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A057059
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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)).
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4
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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
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OFFSET
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1,2
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COMMENTS
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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
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
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LINKS
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FORMULA
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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.)
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)
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EXAMPLE
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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
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
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MATHEMATICA
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Table[If[OddQ@ k, n - (k - 1)/2, k/2], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Aug 28 2016 *)
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PROG
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(PARI) {T(n, k) = if( k<1 | k>n, 0, if( k%2, n - (k - 1) / 2, k / 2))} /* Michael Somos, Apr 21 2011 */
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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