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.

A119480 Numbers n such that the Bernoulli number B_{4n} has denominator 30.

Original entry on oeis.org

1, 2, 17, 19, 31, 38, 47, 59, 61, 62, 71, 94, 101, 103, 107, 109, 118, 122, 137, 149, 151, 157, 167, 181, 197, 206, 211, 218, 223, 227, 229, 241, 257, 263, 269, 271, 283, 289, 302, 311, 313, 314, 317, 331, 334, 337, 347, 349, 353, 361, 362, 367, 379
Offset: 1

Views

Author

Alexander Adamchuk, Jul 26 2006

Keywords

Comments

Most a(n) are primes from A043297(n) except for a(1) = 1 and composite a(n) for n=6,10,12,17,18,26,28,38,39,42,45,50,51, ... a(6) = 38 = 2*19, a(10) = 62 = 2*31, a(12) = 94 = 2*47, a(17) = 118 = 2*59, a(18) = 122 = 2*61, a(26) = 206 = 2*103, a(28) = 218 = 2*109, a(38) = 289 = 17*17, a(39) = 302 = 2*151, a(42) = 314 = 2*157, a(45) = 334 = 2*167, a(50) = 361 = 19*19, a(51) = 362 = 2*181, ... It appears that most composite a(n) are the doubles of some primes from A043297(n) belonging to A081092[n] and A045404[n] - Primes congruent to {3, 4, 5, 6} mod 7. The rest of composite a(n) are the squares of the primes from A043297(n).
Some a(n) are the products of different primes from A043297(n), for example a(77) = 527 = 17*31. a(n) belong to A045402 Primes congruent to {1, 3, 4, 5, 6} mod 7. a(n) is a subset of A053176 Primes p such that 2p+1 is composite, A045979 Bernoulli number B_{2n} has denominator 6, A090863 Numbers n such that F(n+1)*F(n-1)*B(2n) is an integer, where F(k)=k-th Fibonacci number and B(2k)=2k-th Bernoulli number. - Alexander Adamchuk, Jul 27 2006

Links

Crossrefs

Cf. A043297, A051225, A051222, A051230, A045404.
Cf. A045402, A053176, A045979, A090863.

Programs

  • Mathematica
    Select[Range@ 400, Denominator@ BernoulliB[4 #] == 30 &] (* Michael De Vlieger, Aug 09 2017 *)

Formula

a(n) = A051225[n]/2.

A023212 Primes p such that 4*p+1 is also prime.

Original entry on oeis.org

3, 7, 13, 37, 43, 67, 73, 79, 97, 127, 139, 163, 193, 199, 277, 307, 373, 409, 433, 487, 499, 577, 619, 673, 709, 727, 739, 853, 883, 919, 997, 1033, 1039, 1063, 1087, 1093, 1123, 1129, 1297, 1327, 1423, 1429, 1453, 1543, 1549, 1567, 1579, 1597, 1663, 1753
Offset: 1

Views

Author

David W. Wilson

Keywords

Comments

If p > 3 is a Sophie Germain prime (A005384), p cannot be in this sequence, because all Germain primes greater than 3 are of the form 6k - 1, and then 4p + 1 = 3*(8k-1). - Enrique Pérez Herrero, Aug 15 2011
a(n), except 3, is of the form 6k+1. - Enrique Pérez Herrero, Aug 16 2011
According to Beiler: the integer 2 is a primitive root of all primes of the form 4p + 1 with p prime. - Martin Renner, Nov 06 2011
Chebyshev showed that 2 is a primitive root of all primes of the form 4p + 1 with p prime. - Jonathan Sondow, Feb 04 2013
Also solutions to the equation: floor(4/A000005(4*n^2+n)) = 1. - Enrique Pérez Herrero, Jan 12 2013
Prime numbers p such that p^p - 1 is divisible by 4*p + 1. - Gary Detlefs, May 22 2013
It appears that whenever (p^p - 1)/(4*p + 1) is an integer, then this integer is even (see previous comment). - Alexander R. Povolotsky, May 23 2013
4p + 1 does not divide p^n + 1 for any n. - Robin Garcia, Jun 20 2013
Primes in this sequence of the form 4k+1 are listed in A113601. - Gary Detlefs, May 07 2019
There are no numbers with last digit 1 in this list (i.e., members of A030430) because primes p == 1 (mod 10) lead to 5|(4p+1) such that 4p+1 is not prime. - R. J. Mathar, Aug 13 2019

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 102, nr. 5.
  • P. L. Chebyshev, Theory of congruences, Elements of number theory, Chelsea, 1972, p. 306.

Links

Crossrefs

Cf. A001122, A005384, A043297, A088730.
Cf. A005098, A090866.
Cf. A182265, A182434.

