%I
%S 1,2,1,3,1,2,4,1,3,2,5,1,4,2,3,6,1,5,2,4,3,7,1,6,2,5,3,4,8,1,7,2,6,3,
%T 5,4,9,1,8,2,7,3,6,4,5,10,1,9,2,8,3,7,4,6,5,11,1,10,2,9,3,8,4,7,5,6,
%U 12,1,11,2,10,3,9,4,8,5,7,6,13,1,12,2,11,3,10
%N Let R(i,j) be the rectangle with antidiagonals 1; 2,3; 4,5,6; ... Define i(m) and j(m) by R(i(m),j(m)) = m. Then a(n) = j(A057027(n)).
%C Since A057027 is a permutation of the natural numbers, every natural number occurs in this sequence infinitely many times.
%C Triangle of spiral permutations. In the Saclolo reference sigma_n(x) is called a spiral permutation.  _Michael Somos_, Apr 21 2011
%C Second inverse function (numbers of columns) for pairing function A194982.  _Boris Putievskiy_, Jan 10 2013
%C The triangle T(n, k) (see the formula by M. Somos) has in row n a certain permutation of [1, 2, ..., n]. This permutation is useful for the proof of the identity Product_{k=1..n} f(sin(Pi*k/(2*n+1))) = Product_{m=1..n} f(sin(2*Pi*m/(2*n+1))) for any function f, n >= 1 (also for n = 0). The permutation of the arguments of f goes via m = T(n, k), and this is due to sin(Pix) = sin(x). Of course, one can replace the product by a sum in this identity. The product identity is used in a trivial variant of Eisenstein's proof of the quadratic reciprocity law. See the W. Lang Aug 28 2016 comment under A049310.  _Wolfdieter Lang_, Aug 28 2016
%C For the proof of the (slightly extended) conjecture stated in the formula section by _L. Edson Jeffery_ see the W. Lang link.  _Wolfdieter Lang_, Sep 14 2016
%H Wolfdieter Lang, <a href="/A057059/a057059_1.pdf">Proof of a Conjecture Involving Chebyshev Polynomials.</a>
%H Boris Putievskiy, <a href="http://arxiv.org/abs/1212.2732">Transformations Integer Sequences And Pairing Functions</a>, 2012, arXiv:1212.2732 [math.CO], 2012.
%H M. P. Saclolo, <a href="http://www.ams.org/notices/201105/rtx110500682p.pdf">How a Medieval Troubadour Became a Mathematical Figure</a> see p. 684 Equation (1)
%F T(n, k) = k / 2 if k is even, n  (k  1) / 2 if k is odd where 0 < k <= n are integers.  _Michael Somos_, Apr 21 2011
%F (Conjecture) Define the Chebyshev polynomials of the second kind by U_0(t) = 1, U_1(t) = 2*t, and U_r(t) = 2*t*U_(r1)(t)  U_(r2)(t) (r>1). Then T(n,k) = Sum_{j=1..n} U_(k1)(cos((2*j1)*Pi/(2*n+1))), 1<=k<=n.  _L. Edson Jeffery_, Jan 09 2012 (See the Sep 14 2016 comment above.)
%F From _Boris Putievskiy_, Jan 10 2013: (Start)
%F a(n) = (A004736(n)+(A002260(n)1)/2)*((1)^A002260(n)1)/2+(A002260(n)/2)*((1)^A002260(n)+1)/2.
%F a(n) = (j+(i1)/2)*((1)^i1)/2+(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)
%e Formatted as a triangle T(n, k) (see M. Somos' formula):
%e n, 2n+1\k 1 2 3 4 5 6 7 8 9 10 11 12 ..
%e 1, 3: 1
%e 2, 5: 2 1
%e 3, 7: 3 1 2
%e 4, 9: 4 1 3 2
%e 5, 11: 5 1 4 2 3
%e 6, 13: 6 1 5 2 4 3
%e 7, 15: 7 1 6 2 5 3 4
%e 8, 17: 8 1 7 2 6 3 5 4
%e 9, 19: 9 1 8 2 7 3 6 4 5
%e 10, 21: 10 1 9 2 8 3 7 4 6 5
%e 11, 23: 11 1 10 2 9 3 8 4 7 5 6
%e 12, 25: 12 1 11 2 10 3 9 4 8 5 7 6
%e ... formatted by _Wolfdieter Lang_, Aug 28 2016
%e n=4: sin identity: sin(Pi*k/9) = sin(2*Pi*T(4, k))/9), for k =1, ..., n. That is: sin(Pi*1/9) = sin(2*Pi*4/9) = sin(Pi*(1  8/9), sin(Pi*3/9) = sin(2*Pi*3/9) = sin(Pi*(1  6/9)). For even k this is trivial.  _Wolfdieter Lang_, Aug 28 2016
%t Table[If[OddQ@ k, n  (k  1)/2, k/2], {n, 12}, {k, n}] // Flatten (* _Michael De Vlieger_, Aug 28 2016 *)
%o (PARI) {T(n, k) = if( k<1  k>n, 0, if( k%2, n  (k  1) / 2, k / 2))} /* _Michael Somos_, Apr 21 2011 */
%Y Cf. A057058, A194982; related to A141419.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Jul 30 2000
