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A097463
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Let P(i) = i-th prime. To get a(n), factor P(n)-1 as a product of primes, then concatenate the exponents.
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0
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0, 1, 2, 11, 101, 21, 4, 12, 10001, 2001, 111, 22, 301, 1101, 100000001, 200001, 1000000001, 211, 11001, 1011, 32, 110001, 1000000000001, 30001, 51, 202, 1100001, 1000000000000001, 23, 4001, 1201, 101001, 3000001, 110000001, 200000000001
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OFFSET
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1,3
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COMMENTS
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If P(n)-1 = P(1)^a * P(2)^b *....* P(j)^k then a(n) = ab...k.
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LINKS
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EXAMPLE
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3-1=2^1, so a(2)=1.
5-1=2^2, so a(3)=2.
7-1=2^1*3^1, so a(4)=11.
23=(2^1)*(11^1)+1. So a(9) = 10001.
37 = 36 + 1 = 2^2*3^2 + 1, so 37 becomes 22 (a=2,b=2).
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PROG
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(PARI) {forprime(p=2, 150, f=factor(p-1); j=1; q=2; s="0"; while(j<=matsize(f)[1], if(q==f[j, 1], s=concat(s, f[j, 2]); j++, s=concat(s, 0)); q=nextprime(q+1)); print1(eval(s), ", "))} \\ Klaus Brockhaus, Apr 25 2005
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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a(9) corrected by Dennis (tuesdayist(AT)juno.com), Mar 30 2006
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STATUS
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approved
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