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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A238576 Number of odd primes p < 2*n with prime(n*(p-1)/2)^2 - 2 prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 01 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 7, 23, 61, 73.
(ii) For any integer n > 1, there is an odd prime p < 2*n with prime(n*(p+1)/2)^2 - 2 prime.
Clearly, either part of the conjecture implies that there are infinitely many primes of the form p^2 - 2 with p prime.

Examples

			a(2) = 1 since 2 and prime(2*(3-1)/2)^2 - 2 = 3^2 - 2 = 7 are both prime.
a(7) = 1 since 5 and prime(7*(5-1)/2)^2 - 2 = 43^2 - 2 = 1847 are both prime.
a(23) = 1 since 29 and prime(23*(29-1)/2)^2 - 2 = 2137^2 - 2 = 4566767 are both prime.
a(61) = 1 since 43 and prime(61*(43-1)/2)^2 - 2 = 10463^2 - 2 = 109474367 are both prime.
a(73) = 1 since 7 and prime(73*(7-1)/2)^2 - 2 = 1367^2 - 2 = 1868687 are both prime.
		

Crossrefs

Programs

  • Mathematica
    p[k_,n_]:=PrimeQ[Prime[(Prime[k]-1)/2*n]^2-2]
    a[n_]:=Sum[If[p[k,n],1,0],{k,2,PrimePi[2n-1]}]
    Table[a[n],{n,1,80}]

A247011 Numbers n for which A242719(n) = (prime(n) + 2)^2 + 1.

Original entry on oeis.org

5, 7, 13, 17, 26, 33, 64, 81, 98, 140, 171, 176, 190, 201, 215, 225, 318, 332, 336, 444, 469, 475, 495, 551, 558, 563, 577, 601, 636, 671, 828, 849, 862, 870, 948, 1004, 1064, 1074, 1189, 1198, 1230, 1238, 1305, 1328, 1445, 1449, 1528, 1618, 1634, 1642, 1679
Offset: 1

Views

Author

Vladimir Shevelev, Sep 09 2014

Keywords

Comments

(prime(n) + 2)^2 + 1 is the second minimal possible value of A242719(n) after prime(n)^2 + 1. Indeed, by the definition lpf(A242719(n) - 3) > lpf(A242719(n) - 1) >= prime(n), thus after prime(n)^2 + 1 we should consider prime(n)*(prime(n) + 2) + 1. Then prime(n) should be lesser number of twin primes, but then prime(n) + 1 == 0 (mod 3). So, prime(n)*(prime(n) + 2) - 2 == 0 (mod 3). Analogously one can prove that prime(n)*(prime(n) + 4) - 2 == 0 (mod 3).
Note that for the sequence prime(n+1) is in intersection of A006512 and A062326, but prime(n) is not in A062326.

Crossrefs

Formula

If prime(n) is not in A062326, then A242719(n) >= (prime(n)+2)^2 + 1.
Intersection of A247011 and A246824 forms sequence 81, 215, 828, 1189, 1634, ... For these values of n we have A242719(n) - A242720(n) = 2*(prime(n) + 1).

Extensions

More terms from Peter J. C. Moses, Sep 09 2014

A118939 Primes p such that (p^2+3)/4 is prime.

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 29, 31, 41, 43, 67, 83, 101, 109, 139, 151, 157, 179, 181, 199, 211, 223, 239, 263, 277, 283, 307, 311, 337, 347, 353, 379, 389, 419, 431, 463, 491, 557, 577, 587, 619, 659, 673, 739, 757, 797, 809, 811, 829, 853, 907, 911, 953, 991, 1051
Offset: 1

Views

Author

T. D. Noe, May 06 2006

Keywords

Comments

For all primes q>2, we have q=4k+-1 for some k, which makes it easy to show that 4 divides q^2+3. Similar sequences, with p and (p^2+a)/b both prime, are A048161, A062324, A062326, A062718, A109953, A110589, A118915, A118918, A118940, A118941 and A118942.

Programs

  • Mathematica
    Select[Prime[Range[200]],PrimeQ[(#^2+3)/4]&]

A137270 Primes p such that p^2 - 6 is also prime.

