A382438 Numbers k in A024619 such that all residues r (mod k) in row k of A381801 are such that rad(r) divides k, where rad = A007947.

6, 12, 14, 24, 39, 62, 155, 254, 3279, 5219, 16382, 19607, 70643, 97655, 208919, 262142, 363023, 402233, 712979, 1040603, 1048574, 1508597, 2265383, 2391483, 4685519, 5207819, 6728903, 21243689, 25239899, 56328959, 61035155, 67977559, 150508643

Offset: 1

Author

Michael De Vlieger, Mar 27 2025

Keywords

• nonn

Comments

Numbers k in A024619 such that A381804(k) = 0.

Let S(n,p) be the set of distinct power residues r (mod n) beginning with empty product and recursively multiplying by prime p | n. For example, S(10,2) = {1,2,4,8,6}.

This sequence builds on A381750, taking the tensor product T(k) (mod k) of S(k,p), p | k. If all products r (mod k) are such that rad(r) | k, then k is in this sequence. Distinct residues r (mod k) in T(k) are listed in row k of A381801.

Proper subset of A381750.

A139257 is a proper subset since 2^m is congruent to 2 (mod 2^m-2).

Conjecture: 12 and 24 are the only nonsquarefree numbers in this sequence, i.e., in A126706.

Examples

			Table of a(n) for n = 1..10, showing prime decomposition (facs(a(n))), row a(n) of A381801:
                        Row a(n) of A381801
 n    a(n)  facs(a(n))  k (mod a(n)) such that rad(k) | a(n).
-------------------------------------------------------------
 1      6   2 * 3       {0, 1, 2, 3, 4}
 2     12   2^2 * 3     {0, 1, 2, 3, 4, 6, 8, 9}
 3     14   2 * 7       {0, 1, 2, 4, 7, 8}
 4     24   2^3 * 3     {0, 1, 2, 3, 4, 6, 8, 9, 12, 16, 18}
 5     39   3 * 13      {0, 1, 3, 9, 13, 27}
 6     62   2 * 31      {0, 1, 2, 4, 8, 16, 31, 32}
 7    155   5 * 31      {0, 1, 5, 25, 31, 125}
 8    254   2 * 127     {0, 1, 2, 4, 8, 16, 32, 64, 127, 128}
 9   3279   3 * 1093    {0, 1, 3, 9, 27, 81, 243, 729, 1093, 2187}
10   5219   17 * 307    {0, 1, 17, 289, 307, 4913}
Let b = A381750.
a(1) = 6 since T(6) (mod 6) = {1,2,4} X {1,3} = {{1,2,4},{3,0,0}}; all residues r (mod 6) in T(6) (i.e., in row 6 of A381801) are such that rad(r) | 6.
a(2) = 12 since T(12) (mod 12) = {1,2,4,8} X {1,3,9} = {{1,2,4,8},{3,6,0,0},{9,6,0,0}}; all residues r (mod 12) in T(12) are such that rad(r) | 12.
a(3) = 14 since T(14) (mod 14) = {1,2,4,8} X {1,7} = {{1,2,4,8},{7,0,0,0}}; all residues r (mod 14) in T(14) are such that rad(r) | 14.
a(4) = 24 since T(24) (mod 24) = {1,2,4,8,16} X {1,3,9} = {{1,2,4,8,16},{3,6,12,0,0},{9,18,0,0,0}}; all residues r (mod 24) in T(24) are such that rad(r) | 24.
b(6) = 56 is not in the sequence since 49*2 = 98 = 42 (mod 56), rad(42) does not divide 56.
b(8) = 112 is not in the sequence since 49*4 = 196 = 84 (mod 112), rad(84) does not divide 112, etc.
		

Links

• Michael De Vlieger, Efficient Mathematica code for this sequence (2025).

Crossrefs

Cf. A007947, A024619, A381750, A381801.