A246547 -id:A246547 - cp's OEIS frontend

A320700 Odd numbers whose product of prime indices is a nonprime prime power (A246547).

7, 9, 19, 21, 23, 25, 27, 49, 53, 57, 63, 81, 97, 103, 115, 121, 125, 131, 133, 147, 159, 171, 189, 227, 243, 289, 311, 343, 361, 371, 393, 399, 419, 441, 477, 513, 515, 529, 567, 575, 625, 661, 691, 719, 729, 917, 931, 933, 961, 1007, 1009, 1029, 1067, 1083

Offset: 1

Gus Wiseman, Oct 19 2018

nonn

			The sequence of all integer partitions whose Heinz numbers belong to the sequence begins: (4), (2,2), (8), (4,2), (9), (3,3), (2,2,2), (4,4), (16), (8,2), (4,2,2), (2,2,2,2), (25), (27), (9,3), (5,5), (3,3,3), (32), (8,4), (4,4,2), (16,2), (8,2,2), (4,2,2,2), (49), (2,2,2,2,2)

Index entries for sequences computed from indices in prime factorization

Cf. A000720, A000961, A001597, A003963, A056239, A064573, A112798, A246547, A279787, A320322, A320325, A320698, A320699.

Select[Range[1000],With[{x=Times@@Cases[FactorInteger[#],{p_,k_}:>PrimePi[p]^k]},OddQ[#]&&!PrimeQ[x]&&PrimePowerQ[x]]&]

A060847 Difference between a nontrivial prime power (A246547) and the previous prime.

Original entry on oeis.org

1, 1, 2, 3, 2, 4, 1, 2, 3, 2, 8, 12, 1, 2, 2, 5, 6, 6, 2, 3, 6, 6, 2, 2, 8, 3, 4, 2, 12, 2, 9, 8, 18, 2, 2, 6, 4, 12, 2, 3, 6, 4, 2, 6, 12, 8, 2, 6, 2, 1, 6, 8, 2, 2, 14, 4, 6, 2, 6, 2, 3, 20, 2, 12, 2, 2, 8, 14, 10, 18, 8, 6, 2, 2, 2, 12, 12, 19, 2, 6, 6, 20, 2, 2, 2, 8, 8, 2, 2, 8, 20, 12, 15, 2, 4

Offset: 1

Labos Elemer, May 03 2001

nonn

a(n)=1 only for some powers of 2.

			78125=5^7 follows 78121, the difference is 4.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A025475, A000961, A001597, A001694, A007917, A007918, A013632, A013633, A049711, A246547.

N:= 10^5: # to consider prime powers <= N
P:= select(isprime,[2,seq(i,i=3..floor(sqrt(N)),2)]):
PP:= sort([seq(seq(p^k,k=2..ilog[p](N)),p=P)]):
map(t -> t - prevprime(t), PP); # Robert Israel, Nov 13 2024

from sympy import primepi, integer_nthroot, prevprime
def A060847(n):
    def f(x): return int(n+x-sum(primepi(integer_nthroot(x,k)[0]) for k in range(2,x.bit_length())))
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    return (a:=bisection(f,n,n))-prevprime(a) # Chai Wah Wu, Sep 13 2024

a(n) = A246547(n)-prevprime(A246547(n)) = A246547(n)-A049711(A246547(n)).

A282633 Numbers n such that n^2 + 1 is the sum of two proper prime powers (A246547) in more than one way.

Original entry on oeis.org

47, 73, 83, 133, 157, 173, 187, 191, 203, 217, 317, 319, 353, 437, 463, 467, 487, 499, 557, 577, 583, 593, 599, 613, 623, 697, 703, 727, 733, 767, 829, 857, 863, 871, 931, 983, 1013, 1027, 1033, 1067, 1087, 1097, 1123, 1139, 1177, 1267, 1279, 1321, 1327, 1333, 1363, 1403, 1409, 1433, 1453, 1477, 1487, 1493, 1507, 1517, 1543, 1567, 1603, 1607, 1613

Offset: 1

Robert Israel and Altug Alkan, Feb 19 2017

nonn

			83 is a term because 83^2 + 1 = 7^4 + 67^2 = 43^2 + 71^2.

Robert Israel, Table of n, a(n) for n = 1..2672

Cf. A002522, A225103, A246547.

