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

0,18

Links

Karl-Heinz Hofmann, Table of n, a(n) for n = 0..10000

Eric Weisstein's World of Mathematics, Perfect Power.

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 *)

Prog

(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

Crossrefs

Cf. A001597, A072103, A362425.

Sequence in context: A334109 A109527 A373241 * A186715 A331983 A219485

Adjacent sequences: A362421 A362422 A362423 * A362425 A362426 A362427

Keyword

nonn

Author

Ilya Gutkovskiy, Apr 19 2023

Status

approved