This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A271082 #16 Nov 01 2021 16:53:58 %S A271082 1,-3,3,1,-30,5,-15,7,1,2,4,16,-1920,9,1,4,8,16,64,-7680,11,1,2,8, %T A271082 -960,13,1,-120,15,1,2,4,8,-3840,17,1,4,-480,19,1,2,16,32,128,-15360, %U A271082 21,-63,23,1,2,4,-1920,25,1,4,8,64,128,512,-61440 %N A271082 Triangle read by rows, the coefficients of the (3x+1)-polynomials. %C A271082 Definition of the (3x+1)-polynomials. %C A271082 The 3x+1 problem is an exceptional case of the zx + 1 problem (for z real or complex). We associate each odd integer x with a polynomial f(z) whose roots have the same behavior as the integer 3 in the 3x + 1 problem. %C A271082 The polynomial f(z) is called "(3x+1)-polynomials" and the problem zx + 1 generates the same number of iterations as the 3x + 1 problem requires to reach 1. The polynomial f(z) has interesting properties, for instance the study of the roots of f(z)= 0. %C A271082 The following example shows the process. %C A271082 Let’s consider x = 17. The corresponding reduced Collatz trajectory containing only odd numbers (17, 13, 5, 1) is obtained from the following steps: %C A271082 start with x = 17; %C A271082 step 1: (3*17 + 1)/4 = 52/4 = 13; %C A271082 step 2: (3*(3*17 + 1)/4 + 1)/8 = 40/8 = 5; %C A271082 step 3: (3*(3*(3*17 + 1)/4 + 1)/8 + 1)/16 = 16/16 = 1. %C A271082 Step 4: substitute the number 3 by the variable z. So, we obtain the following equation: %C A271082 f(z) = 17z^3 + z^2 + 4z - 480 = (z-3) g(z) = (z-3)(17z^2 + 52z + 160)= 0. %C A271082 We would consider that the polynomial f(z) is associated with the integer 17. %C A271082 The three roots are: %C A271082 z0 = 3; %C A271082 z1 = -1.529411765 + 2.659448131 I; %C A271082 z2 = -1.529411765 - 2.659448131 I. %C A271082 The roots z1 and z2 have the same behavior as the integer z0=3, and the 3*x + 1 problem, z1*x + 1 problem and z2*x + 1 problem are identical for x = 17 : we obtain the same number of iterations of the reduced Collatz function required to yield 1: 17 = 2*9-1 => A075680(9) = 3 iterations. %C A271082 For example, with z1 we obtain the following steps: %C A271082 (17*z1 + 1)/4 = -6.250000001 + 11.30265455*I %C A271082 (z1*(17*z1 + 1)/4 + 1)/8 = -2.437500001 - 4.238495460*I %C A271082 (z1*(z1*(17*z1 + 1)/4 + 1)/8 + 1)/16 = 1. %C A271082 For each number x = 2n-1, if the Collatz conjecture is true, the polynomial f(z) is of the general form : %C A271082 f(z) =(2n-1)*z^p + z^(p-1) + 2^a*z^(n-2) + 2^b*z^(n-3) + ... + 2^w*z + 2^r - 2^s = (z-3) g(z) with the property : degree(f(z)) = p = A075680(n), n>1. %C A271082 s is the number of divisions by 2 at the last step %C A271082 r is the number of divisions by 2 at before the last step %C A271082 a is the number of divisions by 2 at the first step %C A271082 b is the number of divisions by 2 at the second step %H A271082 Michel Lagneau, <a href="/A271082/a271082.pdf">Coefficients</a> %e A271082 Triangle begins: %e A271082 1, -3, %e A271082 