Original entry on oeis.org

3, 5, 7, 13, 17, 23, 47, 53, 67, 73, 83, 97, 107, 113, 167, 193, 197, 263, 293, 317, 367, 373, 383, 457, 463, 467, 487, 503, 557, 593, 607, 643, 647, 673, 677, 683, 773, 787, 797, 823, 827, 857, 877, 887, 947, 1033, 1063, 1087, 1103, 1187, 1193, 1223, 1303
Offset: 1

Views

Author

Ben de la Rosa and Johan Meyer (meyerjh.sci(AT)ufa.ac.za), Mar 13 2008

Keywords

Comments

Each of the primes p = 2,3,5,7,13 has the property that the quadratic polynomial phi(x) = x^2 + x - p^2 takes on only prime values for x = 1,2,...,2p-2; each case giving exactly one repetition, in phi(p-1) = -p and phi(p) = p.
The only common term in A062718 and A137270 is 5. - Zak Seidov, Jun 16 2015

Examples

			The (2 x 7 - 2) -1 = 11 primes given by the polynomial x^2 + x - 7^2 for x = 1, 2, ..., 2 x 7 - 2 are -47, -43, -37, -29, -19, -7, 7, 23, 41, 61, 83, 107.
		

References

  • F. G. Frobenius, Uber quadratische Formen, die viele Primzahlen darstellen, Sitzungsber. d. Konigl. Acad. d. Wiss. zu Berlin, 1912, 966 - 980.

Crossrefs

Programs

  • Magma
    [p: p in PrimesUpTo(1350) | IsPrime(p^2-6)]; // Vincenzo Librandi, Apr 14 2013
  • Maple
    isA028879 := proc(n) isprime(n^2-6) ; end: isA137270 := proc(n) isprime(n) and isA028879(n) ; end: for i from 1 to 300 do if isA137270(ithprime(i)) then printf("%d, ",ithprime(i)) ; fi ; od: # R. J. Mathar, Mar 16 2008
  • Mathematica
    Select[Prime[Range[2,300]],PrimeQ[#^2-6]&] (* Harvey P. Dale, Jul 24 2012 *)

Formula

A000040 INTERSECT A028879. - R. J. Mathar, Mar 16 2008

Extensions

Corrected and extended by R. J. Mathar, Mar 16 2008

A238701 Number of primes p < n with q = floor((n-p)/4) and q^2 - 2 both prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 03 2014

Keywords

Comments

Conjecture: Let m > 0 and n > 2*m + 1 be integers. If m = 1 and 2 | n, or m = 3 and n is not congruent to 1 modulo 6, or m = 2, 4, 5, ..., then there is a prime p < n with q = floor((n-p)/m) and q^2 - 2 both prime.
In the case m = 1, this is a refinement of Goldbach's conjecture. In the case m = 2, this is stronger than Lemoine's conjecture (cf. A046927). The conjecture for m > 2 seems completely new. We view the conjecture as a natural extension of Goldbach's conjecture.

Examples

			a(11) = 2 since 2, floor((11-2)/4)= 2 and 2^2 - 2 are all prime, and 3, floor((11-3)/4) = 2 and 2^2 - 2 are all prime.
		

Crossrefs

Programs

  • Mathematica
    PQ[n_]:=PrimeQ[n]&&PrimeQ[n^2-2]
    p[n_,k_]:=PQ[Floor[(n-Prime[k])/4]]
    a[n_]:=Sum[If[p[n,k],1,0],{k,1,PrimePi[n-1]}]
    Table[a[n],{n,1,80}]

A245363 Numbers n for which A242719(n) = prime(n)*(prime(n)+6) + 1.

Original entry on oeis.org

19, 37, 49, 69, 73, 102, 165, 236, 253, 365, 465, 542, 595, 694, 713, 723, 762, 920, 962, 979, 1119, 1162, 1259, 1334, 1387, 1441, 1706, 1797, 1843, 1906
Offset: 1

Views

Author

Vladimir Shevelev, Sep 09 2014

Keywords

Comments

These are such numbers n for which prime(n), prime(n)+6, prime(n)*(prime(n)+6)-2 are primes, but prime(n) is not in A062326; besides, if prime(n) is in A001359, then also prime(n+1) is not in A062326.
Prime(n)*(prime(n)+6) + 1 is the third minimal possible value of A242719(n) after prime(n)^2 + 1 and (prime(n)+2)^2 + 1 (cf. A247011 and comment there).

Crossrefs

Formula

prime(n) == 7 (mod 10), A242719(n) == 92 (mod 100).

Extensions

More terms from Peter J. C. Moses, Sep 09 2014

A260080 Least positive integer k such that prime(k*n)^2 - 2 = prime(i*n)*prime(j*n) for some integers 0 < i < j.

Original entry on oeis.org

5, 18, 18, 9, 115, 208, 69, 373, 68, 430, 8, 214, 57, 1887, 1255, 295, 880, 542, 5612, 767, 1562, 40, 853, 884, 753, 4332, 4750, 6077, 799, 1394, 639, 5442, 4785, 440, 7417, 1290, 15830, 27745, 3927, 5701, 1891, 22008, 8243, 6031, 9172, 5949, 43286, 20778, 9876, 12472
Offset: 1

Views

Author

Zhi-Wei Sun, Jul 15 2015

Keywords

Comments

Conjecture: a(n) exists for any n > 0.

