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A343771
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Smallest k such that circle centered at the origin and with radius sqrt(k) passes through exactly 6*n integer points in the hexagonal lattice (see A004016).
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7
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1, 7, 49, 91, 2401, 637, 117649, 1729, 8281, 31213, 282475249, 12103, 13841287201, 1529437, 405769, 53599, 33232930569601, 157339, 1628413597910449, 593047, 19882681, 3672178237, 3909821048582988049, 375193, 68574961, 179936733613, 2989441, 29059303, 459986536544739960976801, 7709611
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OFFSET
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1,2
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COMMENTS
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a(n) is the smallest k such that A004016(k) = 6*n.
Also a(n) is the smallest index of n in A002324.
a(n) is the smallest term in A004611 that has exactly n divisors.
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LINKS
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FORMULA
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If the factorization of n into primes is n = Product_{i=1..r} p_i with p_1 >= p_2 >= ... >= p_r, then a(n) <= (q_1)^((p_1)-1) * (q_2)^((p_2)-1) * ... * (q_r)^((p_r)-1), where q_1 < q_2 < ... < q_r are the first r primes congruent to 1 modulo 3. The smallest n such that the equality does not hold is n = 128. [Those n such that the equality does not hold are listed in A355919. - Jianing Song, Jul 21 2022]
a(n) <= 7^(n-1) for all n, where the equality holds if and only if n = 1 or n is a prime.
a(p*q) = 7^(p-1) * 13^(q-1) for primes p >= q.
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EXAMPLE
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91 = 7 * 13 is the smallest number all of whose prime factors are congruent to 1 modulo 3 with exactly 4 divisors, so a(4) = 91.
8281 = 7^2 * 13^2 is the smallest number all of whose prime factors are congruent to 1 modulo 3 with exactly 9 divisors, so a(9) = 8281.
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PROG
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(PARI) primelist(d, r, l) = my(v=vector(l), i=0); if(l>0, forprime(p=2, oo, if(Mod(p, d)==r, i++; v[i]=p; if(i==l, break())))); v
prodR(n, maxf)=my(dfs=divisors(n), a=[], r); for(i=2, #dfs, if( dfs[i]<=maxf, if(dfs[i]==n, a=concat(a, [[n]]), r=prodR(n/dfs[i], min(dfs[i], maxf)); for(j=1, #r, a=concat(a, [concat(dfs[i], r[j])]))))); a
A343771(n)=my(pf=prodR(n, n), a=1, b, v=primelist(3, 1, bigomega(n))); for(i=1, #pf, b=prod(j=1, length(pf[i]), v[j]^(pf[i][j]-1)); if(b<a || i==1, a=b)); a \\ following program for A005179.
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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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