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A107629
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The present sequence depends on the index k of a Gaussian prime a + bi in A103431. Such an index k is a term of this sequence when an integer multiplier m exists such that m*norm(a+bi) lies in an interval of length 1 around the index k of a+bi in A103431: k - 1/2 < m*norm(a+bi) < k + 1/2.
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3
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1, 2, 8, 12, 13, 38, 39, 80, 142, 143, 216, 218, 221, 222, 325, 329, 330, 447, 448, 450, 590, 594, 765, 954, 955, 1156, 1413, 1418, 1419, 1658, 1660, 1661, 1666, 1667, 1958, 2259, 2260, 2590, 2595, 2940, 3340, 3342, 3763, 4209, 4656, 4657, 4662, 4663, 4668
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OFFSET
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1,2
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COMMENTS
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Consider the Gaussian primes a + bi of the first quadrant ordered as a sequence as in A103431. In A103431 and A103432 these primes are ordered first by their norm and if the norms are equal, by the size of the real part a. A prime p == 1 (mod 4) splits into two different Gaussian primes p = -i(a+bi)(b+ai) where a^2 + b^2 = p and these two primes have the same norm. Through this kind of ordering the primes have a well-defined index k in A103431. The present sequence depends on the index k of a Gaussian prime a + bi in A103431. Such an index k is a term of this sequence when an integer multiplier m exists such that m*norm(a+bi) lies in an interval of length 1 around the index k of a+bi in A103431: k - 1/2 < m*norm(a+bi) < k + 1/2. Counting roughly the first 50000000 Gaussian primes of A103431, every integer < 1600 appeared at least once as a multiplier.
As this property depends only on the norm, one could choose for example the Gaussian primes of the 4th quadrant and would get the same results. It is only necessary that no Gaussian primes are included which are multiples of each other and a unit (-1,i,-i). A107630 gives the squares of the norms, which are integers. A107631 gives the multipliers m. Sequence A107632 (cf. also A107633, A107634) is a subsequence of the present sequence where the distance m*norm(a+bi) from index k is smaller than for all previous values, abs(m*norm(a+bi)-k) is minimal up to k.
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LINKS
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EXAMPLE
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The Gaussian prime with index k=80 in sequence A103431 is 1+20i, norm(1+20i)=20.0249..., norm(1+20i)^2=401. With multiplier m = 4, 4*norm(1+20i) = 80.0999375..., which is in the interval with length 1 around 80. So a(8)=80.
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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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