Programs

  • Magma
    [n: n in [0..1000] | IsPrime(n) and IsPrime(4*n+1)]; // Vincenzo Librandi, Nov 20 2010
  • Maple
    isA023212 := proc(n)
    isprime(n) and isprime(4*n+1) ;
end proc:
for n from 1 to 1800 do
    if isA023212(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, May 26 2013
  • Mathematica
    Select[Range[2000], PrimeQ[#] && PrimeQ[4# + 1] &] (* Alonso del Arte, Aug 15 2011 *)
Join[{3}, Select[Range[7, 2000, 6], PrimeQ[#] && PrimeQ[4# + 1] &]] (* Zak Seidov, Jan 21 2012 *)
Select[Prime[Range[300]],PrimeQ[4#+1]&] (* Harvey P. Dale, Oct 17 2021 *)
  • PARI
    forprime(p=2,1800,if(Mod(p,4*p+1)^p==1, print1(p", \n"))) \\ Alexander R. Povolotsky, May 23 2013

Formula

Sum_{n>=1} 1/a(n) is in the interval (0.892962433, 1.1616905) (Wagstaff, 2021). - Amiram Eldar, Nov 04 2021

Extensions

Name edited by Michel Marcus, Nov 27 2020

A175667 Smallest number m such that phi(m) = n*tau(m), with phi=A000010 and tau=A000005; a(n)=0 if no such m exists.

Original entry on oeis.org

1, 5, 7, 34, 11, 13, 58, 17, 19, 55, 23, 65, 106, 29, 31, 85, 0, 37, 0, 41, 43, 115, 47, 119, 125, 53, 133, 145, 59, 61, 0, 388, 67, 274, 71, 73, 298, 0, 79, 187, 83, 203, 346, 89, 209, 235, 0, 97, 394, 101, 103, 169, 107, 109, 253, 113, 458, 295, 0, 287, 0, 0, 127, 514, 131
Offset: 1

Views

Author

Enrique Pérez Herrero, Aug 05 2010

Keywords

Comments

If p = 2*n+1 is a prime, and if n > 1 then a(n)=p.
From R. J. Mathar, Aug 07 2010: (Start)
First column in the array
1,3,8,10,18,24,30: A020488
5,9,15,28,40,72,84,90,120: A062516
7,21,26,56,70,78,108,126,168,210: A063469
34,45,52,102,140,156,252,360,420: A063470
11,33,88,110,198,264,330,
13,35,39,63,76,104,105,130,228,234,280,312,390,504,540,630,840,
58,98,174,294,
17,51,128,136,170,176,224,260,306,384,408,468,510,528,672,780,1260,
19,57,74,135,152,182,190,222,342,456,546,570,756,1080,
55,82,99,124,165,246,308,350,372,440,792,924,990,1050,1320,
23,69,184,230,414,552,690,
65,117,148,195,238,315,364,380,444,520,684,714,864,936,1092,1140,1170,1560,2520,
... (End)

Links

Crossrefs

Cf. A112954, A112955
Cf. A005384, A023212, A043297.

Programs

  • Mathematica
    Table[SelectFirst[Range[10^5], EulerPhi@ # == n DivisorSigma[0, #] &] /.
k_ /; MissingQ@ k -> 0, {n, 120}] (* Michael De Vlieger, Aug 09 2017, Version 10.2 *)

Formula

From Enrique Pérez Herrero, Jan 01 2012: (Start)
If n > 1 then a(n) >= 2*n+1 or a(n)=0.
If p and q = 2*p+1 are both prime, A005384, then a(p) = 2*p+1.
If p > 3 and q = 4*p+1 are both prime, A023212, then a(p) = 8*p + 2 = 2*q.
If p > 2 is prime and both 2*p+1 and 4*p+1 are composite, A043297, then a(n)=0.
(End)

Extensions

More terms from R. J. Mathar, Aug 07 2010
Comment corrected by Enrique Pérez Herrero, Aug 12 2010

A087634 Primes p such that the equation phi(x) = 4p has a solution, where phi is the totient function.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 23, 29, 37, 41, 43, 53, 67, 73, 79, 83, 89, 97, 113, 127, 131, 139, 163, 173, 179, 191, 193, 199, 233, 239, 251, 277, 281, 293, 307, 359, 373, 409, 419, 431, 433, 443, 487, 491, 499, 509, 577, 593, 619, 641, 653, 659, 673, 683, 709, 719, 727
Offset: 1

Views

Author

T. D. Noe, Oct 24 2003

Keywords

Comments

Except for p=2, the complement of A043297. Note that for primes p < 1000, we need to check for solutions x < 18478. The equation phi(x) = 2p has solutions for Sophie Germain primes, A005384
a(n) is also the primes p with 2p+1 or 4p+1 also prime, sequences A005384 and A023212. For the case 2p+1 a trivial solution is phi(6p+3)=4p, and for 4p+1, phi(4p+1)=4p. - Enrique Pérez Herrero, Aug 16 2011

Links

Crossrefs

Cf. A005384, A043297.
Cf. A023212.

Programs

  • Mathematica
    t=Table[EulerPhi[n], {n, 3, 20000}]; Union[Select[t, Mod[ #, 4]==0&&PrimeQ[ #/4]&& #/4<1000&]/4] (* or *)
Select[Prime[Range[100]],PrimeQ[4#+1]||PrimeQ[2#+1]&] (* Enrique Pérez Herrero, Aug 16 2011 *)
Showing 1-4 of 4 results.

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