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July 30, 2015

Nineteenth century non-avian dinosaur discoveries

Filed under: nature,science — ckrao @ 11:05 am

Below is an attempted chronological list of non-avian dinosaur discoveries of the 19th century that today are considered valid genera. There may still be some where there are only scant remains of the fossil (e.g. a tooth or single bone remain). The list came from [1] with some help from [2] and Wikipedia to filter out doubtful names. Many of the best known dinosaurs are listed here and it looks like most of the major groups are covered. Good histories of dinosaur paleontology are in [3] and [4].

 

Genus Discoverer Year Dinosaur type
Megalosaurus Buckland 1824 tetanuran (stiff-tailed) theropod
Iguanodon Mantell 1825 beaked ornithopod
Streptospondylus von Meyer 1830 megalosaurid
Hylaeosaurus Mantell 1833 armoured
Thecodontosaurus Riley & Stutchbury 1836 prosauropod
Plateosaurus von Meyer 1837 prosauropod
Poekilopleuron Eudes-Deslongchamps 1838 megalosaurid
Cardiodon Owen 1841 sauropod
Cetiosaurus Owen 1841 sauropod
Pelorosaurus Mantell 1850 brachiosaur
Aepisaurus Gervais 1852 sauropod
Oplosaurus Gervais 1852 sauropod
Massospondylus Owen 1854 prosauropod
Nuthetes Owen 1854 maniraptoran
Troodon Leidy 1856 raptor
Stenopelix von Meyer 1857 pachycephalosaur
Astrodon Johnston 1858 sauropod
Hadrosaurus Leidy 1858 duckbilled ornithopod
Compsognathus J. A. Wagner 1859 coelurosaur (fuzzy theropod)
Scelidosaurus Owen 1859 armoured
Echinodon Owen 1861 heterodontosaurid (early bird-hipped dinosaur)
Polacanthus Owen vide [Anonymous] 1865 armoured
Calamospondylus Fox 1866 oviraptorosaur
Euskelosaurus Huxley 1866 prosauropod
Acanthopholis Huxley 1867 armoured
Hypselosaurus Matheron 1869 sauropod
Hypsilophodon Huxley 1869 beaked ornithopod
Rhabdodon Matheron 1869 beaked ornithopod
Ornithopsis Seeley 1870 sauropod
Struthiosaurus Bunzel 1870 armoured
Craterosaurus Seeley 1874 stegosaurian
Chondrosteosaurus Owen 1876 sauropod
Macrurosaurus Seeley 1876 sauropod
Allosaurus Marsh 1877 carnosaur
Apatosaurus Marsh 1877 sauropod
Camarasaurus Cope 1877 sauropod
Dryptosaurus Marsh 1877 tyrannosaur
Dystrophaeus Cope 1877 sauropod
Nanosaurus Marsh 1877 early bird-hipped dinosaur
Stegosaurus Marsh 1877 plated dinosaur
Diplodocus Marsh 1878 sauropod
Brontosaurus Marsh 1879 sauropod
Anoplosaurus Seeley 1879 armoured
Coelurus Marsh 1879 coelurosaur (fuzzy theropod)
Mochlodon Seeley 1881 beaked ornithopod
Craspedodon Dollo 1883 horned dinosaur
Ceratosaurus Marsh 1884 theropod
Anchisaurus Marsh 1885 prosauropod
Camptosaurus Marsh 1885 beaked ornithopod
Aristosuchus Seeley 1887 coelurosaur (fuzzy theropod)
Ornithodesmus Seeley 1887 raptor
Cumnoria Seeley 1888 beaked ornithopod
Priconodon Marsh 1888 armoured
Coelophysis Cope 1889 early theropod
Nodosaurus Marsh 1889 armoured
Triceratops Marsh 1889 horned dinosaur
Barosaurus Marsh 1890 sauropod
Claosaurus Marsh 1890 duckbilled ornithopod
Ornithomimus Marsh 1890 ostrich dinosaur
Ammosaurus Marsh 1891 prosauropod
Torosaurus Marsh 1891 horned dinosaur
Argyrosaurus Lydekker 1893 sauropod
Sarcolestes Lydekker 1893 armoured
Dryosaurus Marsh 1894 beaked ornithopod

References

[1] Dinosaur Genera List – http://www.polychora.com/dinolist.html

[2] Genus List for Holtz (2007) Dinosaurshttp://www.geol.umd.edu/~tholtz/dinoappendix/HoltzappendixWinter2011.pdf

[3] Benton, M. J. 2000. A brief history of dinosaur paleontology. Pp. 10-44, in Paul, G. S. (ed.), The Scientific American book of dinosaurs. St Martin’s Press, New York. – http://palaeo.gly.bris.ac.uk/Essays/dinohist.html

[4] Equatorial Minnesota The generic history of dinosaur paleontology 1699 to 1869 – http://equatorialminnesota.blogspot.com.au/2014/06/the-generic-history-of-dinosaur.html

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July 25, 2015

Compositions of positive integers where adjacent parts have different parity

Filed under: mathematics — ckrao @ 5:39 am

Here is an interesting problem based on one I found in the 2008 AIME (Q11).

How many compositions of the positive integer n have adjacent parts with different parity?

For example, a valid composition of n=14 is

14 = 1 + 2 + 5 + 2 + 1 + 2 + 1,

since the 1,2,5,2,1,2,1 alternate between being odd and even.