N:= 10^8: # to get all terms <= sqrt(N-1).
PP:= sort([seq(seq(p^k, k=2..floor(log[p](N))), p = select(isprime, [2, seq(i, i=3..floor(sqrt(N)), 2)]))]):
npp:= nops(PP):
res:= {}: R:= 'R':
for i from 2 to npp do
   for j from 1 to i-1 do
     q:= PP[i]+PP[j];
     if q > N then break fi;
      if issqr(q-1) then
       if assigned(R[q]) then res:= res union {q}
        else R[q]:= 1
      fi fi
od od:
sort(convert(map(t -> sqrt(t-1), res),list));

A287299 Number of ways of writing n as a sum of a proper prime power (A246547) and a nonprime squarefree number (A000469).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 2, 1, 0, 1, 2, 2, 0, 0, 2, 2, 1, 1, 3, 0, 1, 1, 4, 3, 0, 2, 2, 2, 0, 3, 4, 3, 1, 2, 6, 3, 1, 0, 5, 4, 2, 2, 4, 3, 0, 2, 3, 5, 0, 1, 3, 4, 3, 2, 4, 3, 3, 4, 5, 4, 0, 2, 5, 5, 0, 4, 6, 2, 1, 1, 7, 3, 1, 2, 7, 4, 2, 4, 5, 5, 1, 3, 6, 5, 1, 3, 6, 6, 3, 4, 4, 4, 2, 4, 7, 6, 3, 1, 4, 4, 0, 4, 6, 5, 2, 2, 7, 5, 2, 1, 7, 8, 4

Offset: 0

Ilya Gutkovskiy, May 23 2017

nonn

Conjecture: a(n) > 0 for all n > 108.

			a(26) = 3 because we have [25, 1], [22, 4] and [16, 10].

Ilya Gutkovskiy, Extended graphical example
Eric Weisstein's World of Mathematics, Prime Power
Eric Weisstein's World of Mathematics, Squarefree

Cf. A000469, A098983, A246547, A282192, A282290, A282318, A282947.

nmax = 120; CoefficientList[Series[(Sum[Boole[SquareFreeQ[k] && ! PrimeQ[k]] x^k, {k, 1, nmax}]) (Sum[Boole[PrimePowerQ[k] && ! PrimeQ[k]] x^k, {k, 1, nmax}]), {x, 0, nmax}], x]

x='x+O('x^120); concat([0, 0, 0, 0, 0], Vec(sum(k=1, 120, (issquarefree(k) && !isprime(k))*x^k) * sum(k=1, 120, (isprimepower(k) && !isprime(k))*x^k))) \\ Indranil Ghosh, May 23 2017

G.f.: (Sum_{k>=1} x^A246547(k))*(Sum_{k>=1} x^A000469(k)).

A290135 Numbers that are the sum of two proper prime powers (A246547).

Original entry on oeis.org

8, 12, 13, 16, 17, 18, 20, 24, 25, 29, 31, 32, 33, 34, 35, 36, 40, 41, 43, 48, 50, 52, 53, 54, 57, 58, 59, 64, 65, 68, 72, 73, 74, 76, 80, 81, 85, 89, 90, 91, 96, 97, 98, 106, 108, 113, 125, 128, 129, 130, 132, 133, 134, 136, 137, 141, 144, 145, 146, 148, 150, 152, 153, 155, 157, 160, 162, 170, 173, 174, 177, 178

Offset: 1

Ilya Gutkovskiy, Jul 20 2017

nonn

Is 2213 the largest prime term that can be expressed as the sum of two proper prime powers in more than one way? - Altug Alkan, Jul 22 2017

			13 is in the sequence because 13 = 2^2 + 3^2.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A014091, A070049, A071330, A071331, A225102, A225103, A246547.

N:= 1000: # to get all terms <= N
P:= select(isprime, [$2..floor(sqrt(N))]):
PP:= {seq(seq(p^j, j=2..floor(log[p](N))),p=P)}:
A:= select(`<=`,{seq(seq(PP[i]+PP[j],j=1..i),i=1..nops(PP))},N):
sort(convert(A,list)); # Robert Israel, Jul 21 2017

nmax = 180; f[x_] := Sum[Boole[PrimePowerQ[k] && PrimeOmega[k] > 1] x^k, {k, 1, nmax}]^2; Exponent[#, x] & /@ List @@ Normal[Series[f[x], {x, 0, nmax}]]

Exponents in expansion of (Sum_{k>=1} x^A246547(k))^2.

A343123 Primes p such that the sum of A001414(k) for k strictly between p and the following prime is a proper prime power (a term of A246547).

