A289629 -id:A289629 - cp's OEIS frontend

A280302 Smallest k such that (n+1)^k - n^k is divisible by a square > 1.

6, 10, 4, 2, 21, 20, 3, 20, 33, 6, 20, 2, 2, 5, 21, 6, 10, 6, 6, 4, 4, 2, 7, 2, 6, 3, 10, 4, 18, 6, 2, 10, 20, 6, 57, 17, 2, 14, 42, 2, 10, 10, 6, 39, 14, 4, 10, 20, 2, 21, 20, 6, 4, 21, 6, 20, 10, 2, 5, 2, 5, 2, 20, 6, 42, 14, 2, 6, 55, 6, 3, 7, 2, 42, 3, 2

Offset: 1

Juri-Stepan Gerasimov, Dec 31 2016

nonn

a(209) > 70.

a(n) <= p^2 - p, where p = A053670(n). - Jinyuan Wang, May 15 2020

			a(1) = 6 is because (1+1)^6 - 1^6 = 63 is divisible by 9 = 3^2.

Lars Blomberg, Table of n, a(n) for n = 1..208

Cf. A049094, A053670, A237043, A280203, A280208, A280209, A280296, A282176, A282177, A280307.

Cf. A289629, A280547, A289985.

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

More terms from Lars Blomberg, Jan 10 2017

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

Original entry on oeis.org

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

Juri-Stepan Gerasimov, Jan 06 2017

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

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

Cf. A280302, A281996, A282174.

Cf. A289629, A289985.

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 *)

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

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

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

A289985 Smallest positive k such that (n+1)^k + (-n)^k is divisible by a square greater than 1.

Original entry on oeis.org

10, 11, 2, 55, 21, 10, 3, 10, 33, 26, 10, 21, 10, 5, 21, 10, 55, 10, 8, 2, 2, 3, 7, 78, 55, 3, 34, 2, 21, 78, 10, 68, 10, 41, 57, 10, 55, 10, 55, 21, 10

Offset: 1

Juri-Stepan Gerasimov, Sep 02 2017

From Robert Israel, Sep 04 2017: (Start)
If (n+1)^k + (-n)^k is divisible by p^2 then so is (m+1)^k + (-m)^k
for m == n (mod p^2), so a(m) <= k for such m.
For example, a(n) = 2 if n == 3 or 21 (mod 25).
a(24) = 78 because 25^78 + (-24)^78 is divisible by 13^2.
a(42) <= 171 because 43^171 + (-42)^171 is divisible by 19^2.
(End)

			a(1) = 10 because (1+1)^10 + (-1)^10 = 1025 is divisible by 5^2.

Robert Israel, Upper bounds on a(n) for n = 1..2000

Cf. A285929, A289629.

Cf. A280302, A280547.

A289985 := proc(n)
    local k;
    for k from 1 do
        if not issqrfree((n+1)^k+(-n)^k) then
            return k;
        end if;
    end do:
end proc:
for n from 1 do
    printf("%d,\n",A289985(n)) ;
end do: # R. J. Mathar, Sep 04 2017

Table[SelectFirst[Range[10^2], ! SquareFreeQ[(n + 1)^# + (-n)^#] &], {n, 23}] (* Michael De Vlieger, Sep 03 2017 *)

a(24)-a(41) from Giovanni Resta, Sep 04 2017

A349988 a(n) is the smallest k such that n^k + (n+1)^k is divisible by a square > 1.

Original entry on oeis.org

3, 5, 2, 1, 11, 10, 3, 10, 19, 3, 10, 1, 1, 29, 26, 3, 5, 3, 3, 2, 2, 1, 10, 1, 3, 10, 5, 2, 9, 3, 1, 5, 10, 3, 39, 10, 1, 7, 21, 1, 5, 5, 3, 21, 7, 2, 5, 10, 1, 78, 10, 3, 2, 26, 3, 10, 5, 1, 7, 1, 3, 1, 10, 3, 21, 7, 1, 3, 68, 3, 2, 5, 1, 21, 26, 1, 5, 2, 3

Offset: 1

Jon E. Schoenfield, Dec 07 2021

nonn

a(285) <= 111. - Kevin P. Thompson, Jan 13 2022

			a(2) = 5 since the values of 2^k + (2+1)^k for k=1..4 are 5, 13, 35, and 97, none of which are divisible by a square > 1, and 2^5 + (2+1)^5 = 275 which is divisible by 25 = 5^2.

