%I #22 Jul 22 2023 12:38:26
%S 1,0,1,1,2,1,1,2,1,0,1,3,1,1,3,2,0,1,1,0,2,1,1,1,4,1,1,4,3,2,2,3,3,0,
%T 1,1,0,3,2,1,1,1,2,1,1,1,2,2,0,0,1,1,0,0,2,1,1,0,1,1,0,1,1,1,0,0,0,1,
%U 5,1,1,5,4,0,1,1,0,4,3,0,2,2,0,3,3,1,1,1,3,1,1,1,3,2,2,1,2,1,2,1,2,2
%N Irregular triangle T read by rows, giving in row n the nonnegative coefficients of polynomials of height n and degree k (of decreasing powers), for k = 1, 2, ..., n-1, used for Cantor's counting of algebraic numbers, written for m = 1, 2, ..., A364313(n), for n >= 2, and for n = 1 the degree is k = 1.
%C The length of row n is A364313(n). Different orders k are separated by a semicolon in the examples below.
%C The number of polynomials given from row n is A364314(n).
%C The entries for k = n-1 are only present for n >= 2 and A001227(n-1) = 1, that is, n = 2^q + 1 = A000051(q), for q >= 0. This is because otherwise x^(n-1) + 1 and x^(n-1) - 1 are both reducible (factorize over the integers).
%C For Cantor's counting (and determination) of algebraic numbers these polynomials have later to be signed, keeping the positive leading coefficient. See the example for n = 4 below. Complex solutions are omitted if real algebraic numbers are counted.
%C The polynomials with nonnegative coefficients recorded here are sometimes reducible over the integers. But in this case irreducible signed versions exist. E.g., for n = 6 and k = 2 the polynomial x^2 + 3*x + 2 = (x + 1)*(x + 2) is recorded as [1,3,2] (falling powers of x), because x^2 + 3*x - 2 and x^2 - 3*x - 2 are irreducible, each having two real solutions.
%C The number of distinct real solutions of the signed polynomials of degree k and height n is given in A364315(n, k). The total number is A364316(n). Note that no repetition of real solutions already obtained for lower heights can appear due to irreducibility. For the list of all real algebraic numbers for heights 1 to 7 see the W. Lang link.
%C The coefficients of the polynomials are determined from the relative prime compositions of K = n - (k-1). The order is taken from the corresponding partitions, with rising number of parts m, and for given m the order is anti-lexicographic (e.g., [4,1,1], [3,2,1] for K = 6 and m = 3). For each partition the compositions are ordered also anti-lexicographically, not considering the possible 0 parts which are distributed according to decreasing powers of x (e.g., [3,1,0,1], [3,0,1,1], [1,3,0,1], [1,0,3,1], [1,1,0,3], [1,0,1,3]).
%H Georg Cantor, <a href="https://gdz.sub.uni-goettingen.de/id/PPN243919689_0077?tify=%7B%22view%22:%22info%22,%22pages%22:%5B266%5D%7D">Ueber eine Eigenschaft des Inbegriffes aller reellen algebraischen Zahlen</a>, Journal f. d. reine u. angew. Math. 77 (1874), 258-262.
%H Georg Cantor, <a href="https://www.math.byu.edu/~grant/courses/cantor1874.pdf">On a Property of the Class of all Real Algebraic Numbers</a>, (English version).
%H Wolfdieter Lang, <a href="https://arxiv.org/abs/2307.10645">Cantor's List of Real Algebraic Numbers of Heights 1 to 7</a>, arXiv:2307.10645 [math.NT], 2023.
%e The irregular triangle T, with entries T(n, m), begins: (increasing k >= 1 values are separated by ;)
%e n\m 1 2 3 4 5 6 7 8 9 10 11 12 13 ...
%e 1: [1, 0]
%e 2: [1, 1]
%e 3: [2, 1], [1, 2]; [1, 0, 1]
%e 4: [3, 1], [1, 3]; [2, 0, 1], [1, 0, 2], [1, 1, 1]
%e ...