The first terms of the sequence are (OEIS A062200)

1, 1, 3, 2, 6, 6, 11, 16, 22, 37, 49, 80, …

Interestingly the sequence decreases between n=3 and 4 since the only valid compositions of 4 are 4 = 1 + 2 + 1 and 4, while 3 = 1 + 1 + 1 = 2 + 1 = 1 + 2.

Here is the first approach I came up with to solve the problem. Let a_n be the number of such compositions starting with an even number and let b_n be the number of compositions starting with an odd number. We wish to find t_n := a_n + b_n. We identify four cases:

  • If n is odd and the composition starts with an even number 2c, we remove that even number and have a composition of n-2c starting with an odd number. Summing over choices of c,
    a_{2k+1} = b_{2k-1} + b_{2k-3} + \ldots + b_1. \quad (1)
  • If n is odd and the composition starts with an odd number 2c+1, we either have the single term composition n, or we remove that odd number and have a composition of n-(2c+1) starting with an even number. Summing over choices of c,
    b_{2k+1} = 1 + a_{2k} + a_{2k-2} + \ldots + a_2. \quad (2)
  • If n is even  and the composition starts with an even number 2c, we either have the single term composition n, or we remove that even number and have a composition of n-2c starting with an odd number. Summing over choices of c,
    a_{2k} = 1 + b_{2k-2} + b_{2k-4} + \ldots + b_2. \quad (3)
  • If n is even and the composition starts with an odd number 2c+1, we remove that odd number and have a composition of n-(2c+1) starting with an even number. Summing over choices of c,
    b_{2k} = a_{2k-1} + a_{2k-3} + \ldots + a_1.\quad (4)

Recursions (1)-(4) with the initial conditions a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0 allows us to generate the following table.

n 1 2 3 4 5 6 7 8 9 10 11 12
a_n 0 1 1 1 3 2 6 6 11 16 22 37
b_n 1 0 2 1 3 4 5 10 11 21 27 43
t_n 1 1 3 2 6 6 11 16 22 37 49 80

After writing out some terms I noticed the pattern

\displaystyle a_n = a_{n-2} + b_{n-2} \text{ and } b_n = a_{n-1} + b_{n-2}.\quad \quad (5)

I later saw why this is so. Firstly, a composition of n that starts with an even number can be obtained from any composition of n-2 by adding 2 to the first term if even or inserting 2 at the start if odd. Hence a_n = t_{n-2} = a_{n-2} + b_{n-2}. For the second relation, if the composition of n starts with an odd number, it either is 1 plus an even-starting composition of n-1 or it can be obtained from an odd-starting composition of n-2 by adding 2 to the first term. Hence b_n = a_{n-1} + b_{n-2}.

 From (5) and using t_n = a_n + b_n we find that

\begin{aligned}  t_n &= a_n + b_n\\  &= t_{n-2} + (a_{n-1} + b_{n-2})\\  &= t_{n-2} + t_{n-3} + t_{n-2} - a{n-2}\\  &= 2t_{n-2} + t_{n-3} - t_{n-4}.\quad \quad (6)  \end{aligned}

Together with the initial terms t_1 = 1, t_2 = 1, t_3 = 3, t_4 = 2, (6) enables us to generate our desired terms without requiring a_n or b_n.

The OEIS page also gives the following expression for t_n:

\displaystyle t_n = \sum_{k=0}^{n+1} \binom{n-k+1}{3k-n+1}.\quad \quad (7)

Here is how to show this result, with help from the solution at AoPS here. For ease of exposition convert the problem to one of binary strings: we require the number of length-n 0-1 strings so that maximal continguous blocks of 1s have even length and maximal contiguous blocks of 0s have odd length.

Suppose there are b maximal blocks of 0s. All but the last such block must necessarily be followed by a pair of 1s. That is, the string has the form

\displaystyle [....][....0][11....][....0][11....] \ldots [....0][....],

where blocks enclosed by [] contain the same digit (0 or 1) and 0 or more pairs of digits are in place of the sets of four dots. We have shown 2b+1 blocks above but the first and last may be empty.

Let k = (n-b)/2, which is the number of pairs of digits to use after b 0s are used (noting that n-b must be even). We have b 0s used above along with b-1 pairs of 1s. Therefore (n-(b + 2(b-1)))/2 pairs of 0s or 1s need to inserted into the 2b+1 blocks above. The blocks that they end up determine whether they are 0 or 1.

Now x pairs can be inserted into y blocks in \binom{x+y-1}{x} ways, since we write out x+y-1 symbols in a row and choose x of them to be pairs, with the remaining symbols representing separators between blocks. In our case,

x = \frac{n-(b + 2(b-1))}{2}, y = 2b+1 and b = n-2k, so

\begin{aligned}  x &= \frac{n-3b+2}{2}\\  &= \frac{n-3n+6k+2}{2}\\  &= 3k-n+1,\end{aligned}

and x+y-1 = 3k-n+1 + 2(n-2k) + 1 = n-k+1.

Summing over choices of k gives (7) as desired. Note that some of the terms will be 0 as k ranges from 0 to n+1.

For example, when n=11 and 12, we have

\displaystyle t_{11} = \binom{8}{2} + \binom{7}{5} = 28 + 21 = 49,

\displaystyle t_{12} = \binom{9}{1} + \binom{8}{4} + \binom{7}{7} = 9 + 70 + 1= 80,

matching the values given in the table above.

 

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