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 1-10 of 10 results.

A237768 Number of primes p < n with pi(n-p) a Sophie Germain prime, where pi(.) is given by A000720.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 13 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 4, and a(n) = 1 only for n = 5, 12, 20, 21, 26, 27, 30, 31, 32, 60, 61, 68, 69, 80, 81.
This is stronger than part (i) of the conjecture in A237705.
We have verified that a(n) > 0 for all n = 5, ..., 2*10^7.

Examples

			a(5) = 1 since 2, pi(5-2) = pi(3) = 2 and 2*2 + 1 = 5 are all prime.
a(12) = 1 since 7, pi(12-7) = pi(5) = 3 and 2*3 + 1 = 7 are all prime.
a(81) = 1 since 47, pi(81-47) = pi(34) = 11 and 2*11 + 1 = 23 are all prime.

Links

Crossrefs

Cf. A000040, A000720, A005384, A237284, A237705, A237706.

Programs

  • Mathematica
    sg[n_]:=PrimeQ[n]&&PrimeQ[2n+1]
a[n_]:=Sum[If[sg[PrimePi[n-Prime[k]]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]

A237769 Number of primes p < n with pi(n-p) - 1 and pi(n-p) + 1 both prime, where pi(.) is given by A000720.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 13 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 8, and a(n) = 1 only for n = 9, 34, 35.
(ii) For any integer n > 4, there is a prime p < n such that 3*pi(n-p) - 1, 3*pi(n-p) + 1 and 3*pi(n-p) + 5 are all prime. Also, for each integer n > 8, there is a prime p < n such that 3*pi(n-p) - 1, 3*pi(n-p) + 1 and 3*pi(n-p) - 5 are all prime.
(iii) For any integer n > 6, there is a prime p < n such that phi(n-p) - 1 and phi(n-p) + 1 are both prime, where phi(.) is Euler's totient function.

Examples

			a(9) = 1 since 2, pi(9-2) - 1 = 3 and pi(9-2) + 1 = 5 are all prime.
a(34) = 1 since 19, pi(34-19) - 1 = pi(15) - 1 = 5 and pi(34-19) + 1 = pi(15) + 1 = 7 are all prime.
a(35) = 1 since 19, pi(35-19) - 1 = pi(16) - 1 = 5 and pi(35-19) + 1 = pi(16) + 1 = 7 are all prime.

Links

Crossrefs

Cf. A000010, A000040, A000720, A001359, A006512, A014574, A022004, A022005, A237705, A237706, A237768.

Programs

  • Mathematica
    TQ[n_]:=PrimeQ[n-1]&&PrimeQ[n+1]
a[n_]:=Sum[If[TQ[PrimePi[n-Prime[k]]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]

A238277 a(n) = |{0 <= k < n: the number of primes in the interval (k*n, (k+1)*n] is a square}|.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 22 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
We have verified this for n up to 10^5.
See also A238278 and A238281 for related conjectures.

Examples

			a(9) = 1 since the interval (0, 9] contains exactly 2^2 = 4 primes.
a(13) = 1 since the interval (9*13, 10*13] contains exactly 1^2 = 1 prime.

Links

Crossrefs

Cf. A000040, A000290, A237598, A237706, A238278, A238281.

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
d[k_,n_]:=PrimePi[(k+1)*n]-PrimePi[k*n]
a[n_]:=Sum[If[SQ[d[k,n]],1,0],{k,0,n-1}]
Table[a[n],{n,1,80}]

A237710 Least prime p < n with pi(n-p) a square, or 0 if such a prime p does not exist.

Original entry on oeis.org

0, 0, 2, 2, 3, 5, 5, 7, 2, 2, 2, 2, 3, 5, 5, 7, 7, 11, 11, 11, 11, 13, 13, 17, 2, 2, 2, 2, 2, 2, 3, 5, 5, 7, 7, 11, 11, 11, 11, 13, 13, 17, 17, 17, 17, 19, 19, 23, 23, 23, 23, 29, 29, 29, 2, 2, 2, 2, 2, 2, 3, 5, 5, 7, 7, 11, 11, 11, 11, 13
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 12 2014

Keywords

Comments

According to the conjecture in A237706, a(n) should be positive for all n > 2.

