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A112374
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Let T(n) = A000078(n+2), n >= 1; a(n) = smallest k such that n divides T(k).
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0
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1, 3, 6, 4, 6, 9, 8, 5, 9, 13, 20, 9, 10, 8, 6, 10, 53, 9, 48, 28, 18, 20, 35, 18, 76, 10, 9, 8, 7, 68, 20, 15, 20, 53, 30, 9, 58, 48, 78, 28, 19, 18, 63, 20, 68, 35, 28, 18, 46, 108, 76, 10, 158, 9, 52, 8, 87, 133, 18, 68, 51, 20, 46, 35, 78, 20, 17, 138, 35, 30, 230, 20, 72, 58, 76
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OFFSET
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1,2
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COMMENTS
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Rank of apparition of n in the tetranacci numbers. - T. D. Noe, Dec 05 2005
This sequence is well-defined. Proof by T. D. Noe: for every prime p, Brenner proves we can find k(p) such that p divides the k(p)-th term of n-step Fibonacci. Using Brenner's methods, we know that p will also divide every j*k(p)-th term of the sequence for any j>0. We use this last fact to go to the general case: For integer m, we can find a term that m divides as follows: (1) factor m into primes: m = p1^e1 p2^e2...pr^er, (2) let K = m k(p1) k(p2)...k(pr) / (p1 p2 ... pr) (3) then m will divide the K-th term of the sequence. In general, K is much too large. However, it does show that every prime divides a term of every n-step Fibonacci sequence for n>1. - T. D. Noe, Dec 05 2005
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LINKS
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FORMULA
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MATHEMATICA
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n=4; Table[a=Join[{1}, Table[0, {n-1}]]; k=0; While[k++; s=Mod[Plus@@a, i]; a=RotateLeft[a]; a[[n]]=s; s!=0]; k, {i, 100}] (* T. D. Noe, Dec 05 2005 *)
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CROSSREFS
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KEYWORD
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easy,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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