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Search: id:A090849
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| A090849 |
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Smallest positive k such that phi(1+k*2^n) <= phi(k*2^n), where phi is Euler's totient function. |
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3
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| 104, 52, 26, 13, 59, 67, 41, 73, 89, 97, 101, 103, 74, 37, 26, 13, 17, 67, 41, 73, 89, 82, 41, 103, 104, 52, 26, 13, 29, 67, 41, 73, 74, 37, 101, 103, 104, 52, 26, 13, 59, 67, 41, 73, 89, 67, 86, 43, 104, 52, 26, 13, 59, 37, 41, 73, 89, 97, 101, 103, 104, 52, 26, 13, 59, 67
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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Newman proves that k always exists for all n. Surprisingly, it appears that only 19 values of k suffice for all n. Note that a(n) = 26 when n = 2 (mod 12), a(n) = 13 when n = 3 (mod 12), a(n) = 41 when n = 6 (mod 12) and a(n) = 73 when n = 7 (mod 12). Is this sequence periodic?
A091025 shows why this sequence has only a finite number of distinct terms.
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REFERENCES
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D. J. Newman, Euler's phi function on arithmetic progressions, Amer. Math. Monthly, Vol. 104, No. 3 (Mar. 1997), pp. 256-257.
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MATHEMATICA
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Table[k=1; While[EulerPhi[1+k*2^n] > EulerPhi[k*2^n], k++ ]; k, {n, 100}]
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CROSSREFS
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Cf. A090851 (least k such that phi(2n*k+1) < phi(2n*k)).
Sequence in context: A088584 A097014 A106297 this_sequence A091025 A054904 A117845
Adjacent sequences: A090846 A090847 A090848 this_sequence A090850 A090851 A090852
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KEYWORD
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nonn
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AUTHOR
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T. D. Noe (noe(AT)sspectra.com), Dec 09 2003
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