Examples

			a(5) = 3 since pi(5-3) = 1^2, but pi(5-2) = 2 is not a square.

Links

Crossrefs

Cf. A000040, A000290, A000720, A237706.

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
q[n_]:=SQ[PrimePi[n]]
Do[Do[If[q[n-Prime[k]],Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[n-1]}];
Print[n," ",0];Label[aa];Continue,{n,1,100}]
lp[n_]:=Module[{p=2},While[!IntegerQ[Sqrt[PrimePi[n-p]]],p=NextPrime[p]];p]; Join[{0,0},Array[ lp,80,3]] (* Harvey P. Dale, Jan 28 2024 *)

A237815 Number of primes p < n such that the number of Sophie Germain primes among 1, ..., n-p is a Sophie Germain prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 13 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 4.
(ii) For any integer n > 11, there is a prime p < n such that the number of Sophie Germain primes among 1, ..., n-p is a square.
See also A237817 for a similar conjecture involving twin primes.

Examples

			a(5) = 1 since there are exactly two Sophie Germain primes not exceeding 5-2 = 3, and 2 is a Sophie Germain prime.

Links

Crossrefs

Cf. A000040, A005384, A000290, A237705, A237706, A237768, A237817.

Programs

  • Mathematica
    sg[n_]:=PrimeQ[n]&&PrimeQ[2n+1]
sum[n_]:=Sum[If[PrimeQ[2Prime[k]+1],1,0],{k,1,PrimePi[n]}]
a[n_]:=Sum[If[sg[sum[n-Prime[k]]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]

A237817 Number of primes p < n such that r = |{q <= n-p: q and q + 2 are both prime}| and r + 2 are both prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 13 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 12.
(ii) For any integer n > 2, there is a prime p < n such that r = |{q <= n-p: q and q + 2 are both prime}| is a square.
See also A237815 for a similar conjecture involving Sophie Germain primes.

Examples

			a(13) = 1 since {q <= 13 - 2: q and q + 2 are both prime} = {3, 5, 11} has cardinality 3, and {3, 3 + 2} is a twin prime pair.

Links

Crossrefs

Cf. A000040, A000290, A001359, A006512, A237705, A237706, A237769, A237815.

Programs

  • Mathematica
    TQ[n_]:=PrimeQ[n]&&PrimeQ[n+2]
sum[n_]:=Sum[If[PrimeQ[Prime[k]+2],1,0],{k,1,PrimePi[n]}]
a[n_]:=Sum[If[TQ[sum[n-Prime[k]]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]

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.

Links

Crossrefs

Cf. A000290, A001359, A005384, A006512, A237706, A237710, A237578, A237598, A237612, A237769, A237817, A237839, A237879, A237975.

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.

Links

Crossrefs

Cf. A000040, A000290, A237706, A237710, A237721.

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.

Links

Crossrefs

Cf. A000040, A000290, A237705, A237706, A237710, A237720.

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}]

A238732 Number of primes p < n with floor((n-p)/3) a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 03 2014

Keywords

Comments

Conjecture: (i) For any integers m > 2 and n > 2, there is a prime p < n such that floor((n-p)/m) is a square.
(ii) For any integer k > 2, if m and n are sufficiently large, then there is a prime p < n such that floor((n-p)/m) is a k-th power. In particular, for any integer n > 2, there is a prime p < n such that floor((n-p)/5) is a cube.
We have verified part (i) for m = 3, ..., 10 and n = 3, ..., 10^6.

Examples

			a(3) = 1 since 2 is prime with floor((3-2)/3) = 0^2.
a(11) = 1 since 7 is prime with floor((11-7)/3) = 1^2.
a(13) = 1 since 13 is prime with floor((13-11)/3) = 0^2.
a(28) = 1 since 23 is prime with floor((28-23)/3) = 1^2.
a(37) = 1 since 23 is prime with floor((37-23)/3) = 2^2.
a(173) = 1 since 97 is prime with floor((173-97)/3) = 5^2.
It seems that a(n) = 1 only for n = 3, 11, 13, 28, 37, 173.

Links

Crossrefs

Cf. A000040, A000290, A237706, A238701, A238134.

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
s[n_,k_]:=SQ[Floor[(n-Prime[k])/3]]
a[n_]:=Sum[If[s[n,k],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]
Showing 1-10 of 10 results.

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