Kevin P. Thompson, Table of n, a(n) for n = 1..284
Kevin P. Thompson, Factorizations to support a(n) for n = 1..284

Cf. A046092, A280302, A280547, A289629, A289985, A349989.

Table[k=1;While[SquareFreeQ[n^k+(n+1)^k],k++];k,{n,33}] (* Giorgos Kalogeropoulos, Dec 08 2021 *)

a(n) = my(k=1); while(issquarefree(n^k + (n+1)^k), k++); k; \\ Michel Marcus, Dec 08 2021

a(9m-5) = 1 for m >= 1 since a(9m-5) = (9m-5)^1 + (9m-5+1)^1 = 18m-9 which is divisible by 9 = 3^2. - Kevin P. Thompson, Jan 13 2022

a(n) = 1 if n is in A046092; in this case, if n = A046092(m), then n^1 + (n+1)^1 = (2*m+1)^2. - Bernard Schott, Jan 17 2022

Thanks to Hugo Pfoertner for computing terms a(50) and a(68).

A349989 a(n) is the smallest k such that k^n + (k+1)^n is divisible by a square > 1.

Original entry on oeis.org

4, 3, 1, 113, 2, 3, 3, 19, 1, 1, 4, 113, 4, 3, 1, 765, 4, 3, 4, 87, 1, 3, 4, 19, 2, 2, 1, 28, 4, 1, 4, 151, 1, 3, 2, 113, 4, 3, 1, 19, 4, 3, 4, 113, 1, 3, 4, 335, 3, 1, 1, 113, 4, 3, 1, 19, 1, 3, 4, 87, 4, 3, 1, 379, 2, 3, 4, 1, 1, 1, 4, 19, 4, 3, 1, 113, 3, 1, 4

Offset: 1

Jon E. Schoenfield, Dec 07 2021

nonn

a(64) <= 379; a(76) <= 113. Terms a(65)..a(79): 2, 3, 4, 1, 1, 1, 4, 19, 4, 3, 1, a(76), 3, 1, 4.

At k=4, k^n + (k+1)^n = 4^n + 5^n is a multiple of 9 for all odd n, and at k=3, k^n + (k+1)^n = 3^n + 4^n is a multiple of 25 for all n == 2 (mod 4). Thus, a(n) <= 4 if n is not a multiple of 4.

Kevin P. Thompson, Factorizations to support a(n) for n = 1..79

Cf. A280302, A280547, A289629, A289985, A349988.

a(n) = my(k=1); while(issquarefree(k^n + (k+1)^n), k++); k; \\ Michel Marcus, Dec 09 2021

from sympy.ntheory.factor_ import core
def squarefree(n): return core(n, 2) == n
def a(n):
    k = 1
    while squarefree(k**n + (k+1)**n): k += 1
    return k
print([a(n) for n in range(1, 16)]) # Michael S. Branicky, Dec 09 2021

a(n) = A289629(n) if n is even.
a(k) = 1 for k in A049096.
a(n) <= 4 if 4 does not divide n; among terms where 4 divides n, certain terms appear repeatedly. E.g.,
  a(n) <= 113 for n ==  4 (mod 8):   for all such n,  17^2 divides 113^n + 114^n;
  a(n) <=  19 for n ==  8 (mod 16):  for all such n,  17^2 divides  19^n +  20^n;
  a(n) <= 765 for n == 16 (mod 32):  for all such n,  97^2 divides 765^n + 766^n;
  a(n) <=  87 for n == 20 (mod 40):  for all such n,  41^2 divides  87^n +  88^n;
  a(n) <=  28 for n == 68 (mod 136): for all such n,  17^2 divides  28^n +  29^n;
  a(n) <= 151 for n == 32 (mod 64):  for all such n, 257^2 divides 151^n + 152^n;
  a(n) <= 335 for n == 48 (mod 96):  for all such n, 769^2 divides 335^n + 336^n.

a(64)-a(79) from Kevin P. Thompson, Feb 23 2022

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A280302 Smallest k such that (n+1)^k - n^k is divisible by a square > 1.

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

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A289985 Smallest positive k such that (n+1)^k + (-n)^k is divisible by a square greater than 1.

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A349988 a(n) is the smallest k such that n^k + (n+1)^k is divisible by a square > 1.

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A349989 a(n) is the smallest k such that k^n + (k+1)^n is divisible by a square > 1.

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