

A057058


Let R(i,j) be the rectangle with antidiagonals 1; 2,3; 4,5,6; ...; each k is an R(i(k),j(k)) and A057058(n)=i(a(n)), where a=A057027.


5



1, 1, 2, 1, 3, 2, 1, 4, 2, 3, 1, 5, 2, 4, 3, 1, 6, 2, 5, 3, 4, 1, 7, 2, 6, 3, 5, 4, 1, 8, 2, 7, 3, 6, 4, 5, 1, 9, 2, 8, 3, 7, 4, 6, 5, 1, 10, 2, 9, 3, 8, 4, 7, 5, 6, 1, 11, 2, 10, 3, 9, 4, 8, 5, 7, 6, 1, 12, 2, 11, 3, 10, 4, 9, 5, 8, 6, 7, 1, 13, 2, 12, 3, 11, 4
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OFFSET

1,3


COMMENTS

Since A057027 is a permutation of the natural numbers, every natural number occurs in A057058 infinitely many times.
Consider the triangle TN := 1; 1, 2; 1, 3, 2; 1, 4, 2, 3; ... Antidiagonal sums give A129819(n+2). TN arises in studying the equation (E) dy/dx=Q(n,x,y)/P(n,x,y) involving saddlepoints quantities, P and Q are bidimensional polynomials n=2,3,4.. . (E) leads also for instance to the onedimension polynomials in A129326, A129587, A130679.  Paul Curtz, Aug 16 2008
First inverse function (numbers of rows) for pairing function A194982.  Boris Putievskiy, Jan 10 2013


LINKS

Table of n, a(n) for n=1..85.
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, 2012, arXiv:1212.2732 [math.CO].


FORMULA

From Boris Putievskiy, Jan 10 2013: (Start)
a(n) = ((A002260(n)+1)/2)*((1)^A002260(n)1)/2+(A004736(n)+A002260(n)/2)*((1)^A002260(n)+1)/2.
a(n) = ((i+1)/2)*((1)^i1)/2+(j+i/2)*((1)^i+1)/2, where i = nt*(t+1)/2, j = (t*t+3*t+4)/2n, t = floor((1+sqrt(8*n7))/2). (End)


CROSSREFS

Cf. A057059, A194982.
Sequence in context: A294733 A275724 A193278 * A278104 A141672 A141671
Adjacent sequences: A057055 A057056 A057057 * A057059 A057060 A057061


KEYWORD

nonn


AUTHOR

Clark Kimberling, Jul 30 2000


STATUS

approved



