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%I #40 Oct 15 2020 16:35:11
%S 0,1,2,3,5,6,4,9,11,14,5,18,10,7,23,26,29,30,33,35,9,39,41,11,24,50,
%T 51,53,18,14,7,65,34,69,74,15,26,81,83,86,89,90,95,48,98,99,105,37,
%U 113,38,29,119,12,25,8,131,134,135,46,35,47,146,51,155,78,158
%N Order of 4 mod n-th prime: least k such that prime(n) divides 4^k-1, n >= 2.
%C The period of the expansion of 1/p, base N (where N=4), is equivalent to determining for base integer 4, the period of the sequence 1, 4, 4^2, 4^3, ... mod p. Thus the cycle length for base 4, 1/7 = 0.021021021... (cycle length 3).
%C The cycle length, base 4, mod p, is equivalent to "clock cycles", given angle A, then the algebraic identity for the doubling angle, 2A.
%C Examples: Given cos A, f(x) for 2A = 2x^2 - 1, seed 2 Pi/7, i.e., (.623489801 == (arrow), -.222520934... == -.900968867...== .623489801...(cycle length 3). Given 2 cos A, the algebraic identity for 2 cos 2A, f(x) = x^2 - 2; e.g., given seed 2 cos A = 2 Pi/7, the 3 cycle is 1.246979604...== .445041867...== -1.801937736...== back to 1.24697... Likewise, the doubling function given sin^2 A, f(x) for sin^2 2A = 4x(1 - x), the logistic equation; getting cycle length of 3 using the seed sin^2 2 Pi/7. Similarly, the doubling function for tan 2A given tan A, where A = 2 Pi/7 gives 2x/(1 - x^2), cycle length of 3. The doubling function for cot 2A given cot A, with A = 2 Pi/7 gives (x^2 - 1)/2x, cycle length of 3. Note that (x^2 - 1)/2x = sinh(log(x)), and is also generated from using Newton's method on x^2 + 1 = 0.
%C Consider the odd pseudoprimes, composite numbers x such that 2^(x-1) = 1 mod x, that have prime(n) as a factor. It appears that all such x can be factored as prime(n) * (2 a(n) k + 1) for some integer k. For example, the first few pseudoprimes having the factor 31 are 31*11, 31*91, 31*141 and 3*151. The 11th prime is 31 and a(11) = 5. Therefore all the cofactors of 31 should have the form 10k+1, which is clearly true. - _T. D. Noe_, Jun 10 2003
%D Albert H. Beiler, Recreations in the Theory of Numbers, Dover, 1964; Table 48, pages 98-99.
%D John H. Conway & R. K. Guy, The Book of Numbers, Springer-Verlag, 1996, pages 207-208, Periodic Points.
%H T. D. Noe, <a href="/A082654/b082654.txt">Table of n, a(n) for n = 1..1000</a>
%H Wolfdieter Lang, <a href="https://arxiv.org/abs/2008.04300">On the Equivalence of Three Complete Cyclic Systems of Integers</a>, arXiv:2008.04300 [math.NT], 2020.
%H G. Vostrov, R. Opiata, <a href="http://dspace.opu.ua/jspui/bitstream/123456789/7943/1/240-247.pdf">Computer modeling of dynamic processes in analytic number theory</a>, Electrical and Computer Systems (Електротехнічні та комп'ютерні системи) 2018, No. 28, Issue 104, 240-247.
%F a(1) = 0, and a(n) = order(4, prime(n)), also used exp_{prime(n)}(4), that is least exponent k >= 1 for which 4^k is congruent to 1 mod prime(n), for n >= 2. prime(n) = A000040(n). [rewritten by _Wolfdieter Lang_, Apr 10 2020]
%F From _Wolfdieter Lang_, Apr 10 2020: (Start)
%F a(n) = A003558(prime(n)), for n >= 2.
%F a(n) = (1/2)*order(2, 3*prime(n)), for n >= 3. [Proof uses 4^k - 1 = (1+3)^k - 1 == 0 (mod 3), for k >= 0.] (End)
%e 4th prime is 7 and mod 7, 4^3 = 1, but not 4^1 or 4^2, so a(4) = 3.
%e n = 4: prime(4) = 7, 2^6 - 1 = 63 = 3*21 == 0 (mod 21), but not 2^k - 1 for lower exponents k >= 1, therefore ord(2, 3*7) = 6 and a(4) = 3. - _Wolfdieter Lang_, Apr 10 2020
%t Join[{0}, Table[MultiplicativeOrder[4, Prime[n]], {n, 2, 100}]]
%o (PARI) a(n)=if(n>1, znorder(Mod(4,prime(n))), 0) \\ _Charles R Greathouse IV_, Sep 07 2016
%o (GAP) A000040:=Filtered([1..350],IsPrime);;
%o List([1..Length(A000040)],n->OrderMod(4,A000040[n])); # _Muniru A Asiru_, Feb 07 2019
%Y Cf. A000040, A014664, A002326, A036116, A036117.
%K nonn
%O 1,3
%A _Gary W. Adamson_, May 17 2003
%E More terms from _Reinhard Zumkeller_, May 17 2003
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