cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A274660 Irregular triangle read by rows in which row n lists the divisors d of 2*n+1 (A274658), given the sign (-1)^(n + (d-1)/2).

Original entry on oeis.org

1, -1, 3, 1, 5, -1, 7, 1, -3, 9, -1, 11, 1, 13, -1, 3, -5, 15, 1, 17, -1, 19, 1, -3, -7, 21, -1, 23, 1, 5, 25, -1, 3, -9, 27, 1, 29, -1, 31, 1, -3, -11, 33, -1, -5, 7, 35, 1, 37, -1, 3, -13, 39, 1, 41, -1, 43, 1, -3, 5, 9, -15, 45, -1, 47, 1, -7, 49, -1, 3, -17, 51, 1, 53, -1, -5, 11, 55, 1, -3, -19, 57, -1, 59, 1, 61
Offset: 0

Views

Author

Wolfdieter Lang, Jul 27 2016

Keywords

Comments

The length of row n is A099774(n+1).
The unsigned irregular triangle is given in A274658.
The sum of row n gives A228443(n).
The entries of row n appear in the Fourier expansion of Jacobi's elliptic function cn in the rewritten second factor Sum_{n>=0} (q^n/(1+q^(2*n+1))) * cos((2*n+1)*v) as Sum_{n>=0} q^n*Sum_{k=1..A099774(n+1)} sign(a(n,k))*cos(abs(a(n,k))*v). See e.g., the formula in Abramowitz-Stegun, p. 575, 16.23.2.

Examples

			The irregular triangle T(n, k) begins:
n, 2n+1\k  1  2   3   4 ...
0,   1:    1
1,   3:   -1  3
2,   5:    1  5
3,   7:   -1  7
4,   9:    1 -3   9
5,  11:   -1 11
6,  13:    1 13
7,  15:   -1  3  -5  15
8,  17:    1 17
9,  19:   -1 19
10, 21:    1 -3  -7  21
11, 23:   -1 23
12, 25:    1  5  25
13, 27:   -1  3  -9  27
14, 29:    1 29
15, 31:   -1 31
16, 33:    1 -3 -11  33
17, 35:   -1 -5   7  35
18, 37:    1 37
19, 39:   -1  3 -13  39
20, 41:    1 41
...
The above mentioned expansion coefficient of q^4 of the second factor of the cn formula is +cos(1*v) - cos(3*v) + cos(9*v).
		

Crossrefs

Programs

  • Mathematica
    Table[(-1)^(n + (# - 1)/2) # &@ Divisors[2 n + 1], {n, 0, 30}] // Flatten (* Michael De Vlieger, Aug 01 2016 *)

Formula

T(n, k) = (-1)^(n + (d(k)-1)/2)*d(k) with d(k) the k-th divisor of 2*n+1 in increasing order.

A274659 Triangle entry T(n, m) gives the m-th contribution T(n, m)*sin((2*m+1)*v) to the coefficient of q^n in the Fourier expansion of Jacobi's elliptic sn(u|k) function when expressed in the variables v = u/(2*K(k)/Pi) and q, the Jacobi nome, written as series in (k/4)^2. K is the real quarter period of elliptic functions.

Original entry on oeis.org

1, 1, 1, -1, 0, 1, -1, -2, 0, 1, 2, 1, -2, 0, 1, 2, 3, 0, -2, 0, 1, -4, -2, 3, 0, -2, 0, 1, -4, -5, 1, 3, 0, -2, 0, 1, 7, 3, -6, 0, 3, 0, -2, 0, 1, 7, 9, -2, -6, 0, 3, 0, -2, 0, 1, -11, -5, 11, 1, -6, 0, 3, 0, -2, 0, 1, -11, -15, 3, 11, 0, -6, 0, 3, 0, -2, 0, 1, 17, 9, -17, -2, 11, 0, -6, 0, 3, 0, -2, 0, 1
Offset: 0

Views

Author

Wolfdieter Lang, Jul 18 2016

Keywords

Comments

If one takes the row polynomials as R(n, x) = Sum_{m=0..n} T(n, m)*x^(2*m+1), n >= 0, Jacobi's elliptic sn(u|k) function in terms of the new variables v and q becomes sn(u|k) = Sum_{n>=0} R(n, x)*q^n, if one replaces in R(n, x) x^j by sin(j*v).
v=v(u,k^2) and q=q(k^2) are computed with the help of A038534/A056982 for (2/Pi)*K and A002103 for q expanded in powers of (k/4)^2.
A test for sn(u|k) with u = 1, k = sqrt(1/2), that is v approximately 0.8472130848 and q approximately 0.04321389673, with rows n=0..10 (q powers not exceeding 10) gives 0.8030018002 to be compared with sn(1|sqrt(1/2)) approximately 0.8030018249.
For the derivation of the Fourier series formula of sn given in Abramowitz-Stegun (but there the notation sn(u|m=k^2) is used for sn(u|k)) see, e.g., Whittaker and Watson, p. 511 or Armitage and Eberlein, Exercises on p. 55.
For the cn expansion see A274661.
See also the W. Lang link, equations (34) and (35).

Examples

			The triangle T(n, m) begins:
      m  0   1  2  3  4  5  6  7  8  9 10 11
n\ 2m+1  1   3  5  7  9 11 13 15 17 19 21 23
0:       1
1:       1   1
2:      -1   0  1
3:      -1  -2  0  1
4:       2   1 -2  0  1
5:       2   3  0 -2  0  1
6:      -4  -2  3  0 -2  0  1
7:      -4  -5  1  3  0 -2  0  1
8:       7   3 -6  0  3  0 -2  0  1
9:       7   9 -2 -6  0  3  0 -2  0  1
10:    -11  -5 11  1 -6  0  3  0 -2  0  1
11:    -11 -15  3 11  0 -6  0  3  0 -2  0  1
...
T(4, 0) = 2 from the x^1 term in b(0, x)*a(4) + b(2, x)*a(2) + b(4, x)*a(0), that is x^1*3 + x^1*(-2) + x^1*1 = +2*x^1.
n=4: R(4, x) = 2*x^1 + 1*x^3 - 2*x^5 + 0*x^7 + 1*x^9, that is the sn(u|k) contribution of order q^4 in the new variables v and q is (2*sin(1*v) + 1*sin(3*v) - 2*sin(5*v) + 1*sin(9*v))*q^4.
		

References

  • J. V. Armitage and W. F. Eberlein, Elliptic Functions, London Mathematical Society, Student Texts 67, Cambridge University Press, 2006.
  • E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, fourth edition, reprinted, 1958, Cambridge at the University Press.

Crossrefs

Formula

T(n, m) = [x^(2*m+1)]Sum_{j=0..n} b(j, x)*a(n-j), with a(k) = A274621(k/2) if k is even and a(k) = 0 if k is odd, and b(j, x) = Sum_{r | 2*j+1} x^r = Sum_{k=1..A099774(j+1)} x^(A274658(j, k)), for j >= 0.
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