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 A141529 A cyclotomic Binet solution for the first 11 primes: solution of a set of linear equations with cyclotomic polynomial roots; roots: r[i]->(x^11-1): a(n)=Sum[r(i)^n*c(i),{i,1,11}]. 3
 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 31, 29, 23 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS http://en.wikipedia.org/wiki/Terence_Tao http://www.math.ucla.edu/~tao/ Terence Tao and Ben Green proved there are polynomials of arbitrary length that generate Primes. Let us assume that they are right, that the primes are represented in an algebraic polynomial of a recursive sequence: a[n]=Prime[j]; a[n]=Sum[a[n-k]*b[k],{k=1,m}] which has a Binet function solution of: f[n]=Sum[r[i]^n*c[i],{i,1.m}] The classical number theory gives sum of the coefficients at n=0: f=Sum[c[i],{i,1,m}]=2 which is in Fibonacci terms a "Lucas" type. But if the primes are a graph substitution generated sequence with this same generating polynomial then: f=Sum[c[i],{i,1,m}]=0 and the primes are Fibonacci like. The real matter here is the algebraicness and the limit: from Markov type theory of algebraic sequences Limit[a[n+1]/a[n],n->Infinity]= Maxroot[r[i]] Let's call that the algebraic root: Limit[a[n+1]/a[n],n->Infinity]=1+d such that: Prime[n+1]=(1+d)*Prime[n] or Prime[n+1]=Prime[n]+2*Gap[n] which gives: d=2*Gap[n]/Prime[n] and the classical value of the limit is: Limit[d,n->Infinity]=0 That being true the limiting algebraic polynomial of the primes is a p-adic cyclotomic: p[x,Prime[n]]=(x^Prime[n]-1)/(x-1) with the root structure: r[m]=Exp[2*Pi*I*m/Prime[n]] Since the roots are equally spaced on the unit circle in the complex plane: Limit[Sum[Exp[2*Pi*I*m/Prime[n]],{m,1, Prime[n]}],{n->Infinity]=0 So if the polynomial of arbitrary length is algebraic that generates the primes and if their classical ratio of one holds, then the zeroth Binet prime is zero: Limit[Sum[c[m]*Exp[2*Pi*I*m/Prime[n]],{m,1, Prime[n]}],{n->Infinity]=0 The primes are Fibonacci like and not Lucas like. It is possible to "arrange" the c[m] is such a way that: Limit[Sum[c[m]*Exp[2*Pi*I*m/Prime[n]],{m,1,Prime[n]}],{n->Infinity]=Integer but that appears "artificial" from this point of view. I came to this conclusion when studying to learn Ring and scheme theory in algebraic geometry. The ring of the primes is defined with an additive zero in the spec definition: you can't have a ring without it. For a field of primes, a "1" term is also necessary. The classical number theory with 2 as the starting element says that the primes can't be the backbone of the Integer field that they are. That classical definition is here proved wrong. In terms of modern theory of the Grothendieck sort/ algebraic geometry, zero has to be a prime and the zeroth prime. LINKS FORMULA roots:r[i]->(x^11-1): c(i) solved in Mathematica as a linear equation in 11 unknown: (equivalent to a Fourier coefficient / series solution) a(n)=Sum[r(i)^n*c(i),{i,1,11}] MATHEMATICA Clear[p, c, r, f, g] a0 = Prime; p[x_] = (x - 1)*Cyclotomic[a0, x]; c = Array[x, a0]; r[m_] := x /. NSolve[p[x] == 0, x][[m]]; Solve[Flatten[Table[Sum[r[n]^(a0 - m)*c[[n]], {n, 1, a0}] - Prime[m] == 0, {m, 1, a0}]], c]; f[i_] := x[i] /. Flatten[Solve[Flatten[Table[Sum[r[n]^(a0 - m)*c[[n]], {n, 1, a0}] - Prime[m] == 0, {m, 1, a0}]], c]][[i]]; g[n_] := Sum[f[i]*r[i]^n, {i, 1, a0}]; Table[Round[Re[g[n]]], {n, 0, 20}] CROSSREFS Sequence in context: A276992 A008685 A162156 * A022987 A023473 A298379 Adjacent sequences:  A141526 A141527 A141528 * A141530 A141531 A141532 KEYWORD nonn,uned AUTHOR Roger L. Bagula, Aug 11 2008 EXTENSIONS More terms from Robert G. Wilson v, Aug 25 2008 STATUS approved

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Last modified March 19 23:02 EDT 2019. Contains 321343 sequences. (Running on oeis4.)