A083501 -id:A083501 - cp's OEIS frontend

A083500 Smallest k such that n*(n+k) + 1 is a cube.

6, 11, 18, 27, 38, 51, 2, 83, 29, 123, 146, 171, 43, 38, 258, 291, 326, 1, 51, 443, 174, 531, 578, 627, 678, 2, 10, 530, 902, 963, 473, 1091, 1158, 1227, 3, 25, 438, 1523, 66, 1683, 1766, 330, 1042, 2027, 46, 2211, 2306, 2403, 70, 2603, 2706, 417, 2918, 73

Offset: 1

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 03 2003

nonn

For all n, n*(n+k) + 1 is a square for k = 2.

Differs from A102305 at positions listed by A102306.

			a(7) = 2 as 7*9 + 1 = 64 = 4^3, though 66 also qualifies but 2<66.

Cf. A083501.

Do[k = 0; While[i = n(n + k) + 1; !IntegerQ[i^(1/3)], k++ ]; Print[k], {n, 1, 55}]

Edited and extended by Robert G. Wilson v, May 11 2003

A091733 a(n) is the least m > 1 such that m^3 = 1 (mod n).

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 2, 9, 4, 11, 12, 13, 3, 9, 16, 17, 18, 7, 7, 21, 4, 23, 24, 25, 26, 3, 10, 9, 30, 31, 5, 33, 34, 35, 11, 13, 10, 7, 16, 41, 42, 25, 6, 45, 16, 47, 48, 49, 18, 51, 52, 9, 54, 19, 56, 9, 7, 59, 60, 61, 13, 5, 4, 65, 16, 67, 29, 69, 70, 11, 72, 25, 8, 47, 76, 45, 23, 55, 23

Offset: 1

David Wasserman, Mar 05 2004

a(n) <= n + 1; the inequality is strict iff n is divisible by 9 or by a prime congruent to 1 mod 3. - Robert Israel, May 27 2014

			a(7) = 2 because 2^3 is congruent to 1 (mod 7).

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A070667, A076947, A083501.

m = 2; while mod(m^3 - 1, n); m = m + 1; end; m

A:= n -> min(select(t -> type((t^3-1)/n, integer), [$2 .. n+1]));
map(A, [$1 .. 1000]); # Robert Israel, May 27 2014

f[n_] := Block[{x = 2}, While[Mod[x^3 - 1, n] != 0, x++]; x]; Array[f, 79] (* Robert G. Wilson v, Mar 29 2016 *)

a(n) = my(k = 2); while(Mod(k, n)^3 != 1, k++); k; \\ Michel Marcus, Mar 30 2016

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A083500 Smallest k such that n*(n+k) + 1 is a cube.

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A091733 a(n) is the least m > 1 such that m^3 = 1 (mod n).

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