A395299 Number of ways of expressing A001597(n) = A088865(k) for some k.

1, 1, 1, 1, 1, 2, 1, 1, 0, 2, 2, 2, 1, 1, 3, 1, 1, 2, 1, 0, 1, 1, 3, 1, 2, 3, 2, 2, 1, 2, 3, 1, 3, 5, 2, 3, 2, 1, 3, 2, 3, 3, 1, 3, 0, 4, 1, 1, 1, 1, 3, 1, 3, 4, 2, 3, 1, 1, 3, 1, 3, 1, 5, 5, 4, 0, 3, 3, 1, 5, 1, 5, 3, 0, 3, 3, 1, 6, 2, 2, 1, 14, 0, 6, 1, 2, 1

Offset

1,6

Comments

Number of ways of expressing A001597(n) = A008472(k)^A001222(k) for some k, where A008472(k) is the sum of distinct prime factors of k, and A001222(k) = bigomega(k) is the sum of exponents of distinct prime factors of k.

For composite k, A088865(k) >= k, therefore, a(n) is finite (see Neretin's theorem in A088865).

Links

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

Formula

For prime power n = p^m, m > 0, a(p^m) >= 1.

Example

Let s = A001597 and let t = A088865.
Table of n, a(n), s(n) for select n:
   n  a(n)   s(n)  k such that t(k) = s(n).
  -----------------------------------------------
   1    1      1   {1}
   2    1      4   {4}
   3    1      8   {8}
   4    1      9   {9}
   5    1     16   {16}
   6    2     25   {6, 25}
   9    0     36   -
  13    1    100   {21}
  15    3    125   {12, 18, 125}
  20    0    216   -
  23    3    256   {39, 55, 256}
  34    5    625   {24, 36, 46, 54, 625}
  78    6   3600   {371, 611, 731, 779, 851, 899}
.
a(6) = 2 since 6 = 2^1*3^1, thus t(6) = (2+3)^(1+1) = s(6) = 5^2 = 25, and t(25) = s(6) = 5^2 = 25.
a(9) = 0 since s(9) = 36 = 6^2, but 6 is not a sum of 2 distinct primes.
a(13) = 1 since for s(13) = 100 and k <= 100, there exists only 1 number t(k) = 100, and that k = 21 = 3*7; (3+7)^(1+1) = 100.
a(15) = 3 since for k <= 125, we have t(k) = 125 for k = {12, 18, 125}; 12 = 2^2*3 and t(12) = (2+3)^(2+1), 18 = 2*3^2 and t(18) = (2+3)^(1+2), and 125 = 5^3 and t(125) = 5^3.
a(23) = 3 since for k <= 256, we have t(k) = 256 for k = {39, 55, 256}; 39 = 3*13 and t(39) = (3+13)^(1+1), 55 = 5*11 and t(256) = (5+11)^(1+1), and 256 = 2^8 and t(256) = 2^8, etc.

Mathematica

nn = 5000; s = {1}~Join~Union@ Flatten[Table[k^m, {m, 2, Log2[nn]}, {k, 2, Surd[nn, m] } ] ]; Block[{c}, c[_] := 0; c[1] = 1; Do[If[CompositeQ[n], If[# <= nn, c[#]++] &[Power @@ Total@ FactorInteger[n] ] ], {n, nn}]; Table[c[s[[n]] ], {n, LengthWhile[s, # <= nn &] } ] ]

Crossrefs

Cf. A001222, A001597, A008472, A088865.

Sequence in context: A113974 A123331 A235141 * A347386 A331410 A336928

Adjacent sequences: A395296 A395297 A395298 * A395300 A395301 A395302

Keyword

nonn

Author

Michael De Vlieger, Apr 19 2026

Status

approved