2017-02-22, 07:49 | #1 |
"Sam"
Nov 2016
146_{16} Posts |
Polynomial Coefficients Integer Sets
Let S[p] denote a set of integers [n, n_2,..., n_k] such that the corresponding minimal polynomial P:
P(x) = Mod [x-1, x^p-(x-1)^p] where p is a prime number denoted Q(p) (the pth prime number) and the corresponding coefficients n_k of P(x) of terms with degree > 0, written as nx^(p-1)+(n_2)x^(p-2)+....+(n_k)x+1 There is a special relation between each integer in S[p]. The first instance (in set S[n]) is that 1 = n = n_k, and also (p-1)/2 = n_2 = n_(k-1). A second observation is that n_y = n_k if and only if y+k = p+1. A third, and most interesting observation is that the two most inner terms are divisible by Q(p-1), the largest prime number q < p. All other integers in S[p] are divisible by Q(p-1) or a product of primes < p. For the primes < 29, the polynomials and coefficients of terms with degree > 0 listed as integer sets (with the corresponding minimal polynomials P(x)) are: x + 1/2 x^2 + x + 1/3 x^4 + 2*x^3 + 2*x^2 + x + 1/5 x^6 + 3*x^5 + 5*x^4 + 5*x^3 + 3*x^2 + x + 1/7 x^10 + 5*x^9 + 15*x^8 + 30*x^7 + 42*x^6 + 42*x^5 + 30*x^4 + 15*x^3 + 5*x^2 + x + 1/11 x^12 + 6*x^11 + 22*x^10 + 55*x^9 + 99*x^8 + 132*x^7 + 132*x^6 + 99*x^5 + 55*x^4 + 22*x^3 + 6*x^2 + x + 1/13 x^16 + 8*x^15 + 40*x^14 + 140*x^13 + 364*x^12 + 728*x^11 + 1144*x^10 + 1430*x^9 + 1430*x^8 + 1144*x^7 + 728*x^6 + 364*x^5 + 140*x^4 + 40*x^3 + 8*x^2 + x + 1/17 x^18 + 9*x^17 + 51*x^16 + 204*x^15 + 612*x^14 + 1428*x^13 + 2652*x^12 + 3978*x^11 + 4862*x^10 + 4862*x^9 + 3978*x^8 + 2652*x^7 + 1428*x^6 + 612*x^5 + 204*x^4 + 51*x^3 + 9*x^2 + x + 1/19 x^22 + 11*x^21 + 77*x^20 + 385*x^19 + 1463*x^18 + 4389*x^17 + 10659*x^16 + 21318*x^15 + 35530*x^14 + 49742*x^13 + 58786*x^12 + 58786*x^11 + 49742*x^10 + 35530*x^9 + 21318*x^8 + 10659*x^7 + 4389*x^6 + 1463*x^5 + 385*x^4 + 77*x^3 + 11*x^2 + x + 1/23 x^28 + 14*x^27 + 126*x^26 + 819*x^25 + 4095*x^24 + 16380*x^23 + 53820*x^22 + 148005*x^21 + 345345*x^20 + 690690*x^19 + 1193010*x^18 + 1789515*x^17 + 2340135*x^16 + 2674440*x^15 + 2674440*x^14 + 2340135*x^13 + 1789515*x^12 + 1193010*x^11 + 690690*x^10 + 345345*x^9 + 148005*x^8 + 53820*x^7 + 16380*x^6 + 4095*x^5 + 819*x^4 + 126*x^3 + 14*x^2 + x + 1/29 The corresponding integer Sets are: S[2] = {1} S[3] = {1, 1} S[5] = {1, 2, 2, 1} S[7] = {1, 3, 5, 5, 3, 1} S[11] = {1, 5, 15, 30, 42, 42, 30, 15, 5, 1} S[13] = {1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1} S[17] = {1, 8, 40, 140, 364, 728, 1144, 1430, 1430, 1144, 728, 364, 140, 40, 8, 1} S[19] = {1, 9, 51, 204, 612, 1428, 2652, 3978, 4862, 4862, 3978, 2652, 1428, 612, 204, 51, 9, 1} S[23] = {1, 11, 77, 385, 1463, 4389, 10659, 21318, 35530, 49742, 58786, 58786, 49742, 35530, 21318, 10659, 4389, 1463, 385, 77, 11, 1} S[29] = {1, 14, 126, 819, 4095, 16380, 53820, 148005, 345345, 690690, 1193010, 1789515, 2340135, 2674440, 2674440, 2340135, 1789515, 1193010, 690690, 345345, 148005, 53820, 16380, 4095, 819, 126, 14, 1} Can any of these observations be proven? Also, why is this the case? |
2017-02-22, 08:37 | #2 |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{3}×7^{3} Posts |
You seem to be describing P(x) = ((x+1)^{p}-x^{p}-1)/p
(only you have "+1/p" still dangling at the end. I will subtract it for symmetry). After binomial expansion: It is obvious that for p>=7, the middle terms will have a factor precprime(p); it follows from , where k = (p-1)/2. This is because precprime(p) will be present in the denominator, but not in the numerator. Note that your own observation that "the two most inner terms are divisible by Q(p-1)" is not holding for p<7 (e.g. for p=5, "the two most inner terms" are not divisible by 3), and the reason for this is obvious. For p>=7, it will hold, - as indeed shown above. |
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