

A000905


Hamilton numbers.
(Formerly M0736 N0275)


5



2, 3, 5, 11, 47, 923, 409619, 83763206255, 3508125906290858798171, 6153473687096578758448522809275077520433167, 18932619208894981833333582059033329370801266249535902023330546944758507753065602135843
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OFFSET

1,1


COMMENTS

a(n) = minimal degree of an equation from which n successive terms after the first can be removed (by a series of transformation comparable to Tschirnhaus') without requiring the solution of an equation of degree greater than n (and excluding cases where an equation of degree greater than n is needed but is in fact factorizable into several equations of degree all less than n). Hamilton computed the first six terms of this sequence (see reference). That is the reason Sylvester and Hammond named them "Hamilton numbers".  Olivier Gérard, Oct 17 2007


REFERENCES

W. R. Hamilton, Sixth Report of the British Association for the Advancement of Science, London, 183i, 295348.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

Table of n, a(n) for n=1..11.
Alexander Chen, Y. H. He, J. McKay, Erland Samuel Bring's "Transformation of Algebraic Equations", arXiv preprint arXiv:1711.09253 [math.HO], 2017. See page 6.
Raymond Garver, The Tschirnhaus transformation, The Annals of Mathematics, 2nd Ser., Vol. 29, No. 1/4. (1927  1928), pp. 329.
E. Lucas, Théorie des Nombres, GauthierVillars, Paris, 1891, Vol. 1, p. 496.
E. Lucas, Théorie des Nombres, GauthierVillars, Paris, 1891, Vol. 1. [Annotated scan of pages 488499 only]
J. J. Sylvester and M. J. Hammond, On Hamilton's numbers, Phil. Trans. Roy. Soc., 178 (1887), 285312.


EXAMPLE

a(1)=2 is the familiar fact than one can always remove the linear term of a quadratic equation.
a(2)=3 because one can put any cubic equation in the form x^3a=0 by a Tschirnhaus transformation based on the solutions of a quadratic equation.
a(4)=11 because one can remove the 4 terms after the first term in a polynomial of degree 11 without having to solve a quintic.


MAPLE

A000905 := proc(n) option remember; local i; if n=1 then 2 else 2+add((1)^(i+1)*binomial(A000905(ni), i+1), i=1..n1); fi; end;


MATHEMATICA

a[1]=2; a[n_] := a[n] = 2+Sum[(1)^(i+1)*Product[a[ni]  k, {k, 0, i}]/(i+1)!, {i, 1, n1}]; Table[a[n], {n, 1, 11}] (* JeanFrançois Alcover, May 17 2011, after Maple prog. *)


CROSSREFS

Cf. A001660.
Equals A006719(n)  1.
Cf. A134294.
Sequence in context: A003686 A086506 A109462 * A065296 A114895 A083685
Adjacent sequences: A000902 A000903 A000904 * A000906 A000907 A000908


KEYWORD

nonn,nice,easy


AUTHOR

N. J. A. Sloane


EXTENSIONS

The formula given by Lucas on p. 498 is slightly in error  see Maple program given here.


STATUS

approved



