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A350341
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a(n) is the smallest abelian order with precisely 2^n groups.
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4
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1, 4, 1225, 354025, 187279225, 256385259025, 720186192601225, 2898439345541083201, 6402652514300252791009, 32275771324587574319476369, 171997585388727183548489570401, 1618325280922534070007738367903009, 18528206141282092567518596574121550041, 298841436852738871021507444144006480611289
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OFFSET
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0,2
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COMMENTS
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If m is an abelian order, then m = (p_1)^2 * (p_2)^2 * ... * (p_r)^2 * q_1 * q_2 * ... * q_s, where p_1, p_2, ... p_r, q_1, q_2, ..., q_s are distinct primes such that (p_i)^2 !== 1 (mod p_j) for i != j, (p_i)^2 !== 1 (mod q_j), q_i !== 1 (mod p_j), q_i !== 1 (mod q_j) for i != j. In this case there are 2^r groups of order m.
Note that the smallest abelian order with precisely 2^n groups must be the square of a squarefree number.
a(n) is the smallest square number with n distinct prime factors that is an abelian order.
a(n) is the smallest number of the form (p_1*p_2*...*p_n)^2 where the p_i are distinct primes and no (p_j)^2-1 is divisible by any p_i.
a(n) exists for all n.
Except for a(1) = 4, no term can be divisible by 2 or 3. Conjecture: lpf(a(n+1)) >= lpf(a(n)) for all n, where lpf = least prime factor.
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LINKS
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FORMULA
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EXAMPLE
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a(2) = 1225 = 35^2 since the smallest square number with 2 distinct prime factors that is an abelian order is 1225.
a(3) = 354025 = 595^2 since the smallest square number with 3 distinct prime factors that is an abelian order is 354025.
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PROG
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(PARI) isA051532(n) = my(f=factor(n), v=vector(#f[, 1])); for(i=1, #v, if(f[i, 2]>2, return(0), v[i]=f[i, 1]^f[i, 2])); for(i=1, #v, for(j=i+1, #v, if(v[i]%f[j, 1]==1 || v[j]%f[i, 1]==1, return(0)))); 1 \\ Charles R Greathouse IV's program for A051532
a(n) = for(k=1, oo, if(issquarefree(k) && omega(k)==n && isA051532(k^2), return(k^2)))
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CROSSREFS
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KEYWORD
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nonn,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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