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A261117
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Smallest positive integer b such that b^(2^n)+1 is divisible by the square of A035089(n+1).
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2
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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
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OFFSET
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0,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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).
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PROG
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(PARI) 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<<n)==-1&return(b))
(PARI) 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<<n, m=min(m, lift(h)); h*=g); m
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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