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A215197
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Numbers k such that k and k + 1 are both of the form p*q^4 where p and q are distinct primes.
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5
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2511, 7856, 10287, 15471, 15632, 18063, 20816, 28592, 36368, 40816, 54512, 75248, 88047, 93231, 101168, 126927, 134703, 160624, 163376, 170991, 178767, 210032, 215216, 217808, 220624, 254096, 256527, 274671, 280624, 292976, 334448, 347408, 443151, 482192
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OFFSET
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1,1
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COMMENTS
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These are numbers n such that n and n+1 both have 10 divisors. Proof: clearly n and n+1 cannot both be of the form p^9, so for contradiction assume either n and n+1 is of the form p*q^4 and the other is of the form r^9 where p, q, and r are prime. So p*q^4 is either r^9 - 1 = (r-1)(r^2+r+1)(r^6+r^3+1) or r^9 + 1 = (r+1)(r^2-r+1)(r^6-r^3+1). But these factors are relatively prime and so cannot represent p*q^4 unless one or more factors are units. But this does not happen for r > 2, and the case r = 2 does not work since neither 511 not 513 is of the form p*q^4. - Charles R Greathouse IV, Jun 19 2016
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LINKS
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EXAMPLE
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2511 is a member as 2511 = 31*3^4 and 2512 = 157*2^4.
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MAPLE
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with(numtheory):for n from 3 to 500000 do:x:=factorset(n):y:=factorset(n+1):n1:=nops(x):n2:=nops(y):if n1=2 and n2=2 then xx1:=x[1]*x[2]^4 : xx2:=x[2]*x[1]^4:yy1:=y[1]*y[2]^4: yy2:=y[2]*y[1]^4:if (xx1=n or xx2=n) and (yy1=n+1 or yy2=n+1) then printf("%a, ", n):else fi:fi:od:
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MATHEMATICA
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lst={}; Do[f1=FactorInteger[n]; If[Sort[Transpose[f1][[2]]]=={1, 4}, f2=FactorInteger[n+1]; If[Sort[Transpose[f2][[2]]]=={1, 4}, AppendTo[lst, n]]], {n, 3, 55000}]; lst
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PROG
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(PARI) is(n)=vecsort(factor(n)[, 2])==[1, 4]~ && vecsort(factor(n+1)[, 2])==[1, 4]~ \\ Charles R Greathouse IV, Jun 19 2016
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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