A095841 Prime powers having exactly one partition into two prime powers.
2, 3, 127, 163, 179, 191, 193, 223, 239, 251, 269, 311, 337, 343, 389, 419, 431, 457, 491, 547, 557, 569, 599, 613, 653, 659, 673, 683, 719, 739, 787, 821, 839, 853, 883, 911, 929, 953, 967, 977, 1117, 1123, 1201, 1229, 1249, 1283, 1289, 1297, 1303, 1327, 1381, 1409, 1423, 1439, 1451, 1471, 1481, 1499, 1607, 1663, 1681
Offset: 1
Keywords
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Programs
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Haskell
a095841 n = a095841_list !! (n-1) a095841_list = filter ((== 1) . a071330) a000961_list -- Reinhard Zumkeller, Jan 11 2013
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Maple
N:= 10^4: # to get all terms <= N primepows:= {1,seq(seq(p^n, n=1..floor(log[p](N))), p=select(isprime,[2,seq(2*k+1,k=1..(N-1)/2)]))}: npp:= nops(primepows): B:= Vector(N,datatype=integer[4]): for n from 1 to npp do for m from n to npp do j:= primepows[n]+primepows[m]; if j <= N then B[j]:= B[j]+1 fi; od od: select(t -> B[t] = 1, primepows); # Robert Israel, Nov 21 2014 -
Mathematica
max = 2000; ppQ[n_] := n == 1 || PrimePowerQ[n]; pp = Select[Range[max], ppQ]; lp = Length[pp]; Table[pp[[i]] + pp[[j]], {i, 1, lp}, {j, i, lp}] // Flatten // Select[#, ppQ[#] && # <= max&]& // Sort // Split // Select[#, Length[#] == 1&]& // Flatten (* Jean-François Alcover, Mar 04 2019 *) -
PARI
is(n)=if(n<127,return(n==2||n==3)); isprimepower(n) && sum(i=2,n\2,isprimepower(i)&&isprimepower(n-i))==1 \\ naive; Charles R Greathouse IV, Nov 21 2014
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PARI
is(n)=if(!isprimepower(n), return(0)); my(s); forprime(p=2, n\2, if(isprimepower(n-p) && s++>1, return(0))); for(e=2, log(n)\log(2), forprime(p=2, sqrtnint(n\2, e), if(isprimepower(n-p^e) && s++>1, return(0)))); s+(!!isprimepower(n-1))==1 || n==2 \\ faster; Charles R Greathouse IV, Nov 21 2014
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PARI
has(n)=my(s); forprime(p=2, n\2, if(isprimepower(n-p) && s++>1, return(0))); for(e=2, log(n)\log(2), forprime(p=2, sqrtnint(n\2, e), if(isprimepower(n-p^e) && s++>1, return(0)))); s+(!!isprimepower(n-1))==1 list(lim)=my(v=List([2])); forprime(p=2,lim,if(has(p), listput(v,p))); for(e=2,log(lim)\log(2), forprime(p=2,lim^(1/e), if(has(p^e), listput(v,p^e)))); Set(v) \\ Charles R Greathouse IV, Nov 21 2014
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