[math-fun] clarification, trapezohedra
I said, Recall the odd (mod 3) squarefree sequence d(k) = 0 1 2 1 0 1 2 0 2 1 0 2 0 1 2 1 0 1 2 0 1 0 2 1 2 0 2 1 0 ... defined as the left-to-right alternating sum of the nonzero trits of k. In *balanced* ternary. (Thanks, Neil!) Theorem: if any trapezohedron be scaled to inscribe in a sphere, its faces are "right kites", i.e., they have two square vertices. Mathworld's trapezohedron article just lists areas and volumes of a few special cases (scaled inconsistently). For an n-trapezohedron inscribed in a spheroid with polar radius a, equatorial radius b, 2 2 2 2 2 2 8 c (b c - b + 2 a ) 2 sqrt(c (a c + b )) s = sqrt(4 a - ----------------------), S = ---------------------, 2 c + 1 (c + 1) 2 2 2 %pi 8 b sqrt(b c + a c) sin(---) n a b c n inradius = ------------------, area = ---------------------------------, 2 2 2 2 sqrt(b c + a c) (c + 1) 2 %pi 8 a b c sin(---) n n 2 b sqrt(c) volume = -------------------, tradius = -----------, 2 c + 1 3 (c + 1) 2 2 2 2 2 2 (c + 1) ((2 - 2 c) S - s ) (c + 1) ((c - 1) S - c s ) sqrt(---------------------------) sqrt(-----------------------------) (1 - c) (2 c - 1) (c - 1) c (2 c - 1) a = ---------------------------------, b = -----------------------------------, 2 2 2 2 %pi 2 b cos(-----) + a c n angles = [acos(--------------------), 2 2 a c + b 2 2 (b - a ) sqrt(1 - c) sqrt(c) acos(-----------------------------------------), 2 2 2 2 sqrt(a c + b ) sqrt(2 b c + a (1 - c)) 2 2 2 2 b c - a (1 - c) %pi - acos(--------------------)], 2 2 2 b c + a (1 - c) 2 3/2 b c --------------------------- + a 2 2 sqrt(a c + b ) + a sqrt(c) apex_solid_angle = 2 acos(-------------------------------) n, 2 2 sqrt(b c + a ) 2 2 %pi 2 girdpt_solid_angle = acos(((a cos(-----) + b c) n 4 2 %pi 4 2 2 2 (a cos(-----) + b c - 2 a b c (2 c + 1)) n 2 2 2 3/2 2 2 + 4 a (b - a ) sqrt(1 - c) c (c + 1) sqrt(a c + b ) 2 2 2 2 3 sqrt(2 b c + a (1 - c)))/(b c + a ) ), c + 1 natural_a = --------------------------------------------, %pi 2 sqrt(2) sqrt(1 - c) sqrt(2 c + 1) sin(---) n c + 1 natural_b = ------------------, %pi 4 sqrt(c) sin(---) n where s:=short side, S:=long side, c:=cos(%pi/n), tradius := the "tubular" or girth radius, and natural_a and natural_b are the spheroid radii of the trapezohedron which is the dual of the antiprism with unit edges. The face angle order is [apex, upper girdle, lower girdle]. Note vanishing cosine of upper girdle when a=b. Maybe somebody can simplify the girdle vertex solid angle. When a=b, it boils all the way down to acos(1 - 2 c). (This all started when I added prisms, pyramids, etc. to Macsyma's rather meager PLOT_3D_FIGURE for a geomety lesson.) --rwg PS, while the cube is the only regularhedron whose vertex solid angle is a rational fraction (1/8) of the sphere, the tetrahedral and octahedral vertex solid angles are 2 atan(sqrt(2)/5) and 4 atan(sqrt(2)/4). Since 3 atan(sqrt(2)/4) + 2 atan(sqrt(2)/5) = pi/2, six octahedra and eight tetrahdra come together at each vertex in the octahedral-tetrahedral tessellation of 3-space. This is a (rather lame) example of a pi arctan formula a la Machin's, but with sqrt(rational) series multipliers, which are considered kosher in the game of maximizing digits/term of hypergeometric pi series. So, with sqrts, (but not surds), can we beat Machin?
