A282550 -id:A282550 - cp's OEIS frontend

A285438 Perfect powers that are also the sum of two powers of a prime p.

4, 8, 9, 16, 32, 36, 64, 128, 144, 256, 324, 512, 576, 1024, 2048, 2304, 2744, 2916, 4096, 8192, 9216, 16384, 26244, 32768, 36864, 65536, 131072, 147456, 236196, 262144, 524288, 589824, 941192, 1048576, 2097152, 2125764, 2359296, 4194304, 8388608, 9437184

Offset: 1

Michael Josephy, Apr 18 2017

nonn

Integers n such that there exist integers i, j, k, m, p with i, j >= 0, m, k >= 2 and p prime, such that n = m^k = p^i + p^j.
These are numbers of the form 2^r = 2^(r-1) + 2^(r-1) when r >= 2, numbers of the form (3*2^r)^2 = 2^(2*r) + 2^(2*r+3) and numbers of the form (2*p^r)^k = p^(r*k) + p^(r*k+1) when p = 2^k - 1 is a Mersenne prime. [Edited by Jinyuan Wang, Nov 30 2019]
If n = p^i + p^j is a term with exactly two sets of integer solutions (p, i, j), where i <= j, then n must be 36 = 6^2 = 2^2 + 2^5 = 3^2 + 3^3 or of the form 2^k = 2^(k-1) + 2^(k-1) = p^0 + p^1 where p = 2^k - 1 is a Mersenne prime. There is no n = p^i + p^j in this sequence with at least three sets of integer solutions (p, i, j), where i <= j. - Jinyuan Wang, Nov 30 2019

			324 = 18^2 = 3^4 + 3^5.

Robert Israel, Table of n, a(n) for n = 1..6655
W. Weakley, Problem 11936, Amer. Math. Monthly, 123 (2016), 941.

Cf. A000668, A048645, A072103, A161792, A282550.

N:= 10^9: # to get all terms <= N
R1:= {seq(2^i,i=2..ilog2(N))}:
R2:= {seq(9*2^(2*r), r=0..ilog2(floor(N/9))/2)}:
R3:= {seq(seq(2^k*(2^k-1)^(r*k),r=1..floor(log[2^k-1](N/2^k)/k)),k=select(t -> isprime(2^t-1),[$2..ilog2(N)]))}:
sort(convert(R1 union R2 union R3, list)); # Robert Israel, Apr 25 2017

upto(nn) = {my(v=List([]), k=1); for(r=2, logint(nn, 2), listput(v, 2^r)); for(r=0, logint(nn\9, 4), listput(v, 9*4^r)); while((2*2^k-2)^kJinyuan Wang, Nov 30 2019

a(19)-a(40) from Robert Israel, Apr 25 2017

A304432 Numbers n such that n^2 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).

Original entry on oeis.org

5, 6, 9, 10, 12, 13, 14, 15, 17, 18, 20, 21, 24, 25, 26, 28, 29, 30, 33, 34, 35, 36, 37, 39, 40, 41, 42, 43, 45, 48, 50, 51, 52, 53, 54, 55, 57, 58, 60, 61, 62, 63, 65, 66, 68, 70, 71, 72, 73, 74, 75, 76, 78, 80, 81, 82, 85, 87, 89, 90, 91, 95, 96, 97, 98, 99, 100

Offset: 1

M. F. Hasler, May 13 2018

nonn

			5^2 = 25 = 3^2 + 4^2; 6^2 = 3^2 + 3^3; 9^2 = 2^5 + 7^2, ...

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

Cf. A001597 (perfect powers), A282550.

n = 120; i = 1;
s = Select[Union@ Flatten@ Table[a^2*b^3, {b, Surd[nn, 3]}, {a, Sqrt[nn/b^3]}],
  GCD @@ FactorInteger[#][[All, -1]] > 1 &];
m = Length[s];
Union@ Reap[
  While[i <= m, j = i + 1;
    While[k = s[[i]] + s[[j]]; k <= nn,
      If[And[IntegerQ@ Sqrt[k], i != j],
      Sow[Sqrt[k]]]; j++];
i++] ][[-1, 1]] (* Michael De Vlieger, Dec 02 2024 *)

is(n)=for(i=2,(n^2-1)\2, ispower(i)&&ispower(n^2-i)&&return(i)) \\ For more efficiency, loop over elements of precomputed A001597\{1}.

L=100; PP=List(); a=Set(); for(n=1,L^2, ispower(n)||next; for(i=1,#PP, issquare(n+PP[i],&m)&& m<=L&&  a=setunion(a,[m])); listput(PP,n)); a

A282578 Least k such that k^n is the sum of two distinct proper prime powers (A246547), or 0 if no such k exists.

Original entry on oeis.org

12, 5, 5, 3, 62

Offset: 1

Altug Alkan, Feb 20 2017

Corresponding values of k^n are 12, 25, 125, 81, 916132832, ...

			a(1) = 12 because 12 = 2^2 + 2^3.
a(2) = 5 because 5^2 = 2^4 + 3^2.
a(3) = 5 because 5^3 = 2^2 + 11^2.
a(4) = 3 because 3^4 = 2^5 + 7^2.
a(5) = 62 because 62^5 = 31^5 + 31^6.
a(9) = 2 because 2^9 = 7^3 + 13^2.

Wikipedia, Beal's conjecture

Cf. A001597, A225102, A246547, A282550.

from sympy import nextprime, perfect_power
def ppupto(limit): # distinct proper prime powers <= limit
    p = 2; p2 = pk = p*p; pklist = []
    while p2 <= limit:
        while pk <= limit: pklist.append(pk); pk *= p
        p = nextprime(p); p2 = pk = p*p
    return sorted(pklist)
def sum_of_pp(n):
    pp = ppupto(n); ppset = set(pp)
    for p in pp:
        if p > n//2: break
        if n - p in ppset and n - p != p: return True
    return False
def a(n):
    k = 2
    while not sum_of_pp(k**n): k += 1
    return k
print([a(n) for n in range(1, 6)]) # Michael S. Branicky, Dec 05 2021

a(p) <= 2 * (2^p - 1) where p is in A000043 since (2^p - 1)^p + (2^p - 1)^(p + 1) = (2 * (2^p - 1))^p.

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A285438 Perfect powers that are also the sum of two powers of a prime p.

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A304432 Numbers n such that n^2 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).

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A282578 Least k such that k^n is the sum of two distinct proper prime powers (A246547), or 0 if no such k exists.

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