A362424 Number of partitions of n into 2 distinct perfect powers (A001597).

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 2, 1, 1, 2, 1, 0, 0, 2, 2, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 2, 1, 0, 0, 0, 2, 1, 1, 0, 1, 0, 1, 0, 2, 0, 0, 2, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 2, 0, 0, 0, 2, 1, 1

Offset: 0

Author

Ilya Gutkovskiy, Apr 19 2023

Keywords

• nonn

Links

• Karl-Heinz Hofmann, Table of n, a(n) for n = 0..10000
• Eric Weisstein's World of Mathematics, Perfect Power.

Crossrefs

Cf. A001597, A072103, A362425.

Programs

Mathematica

perfectPowerQ[n_] := n == 1 || GCD @@ FactorInteger[n][[;; , 2]] > 1; a[n_] := Count[IntegerPartitions[n, {2}], ?(AllTrue[#, perfectPowerQ] && UnsameQ @@ # &)]; Array[a, 100, 0] (* _Amiram Eldar, May 05 2023 *)

Python

import numpy
from math import isqrt
A072103 = []
for m in range(2,isqrt(10001)+1):
    k = 2
    while m**k < 10001:
        A072103.append(m**k)
        k += 1
A001597 = sorted(set(A072103)) # eliminates multiples and sorting
A362424 = numpy.zeros(10001+1, dtype="i4")
A001597 = [1] + A001597 # we need a "1" in front of A001597
a = 0
while A001597[a] < 10001 // 2:
    b = a + 1
    while b < len(A001597) and A001597[a] + A001597[b] < 10001:
        A362424[A001597[a] + A001597[b]] += 1
        b += 1
    a += 1
print(list(A362424[0:92])) # Karl-Heinz Hofmann, Sep 16 2023