A280547 Smallest number k such that (k+1)^n - k^n is divisible by a square greater than 1.

4, 7, 3, 14, 1, 23, 3, 7, 2, 2, 1, 75, 3, 7, 3, 36, 1, 2476, 1, 1, 2, 165, 1, 14, 4, 7, 3, 149, 1, 2972, 3, 2, 4, 14, 1, 977, 4, 5, 1, 34, 1, 135, 2, 7, 4, 136, 1, 23, 2, 7, 2, 11, 1, 2, 3, 2, 4

Offset: 2

Author

Juri-Stepan Gerasimov, Jan 06 2017

Keywords

• nonn
• more

Comments

a(31) > 2882.

a(31) <= 2972 (since (2972+1)^31 - 2972^31 is divisible by 1489^2). - Jon E. Schoenfield, Jan 20 2017

From Jon E. Schoenfield, Jan 22 2017: (Start)

Observation: let f(n,k) = (k+1)^n - k^n; then, for each n <= 100, there exists a value of k <= 3735 and a prime p <= 1489 such that p^2 | f(n,k), and the smallest value of k (for a given n) at which any prime-squared divisor p^2 of f(n,k) is found establishes an upper bound on a(n). For n <= 100, there is no prime p in [1490..10^6] whose square divides f(n,k) at a smaller value of k than the upper bound described above; values of that upper bound, for n = 31..100, are 2972, 3, 2, 4, 14, 1, 977, 4, 5, 1, 34, 1, 135, 2, 7, 4, 136, 1, 23, 2, 7, 2, 11, 1, 2, 3, 2, 4, 1155, 1, 3735, 4, 1, 3, 14, 1, 1068, 3, 7, 2, 715, 1, 415, 4, 7, 3, 2, 1, 533, 1, 7, 4, 509, 1, 14, 4, 7, 2, 23, 1, 23, 3, 7, 4, 14, 1, 1550, 3, 2, 1.

Conjecture: for n <= 100, there is no square greater than (10^6)^2 = 10^12 that divides (k+1)^n - k^n at any value of k lower than the upper bound described above; i.e., the upper bound described above is equal to a(n) for each n <= 100. (End)

Confirmed Jon E. Schoenfield's conjecture through a(58). - Robert Price, Feb 04 2017

If p is a prime that divides (k+1)^n - k^n for some k but does not divide n, then by Hensel's lemma there is some k for which p^2 divides (k+1)^n - k^n. In particular, all terms exist. - Robert Israel, Feb 08 2017

Examples

			a(2) = 4 because (4+1)^2 - 4^2 = 9 is a square.
		

Crossrefs

Cf. A280302, A281996, A282174.

Cf. A289629, A289985.

Programs

Mathematica

A280547 = {};
For[n = 2, n < 11, n++,
    k = 0;
    While[SquareFreeQ[(k + 1)^n - k^n], k++];
    AppendTo[A280547, k]];
A280547 (* Robert Price, Feb 04 2017 *)

PARI

a(n) = {my(k = 1); while (issquarefree((k+1)^n - k^n), k++); k;} \\ Michel Marcus, Jan 14 2017

Extensions

a(19)-a(30) from Lars Blomberg, Jan 14 2017

a(31)-a(58) from Robert Price, Feb 04 2017