Search: id:A074279
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%I A074279
%S A074279 1,2,2,2,2,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,
%T A074279 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5
%N A074279 n appears n^2 times.
%C A074279 Since the last occurrence of n comes one before the first occurrence 
               of n+1 and the former is at SUM[i=0..n](i^2) = A000330(n), we have 
               a(A000330(n)) = a(n*(n+1)*(2n+1)/6) = n and a(1+A000330(n)) = a(1+(n*(n+1)*(2n+1)/
               6)) = n+1. The current sequence is, loosely, the inverse function 
               of the square pyramidal sequence. See also: A000330 Square pyramidal 
               numbers: 0^2+1^2+2^2+...+n^2 = n(n+1)(2n+1)/6. A000330 has many alternative 
               formulae, thus yielding many alternative formulae for the current 
               sequence. - Jonathan Vos Post (jvospost3(AT)gmail.com), Mar 18 2006
%H A074279 Y.-F. S. Petermann, J.-L. Remy and I. Vardi, <a href="http://www.lix.polytechnique.fr/
               ~ilan/discrete_derivatives.ps">Discrete derivatives of sequences</
               a>, Adv. in Appl. Math. 27 (2001), 562-84.
%Y A074279 Cf. A000217, A000330, A006331, A050446, A050447, A000537, A006003, A005900.
%Y A074279 Sequence in context: A156875 A066339 A052375 this_sequence A072750 A029835 
               A074280
%Y A074279 Adjacent sequences: A074276 A074277 A074278 this_sequence A074280 A074281 
               A074282
%K A074279 nonn
%O A074279 0,2
%A A074279 Jon Perry (perry(AT)globalnet.co.uk), Sep 21 2002

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