A395721 Largest k < m^2 such that both rad(k) and rad(k+1) divide m, where rad = A007947 and m = A005117(n), or -1 if such k does not exist.

-1, 1, -1, -1, 8, -1, 4, -1, -1, 7, -1, -1, -1, -1, 1, -1, 1, -1, 80, -1, -1, 16, -1, -1, 1, -1, -1, 63, -1, 1, -1, -1, -1, -1, -1, 1, -1, -1, 31, -1, 242, -1, -1, 49, -1, -1, 1, -1, 26, -1, 1, -1, -1, 1, -1, -1, -1, -1, 1, -1, -1, -1, 288, -1, -1, 1, -1, -1, 10

Offset

1,5

Comments

Largest k < A062503(n) such that both A007947(k) and A007947(k+1) divide A005117(n), or -1 if such k does not exist.

Conjecture: for odd A005117(n), a(n) = -1.

Links

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Formula

For A005117(n) = prime p (i.e., for n in A071403), a(n) = -1 except for a(2) = 1.

For even A005117(n) = 2*j (i.e., n in A383008), a(n) >= 1, since rad(1) | 2*j and rad(2) | 2*j.

a(A365435(i)) = A002072(i).

Example

Let s = A005117.
Table of n, s(n), a(n), a(n)+1 for select n:
    n   s(n)                    a(n)              a(n)+1
  -----------------------------------------------------------------
    2     2 = 2                   1                  2
    5     6 = 2 * 3               8 = 2^3            9  = 3^2
    7    10 = 2 * 5               4 = 2^2            5
   10    14 = 2 * 7               7                  8  = 2^3
   19    30 = 2 * 3 * 5          80 = 2^4 * 5       81  = 3^4
   22    34 = 2 * 17             16 = 2^4           17
   28    42 = 2 * 3 * 7          63 = 3^2 * 7       64  = 2^6
   41    66 = 2 * 3 * 11        242 = 2 * 11^2     243  = 3^5
   63   102 = 2 * 3 * 17        288 = 2^5 * 3^2    289  = 17^2
   72   114 = 2 * 3 * 19        512 = 2^9          513  = 3^3 * 19
  129   210 = 2 * 3 * 5 * 7    4374 = 2 * 3^7     4375  = 5^4 * 7
  437   714 = 2 * 3 * 7 * 17   5831 = 7^3 * 17    5832  = 2^3 * 3^6
.
a(3) = -1 since s(3) = 3, and there are no terms k in {1, 3, 9} such that rad(k+1) | 3. Generally, for odd prime p, there are no terms k in {1, p, p^2} such that rad(k+1) | p.
a(11) = -1 since s(11) = 15, and there exist no k in {1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 125, 135, 225} such that rad(k+1) | 15, etc.

Mathematica

s = Select[Range[110], SquareFreeQ]; rad[x_] := Times @@ FactorInteger[x][[All, 1]]; f[x_] := Select[Range[x^2, 1, -1], Divisible[x, rad[#] ] &]; Table[If[ListQ[#], #[[-1]], -1] &@ SelectFirst[Partition[f[s[[n]] ], 2, 1], #1 - #2 == 1 & @@ # &], {n, Length[s]}]

Crossrefs

Cf. A002072, A005117, A007947, A039956, A062503, A071403, A365435, A383008, A395754.

Sequence in context: A176457 A110194 A199283 * A079359 A010156 A197590

Adjacent sequences: A395716 A395717 A395720 * A395723 A395725 A395726

Keyword

sign

Author

Michael De Vlieger, May 04 2026

Status

approved