Examples

			a(1) = 5 since prime(5*1)^2-2 = 11^2-2 = 119 = 7*17 = prime(4*1)*prime(7*1).
a(66) = 149073 since prime(149073*66)^2-2 = 176365951^2-2 = 31104948672134399 = 3160879*9840600881 = prime(3448*66)*prime(9840600881*66).
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    Dv[n_]:=Divisors[Prime[n]^2-2]
    L[n_]:=Length[Dv[n]]
    P[k_,n_]:=L[k*n]==4&&PrimeQ[Part[Dv[k*n],2]]&&Mod[PrimePi[Part[Dv[k*n],2]],n]==0&&PrimeQ[Part[Dv[k*n],3]]&&Mod[PrimePi[Part[Dv[k*n],3]],n]==0
    Do[k=0;Label[bb];k=k+1;If[P[k,n],Goto[aa]];Goto[bb];Label[aa];Print[n," ", k];Continue,{n,1,50}]

A261354 Primes p such that prime(p)^2 - 2 = prime(q) for some prime q.

Original entry on oeis.org

31, 191, 541, 809, 1153, 1301, 2221, 3037, 3847, 4049, 4159, 5441, 8243, 10177, 12277, 13681, 14783, 15619, 17903, 19463, 20897, 22697, 24517, 25163, 25847, 25849, 26633, 26647, 27329, 27407, 28051, 32653, 35059, 35747, 36341, 36527, 37369, 37811, 38609, 40949, 42737, 46679, 51061, 51607, 54443, 54679, 56113, 57637, 60887, 61493
Offset: 1

Views

Author

Zhi-Wei Sun, Aug 15 2015

Keywords

Comments

Conjecture: The sequence has infinitely many terms. In general, for any integers a,b,c with a>0 and gcd(a,b,c)=1, if b^2-4*a*c is not a square, a+b+c is odd, and gcd(b,a+c) is not divisible by 3, then there are infinitely many prime pairs {p,q} such that a*prime(p)^2+b*prime(p)+c = prime(q).

Examples

			a(1) = 31 since 31 is a prime, and prime(31)^2-2 = 127^2-2 = 16127 = prime(1877) with 1877 prime.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    PQ[n_]:=PrimeQ[n]&&PrimeQ[PrimePi[n]]
    f[k_]:=Prime[Prime[k]]^2-2
    n=0;Do[If[PQ[f[k]],n=n+1;Print[n," ",Prime[k]]],{k,1,6200}]

A118941 Primes p such that (p^2-5)/4 is prime.

Original entry on oeis.org

5, 7, 11, 13, 17, 19, 23, 31, 41, 43, 53, 61, 71, 79, 83, 89, 97, 101, 107, 109, 113, 131, 137, 167, 173, 179, 193, 229, 241, 251, 263, 269, 277, 281, 283, 307, 311, 317, 349, 353, 373, 383, 419, 431, 439, 461, 463, 467, 563, 571, 577, 593, 607, 613, 619, 647
Offset: 1

Views

Author

T. D. Noe, May 06 2006

Keywords

Comments

For all primes q>2, we have q=4k+-1 for some k, which makes it easy to show that 4 divides q^2-5. Similar sequences, with p and (p^2+a)/b both prime, are A048161, A062324, A062326, A062718, A109953, A110589, A118915, A118918, A118939, A118940 and A118942.

Programs

  • Mathematica
    Select[Prime[Range[200]],PrimeQ[(#^2-5)/4]&]

A118942 Primes p such that (p^2-13)/12 is prime.

Original entry on oeis.org

7, 13, 17, 19, 23, 31, 37, 41, 53, 67, 71, 73, 89, 103, 107, 113, 131, 139, 157, 163, 181, 199, 211, 233, 239, 257, 269, 283, 307, 311, 337, 359, 373, 379, 401, 419, 463, 487, 491, 499, 509, 521, 577, 593, 607, 617, 631, 647, 653, 683, 701, 733, 761, 769, 787
Offset: 1

Views

Author

T. D. Noe, May 06 2006

Keywords

Comments

For all primes q>3, we have q=6k+-1 for some k, which makes it easy to show that 12 divides q^2-13. Similar sequences, with p and (p^2+a)/b both prime, are A048161, A062324, A062326, A062718, A109953, A110589, A118915, A118918, A118939, A118940 and A118941.

Programs

  • Mathematica
    Select[Prime[Range[200]],PrimeQ[(#^2-13)/12]&]
Previous Showing 11-20 of 42 results. Next