A395299 - OEIS

%I #5 Apr 26 2026 22:47:18

%S 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,

%T 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,

%U 1,5,1,5,3,0,3,3,1,6,2,2,1,14,0,6,1,2,1

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

%C 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.

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

%H Michael De Vlieger, <a href="/A395299/b395299.txt">Table of n, a(n) for n = 1..10000</a>

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

%e Let s = A001597 and let t = A088865.

%e Table of n, a(n), s(n) for select n:

%e    n  a(n)   s(n)  k such that t(k) = s(n).

%e   -----------------------------------------------

%e    1    1      1   {1}

%e    2    1      4   {4}

%e    3    1      8   {8}

%e    4    1      9   {9}

%e    5    1     16   {16}

%e    6    2     25   {6, 25}

%e    9    0     36   -

%e   13    1    100   {21}

%e   15    3    125   {12, 18, 125}

%e   20    0    216   -

%e   23    3    256   {39, 55, 256}

%e   34    5    625   {24, 36, 46, 54, 625}

%e   78    6   3600   {371, 611, 731, 779, 851, 899}

%e .

%e 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.

%e a(9) = 0 since s(9) = 36 = 6^2, but 6 is not a sum of 2 distinct primes.

%e 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.

%e 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.

%e 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.

%t 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 &] } ] ]

%Y Cf. A001222, A001597, A008472, A088865.

%K nonn

%O 1,6

%A _Michael De Vlieger_, Apr 19 2026