Original entry on oeis.org

3, 13, 17, 19, 239, 269, 457, 751, 1091, 1319, 1871, 2129, 2141, 2341, 2549, 2683, 2969, 3167, 3359, 3671, 3821, 4091, 4799, 5437, 5843, 6299, 6551, 6779, 7559, 8387, 8999, 9239, 9419, 10529, 11057, 11717, 11777, 12071, 13309, 13901, 17027, 17203, 18047, 18311, 18521, 21139, 23831, 26249, 26861

Offset: 1

J. M. Bergot and Robert Israel, Apr 05 2021

nonn

Primes prime(k) such that Sum_{prime(k) < j < prime(k+1)} A001414(j) is in A246547.

			a(4) = 19 is a term because 19 and 23 are consecutive primes with Sum_{19 < j < 23} A001414(j) = 9+10+13 = 32 = 2^5.

Robert Israel, Table of n, a(n) for n = 1..3000

Cf. A001414, A246547.

spf:= proc(n) local t; add(t[1]*t[2],t=ifactors(n)[2]) end proc:
R:= NULL: count:= 0: p:= 2:
while count < 100 do
  q:= p; p:= nextprime(p);
  L:= ifactors(add(spf(i),i=q+1..p-1))[2];
  if nops(L) = 1 and L[1][2]>1 then
    count:= count+1; R:= R, q;
  fi
od:
R;

A282533 Primes that are the sum of two proper prime powers (A246547) in more than one way.

Original entry on oeis.org

41, 89, 113, 137, 593, 857, 2213

Offset: 1

Altug Alkan, Feb 17 2017

Primes of the form 2^k + p^e in more than one way where p is an odd prime (e > 1, k > 1).
Prime terms in A225103.
29 = 2^4 + 5^2 = 2 + 3^3 is a border case not included in this sequence - Olivier Gérard, Feb 25 2019
a(8) > 10^8 if it exists. - Robert Israel, Feb 17 2017
a(8) > 10^18 if it exists. - Charles R Greathouse IV, Feb 19 2017

			41 = 2^4 + 5^2 = 2^5 + 3^2.
89 = 2^3 + 3^4 = 2^6 + 5^2.
113 = 2^5 + 3^4 = 2^6 + 7^2.
137 = 2^7 + 3^2 = 2^4 + 11^2.
593 = 2^9 + 3^4 = 2^6 + 23^2.
857 = 2^7 + 3^6 = 2^4 + 29^2.
2213 = 2^4 + 13^3 = 2^2 + 47^2.

Cf. A225099, A225102, A225103, A246547.

Cf. A115231 (prime numbers which cannot be written as 2^a + p^b, b>=0)

N = 10^8; % to get all terms <= N
C = sparse(1,N);
for p = primes(sqrt(N))
  C(p .^ [2:floor(log(N)/log(p))]) = 1;
end
R = zeros(1,N);
for k = 2: floor(log2(N))
  R((2^k+1):N) = R((2^k+1):N) + C(1:(N-2^k));
end
P = primes(N);
P(R(P) > 1.5) % Robert Israel, Feb 17 2017

N:= 10^6: # to get all terms <= N
B:= Vector(N):
C:= Vector(N):
for k from 2 to ilog2(N) do B[2^k]:= 1 od:
p:= 2:
do
  p:= nextprime(p);
  if p^2 > N then break fi;
  for k from 2 to floor(log[p](N)) do C[p^k]:= 1 od:
od:
R:= SignalProcessing:-Convolution(B,C):
select(t -> isprime(t) and R[t-1] > 1.5, [seq(i,i=3..N,2)]); # Robert Israel, Feb 17 2017

Select[Prime@ Range[10^3], Function[n, Count[Transpose@{n - #, #}, w_ /; Times @@ Boole@ Map[And[PrimePowerQ@ #, ! PrimeQ@ #] &, w] > 0] >= 2 &@ Range[4, Floor[n/2]]]] (* or *)
With[{n = 10^8}, Keys@ Select[#, Length@ # > 1 &] &@ GroupBy[#, First] &@ SortBy[Transpose@ {Map[Total, #], #}, First] &@ Select[Union@ Map[Sort, Tuples[#, 2]], PrimeQ@ Total@ # &] &@ Flatten@ Map[#^Range[2, Log[#, Prime@ n]] &, Array[Prime@ # &, Floor@ Sqrt@ n]]] (* Michael De Vlieger, Feb 19 2017, latter program Version 10 *)

is(n) = if(!ispseudoprime(n), return(0), my(x=n-1, y=1, i=0); while(y < x, if(isprimepower(x) > 1 && isprimepower(y) > 1, if(i==0, i++, return(1))); y++; x--)); 0 \\ Felix Fröhlich, Feb 18 2017

has(p)=my(t,q); p>40 && sum(k=2,logint(p-9,2), t=2^k; sum(e=2,logint(p-t,3), ispower(p-t,e,&q) && isprime(q)))>1
list(lim)=my(v=List(),t,q); lim\=1; if(lim<9,lim=9); for(k=2,logint(lim-9,2), t=2^k; for(e=2,logint(lim-t,3), forprime(p=3,sqrtnint(lim-t,e), q=t+p^e; if(isprime(q) && has(q), listput(v,q))))); Set(v) \\ Charles R Greathouse IV, Feb 18 2017

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.

