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

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

Offset: 1

Author

Altug Alkan, Feb 20 2017

Keywords

• nonn
• more

Comments

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

Examples

			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
		

Crossrefs

Cf. A225102, A246547.

Programs

Maple

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

Mathematica

(* 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 *)

PARI

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

PARI

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

Extensions

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