A260824 -id:A260824 - cp's OEIS frontend

A248214 Least integer b > 0 such that b^n + 1 is not squarefree.

3, 7, 2, 110, 3, 7, 3, 40, 2, 2, 3, 110, 3, 7, 2, 392, 3, 7, 3, 110, 2, 7, 3, 40, 3, 5, 2, 110, 3, 2, 3, 894, 2, 4, 3, 110, 3, 7, 2, 40, 3, 7, 3, 110, 2, 7, 3, 107, 3, 2, 2, 110, 3, 7, 2, 40, 2, 7, 3, 110, 3, 7, 2, 315, 3, 7, 3, 2, 2, 2, 3, 40, 3, 6, 2, 110, 3, 2

Offset: 1

Jeppe Stig Nielsen, Oct 04 2014

nonn

If m is an odd multiple of n, so m=(2k+1)n, then a(m)=a((2k+1)n)<=a(n). This follows from raising the congruence b^n == -1 (mod p^2) to the (2k+1)th power. Because of this, for all k, a(2k+1) <= a(1)=3, a(2*(2k+1)) <= a(2)=7, a(4*(2k+1)) <= a(4)=110, a(8*(2k+1)) <= a(8)=40, a(16(2k+1)) <= a(16)=392, etc. Also a(3(2k+1)) <= a(3)=2.
To show that a(n) is finite for all n, it suffices to show that a(2^j) is finite for all j.
Correspondingly (to the first comment), large & in particular record values are obtained for powers of 2: a(1)=3, a(2)=7, a(4)=110, a(16)=392, a(32)=894, ... - M. F. Hasler, Oct 08 2014
See A248576 for the least prime p such that p^2 divides b^n+1. - M. F. Hasler, Oct 08 2014
For a criterion for a(n) to be finite when n is a power of two, see A261117. - Jeppe Stig Nielsen, Aug 08 2015

			For n = 12, we have that 110^12 + 1 is divisible by a (nonunit) square (namely by 5^2), and since 110 is minimal with this property, a(12) = 110.
For n=32, we have that 894^32 + 1 is divisible by 193^2, and there is no b < 894 such that b^32 + 1 would be divisible by a square > 1. (Conjectural: no factor p^2 with p < 10^6 for any b < 894.) - _M. F. Hasler_, Oct 08 2014

Cf. A248576, A260824, A261117.

for(n=1,1000,b=1;while(issquarefree(b^n+1),b++);print1(b,","))

a(n,bound=b->n*b*20)=for(b=1,9e9,forprime(p=1,bound(b),Mod(b,p^2)^n+1||return(b))) \\ The given default search bound is experimental; might yield only an upper bound as result. You may use, e.g., a(n,b->10^5), for a constant bound. - M. F. Hasler, Oct 08 2014

More terms from M. F. Hasler, Oct 08 2014

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<

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A248214 Least integer b > 0 such that b^n + 1 is not squarefree.

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

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