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 a(n)=i(A057027(n)).

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

Offset: 1

Author

Clark Kimberling, Jul 30 2000

Keywords

• nonn

Comments

Since A057027 is a permutation of the natural numbers, every natural number occurs infinitely many times in this sequence.

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 saddle-points quantities, P and Q are bidimensional polynomials n=2,3,4.. . (E) leads also for instance to the one-dimension 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

• Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Crossrefs

Cf. A057059, A194982.

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)^i-1)/2+(j+i/2)*((-1)^i+1)/2, where i = n-t*(t+1)/2, j = (t*t+3*t+4)/2-n, t = floor((-1+sqrt(8*n-7))/2). (End)