 2014

Keywords

Comments

According to the conjecture in A236417, this sequence should have infinitely many terms.

Examples

			a(1) = 13 with 13 and A047967(13) = 83 both prime.
		

Crossrefs

Programs

  • Mathematica
    pq[n_]:=PrimeQ[n]&&PrimeQ[PartitionsP[n]-PartitionsQ[n]]
    n=0;Do[If[pq[m],n=n+1;Print[n," ",m]],{m,1,10000}]
    Select[Prime[Range[20000]],PrimeQ[PartitionsP[#]-PartitionsQ[#]]&] (* Harvey P. Dale, Jan 02 2022 *)

A236439 a(n) = |{0 < k < n-2: A000009(m)^2 + A047967(m)^2 is prime with m = k + phi(n-k)/2}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 25 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
We have verified this for n up to 50000.
The conjecture implies that there are infinitely many positive integers m with A000009(m)^2 + A047967(m)^2 prime. See A236440 for such numbers m.

Examples

			a(14) = 1 since 2 + phi(12)/2 = 4 with A000009(4)^2 + A047967(4)^2 = 2^2 + 3^2 = 13 prime.
a(17) = 1 since 10 + phi(7)/2 = 13 with A000009(13)^2 + A047967(13)^2 = 18^2 + 83^2 = 7213 prime.
		

Crossrefs

Programs

  • Mathematica
    p[n_]:=PrimeQ[PartitionsQ[n]^2+(PartitionsP[n]-PartitionsQ[n])^2]
    a[n_]:=Sum[If[p[k+EulerPhi[n-k]/2],1,0],{k,1,n-3}]
    Table[a[n],{n,1,100}]
Showing 1-4 of 4 results.