cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A281996 The smallest square referenced in A280547 (Smallest number k such that (k+1)^n - k^n is divisible by a square > 1).

Original entry on oeis.org

9, 169, 25, 121, 9, 1849, 25, 169, 25, 529, 9, 2809, 49, 169, 25, 18769, 9, 52441, 25, 49, 529, 2209, 9, 121, 9, 169, 25, 3481, 9, 2217121, 25, 529, 9, 121, 9, 22201, 9, 169, 25, 6889, 9, 29929, 529, 169, 9, 80089, 9, 1849, 25, 169, 169, 11449, 9, 529, 25, 361, 9
Offset: 2

Views

Author

Robert Price, Feb 04 2017

Keywords

Comments

The square roots of a(n) are given in A282174.

Crossrefs

Cf. A280547, A282174.

A282174 Square root of the smallest square referenced in A280547 (Smallest number k such that (k+1)^n - k^n is divisible by a square > 1).

Original entry on oeis.org

3, 13, 5, 11, 3, 43, 5, 13, 5, 23, 3, 53, 7, 13, 5, 137, 3, 229, 5, 7, 23, 47, 3, 11, 3, 13, 5, 59, 3, 1489, 5, 23, 3, 11, 3, 149, 3, 13, 5, 83, 3, 173, 23, 13, 3, 283, 3, 43, 5, 13, 13, 107, 3, 23, 5, 19, 3
Offset: 2

Views

Author

Robert Price, Feb 07 2017

Keywords

Crossrefs

Cf. A280547, A281996.

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

Original entry on oeis.org

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

Views

Author

Juri-Stepan Gerasimov, Dec 31 2016

Keywords

Comments

a(209) > 70.
a(n) <= p^2 - p, where p = A053670(n). - Jinyuan Wang, May 15 2020

Examples

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

Links

Crossrefs

Cf. A049094, A053670, A237043, A280203, A280208, A280209, A280296, A282176, A282177, A280307.
Cf. A289629, A280547, A289985.

Programs

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

Extensions

More terms from Lars Blomberg, Jan 10 2017

A289629 Smallest positive k such that (k+1)^n + (-k)^n is divisible by a square greater than 1.

Original entry on oeis.org

3, 7, 113, 14, 3, 23, 19, 7, 1, 2, 113, 75, 3, 7, 765, 36, 3, 2476, 87, 1, 3, 165, 19, 14, 2, 7, 28, 149, 1, 2972, 151, 2, 3, 14, 113, 977, 3, 5, 19, 34, 3, 135, 113, 7, 3, 136, 335, 23, 1, 7, 113, 11, 3, 2, 19, 2, 3
Offset: 2

Views

Author

Juri-Stepan Gerasimov, Sep 02 2017

Keywords

Comments

From Kevin P. Thompson, Dec 03 2021: (Start)
Most of the factorizations to prove known terms of this sequence have been uploaded to factordb.com.
For known terms up to n = 100, the square that divides (k+1)^n + (-k)^n is very small, less than 1500^2, with only one example greater than 1000^2.
a(20m+10) = 1 for m >= 0. With k = 1 and starting at n = 20*0 + 10 = 10, (k+1)^n + (-k)^n = 2^10 + 1 = 1024 + 1 = 1025 which is divisible by 5^2. Since the last two digits of 2^n repeat in a cycle of length 20, (k+1)^n + (-k)^n will always be divisible by 5^2 for n = 20m + 10.
Conjecture: (k+1)^n + (-k)^n is not squarefree for the following (n, k) patterns, with m >= 1: (22m-11, 2), (20m-6, 3), (20m-2, 3), (3^m, 7), (15m, 7), (20m-15, 14), (16m-8, 19), (42m-35, 23), and (8m-4, 113). In each case, the value of a(n) in this sequence is usually equal to the value specified for k, but sometimes this value is not the smallest such k. For example, in the (n, k) = (20m-2, 3) case, a(20m-2) = 3 for m = 1..3, but at m = 4, a(20*4-2) = a(78) = 1.
(End)

Examples

			a(2) = 3 because (1+1)^2 + (-1)^2 = 5 is squarefree, (2+1)^2 + (-2)^2 = 13 is squarefree, and (3+1)^2 + (-3)^2 = 25 is divisible by 5^2.

Links

  • Kevin P. Thompson, Table of n, a(n) for n = 2..150 with unconfirmed terms (term 31 confirmed by _Hugo Pfoertner_ and term 48 confirmed by _Jon E. Schoenfield_ and _Hugo Pfoertner_)

Crossrefs

Cf. A280547, A285929.
Cf. A280302, A289985.

Programs

  • Mathematica
    Table[k=1;While[SquareFreeQ[(k+1)^n+(-k)^n],k++];k,{n,2,15}] (* Giorgos Kalogeropoulos, Dec 03 2021 *)
  • PARI
    a(n) = my(k=1); while (issquarefree((k+1)^n + (-k)^n), k++); k; \\ Michel Marcus, Dec 04 2021

Formula

a(2n+1) = A280547(2n+1).

Extensions

Offset corrected; a(16), a(32), a(36), a(44), and a(48) corrected; and a(50)-a(58) added by Kevin P. Thompson, Dec 05 2021.

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

Views

Author

Juri-Stepan Gerasimov, Sep 02 2017

Keywords

Comments

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)

Examples

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

Links

Crossrefs

Cf. A285929, A289629.
Cf. A280302, A280547.

Programs

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

Extensions

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

Views

Author

Jon E. Schoenfield, Dec 07 2021

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

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

Programs

  • Mathematica
    Table[k=1;While[SquareFreeQ[n^k+(n+1)^k],k++];k,{n,33}] (* Giorgos Kalogeropoulos, Dec 08 2021 *)
  • PARI
    a(n) = my(k=1); while(issquarefree(n^k + (n+1)^k), k++); k; \\ Michel Marcus, Dec 08 2021

Formula

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

Extensions

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

Views

Author

Jon E. Schoenfield, Dec 07 2021

Keywords

Comments

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.

Links

Crossrefs

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

Programs

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

Formula

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.

Extensions

a(64)-a(79) from Kevin P. Thompson, Feb 23 2022
Showing 1-7 of 7 results.

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