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A318135
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The 10-adic integer a = ...1588948901 satisfying a^2 + 1 = b, b^2 + 1 = c, c^2 + 1 = d, d^2 + 1 = e, e^2 + 1 = f, and f^2 + 1 = a.
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8
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1, 0, 9, 8, 4, 9, 8, 8, 5, 1, 0, 2, 0, 1, 1, 9, 3, 5, 1, 0, 7, 9, 3, 2, 1, 8, 0, 0, 1, 2, 2, 4, 8, 5, 9, 2, 2, 4, 6, 7, 7, 1, 3, 3, 2, 7, 7, 4, 8, 2, 8, 5, 6, 0, 8, 5, 7, 1, 6, 6, 7, 4, 8, 0, 0, 5, 1, 4, 9, 8, 8, 1, 1, 4, 6, 4, 7, 4, 4, 4, 9, 5, 8, 8, 7, 0, 3, 1, 3, 3, 2, 5, 8, 4, 6, 7, 2, 4, 0, 9, 8, 0, 0, 0, 4, 1, 7, 5, 8, 7, 0, 1, 4, 5, 9, 4, 0, 9, 4, 5, 3, 3, 5, 8, 0, 8, 2, 5, 9, 5, 9, 8, 2, 3, 1, 0, 4, 7, 7, 6, 6, 4, 4, 0, 7, 3, 1, 1, 7, 6
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OFFSET
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0,3
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COMMENTS
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Data generated using MATLAB.
Conjecture: Let r(k) = the smallest positive residue of A003095(6*k+1) mod 10^(6*k+1). Then the first 2*k + 2 digits of r(k), reading from right to left, give the first 2*k + 2 digits of this 10-adic number. For example with k = 5, r(k) = 2121286728960294(201588948901) gives the first 12 digits correctly. - Peter Bala, Nov 14 2022
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LINKS
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EXAMPLE
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901^2 + 1 == 802 (mod 10^3), 802^2 + 1 == 205 (mod 10^3), 205^2 + 1 == 26 (mod 10^3), 26^2 + 1 == 677 (mod 10^3), 677^2 + 1 == 330 (mod 10^3), and 330^2 + 1 == 901 (mod 10^3), so 1 0 9 comprise the sequence's first three terms.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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