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A085866
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a(1) = 3, a(n+1) = a(n)*phi(a(n)), where phi(n) is Euler's totient function.
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2
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3, 6, 12, 48, 768, 196608, 12884901888, 55340232221128654848, 1020847100762815390390123822295304634368, 347376267711948586270712955026063723559809953996921692118372752023739388919808
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OFFSET
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1,1
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COMMENTS
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a(1) = 1, a(n+1) = a(n) + phi(a(n)) gives A074693.
For n > 1, a(n)/3 is 2^(2^(n-2)). This sequence is 2, 4, 16, 256, ..., which is phi(a(n-1)).
The Harris 1935 problem is to show 1 + sqrt(13) = sqrt(12 + sqrt(48 + sqrt( 768 + ...))). - Michael Somos, Jun 18 2018
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LINKS
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V. C. Harris, Problem 78, National Mathematics Magazine 9, no.6 (1935), p. 180.
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FORMULA
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a(n) = 3*2^(2^(n-2)).
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EXAMPLE
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a(3) = 12 and phi(12)= 4, hence a(4) = 12*4 = 48.
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MATHEMATICA
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RecurrenceTable[{a[1]==3, a[n+1]==a[n] EulerPhi [a[n]]}, a, {n, 20}] (* Vincenzo Librandi, Jun 19 2018 *
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PROG
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(PARI) for(n=1, 11, if(n==1, a=3, a*=eulerphi(a)); print1(a, ", "); )
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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