A248214 -id:A248214 - cp's OEIS frontend

A248576 Least prime p such that m^n+1 is divisible by p^2, where m = min{ b>0 | b^n+1 not squarefree} = A248214(n).

2, 5, 3, 17, 2, 5, 2, 17, 3, 5, 2, 17, 2, 5, 3, 769, 2, 5, 2, 17, 3, 5, 2, 17, 2, 13, 3, 17, 2, 5, 2, 193, 3, 17, 2, 17, 2, 5, 3, 17, 2, 5, 2, 17, 3, 5, 2, 97, 2, 5, 3, 17, 2, 5, 11, 17, 3, 5, 2, 17, 2, 5, 3, 257, 2, 5, 2, 17, 3, 5, 2, 17, 2, 37, 3, 17, 2, 13, 2, 769, 3, 5, 2

Offset: 1

M. F. Hasler, Oct 08 2014

nonn

See the main entry A248214 for all further information.

Cf. A248214.

a(n,bound=b->n*b*20)=for(b=1,9e9,forprime(p=1,bound(b),Mod(b,p^2)^n+1||return(p))) \\ The given default bound is experimental. You may use, e.g., a(n,b->10^5) for a fixed bound. Especially for n = 2^k >= 32, there might be a larger p leading to a smaller b, than the one found with this bound.

A283452 The smallest square referenced in A248214 (Least integer b > 0 such that b^n + 1 is not squarefree).

Original entry on oeis.org

4, 25, 9, 289, 4, 25, 4, 289, 9, 25, 4, 289, 4, 25, 9, 591361, 4, 25, 4, 289, 9, 25, 4, 289, 4, 169, 9, 289, 4, 25, 4, 37249, 9, 289, 4, 289, 4, 25, 9, 289, 4, 25, 4, 289, 9, 25, 4, 9409, 4, 25, 9, 289, 4, 25, 121, 289, 9, 25, 4, 289, 4, 25, 9, 66049, 4, 25, 4, 289, 9, 25, 4, 289, 4, 1369, 9, 289, 4, 169

Offset: 1

Robert Price, Mar 07 2017

nonn

			A248214(2)=7, 7^2 + 1 = 50 = 2*5*5, so 50 is not squarefree, the square being 5*5 = 25.

Cf. A248214, A248576.

A260824 Least positive integer b such that b^(2^n)+1 is not squarefree.

Original entry on oeis.org

3, 7, 110, 40, 392, 894, 315, 48

Offset: 0

Jeppe Stig Nielsen, Aug 04 2015

For any n, a(n) <= A261117(n).
The smallest square in the factors of b^(2^n)+1 are 2^2, 5^2, 17^2, 17^2, 769^2. - Robert Price, Mar 07 2017; edited by Jeffrey Shallit, May 10 2017
a(8) <= 50104 (corresponding square 10753^2). - Jeffrey Shallit, May 10 2017
Some better bounds than A261117(n): a(9) <= 65863 (factor 13313^2), a(12) <= 265801 (factor 65537^2), a(16) <= 1493667 (factor 1179649^2), a(18) <= 15834352 (factor 7340033^2), a(19) <= 15786037 (factor 23068673^2), a(21) <= 78597313 (factor 230686721^2), a(22) <= 13753565041 (factor 469762049^2), a(23) <= 6276931961 (factor 469762049^2). - Max Alekseyev, Feb 20 2018

			a(1) = A049532(1) = 7.
For n=4, we consider b^16+1. The first time it is not squarefree is for b=392, where 392^16+1 is divisible by 769^2. So a(4)=392.

Subsequence of A248214.

Cf. A261117, A049532.

a(n) = for(b=1,10^42, !issquarefree(b^(2^n)+1) & return(b) );

from sympy.ntheory.factor_ import core
def a(n):
  b, pow2, t = 1, 2**n, 2
  while core(t, 2) == t:
    b += 1
    t = b**(pow2) + 1
  return b
print([a(n) for n in range(4)]) # Michael S. Branicky, Mar 07 2021

a(n) = A248214(2^n).

Edited and a(5)-a(7) added by Max Alekseyev, Feb 20 2018

A261117 Smallest positive integer b such that b^(2^n)+1 is divisible by the square of A035089(n+1).

Original entry on oeis.org

8, 7, 110, 40, 1497, 894, 315, 48, 166107, 95853, 63609, 71589, 492348, 209628, 388440, 48853, 6118793, 2684186, 25787045, 49643800, 54302036, 3969770538, 17592956651, 7347360617, 991255542, 8249087392, 11518171450, 51385581002, 2268777293, 21252616802, 2822082710511

Offset: 0

Jeppe Stig Nielsen, Aug 08 2015

nonn

For given n, if A035089(n+1) exists (which is true by Dirichlet's theorem on arithmetic progressions), then a(n) exists. Proof: p := A035089(n+1) is a prime of the form p=k*2^(n+1)+1, then the group (Z/(p^2)Z)* is cyclic of order p*(p-1) = p*k*2^(n+1). It therefore has an element b of order exactly 2^(n+1). For that b we have then b^(2^n) == -1 (mod p^2).

For given n, a(n) is not necessarily the smallest b such that b^(2^n)+1 is nonsquarefree; see A260824.

			Consider n=4, hence generalized Fermat numbers b^16+1. The first prime (A035089(4+1)) of the form 32*k+1 is 97. It follows that 97 is the smallest prime whose square divides a number of the form b^16+1. The first time 97^2 divides b^16+1 is for b=1497. Hence a(4)=1497. However, A260824(4) is smaller, A260824(4)=392. This is because already 392^16+1 is nonsquarefree (but the prime with a square dividing it, 769, exceeds 97).

Cf. A260824, A248214, A035089.

a(n)=for(k=1,10^10,p=(k<<(n+1))+1;if(isprime(p),break()));for(b=1,p^2,b%p!=0&Mod(b,p^2)^(1<

a(n)=for(k=1, 10^10, p=(k<<(n+1))+1; if(isprime(p), break())); e=p*(p-1)/(1<<(n+1)); h=znprimroot(p^2)^e; g=h^2; m=p^2; for(i=1,1<

cp's OEIS Frontend

A248576 Least prime p such that m^n+1 is divisible by p^2, where m = min{ b>0 | b^n+1 not squarefree} = A248214(n).

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

A283452 The smallest square referenced in A248214 (Least integer b > 0 such that b^n + 1 is not squarefree).

Views

Author

Keywords

Examples

Crossrefs

A260824 Least positive integer b such that b^(2^n)+1 is not squarefree.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Python

Formula

Extensions

A261117 Smallest positive integer b such that b^(2^n)+1 is divisible by the square of A035089(n+1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

PARI