A282686 Least sum of two proper prime powers (A246547) that is the product of n distinct primes.

Original entry on oeis.org

13, 33, 130, 966, 14322, 81510, 3530730, 117535110, 2211297270, 131031070170, 1295080356570, 163411918786830, 3389900689405230, 414524121952915590, 2951531806477464210, 754260388389042905370

Offset: 1

Altug Alkan, Feb 20 2017

Least value of A225102 that is the product of n distinct primes.
From Jon E. Schoenfield, Mar 18 2017: (Start)
For each n, we can write a(n) = p^j + q^k where p and q are prime and 2 <= j <= k; since a(n) is squarefree, p and q are distinct.
Suppose j and k are both even. Then a(n) cannot have any prime factor f such that f == 3 (mod 4) (see A002145). Thus, a(n) is the product of n distinct terms of {2, 5, 13, 17, 29, 37, 41, ...} = A002313, so a(n) >= Product_{i=1..n} A002313(i) = A185952(n).
In fact, however, a(n) < A185952(n) for n = 4..15, and it seems nearly certain that this holds for all n > 3. In any case, if we search for a(n) by generating products of n distinct primes and, for each such product P, testing whether there exists a solution for P = p^j + q^k, then we need not consider solutions in which both j and k are even unless P >= A185952(n).
Additionally, since the sum of any two cubes that is divisible by 3 is also divisible by 9 (hence nonsquarefree), any P that is divisible by 3 cannot be the sum of two cubes, so the exponents j and k cannot both be divisible by 3. (Every P < 2*5*7*11*...*prime(n+1) = A002110(n+1)/3 is divisible by 3.) Thus, for every P that is divisible by 3 and < A185292(n), we can rule out every ordered pair (j,k) except (2,3) and (3,4) (which could be tested together by computing t = P - r^3 for each prime r < P^(1/3) and, if t is square, checking whether sqrt(t) is a prime or the square of a prime) and those with k >= 5 (which could be tested by checking whether t = P - q^k is a prime power for each prime power q^k that is less than P and has k >= 5). (End)
a(17) <= 63985284333636413237490 = 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 29 * 37 * 41 * 43 * 59 * 61 * 103 * 409 = 10461281^3 + 250679912393^2. - Jon E. Schoenfield, Mar 31 2017

			a(1) = 13 = 2^2 + 3^2.
a(2) = 33 = 5^2 + 2^3 = 3 * 11.
a(3) = 130 = 3^2 + 11^2 = 2 * 5 * 13.
a(4) = 966 = 5^3 + 29^2 = 2 * 3 * 7 * 23.
a(5) = 14322 = 17^3 + 97^2 = 2 * 3 * 7 * 11 * 31.
a(6) = 81510 = 29^3 + 239^2 = 2 * 3 * 5 * 11 * 13 * 19.
a(7) = 3530730 = 41^4 + 89^3 = 2 * 3 * 5 * 7 * 17 * 23 * 43.
a(8) = 117535110 = 461^3 + 4423^2 = 2 * 3 * 5 * 7 * 11 * 17 * 41 * 73.
From _Jon E. Schoenfield_, Mar 14 2017: (Start)
a(9) = 2211297270 = 1301^3 + 3037^2 = 2 * 3 * 5 * 7 * 13 * 17 * 29 * 31 * 53.
a(10) = 131031070170 = 1361^3 + 358483^2 = 2 * 3 * 5 * 7 * 11 * 13 * 17 * 43 * 47 * 127. (End)
From _Giovanni Resta_, Mar 14 2017: (Start)
a(11) = 810571^2 + 8609^3,
a(12) = 12694849^2 + 13109^3. (End)
From _Jon E. Schoenfield_, Mar 18 2017: (Start)
a(13) = 24537703^2 + 140741^3.
a(14) = 639414679^2 + 178349^3.
a(15) = 1632727069^2 + 658649^3. (End)
a(16) = 1472015189^2 + 9094049^3. - _Jon E. Schoenfield_, Mar 19 2017

Cf. A225102, A246547.

