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A251731
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Least k such that k^3 + q is divisible by 3^n where q is the n-th number congruent to 1 or -1 (mod 18).
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1
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2, 1, 2, 16, 32, 145, 62, 1363, 3458, 19492, 58928, 89308, 70028, 1594318, 1890551, 189871, 31401806, 47918575, 190704887, 163454602, 502048577, 9481323661, 11627845304, 34656488290, 115450061084, 286130228125, 2303721331049, 1569269836240, 22013516320412
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OFFSET
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1,1
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COMMENTS
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It is known that k always exists if q is congruent to +-1 mod 18.
The numbers congruent to 1 or -1 (mod 18) are 1, 17, 19, 35, 37, ... = {A161705} UNION {A239129}.
For n >= 2, k^3 == (9 - 18*n - 7*(-1)^n)/2 (mod 3^n) if and only if k - a(n) is divisible by 3^(n-1). - Jinyuan Wang, Feb 13 2020
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LINKS
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EXAMPLE
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a(1) = 2 because the first number of the form +-1 (mod 18) is 1, and 2^3 + 1 = 9 = 3*3^1;
a(2) = 1 because the second number of the form +-1 (mod 18) is 17, and 1^3 + 17 = 18 = 2*3^2;
a(3) = 2 because the third number of the form +-1 (mod 18) is 19, and 2^3 + 19 = 27 = 3^3;
a(4)= 16 because the fourth number of the form +-1 (mod 18) is 35, and 16^3 + 35 = 4131 = 51*3^4.
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MAPLE
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f:= proc(n) local q, R, k;
if n::odd then q:= 9*n-8 else q:= 9*n-1 fi;
min(map(subs, [msolve(k^3+q, 3^n)], k))
end proc:
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MATHEMATICA
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lst1={1}; Do[lst1=Union[lst1, Union[{18*n+1}, {18*n-1}]], {n, 1, 10}]; lst={}; Do[k=1; While[Mod[k^3+lst1[[n]], 3^n]!=0, k++]; Print[n, " ", k], {n, 1, 10}]; lst
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PROG
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(PARI) a(n) = {if (n % 2, q = 9*(n-1)+1, q = 9*n-1); m = 3^n; k = 1; while ((k^3+q) % m, k++); k; } \\ Michel Marcus, Jan 07 2015
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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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