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.

A237706 Number of primes p < n with pi(n-p) a square, where pi(.) is given by A000720.

Original entry on oeis.org

0, 0, 1, 2, 1, 1, 1, 1, 2, 2, 2, 4, 3, 3, 3, 1, 1, 2, 2, 3, 3, 1, 1, 2, 3, 4, 4, 4, 4, 6, 5, 4, 4, 2, 2, 3, 3, 5, 5, 3, 3, 3, 3, 4, 4, 3, 3, 3, 3, 3, 3, 2, 2, 4, 5, 5, 5, 4, 4, 7, 6, 5, 5, 4, 4, 5, 5, 7, 7, 5
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 11 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3, 5, 6, 7, 8, 16, 17, 22, 23, 148, 149.
(ii) For any integer n > 2, there is a prime p < n with pi(n-p) a triangular number.
We have verified that a(n) > 0 for every n = 3, ..., 1.5*10^7. See A237710 for the least prime p < n with pi(n-p) a square.
See also A237705, A237720 and A237721 for similar conjectures.

Examples

			a(8) = 1 since 7 is prime with pi(8-7) = 0^2.
a(16) = 1 since 7 is prime with pi(16-7) = 2^2.
a(149) = 1 since 139 is prime with pi(149-139) = pi(10) = 2^2.
a(637) = 2 since 409 is prime with pi(637-409) = pi(228) = 7^2, and 613 is prime with pi(637-613) = pi(24) = 3^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
    q[n_]:=SQ[PrimePi[n]]
    a[n_]:=Sum[If[q[n-Prime[k]],1,0],{k,1,PrimePi[n-1]}]
    Table[a[n],{n,1,70}]

A237840 a(n) = |{0 < k <= n: the number of twin prime pairs not exceeding k*n is a square}|.

Original entry on oeis.org

1, 2, 2, 1, 3, 3, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 3, 1, 3, 4, 3, 4, 4, 3, 3, 4, 2, 2, 4, 2, 3, 2, 1, 3, 2, 2, 2, 1, 2, 1, 2, 2, 2, 4, 3, 2, 2, 1, 3, 4, 3, 1, 3, 1, 2, 4, 2, 5, 2, 3, 2, 3, 1, 3, 2, 4, 4, 1, 3, 2, 4, 2, 4, 4, 4, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 14 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 for no n > 159.
(ii) For every n = 1, 2, 3, ..., there is a positive integer k <= n such that the number |{{p, 2*p+1}: both p and 2*p + 1 are primes not exceeding k*n}| is a square.
We have verified that a(n) > 0 for all n = 1, ..., 22000.
See also A237879 for the least k among 1, ..., n such that the number of twin prime pairs not exceeding k*n is a square.

Examples

			a(4) = 1 since the number of twin prime pairs not exceeding 1*4 = 4 is 0^2.
a(9) = 1 since there are exactly 2^2 twin prime pairs not exceeding 3*9 = 27 (namely, they are {3, 5}, {5, 7}, {11, 13} and {17, 19}).
a(18055) > 0 since there are exactly 675^2 = 455625 twin prime pairs not exceeding 5758*18055 = 103960690.
a(18120) > 0 since there are exactly 729^2 = 531441 twin prime pairs not exceeding 6827*18120 = 123705240.
a(18307) > 0 since there are exactly 681^2 = 463761 twin prime pairs not exceeding 5792*18307 = 106034144.
a(18670) > 0 since there are exactly 683^2 = 466489 twin prime pairs not exceeding 5716*18670 = 106717720.
a(19022) > 0 since there are exactly 737^2 = 543169 twin prime pairs not exceeding 6666*19022 = 126800652.
a(19030) > 0 since there are exactly 706^2 = 498436 twin prime pairs not exceeding 6045*19030 = 115036350.
a(19805) > 0 since there are exactly 717^2 = 514089 twin prime pairs not exceeding 6015*19805 = 119127075.
a(19939) > 0 since there are exactly 1000^2 = 10^6 twin prime pairs not exceeding 12660*19939 = 252427740.
a(20852) > 0 since there are exactly 747^2 = 558009 twin prime pairs not exceeding 6268*20852 = 130700336.
a(21642) > 0 since there are exactly 724^2 = 524176 twin prime pairs not exceeding 5628*21642 = 121801176.
		

Crossrefs

Programs

  • Mathematica
    tw[0]:=0
    tw[n_]:=tw[n-1]+If[PrimeQ[Prime[n]+2],1,0]
    SQ[n_]:=IntegerQ[Sqrt[tw[PrimePi[n]]]]
    a[n_]:=Sum[If[SQ[k*n-2],1,0],{k,1,n}]
    Table[a[n],{n,1,80}]

A237720 Number of primes p <= (n+1)/2 with floor( sqrt(n-p) ) prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 12 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 5, and a(n) = 1 only for n = 6, 23, 24, 111, 112, ..., 120.
(ii) For any integer n > 2, there is a prime p < n with floor(sqrt(n+p)) prime.
Note that floor(sqrt(n)) is the number of squares among 1, ..., n.
See also A237705, A237706 and A237721 for similar conjectures.

Examples

			a(6) = 1 since 2 and floor(sqrt(6-2)) = 2 are both prime.
a(23) = 1 since 11 and floor(sqrt(23-11)) = 3 are both prime.
a(24) = 1 since 11 and floor(sqrt(24-11)) = 3 are both prime.
a(27) = 2 since 2 and floor(sqrt(27-2)) = 5 are both prime, and 13 and floor(sqrt(27-13)) = 3 are both prime.
a(n) = 1 for n = 111, ..., 116 since 53 and floor(sqrt(n-53)) = 7 are both prime.
a(n) = 1 for n = 117, 118, 119, 120 since 59 and floor(sqrt(n-59)) = 7 are both prime.
		

Crossrefs

Programs

  • Mathematica
    q[n_]:=PrimeQ[Floor[Sqrt[n]]]
    a[n_]:=Sum[If[q[n-Prime[k]],1,0],{k,1,PrimePi[(n+1)/2]}]
    Table[a[n],{n,1,70}]

A237721 Number of primes p <= n with floor( sqrt(n-p) ) a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 12 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 9, 10, 11, 12, 15, 16, 17.
We have verified this for n up to 10^6.
See also A237705, A237706 and A237720 for similar conjectures.

Examples

			a(2) = 1 since 2 is prime with floor(sqrt(2-2)) = 0^2.
a(3) = 2 since 2 is prime with floor(sqrt(3-2)) = 1^2, and 3 is prime with floor(sqrt(3-3)) = 0^2.
a(9) = a(10) = 1 since 7 is prime with floor(sqrt(9-7)) = floor(sqrt(10-7)) = 1^2.
a(11) = 1 since 11 is prime with floor(sqrt(11-11)) = 0^2.
a(12) = 1 since 11 is prime with floor(sqrt(12-11)) = 1^2.
a(15) = a(16) = 1 since 13 is prime with floor(sqrt(15-13)) = floor(sqrt(16-13)) = 1^2.
a(17) = 1 since 17 is prime with floor(sqrt(17-17)) = 0^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
    q[n_]:=SQ[Floor[Sqrt[n]]]
    a[n_]:=Sum[If[q[n-Prime[k]],1,0],{k,1,PrimePi[n]}]
    Table[a[n],{n,1,70}]
Showing 1-4 of 4 results.