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.

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A233360 Primes of the form L(k) + q(m) with k > 0 and m > 0, where L(k) is the k-th Lucas number (A000204), and q(.) is the strict partition function (A000009).

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 79, 83, 101, 103, 107, 127, 131, 149, 151, 193, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 307, 337, 347, 349, 379, 397, 401, 419, 421, 449, 463, 487, 523, 541, 571, 643, 647, 661
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 08 2013

Keywords

Comments

Conjecture: The sequence has infinitely many terms.
This follows from the conjecture in A233359.

Examples

			a(1) = 2 since L(1) + q(1) = 1 + 1 = 2.
a(2) = 3 since L(1) + q(3) = 1 + 2 = 3.
a(3) = 5 since L(2) + q(3) = 3 + 2 = 5.
		

Crossrefs

Programs

  • Mathematica
    n=0
    Do[Do[If[LucasL[j]>=Prime[m],Goto[aa],
    Do[If[PartitionsQ[k]==Prime[m]-LucasL[j],
    n=n+1;Print[n," ",Prime[m]];Goto[aa]];If[PartitionsQ[k]>Prime[m]-LucasL[j],Goto[bb]];Continue,{k,1,2*(Prime[m]-LucasL[j])}]];
    Label[bb];Continue,{j,1,2*Log[2,Prime[m]]}];
    Label[aa];Continue,{m,1,125}]

A233439 a(n) = |{0 < k < n: prime(k)^2 + 4*prime(n-k)^2 is prime}|.

Original entry on oeis.org

0, 0, 0, 1, 2, 1, 2, 1, 3, 4, 4, 8, 4, 6, 3, 1, 7, 3, 8, 5, 2, 9, 2, 11, 8, 7, 5, 4, 8, 7, 8, 8, 8, 7, 5, 9, 5, 10, 9, 7, 13, 9, 11, 10, 14, 5, 11, 10, 10, 11, 12, 7, 13, 10, 10, 8, 15, 11, 12, 11, 13, 14, 6, 12, 11, 22, 21, 5, 15, 7, 13, 15, 17, 15, 10, 16, 11, 13, 14, 12, 17, 12, 16, 16, 19, 22, 17, 12, 19, 17, 19, 17, 16, 17, 18, 20, 19, 17, 10, 16
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 09 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 3.
(ii) For any integer n > 10, prime(j)^3 + 2*prime(n-j)^2 is prime for some 0 < j < n, and prime(k)^3 + 2*prime(n-k)^3 is prime for some 0 < k < n.
(iii) If n > 5, then prime(k)^3 + 2*p(n-k)^3 is prime for some 0 < k < n, where p(.) is the partition function (A000041). If n > 2, then prime(k)^3 + 2*q(n-k)^3 is prime for some 0 < k < n, where q(.) is the strict partition function (A000009).

Examples

			a(4) = 1 since prime(3)^2 + 4*prime(1)^2 = 5^2 + 4*2^2 = 41 is prime.
a(6) = 1 since prime(5)^2 + 4*prime(1)^2 = 11^2 + 4*2^2 = 137 is prime.
a(8) = 1 since prime(3)^2 + 4*prime(5)^2 = 5^2 + 4*11^2 = 509 is prime.
a(16) = 1 since prime(6)^2 + 4*prime(10)^2 = 13^2 + 4*29^2 = 3533 is prime.
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[Prime[k]^2+4*Prime[n-k]^2],1,0],{k,1,n-1}]
    Table[a[n],{n,1,100}]

A229835 Number of ways to write n = (p - 1)/6 + q, where p is a prime, and q is a term of the sequence A000009.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 19 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. Also, any integer n > 1 can be written as (p + 1)/6 + q, where p is a prime and q is a term of A000009.
We have verified this for n up to 2*10^8. Note that 26128189 cannot be written as (p - 1)/4 + q with p a prime and q a term of A000009. Also, 65152682 cannot be written as (p + 1)/4 + q with p a prime and q a term of A000009.

Examples

			a(2) = 1 since 2 = (7 - 1)/ 6 + 1 with 7 prime, and 1 = A000009(i) for i = 0, 1, 2.
a(3) = 2 since 3 = (7 - 1 )/6 + 2 with 7 prime and 2 = A000009(3) = A000009(4), and 3 = (13 - 1 )/6 + 1 with 13 prime and 1 = A000009(i) for i = 0, 1, 2.
		

Crossrefs

Programs

  • Mathematica
    Do[m=0;Do[If[PartitionsQ[k]>=n,Goto[aa]];If[k>1&&PartitionsQ[k]==PartitionsQ[k-1],Goto[bb]];
    If[PrimeQ[6(n-PartitionsQ[k])+1],m=m+1];Label[bb];Continue,{k,1,2n}];
    Label[aa];Print[n," ",m];Continue,{n,1,100}]

A236442 Number of ordered ways to write n = k + m with k > 0 and m > 0 such that A000009(k) + A047967(m) is prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 26 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 2.
(ii) If n > 2 is neither 18 nor 30, then n can be written as k + m with k > 0 and m > 0 such that A000009(k)^2 + A047967(m)^2 is prime.
(iii) Any integer n > 4 can be written as k + m with k > 0 and m > 0 such that A000009(k)*A047967(m) - 1 (or A000009(k)*A047967(m) + 1) is prime.

Examples

			a(15) = 1 since 15 = 13 + 2 with A000009(13) + A047967(13) = 18 + 1 = 19 prime.
a(18) = 1 since 18 = 3 + 15 with A000009(3) + A047967(15) = 2 + 149 = 151 prime.
		

Crossrefs

Programs

  • Mathematica
    p[n_,k_]:=PrimeQ[PartitionsQ[k]+(PartitionsP[n-k]-PartitionsQ[n-k])]
    a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
    Table[a[n],{n,1,100}]
Previous Showing 11-14 of 14 results.