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A085331
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Numbers n such that phi(rev(n))=n.
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6
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1, 12, 36, 192, 1992, 2016, 31067664, 39206496, 1564356432, 3937403136, 15600000432, 22871605008, 156043560432, 156439956432, 1560000000432, 1956000004392
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OFFSET
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1,2
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COMMENTS
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rev(2*(10^k-4)) = 3*(10^k-3). If 10^k-3 is prime, then phi(3*(10^k-3)) = 2*(10^k-4), so 2*(10^k-4) is a term. 10^1-3=7 is prime, so 2*(10^1-4)=12 is a term, a(2). 10^2-3=97 is prime, so 2*(10^2-4)=192 is a term, a(4). 10^3-3=997 is prime, so 2*(10^3-4)=1992 is a term, a(5). 10^17-3 is prime, so 2*(10^17-4)=199999999999999992 is a term. 10^140-3 is prime, so 2*(10^140-4) is a term. 10^990-3 is prime, so 2*(10^990-4) is a term. Conjecture: sequence is infinite. - Ray Chandler, Jul 20 2003
Let f(m,n,r,t)=((9).(m).78.(0)(n).21.(9)(m))(r).(9)(t).7 where m, n, r & t are nonnegative integers; dot between numbers means concatenation and "(m)(n)" means number of m's is n. If r*t=0 & p=f(m,n,r,t) is prime then reversal(3*p) = 1.((9)(m).56.(0)(n).43.(9)(m))(r).(9)(t).2 is in the sequence. For example p1=f(0,0,0,0)=7 so reversal(3*p1) = 12 is in the sequence, p2=f(0,0,2,0)=(7821)(2).7=782178217 so reversal(3*p2) = 1.(5643)(2).2 = 1564356432 is in the sequence & p3=f(0,0,674,0) so reversal(3*p3) = 1.(5643)(674).2 is in the sequence. Primes of the form f(m,n,r,t) are a generalized form of primes of the form 10^j-3 that were already related to this sequence by Ray Chandler. For all n, A085331(n) = reversal(A072395(n)). - Farideh Firoozbakht, Jan 08 2005
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LINKS
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EXAMPLE
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phi[{1,21,63,291,2991,6102}] = {1,12,36,192,1992,2016}
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MATHEMATICA
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v = {1}; Do[ If[ n == EulerPhi[ FromDigits[ Reverse[ IntegerDigits [ n ] ] ] ], v = Append[ v, n ]; Print[ v ], If[ Mod[ n, 1000000 ] == 0, Print[ -n ] ] ], {n, 2, 2050000000, 2} ] (Firoozbakht)
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CROSSREFS
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KEYWORD
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nonn,base,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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