A076481 -id:A076481 - cp's OEIS frontend

A179625 Legal generalized repunit prime numbers.

5, 7, 13, 31, 43, 73, 157, 211, 241, 1093, 2801, 19531, 22621, 30941, 55987, 88741, 245411, 292561, 346201, 797161, 5229043, 8108731, 12207031, 25646167, 305175781, 321272407, 917087137, 16148168401, 2141993519227, 10778947368421, 17513875027111, 610851724137931, 50544702849929377

Offset: 1

Tim Johannes Ohrtmann, Jan 09 2011

nonn

In Chris Caldwell's sense, a legal generalized repunit prime is a prime number of the form (b^p-1)/(b-1) such that 3 <= b <= 5*p, b != 10, and p prime.

Charles R Greathouse IV, Table of n, a(n) for n = 1..640, replacing an earlier file of T. D. Noe
Chris Caldwell, The Top Twenty: Generalized Repunit
Robert Granger and Andrew Moss, Generalised Mersenne numbers revisited, arXiv:1108.3054 [math.NT], 2011-2012.

Cf. A076481, A086122, A165210, A102170 (repunit primes in bases 3, 5, 6, and 7)

This sequence except for the term 5 is subsequence of A085104.

lim=10^17; n=1; Sort[Reap[While[p=Prime[n]; b=3; While[num=Cyclotomic[p,b]; b<=5p && num<=lim, If[b!=10 && PrimeQ[num], Sow[num]]; b++]; b>3, n++]][[2,1]]]

upTo(lim)=my(v=List(),t);forprime(p=2,log(2*lim+1)\log(3),for(b=3,5*p,if(b==10,next);t=(b^p-1)/(b-1);if(t>lim,break);if(isprime(t),listput(v,t))));vecsort(Vec(v)) \\ Charles R Greathouse IV, Aug 21 2011

A255510 Numbers n of the form 3^k such that sigma(n) is a prime p.

Original entry on oeis.org

9, 729, 531441, 2503155504993241601315571986085849, 4638397686588101979328150167890591454318967698009

Offset: 1

Jaroslav Krizek, Mar 25 2015

nonn

Powers of 3 from A023194 (numbers n such that sigma(n) is a prime).

Jaroslav Krizek, Table of n, a(n) for n = 1..10

Cf. A000203 (sigma), A023194 (sigma(n) is prime).

Cf. A003462 (sigma(3^n)), A028491 (sigma(3^n) is prime) , A076481.

[(3^n): n in [1..1000] | IsPrime((SumOfDivisors(3^n)))]

Select[3^Range[0,110],PrimeQ[DivisorSigma[1,#]]&] (* Harvey P. Dale, Mar 29 2015 *)

a(n) = 3^(A028491(n) - 1).

sigma(a(n)) = A076481(n).

A308442 Primes of the form (p^k+1)/2 where p is prime and k > 1.

Original entry on oeis.org

5, 13, 41, 61, 181, 313, 421, 1201, 1741, 1861, 2521, 3121, 5101, 7321, 8581, 9661, 14281, 16381, 19801, 36721, 41761, 60901, 71821, 83641, 100801, 106261, 135721, 139921, 161881, 163021, 199081, 205441, 218461, 273061, 282001, 337021, 353641, 388081, 431521, 491041, 531481, 539761, 552301, 571381

Offset: 1

J. M. Bergot and Robert Israel, May 27 2019

nonn

The only primes of the form (p^k-1)/2 are A076481, since (p^k-1)/2 is divisible by (p-1)/2.

k must be a power of 2, since if k has an odd divisor d>1, (p^k+1)/2 is divisible by (p^(k/d)+1)/2.

			a(3) = 41 is in the sequence because 41 = (3^4 + 1)/2.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A076481.

Contains A067756.

N:= 10^6: # to get terms <= N
p:= 2:
Res:= NULL:
do
  p:= nextprime(p);
  if p^2 >= 2*N then break fi;
  pk:= p;
  do
    pk:= pk^2;
    if pk >= 2*N then break fi;
    v:= (pk+1)/2;
    if isprime(v) then Res:= Res, v;
    fi;
  od
od:
sort([Res]); # Robert Israel, May 27 2019

A173759 Prime numbers p such that 1/p is in the Cantor set.

Original entry on oeis.org

3, 13, 757, 1093, 797161

Offset: 1

T. D. Noe, Feb 23 2010

These are the prime numbers in A121153. Some of these primes are in A076481, the base-3 repunit primes, which have the form (3^n-1)/2. However, 757 has base-3 representation 1001001 and a base-27 representation 111. Are more exceptions?

Contains A076481 as a subsequence, implying that a(6) <= A076481(4) = 3754733257489862401973357979128773. - Max Alekseyev, Aug 17 2013

			3 is here because 1/3 can be written 0.02222222...

Christian Salas, On Prime Reciprocals in the Cantor Set, arXiv:0906.0465 [math.NT]

A385326 The number of positive k <= 2n + 1 such that 2n + 1 divides (2^k + 2*n + 1)^2 - 1.

Original entry on oeis.org

1, 3, 2, 2, 3, 2, 2, 7, 4, 2, 7, 2, 2, 3, 2, 6, 6, 5, 2, 6, 4, 6, 7, 2, 2, 12, 2, 5, 6, 2, 2, 21, 10, 2, 6, 2, 8, 7, 5, 2, 3, 2, 21, 6, 8, 15, 18, 5, 4, 6, 2, 2, 17, 2, 6, 6, 8, 5, 19, 9, 2, 12, 2, 18, 18, 2, 14, 7, 4, 2, 6, 4, 10, 7, 2, 10, 12, 15, 6, 6, 4, 2, 16, 2, 2, 19, 2, 5, 6, 2, 2, 6, 10, 9, 21, 2, 4, 32, 2, 2, 6

Offset: 0

Juri-Stepan Gerasimov, Jun 25 2025

nonn

			1 is the term because 2*0 + 1 = 1 is divisor of (2^1 + 2*0 + 1)^2 - 1 = 3^2 - 1 = 8.

Cf. A003462 (numbers m > 0 such that a(m) = 3), A005384 (primes p such that a(p) = 2), A005408 (odd numbers), A076481 (primes q such that a(q) = 3), A081858 (numbers k numbers k >= 0 such that 2k + 1 divides 2^k - 1), A102781 (numbers k such that 2k + 1 divides (2^k + 2*k + 1)^2 - 1), A224486 (numbers k such that 2k + 1 divides 2^k + 1).

[#[k: k in [1..2*n+1] | ((2^k+2*n+1)^2 - 1) mod (2*n + 1) eq 0]: n in [0..100]];

a[n_]:=Length[Select[Range[2n+1],Divisible[(2^#+2n+1)^2-1,2n+1] &]]; Array[a,101,0] (* Stefano Spezia, Jun 25 2025 *)

a(n) = sum(k=1, 2*n+1, !Mod((2^k + 2*n + 1)^2 - 1, 2*n + 1)); \\ Michel Marcus, Jun 25 2025

cp's OEIS Frontend

A179625 Legal generalized repunit prime numbers.

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A255510 Numbers n of the form 3^k such that sigma(n) is a prime p.

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A308442 Primes of the form (p^k+1)/2 where p is prime and k > 1.

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A173759 Prime numbers p such that 1/p is in the Cantor set.

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A385326 The number of positive k <= 2*n + 1 such that 2*n + 1 divides (2^k + 2*n + 1)^2 - 1.

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Magma

Mathematica

PARI

A385326 The number of positive k <= 2n + 1 such that 2n + 1 divides (2^k + 2*n + 1)^2 - 1.