I pontificated: Theorem: if any trapezohedron be scaled to inscribe in a sphere, its faces are "right kites", i.e., they have two square vertices. If a planar kite lies on a sphere, it lies on a circle, with diameter = main diagonal. --rwg
rcs>RWG: Can your matrices of Q-products explain why Unequal-Partitions(N) has so many powers of 2 divisors? Probably not. The most egregious values (from A&S table 24.5 on page 836) are Q(20) = 64, Q(34) = 512, Q(45) = 2048. And Q(29) = 256. There are occasional odd Q(N), but no apparent pattern. Q(n) = Q_even(n) + Q_odd(n). In generating functions, n (n + 1) inf --------- inf ==== 2 /===\ \ q | | n > ---------- = | | (q + 1), / (q; q) | | ==== n n = 1 n = 0 with alternate terms of the sum generating Q_even and Q_odd. But Q_even(n) - Q_odd(n) = 1, - 1, - 1, 0, 0, 1, 0, 1, 0, 0, 0 ... = A010815, the coeffs of n (n + 1) inf --------- inf ==== n 2 /===\ \ (- 1) q | | n > ----------------- = | | (1 - q ) / (q; q) | | ==== n n = 1 n = 0 2 5 7 12 15 22 26 35 40 A001318 = 1 - q - q + q + q - q - q + q + q - q - q + . . . = q . So this explains one power of 2, since Q_even(n) and Q_odd(n) are usually equal (and never differ by > 1). Note that Q_odd(51) = 2048, but Q(51) = 4097, since 51 is a generalized pentagonal number. Further bisecting the generating series into oddly evenly many parts, etc, failed to produce the near-equalities that might explain one more factor of 2. But before we can explain this phenomenon, we need to quantify it, and with larger n, it seems to taper off. Based on n<2049, I conjecture only only finitely many Q(n) < T(n)^4, where T(n):= the largest power of 2 dividing Q(n). So we need a more modest estimate for T(n), which, for large enough n, might wind up completely unremarkable. Finally, while [ inf inf ] inf [ ==== ==== ] /===\ [ k ] [ \ n \ n ] | | [ 1 q ] [ > Q (n) q > Q (n) q ] | | [ ] = [ / even / odd ] | | [ k ] [ ==== ==== ] k = 1 [ q 1 ] [ n = 0 n = 0 ] [ ] [ vice versa ] computes both g.f.s at once, you need 1000 terms, e.g., for Q(1000), whereas you need only 22 for [ inf ] inf [ 4 n + 1 ] [ ==== ] /===\ [ q q ] [ \ n ] | | [ ------------------------- ----- ] [ 0 > Q (n) q ] | | [ 2 n 2 n + 1 1 - q ] = [ / odd ]. | | [ (1 - q ) (1 - q ) ] [ ==== ] n = 1 [ ] [ n = 0 ] [ 0 1 ] [ ] [ 0 1 ] --rwg PS, Correction: in an unsuccessful attempt to confuse him, I said Neil, you might want to add these G.f.s (and yesterday's A027193,A027193) meaning ...193,A027187. PPS, did RSA200 protect any squeamish ossifrage type sentiments?
