[seqfan] Strange thing in A005408: The odd numbers: a(n) = 2n+1.

Joerg Arndt arndt at jjj.de
Thu Mar 4 10:45:16 CET 2010

Dear Vincenzo,
in seq A005408 you give the following relation:

 Except for the first term, numbers n such that if
 A=(7*n^4-3*n^3-3*n^2-n)/8; B=(7*n^4+4*n^3-6*n^2-4*n-1)/16;
 C=(17*n^4+20*n^3+6*n^2+4*n+1)/16; D=(5*n^4+3*n^3+3*n^2+n)/4; then
 A^3+B^3+C^3=D^3 [From Vincenzo Librandi
 (vincenzo.librandi(AT)tin.it), Sep 08 2009]

This Diophantine equation indeed holds for _all_ n:


t=a^3+b^3+c^3-d^3  \\ == zero

While the equality may be of some interest there
is no connection with the seq at hand.

Also note that you added the following 'example':

  For n=3, A=57, B=38, C=124, D=129, and 57^3+38^3+124^3=129^3; For
  n=5, A=490, B=294, C=831, D=895, and 490^3+294^3+831^3=895^3 [From
  Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Sep 08 2009]

This usage of the 'example' entry strikes me as unusual.

Please remove both from the seq.

I strongly suggest you also remove the other example

  For n=2, a(2)=4*2-1-4=3; n=3, a(3)=4*3-3-4=5; n=4, a(4)=4*4-5-4=7
  [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 22 2009]

The corresponding formula has a typo but may IMHO be left (after correction):

  Let n>1; if a=n, b=(n^2-1)/2, c=(n^2+1)/2, then a^2+b^2=c^2 is alwais
  verified. [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Mar
  27 2009]

regards,   jj

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