A380870 a(n) = A381798(n) - A361373(n) - 1.

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 3, 0, 0, 2, 0, 1, 4, 6, 0, 0, 0, 8, 0, 1, 0, 1, 0, 0, 3, 3, 7, 2, 0, 13, 0, 1, 0, 4, 0, 7, 6, 6, 0, 1, 0, 15, 14, 8, 0, 13, 3, 0, 15, 23, 0, 1, 0, 0, 5, 0, 5, 7, 0, 3, 9, 12, 0, 2, 0, 30, 18, 14, 10, 6, 0, 3, 0, 14, 0

Offset: 1

Author

Michael De Vlieger, Apr 08 2025

Keywords

• nonn

Comments

a(n) = cardinality of the intersection of A024619 and row n of A381799.

Let S(n,p) = {p^m : p | n, m = 1..floor(log_p n)}. Therefore S(10,2) = {1,2,4,8} and S(30,3) = {1,3,9,27}. Then U({S(n,p) : p|n}) = row n of A377485.

Let T(n,p) = {p^m (mod n) : p | n} the set of prime divisor power residues r (mod n) == p^m, p | n. Thus T(10,2) = {1,2,4,8,6} and T(30,3) = {1,3,9,27,21}. Then U({T(n,p) : p|n}) = row n of A381799.

Examples

			Table of n, a(n), and H(n) = intersection of row n of A381799 with A024619.
 n   facs(n)   a(n)  H(n)
--------------------------------------------
 6   2 * 3       0   -
10   2 * 5       1   {6}
12   2^2 * 3     0   -
14   2 * 7       0   -
15   3 * 5       3   {6, 10, 12}
18   2 * 3^2     2   {10, 14}
20   2^2 * 5     1   {12}
21   3 * 7       4   {6, 12, 15, 18}
22   2 * 11      6   {6, 10, 12, 14, 18, 20}
24   2^3 * 3     0   -
30   2 * 3 * 5   1   {21}
.
a(6) = 0 since Q(6) = R(6) = {1,2,3,4}, i.e., all terms in row 6 of A381799 are powers of primes.
a(10) = 1 since Q(10) = {1,2,4,5,8} but R(10) = {1,2,4,5,6,8}; the latter set contains 1 term (i.e., 6) that is not a member of the former set.
a(14) = 0 since R(14) = {1,2,4,7,8} are all powers of primes.
a(15) = 3 since R(15) = {1,3,5,6,9,10,12} has 3 terms {6,10,12} that are not powers of primes.
a(18) = 2 since R(18) = {1,2,3,4,8,9,10,14,16} has 2 terms {6,10} that are not powers of primes, etc.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..16384

Crossrefs

Cf. A000961, A024619, A361373, A377485, A381750, A381798, A381799.

Programs

Mathematica

f[x_, p_] := Block[{m = 2, r, c},
  Which[
    PrimePowerQ[x],
    Join[{0}, #1^Range[0, #2 - 1]] & @@ FactorInteger[x][[1]],
    PowerMod[p, m, x] == p, {1, p},
    True, c[_] := False;
    c[1] = c[p] = True; {1, p}~Join~
      Reap[While[r = PowerMod[p, m, x]; ! c[r], Sow[r];
        c[r] = True; m++] ][[-1, 1]] ] ]
Table[Count[Union@ Flatten@ Map[f[n, #] &, FactorInteger[n][[All, 1]] ], _?(And[# > 1, ! PrimePowerQ[#]] &)], {n, 120}]

Formula

Let Q(n) = {1} joined to row n > 1 of A377485 and let R(n) = row n of A381799.

a(n) = card(U(Q(n) \ R(n))).

a(p^m) = 0 for prime power p^m, m >= 0.

a(n) = 0 for n in A381750.