A057059 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)).

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

Offset: 1

Author

Clark Kimberling, Jul 30 2000

Keywords

• nonn
• easy

Comments

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

Triangle of spiral permutations. In the Saclolo reference sigma_n(x) is called a spiral permutation. - Michael Somos, Apr 21 2011

Second inverse function (numbers of columns) for pairing function A194982. - Boris Putievskiy, Jan 10 2013

The triangle T(n, k) (see the formula by Michael 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(Pi-x) = 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

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

Examples

			Formatted as a triangle T(n, k) (see Michael Somos' formula):
n, 2n+1\k 1 2  3 4  5 6  7 8  9 10 11 12 ..
1,   3:   1
2,   5:   2 1
3,   7:   3 1  2
4,   9:   4 1  3 2
5,  11:   5 1  4 2  3
6,  13:   6 1  5 2  4 3
7,  15:   7 1  6 2  5 3  4
8,  17:   8 1  7 2  6 3  5 4
9,  19:   9 1  8 2  7 3  6 4  5
10, 21:  10 1  9 2  8 3  7 4  6  5
11, 23:  11 1 10 2  9 3  8 4  7  5  6
12, 25:  12 1 11 2 10 3  9 4  8  5  7  6
... formatted by _Wolfdieter Lang_, Aug 28 2016
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
		

Links

• Wolfdieter Lang, Proof of a Conjecture Involving Chebyshev Polynomials.
• Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.
• M. P. Saclolo, How a Medieval Troubadour Became a Mathematical Figure, Notices Amer. Math. Soc. 58 (2011), no. 5, 682-687. See p. 684 Equation (1).

Crossrefs

Cf. A057058, A194982; related to A141419.

Programs

Mathematica

Table[If[OddQ@ k, n - (k - 1)/2, k/2], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Aug 28 2016 *)

PARI

{T(n, k) = if( k<1 || k>n, 0, if( k%2, n - (k - 1) / 2, k / 2))} /* Michael Somos, Apr 21 2011 */

Formula

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

(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_(r-1)(t) - U_(r-2)(t) (r>1). Then T(n,k) = Sum_{j=1..n} U_(k-1)(cos((2*j-1)*Pi/(2*n+1))), 1<=k<=n. - L. Edson Jeffery, Jan 09 2012 (See the Sep 14 2016 comment above.)

From Boris Putievskiy, Jan 10 2013: (Start)

a(n) = -(A004736(n)+(A002260(n)-1)/2)*((-1)^A002260(n)-1)/2+(A002260(n)/2)*((-1)^A002260(n)+1)/2.

a(n) = -(j+(i-1)/2)*((-1)^i-1)/2+(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)