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 1-7 of 7 results.

A234514 Number of ways to write n = k + m with k > 0 and m > 0 such that p = k + phi(m)/2 and q(p) + 1 are both prime, where phi(.) is Euler's totient function, and q(.) is the strict partition function (A000009).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 27 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 12.
(ii) For any integer n > 4, there is a prime p < n - 2 such that q(p + phi(n-p)/2) + 1 is prime.
Clearly, part (i) of the conjecture implies that there are infinitely many primes p with q(p) + 1 prime (cf. A234530).
We have verified part (i) for n up to 10^5.

Examples

			a(11) = 1 since 11 = 1 + 10 with 1 + phi(10)/2 = 3 and q(3) + 1 = 3 both prime.
a(27) = 1 since 27 = 7 + 20 with 7 + phi(20)/2 = 11 and q(11) + 1 = 13 both prime.
a(30) = 1 since 30 = 8 + 22 with 8 + phi(22)/2 = 13 and q(13) + 1 = 19 both prime.
a(38) = 1 since 38 = 21 + 17 with 21 + phi(17)/2 = 29 and q(29) + 1 = 257 both prime.
a(572) = 1 since 572 = 77 + 495 with 77 + phi(495)/2 = 197 and q(197) + 1 = 406072423 both prime.
a(860) = 1 since 860 = 523 + 337 with 523 + phi(337)/2 = 691 and q(691) + 1 = 712827068077888961 both prime.
		

Crossrefs

Programs

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

A234309 a(n) = |{2 < k <= n/2: 2^{phi(k)} + 2^{phi(n-k)} - 1 is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 23 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 5.
(ii) For any integer n > 1, 2^k +2^{phi(n-k)} - 1 is prime for some 0 < k < n, and 2^{sigma(j)} + 2^{phi(n-j)} - 1 is prime for some 0 < j < n, where sigma(j) is the sum of all positive divisors of j.
As phi(k) is even for any k > 2, part (i) of the conjecture implies that there are infinitely many primes of the form 4^a + 4^b - 1 with a and b positive integers (cf. A234310). Note that any Mersenne prime greater than 3 has the form 2^{2a+1} - 1 = 4^a + 4^a - 1.

Examples

			a(6) = 1 since 2^{phi(3)} + 2^{phi(3)} - 1 = 2^2 + 2^2 - 1 = 7 is prime.
a(7) = 1 since 2^{phi(3)} + 2^{phi(4)} - 1 = 2^2 + 2^2 - 1 = 7 is prime.
a(8) = 2 since 2^{phi(3)} + 2^{phi(5)} - 1 = 2^2 + 2^4 - 1 = 19 and 2^{phi(4)} + 2^{phi(4)} - 1 = 2^2 + 2^2 - 1 = 7 are both prime.
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[2^(EulerPhi[k])+2^(EulerPhi[n-k])-1],1,0],{k,3,n/2}]
    Table[a[n],{n,1,100}]

A234475 Number of ways to write n = k + m with 2 < k <= m such that q(phi(k)*phi(m)/4) + 1 is prime, where phi(.) is Euler's totient function and q(.) is the strict partition function (A000009).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 26 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5.
This implies that there are infinitely many primes p with p - 1 a term of A000009.

Examples

			a(6) = 1 since 6 = 3 + 3 with q(phi(3)*phi(3)/4) + 1 = q(1) + 1 = 2 prime.
a(76) = 1 since 76 = 18 + 58 with q(phi(18)*phi(58)/4) + 1 = q(42) + 1 = 1427 prime.
a(197) = 1 since 197 = 4 + 193 with q(phi(4)*phi(193)/4) + 1 = q(96) + 1 = 317789.
a(356) = 1 since 356 = 88 + 268 with q(phi(88)*phi(268)/4) + 1 = q(1320) + 1 = 35940172290335689735986241 prime.
		

Crossrefs

Programs

  • Mathematica
    f[n_,k_]:=PartitionsQ[EulerPhi[k]*EulerPhi[n-k]/4]+1
    a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,3,n/2}]
    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}]

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}]
Showing 1-7 of 7 results.