|
| |
|
|
A282686
|
|
Least sum of two proper prime powers (A246547) that is the product of n distinct primes.
|
|
0
|
|
|
|
13, 33, 130, 966, 14322, 81510, 3530730, 117535110, 2211297270, 131031070170, 1295080356570, 163411918786830, 3389900689405230, 414524121952915590, 2951531806477464210, 754260388389042905370
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Least value of A225102 that is the product of n distinct primes.
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
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
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.
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)
a(11) = 810571^2 + 8609^3,
a(12) = 12694849^2 + 13109^3. (End)
a(13) = 24537703^2 + 140741^3.
a(14) = 639414679^2 + 178349^3.
a(15) = 1632727069^2 + 658649^3. (End)
|
|
|
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 *)
|
|
|
PROG
|
(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 && t<v[#f], v[#f]=t))); apply(k->if(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 && t<v[#f], v[#f]=t))); forprime(p=2, sqrtint(lim), p2=p^2; for(i=1, #u, t=u[i]+p2; if(t>lim, break); f=factor(t)[, 2]; if(vecmax(f)==1 && t<v[#f], v[#f]=t))); forprime(p=2, sqrtint(lim), p2=p^2; forprime(q=2, min(sqrtint(lim-p2), p), t=p2+q^2; if(t>lim, break); f=factor(t)[, 2]; if(vecmax(f)==1 && t<v[#f], v[#f]=t))); apply(k->if(k==oo, "?", k), v) \\ Charles R Greathouse IV, Mar 19 2017
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,more
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
