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 1-5 of 5 results.

A233390 a(n) = |{0 < k < n: 2^k - 1 + q(n-k) is prime}|, where q(.) is the strict partition function (A000009).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 08 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
We have verified this for n up to 150000. For n = 124669, the least positive integer k with 2^k - 1 + q(n-k) prime is 13413.

Examples

			a(6) = 1 since 2^2 - 1 + q(4) = 3 + 2 = 5 is prime.
a(10) = 1 since 2^4 - 1 + q(6) = 15 + 4 = 19 is prime.
a(41) = 1 since 2^{16} - 1 + q(25) = 65535 + 142 = 65677 is prime.
a(127) = 1 since 2^{21} - 1 + q(106) = 2097151 + 728260 = 2825411 is prime.
a(153) = 1 since 2^{70} - 1 + q(83) = 1180591620717411303423 + 101698 = 1180591620717411405121 is prime.
a(164) = 1 since 2^{26} - 1 + q(138) = 67108863 + 8334326 = 75443189 is prime.
		

Crossrefs

Programs

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

A233393 Primes of the form 2^k - 1 + q(m) with k > 0 and m > 0, where q(.) is the strict partition function (A000009).

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 37, 41, 43, 47, 53, 61, 67, 71, 73, 79, 83, 101, 107, 109, 127, 131, 137, 139, 149, 157, 167, 173, 181, 191, 193, 199, 223, 229, 257, 263, 269, 271, 277, 293, 311, 331, 347, 349, 359, 383, 397, 421, 449, 463, 467, 479, 521, 523, 557, 587
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 A233390.

Examples

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

Crossrefs

Programs

  • Mathematica
    Pow[n_]:=Pow[n]=Mod[n,2]==0&&2^(IntegerExponent[n,2])==n
    n=0
    Do[Do[If[Pow[Prime[m]-PartitionsQ[k]+1],
    n=n+1;Print[n," ",Prime[m]];Goto[aa]];If[PartitionsQ[k]>=Prime[m],Goto[aa]];Continue,{k,1,2*Prime[m]}];
    Label[aa];Continue,{m,1,110}]

A233417 a(n) = |{0 < k <= n/2: q(k)*q(n-k) + 1 is prime}|, where q(.) is the strict partition function (A000009).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 2, 3, 2, 2, 4, 5, 3, 1, 5, 7, 1, 3, 4, 4, 3, 2, 5, 3, 6, 6, 1, 6, 8, 6, 6, 4, 7, 7, 3, 5, 5, 6, 6, 5, 5, 3, 7, 8, 7, 7, 8, 8, 6, 4, 8, 8, 5, 3, 8, 8, 5, 15, 6, 8, 3, 9, 5, 6, 7, 9, 4, 6, 8, 9, 5, 4, 7, 8, 7, 6, 10, 9, 9, 8, 6, 6, 9, 9, 7, 12, 5, 10, 7, 7, 5, 3, 8, 10, 7, 5, 9, 7, 4, 5
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 09 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1. Similarly, for any integer n > 5, there is a positive integer k < n with q(k)*q(n-k) - 1 prime.
(ii) Let n > 1 be an integer. Then p(k) + q(n-k)^2 is prime for some 0 < k < n, where p(.) is the partition function (A000041). If n is not equal to 8, then k^3 + q(n-k)^2 is prime for some 0 < k < n.

Examples

			a(14) = 1 since q(1)*q(13) + 1 = 1*18 + 1 = 19 is prime.
a(17) = 1 since q(4)*q(13) + 1 = 2*18 + 1 = 37 is prime.
a(27) = 1 since q(13)*q(14) + 1 = 18*22 + 1 = 397 is prime.
		

Crossrefs

Programs

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

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}]

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}]
Showing 1-5 of 5 results.