A082654 Order of 4 mod n-th prime: least k such that prime(n) divides 4^k-1, n >= 2.

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, 51, 53, 18, 14, 7, 65, 34, 69, 74, 15, 26, 81, 83, 86, 89, 90, 95, 48, 98, 99, 105, 37, 113, 38, 29, 119, 12, 25, 8, 131, 134, 135, 46, 35, 47, 146, 51, 155, 78, 158

Offset: 1

Author

Gary W. Adamson, May 17 2003

Keywords

• nonn
• easy

Comments

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).

The cycle length, base 4, mod p, is equivalent to "clock cycles", given angle A, then the algebraic identity for the doubling angle, 2A.

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.

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

Examples

			4th prime is 7 and mod 7, 4^3 = 1, but not 4^1 or 4^2, so a(4) = 3.
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
		

References

• Albert H. Beiler, Recreations in the Theory of Numbers, Dover, 1964; Table 48, pages 98-99.
• John H. Conway & R. K. Guy, The Book of Numbers, Springer-Verlag, 1996, pages 207-208, Periodic Points.

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• Wolfdieter Lang, On the Equivalence of Three Complete Cyclic Systems of Integers, arXiv:2008.04300 [math.NT], 2020.
• G. Vostrov, R. Opiata, Computer modeling of dynamic processes in analytic number theory, Electrical and Computer Systems (Електротехнічні та комп'ютерні системи) 2018, No. 28, Issue 104, 240-247.

Crossrefs

Cf. A053447 (order of 4 mod 2n+1), A216371.

In other bases: A014664, A062117, A211241, A211242, A211243, A211244, A211245, A002371, A372801.

Programs

GAP

A000040:=Filtered([1..350],IsPrime);;
List([1..Length(A000040)],n->OrderMod(4,A000040[n])); # Muniru A Asiru, Feb 07 2019

Mathematica

Join[{0}, Table[MultiplicativeOrder[4, Prime[n]], {n, 2, 100}]]

PARI

a(n)=if(n>1, znorder(Mod(4,prime(n))), 0) \\ Charles R Greathouse IV, Sep 07 2016

Formula

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]

From Wolfdieter Lang, Apr 10 2020: (Start)

a(n) = A003558(prime(n)), for n >= 2.

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)

From Jianing Song, May 13 2024: (Start)

a(n) = A014664(n)/gcd(2, A014664(n)).

a(n) <= (prime(n) - 1)/2. Those prime(n) for which a(n) = (prime(n) - 1)/2 are listed in A216371. (End)

Extensions

More terms from Reinhard Zumkeller, May 17 2003