%e n = 5: [4, 1], [1, 4], [3, 2], [2, 3]; [3, 0, 1], [1, 0, 3], [2, 1, 1], [1, 2, 1], [1, 1, 2]; [2, 0, 0, 1], [1, 0, 0, 2], [1, 1, 0, 1], [1, 0, 1, 1]; [1, 0, 0, 0, 1]
%e ---------
%e n = 6: [5, 1], [1, 5]; [4, 0, 1], [1, 0, 4], [3, 0, 2], [2, 0, 3], [3, 1, 1], [1, 3, 1], [1, 1, 3], [2, 2, 1], [2, 1, 2], [1, 2, 2]; [3, 0, 0, 1], [1, 0, 0, 3], [2, 1, 0, 1], [2, 0, 1, 1], [1, 2, 0, 1], [1, 0, 2, 1], [1, 1, 0, 2], [1, 0, 1, 2], [1, 1, 1, 1]; [2, 0, 0, 0, 1], [1, 0, 0, 0, 2], [1, 1, 0, 0, 1], [1, 0, 1, 0, 1], [1, 0, 0, 1, 1]
%e ---------
%e n = 7: [6, 1], [1, 6], [5, 2], [2, 5], [4, 3], [3, 4]; [5, 0, 1], [1, 0, 5], [4, 1, 1], [1, 4, 1], [1, 1, 4], [3, 2, 1], [3, 1, 2], [2, 3, 1], [2, 1, 3], [1, 3, 2], [1, 2, 3]; [4, 0, 0, 1], [1, 0, 0, 4], [3, 0, 0, 2], [2, 0, 0, 3], [3, 1, 0, 1], [3, 0, 1, 1], [1, 3, 0, 1], [1, 0, 3, 1], [1, 1, 0, 3], [1, 0, 1, 3], [2, 2, 0, 1], [2, 0, 2, 1], [2, 1, 0, 2], [2, 0, 1, 2], [1, 2, 0, 2], [1, 0, 2, 2], [2, 1, 1, 1], [1, 2, 1, 1], [1, 1, 2, 1], [1, 1, 1, 2]; [3, 0, 0, 0, 1], [1, 0, 0, 0, 3], [2, 1, 0, 0, 1], [2, 0, 1, 0, 1], [2, 0, 0, 1, 1], [1, 2, 0, 0, 1], [1, 0, 2, 0, 1], [1, 0, 0, 2, 1], [1, 1, 0, 0, 2], [1, 0, 1, 0, 2], [1, 0, 0, 1, 2], [1, 1, 1, 0, 1], [1, 1, 0, 1, 1], [1, 0, 1, 1, 1]; [2, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 2], [1, 1, 0, 0, 0, 1], [1, 0, 1, 0, 0, 1], [1, 0, 0, 1, 0, 1], [1, 0, 0, 0, 1, 1]
%e -------
%e x^6 + 1 = (x^2 + 1)*(x^4 - x^2 + 1), hence no [1, 0, 0, 0, 0, 0, 1] recorded, because x^6 - 1 also factorizes.
%e ...
%e ---------------------------------------------------------------------------
%e Polynomials: n = 4, degree k = 1: 3*x + 1, x + 3; k = 2: 2*x^2 + 1, x^2 + 2, x^2 + x + 1; k = 3: no entry [1, 0, 0, 1], because x^3 + 1 factorizes, as well as x^3 - 1.
%e ---------------------------------------------------------------------------
%e Height n = 4, degree k = 2, with signed polynomials:
%e [2, 0, 1] for 2*x^2 + 1, 2*x^2 - 1, [1, 0, 2] for x^2 + 2, x^2 - 2, and [1, 1, 1] for x^2 + x + 1, x^2 + x - 1, x^2 - x + 1, x^2 - x - 1. The corresponding real algebraic numbers come in signed pairs only from 2*x^2 - 1, x^2 - 2, x^2 + x - 1, and x^2 - x - 1, namely, -sqrt(1/2), +sqrt(1/2), -sqrt(2), +sqrt(2), -phi = -A001622, phi - 1, and -(phi - 1), phi. So Cantor's phi (our Phi) is Phi(4, 2) = 8. Together with the four real k = 1 roots from the signed polynomials for [3, 1] and [1, 3] one finds Phi(4) = 12. See A362364.
%Y Cf. A000051, A001227, A001622, A005409, A364313, A364314, A364315, A364316.
%K nonn,tabf
%O 1,5
%A _Wolfdieter Lang_, Jul 19 2023
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