N:= 1.2*10^8: # to get all terms <= N
PP:= {seq(seq(p^k,k=2..floor(log[p](N))), p = select(isprime, [2,seq(i,i=3..floor(sqrt(N)),2)]))}:
PP:= sort(convert(PP,list)):
A:= 'A':
for i from 1 to nops(PP) do
  for j from 1 to i do
     Q:= PP[i]+PP[j];
     if Q > N then break fi;
     F:= ifactors(Q)[2];
     if max(seq(f[2],f=F))>1 then next fi;
     m:= nops(F);
     if not assigned(A[m]) or A[m] > Q then A[m]:= Q fi
od od:
seq(A[i],i=1..max(map(op,[indices(A)]))); # Robert Israel, Mar 01 2017

(* first 8 terms *) mx = 1.2*^8; a = 0 Range[8] + mx; p = Sort@ Flatten@ Table[ p^Range[2, Log[p, mx]], {p, Prime@ Range@ PrimePi@ Sqrt@ mx}]; Do[ j=1; While[j <= i && (v = p[[i]] + p[[j]]) < mx, f = FactorInteger@v; If[Max[Last /@ f] == 1, c = Length@f; If[c < 9 && v < a[[c]], a[[c]] = v]]; j++], {i, Length@p}]; a (* Giovanni Resta, Mar 19 2017 *)

do(lim)=my(v=List(),u=v,t,f); t=1; for(i=1,lim, t*=prime(i); if(t>lim,break); listput(v, oo)); v=Vec(v); for(e=2,logint(lim\=1,2), forprime(p=2,sqrtnint(lim-4,e), listput(u,p^e))); u=Set(u); for(i=1,#u, for(j=1,i, t=u[i]+u[j]; if(t>lim, break); f=factor(t)[,2]; if(vecmax(f)==1 && tif(k==oo,"?",k), v) \\ Charles R Greathouse IV, Mar 19 2017

do(lim)=my(v=List(),u=v,t,f,p2); t=1; for(i=1,lim, t*=prime(i); if(t>lim,break); listput(v, oo)); v=Vec(v); for(e=3,logint(lim\=1,2), forprime(p=2,sqrtnint(lim-4,e), listput(u,p^e))); u=Set(u); for(i=1,#u, for(j=1,i, t=u[i]+u[j]; if(t>lim, break); f=factor(t)[,2]; if(vecmax(f)==1 && tlim, break); f=factor(t)[,2]; if(vecmax(f)==1 && tlim, break); f=factor(t)[,2]; if(vecmax(f)==1 && tif(k==oo,"?",k), v) \\ Charles R Greathouse IV, Mar 19 2017

a(7)-a(8) from Giovanni Resta, Feb 21 2017
a(9)-a(10) from Jon E. Schoenfield, Mar 14 2017
a(11)-a(12) from Giovanni Resta, Mar 14 2017
a(13)-a(15) from Jon E. Schoenfield, Mar 18 2017
a(16) from Jon E. Schoenfield, Mar 19 2017

A301296 Smallest distance from n to a prime power (as defined in A246547).

Original entry on oeis.org

3, 2, 1, 0, 1, 2, 1, 0, 0, 1, 2, 3, 3, 2, 1, 0, 1, 2, 3, 4, 4, 3, 2, 1, 0, 1, 0, 1, 2, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4

Offset: 1

N. J. A. Sloane, Mar 24 2018

nonn

There are four different sequences which may legitimately be called "prime powers": A000961 (p^k, k >= 0), A246655 (p^k, k >= 1), A246547 (p^k, k >= 2), A025475 (p^k, k=0 and k >= 2).

Cf. A061670, A080732, A301295.

cp's OEIS Frontend

A320700 Odd numbers whose product of prime indices is a nonprime prime power (A246547).

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A060847 Difference between a nontrivial prime power (A246547) and the previous prime.

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A282633 Numbers n such that n^2 + 1 is the sum of two proper prime powers (A246547) in more than one way.

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A287299 Number of ways of writing n as a sum of a proper prime power (A246547) and a nonprime squarefree number (A000469).

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A290135 Numbers that are the sum of two proper prime powers (A246547).

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A343123 Primes p such that the sum of A001414(k) for k strictly between p and the following prime is a proper prime power (a term of A246547).

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A282533 Primes that are the sum of two proper prime powers (A246547) in more than one way.

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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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