3, 1, -30, %e A271082 5, -15, %e A271082 7, 1, 2, 4, 16, -1920, %e A271082 9, 1, 4, 8, 16, 64, -7680, %e A271082 11, 1, 2, 8, -960, %e A271082 13, 1, -120, %e A271082 15, 1, 2, 4, 8, -3840, %e A271082 17, 1, 4, -480, %e A271082 19, 1, 2, 16, 32, 128, -15360, %e A271082 21, -63, %e A271082 23, 1, 2, 4, -1920, %e A271082 25, 1, 4, 8, 64, 128, 512, -61440, %e A271082 The corresponding polynomials are: %e A271082 +----+-----------------------------------------------------------+ %e A271082 | x | Polynomials f(z) including the factor (z - 3) | %e A271082 +----+-----------------------------------------------------------+ %e A271082 | 1 | z - 3 | %e A271082 | 3 | 3z^2 + z - 30 | %e A271082 | 5 | 5z - 15 | %e A271082 | 7 | 7z^5 + z^4 + 2z^3 + 4z^2 + 16^z - 1920 | %e A271082 | 9 | 9z^6 + z^5 + 4z^4 + 8z^3 + 16z^2 + 64z - 7680 | %e A271082 | 11 | 11z^4 + z^3 + 2z^2 + 8z - 960 | %e A271082 | 13 | 13z^2 + z -120 | %e A271082 | 15 | 15z^5 + z^4 + 2z^3 + 4z^2 + 8z - 3840 | %e A271082 | 17 | 17z^3 + z^2 + 4z - 480 | %e A271082 | 19 | 19z^6 + z^5 + 2z^4 + 16z^3 + 32z^2 + 128z - 15360 | %e A271082 | 21 | 21z - 63 | %e A271082 | 23 | 23z^4 + z^3 + 2z^2 + 4z - 1920 | %e A271082 +----+-----------------------------------------------------------+ %e A271082 +----+-----------------------------------------------------------+ %e A271082 | x | Polynomials f(z)/(z - 3) | %e A271082 +----+-----------------------------------------------------------+ %e A271082 | 1 | 1 | %e A271082 | 3 | 3z + 10 | %e A271082 | 5 | 5 | %e A271082 | 7 | 7z^4 + 22z^3 + 68z^2 + 208z +640 | %e A271082 | 9 | 9z^5 + 28z^4 + 88z^3 + 272z^2 + 832z + 2560 | %e A271082 | 11 | 11z^3 + 34z^2 + 104z + 320 | %e A271082 | 13 | 13z + 40 | %e A271082 | 15 | 15z^4 + 46z^3 + 140z^2 + 424z + 1280 | %e A271082 | 17 | 17z^2 + 52z + 160 | %e A271082 | 19 | 19z^5 + 58z^4 + 176z^3 + 544z^2 + 1664z + 5120 | %e A271082 | 21 | 21 | %e A271082 | 23 | 23z^3 + 70 z^2 + 212z + 640 | %e A271082 +----+-----------------------------------------------------------+ %p A271082 for m from 1 by 2 to 27 do: T:=array(1..50,[0$50]):U:=array(1..50,[0$50]): %p A271082 n:=m:ii:=2:xx1:=2:pp1:=0:s:=0:U[1]:=n:U[2]:=1: %p A271082 for q from 1 to 100 while(xx1<>1)do: %p A271082 n1:=3*n+1: %p A271082 for p from 1 to 50 do: %p A271082 p1:=2^p:x1:=floor(n1/p1):x0:=irem(n1,p1): %p A271082 if x0=0 and xx1<> 1 %p A271082 then %p A271082 pp1:=p:xx1:=x1: %p A271082 else %p A271082 fi: %p A271082 od: %p A271082 T[ii]:=pp1:n1:=x1:n:=xx1:ii:=ii+1:od:s:=0: %p A271082 for j from 1 to ii-3 do: %p A271082 s:=s+T[j]:U[j+2]:=2^s: %p A271082 od: %p A271082 s:=s+T[ii-2]:s1:=2^s:s:=s+T[ii-1]: %p A271082 s2:=2^s:U[ii]:=s1-s2: %p A271082 W:=array(1..ii-1,[0$ii-1]): %p A271082 W[1]:=U[1]: %p A271082 for l from 2 to ii-1 do: %p A271082 W[l]:=U[l+1]: %p A271082 od: %p A271082 print(m): %p A271082 print(W): %p A271082 od: %Y A271082 Cf. A075680, A171870. %K A271082 sign,tabf %O A271082 1,2 %A A271082 _Michel Lagneau_, Mar 30 2016