George Andrews kindly informs me that any fixed power of 2 divides almost all Q(n) (usually written q(n), to confuse you with the nome). This is the p=2, k=2 case of the more general result (www.math.wisc.edu/~ono/reprints/018.pdf) that as N ->oo, there is a positive alpha(k,p,j) such that for n<N, p^j fails to divide at most N/(log N)^alpha of the partitions of n into nonmultiples of k, provided that there is an integer a such that p^a|k and p^(2a) > k. While impressive, this can't be sharp, at least for k=2, j=1, p=2, where the density is ~ sqrt(2N/3). On the other hand, for j=2, 325 of the first 801 Q(n) are nonzero mod 4. But even knowing alpha won't tell us much about early appearances of large 2^j. This article (by B. Gordon and K. Ono) cites K. Alladi, www.ams.org/tran/1997-349-12/ S0002-9947-97-01831-X/S0002-9947-97-01831-X.pdf which apparently gives combinatorial arguments for the first few j. He also claims eventual evenness for partitions with even parts distinct. But J. McKay wonders about an odd # of even parts. I get g.f.s 2 3 4 5 6 7 8 9 10 11 12 q + q + 2 q + 3 q + 5 q + 7 q + 10 q + 14 q + 20 q + 27 q + 37 q 13 14 15 16 17 18 19 20 + 49 q + 66 q + 86 q + 113 q + 146 q + 190 q + 242 q + 310 q 21 22 23 24 25 26 27 + 392 q + 497 q + 623 q + 782 q + 973 q + 1212 q + 1498 q 28 29 30 31 32 33 34 + 1851 q + 2274 q + 2793 q + 3411 q + 4163 q + 5059 q + 6142 q 35 36 37 38 39 40 + 7427 q + 8972 q + 10801 q + 12989 q + 15572 q + 18646 q 41 42 43 44 45 46 + 22267 q + 26561 q + 31602 q + 37556 q + 44533 q + 52743 q 47 48 49 50 + 62338 q + 73593 q + 86716 q + 102064 q + . . . 1 1 1 2 -------- - ---------- -------- - (- q; q ) (q; q) (q; - q) (q; q) oo oo oo oo = ----------------------- = ---------------------- 2 2 =(unrestricted - distinct odd)/2 oo oo 2 ==== k ==== k \ q \ q > ------- - > --------- / (q; q) / 2 2 ==== k ==== (q ; q ) k = 0 k = 0 k = ------------------------------- 2 2 8 18 32 50 q - q + q - q + q + . . . = ----------------------------------- (q; q) oo = (electron shell #s = (theta_4(q^2)-1)/2) * unrestricted . Since zilch is known about unrestricted mod 2, this quashes oddevens(n). Confession: I had actually downloaded the Ono article prior to the first even Q(n) message, but suffered mental indigestion due to a defishency. --rwg
Apropos my last, EIS gives for A000701, Name: One half of number of non-self-conjugate partitions; also half of number of asymmetric Ferrers graphs with n nodes. Comments: Also number of cycle types of odd permutations. Also number of partitions of n with an odd number of even parts. There is no restriction on the odd parts. but no generating function. This is no problem, since A000701 = (A000041-A000700)/2, but in case Neil wants to add some, here are the expressions I gave last time plus a couple more. oo ==== \ n + 2 2 3 4 5 6 7 8
A000701(n) q = q + q + 2 q + 3 q + 5 q + 7 q + 10 q / ==== n = 0
9 10 11 12 13 14 15 + 14 q + 20 q + 27 q + 37 q + 49 q + 66 q + 86 q + . . . oo ==== 2 \ 2 n > (- q ) / oo ==== ==== 2 (2 n - 1) n = 1 \ q = - ----------------------------- = (- q; q) > --------------- oo n (3 n - 1) oo / 2 2 ==== ----------- ==== (q ; q ) \ n 2 n = 1 2 n - 1 > (- 1) q / ==== n = - oo 1 1 1 2 -------- - ---------- -------- - (- q; q ) (q; q) (q; - q) (q; q) oo oo oo oo = ----------------------- = ---------------------- 2 2 oo 2 ==== k \ 1 1 q = > (------- - ---------) --- , / 2 2 2 2 ==== (q; q) (q ; q ) k = 0 k k n-1 where "q-pochhammer" notation (a;q)_n := prod 1-a*q^k . k=0 The this last sum and the ratio of sums are perhaps most efficient, but among the seemingly endless variations I can't find a simple monomial product. --rwg GLISSADE <-> SLIDAGES
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R. William Gosper