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Numbers n for which A276711(n) = 1
0

%I #11 Sep 28 2016 05:23:30

%S 2,4,5,8,9,10,16,22,24,25,26,36,58,64,76,82,85,120,170,196,202,214,

%T 324,328,370,412,505,562,676,706,730,799,841,1024,1204,1243,1549,1681,

%U 1849,2146,2986,10404,46656,52900,112896,122500

%N Numbers n for which A276711(n) = 1

%C Is the sequence finite?

%C a(47) > 10^7 if it exists.

%e 122500 is in the sequence because 122500 = 121169 + 11^3 is the only way to write 122500 as the sum of a prime and a nonnegative perfect power.

%p N:= 10^6: # to get all terms <= N

%p Primes:= select(isprime, [2,seq(i,i=3..N,2)]):

%p Pows:= {0,1,seq(seq(b^k,k=2..floor(log[b](N))),b=2..floor(sqrt(N)))}:

%p G:= expand(add(x^p,p=Primes)*add(x^r,r=Pows)):

%p B:= [seq(coeff(G,x,i),i=1..N)]:

%p select(t -> B[t]=1, [$1..N]);

%Y Cf. A276711.

%K nonn,more

%O 1,1

%A _Robert Israel_, Sep 27 2016