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A136475
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Irregular triangle read by rows: row n gives prime factors of (2^(3^(n+1))+1)/(2^(3^n)+1).
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3
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3, 3, 19, 3, 87211, 3, 163, 135433, 272010961, 3, 1459, 139483, 10429407431911334611, 918125051602568899753, 3, 227862073, 3110690934667, 216892513252489863991753, 1102099161075964924744009, 393063301203384521164229656203691748263012766081190297429488962985651210769817
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OFFSET
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0,1
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COMMENTS
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1. The motivation for this sequence is to quickly generate integers n such that n divides 2^n+1 (sequences A006521, A136473). From the link, it is known that if 3^k||n with n|2^n+1 and n not a power of 3, then n is divisible by a prime p dividing 2^(3^k)+1. Thus for any fixed k every n with n|2^n+1 not a power of 3 is divisible by one of the following numbers: 3^k or some 3^j.p, where p>3 is a prime in A136475 before the k-th '3' and j is the number of '3's before p in the sequence.
2. Note: (2^(3^(k+1)+1)/(2^(3^k)+1) = 2^(2.3^k) - 2^(3^k) + 1.
3. For the primes dividing 2^(3^k)+1 for some k see A136474.
4. Are these numbers always squarefree?
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LINKS
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FORMULA
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The prime factors of (2^(3^(k+1)+1)/(2^(3^k)+1) are given in ascending order *for each k*. For each new value of k the factorization starts with a '3', thus delimiting the different factorizations.
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EXAMPLE
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1. 2^(3^4)+1)/(2^(3^3)+1)=3.163.135433.272010961, the factorization starting at the 4th '3' and ending just before the 5th '3'.
2. From Comment 1 below and k=5, we see that every n not a power of 3 satisfying n|2^n+1 (sequences A006521, A136473) is divisible by 3^5 or 3^2.19 or 3^3.87211 or 3^4.163 or 3^4.135433 or 3^4.272010961.
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MAPLE
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S:=[]; for k from 0 to 4 do f:=op(2, ifactors((2^(3^(k+1))+1)/(2^(3^k)+1))); T:=[]; for j to nops(f) do T:=[op(T), op(1, op(j, f))]; od; S:=[op(S), op(sort(T))]; od; op(S);
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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