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.

Showing 1-4 of 4 results.

A273166 Row sums of triangle A274659.

Original entry on oeis.org

1, 2, 0, -2, 2, 4, -4, -6, 6, 10, -8, -16, 14, 24, -24, -36, 36, 56, -52, -82, 78, 116, -116, -168, 166, 242, -232, -338, 328, 468, -464, -648, 640, 892, -872, -1214, 1192, 1636, -1620, -2198, 2176
Offset: 0

Views

Author

Wolfdieter Lang, Aug 04 2016

Keywords

Crossrefs

Cf. A274659.

Formula

a(n) = Sum_{m=0..n} A274659(n, m), n >= 0.

A274658 Irregular triangle which lists in row n the divisors of 2*n+1.

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
Offset: 0

Views

Author

Wolfdieter Lang, Jul 18 2016

Keywords

Comments

The length of row n is A099774(n+1).
This gives the odd numbered rows of the irregular triangle A027750.
The row sums are given in A008438.
The entries of row n appear, for instance, as arguments of sin in the Fourier expansion of Jacobi's elliptic function sn in the second factor Sum_{n>=0} (q^n/(1-q^(2*n+1)))*sin((2*n+1)*v) as coefficients of q^n. See e.g., the formula in Abramowitz-Stegun, p. 575, 16.23.1 (or 16.23.2 for cn but with signs). See also A274659.

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 second factor in the sn formula has as q^4 coefficient: sin(1*v) + sin(3*v) + sin(9*v).

Links

Crossrefs

Cf. A008438 (row sums), A027750, A099774 (row lengths), A274659.

Programs

  • Mathematica
    Table[Divisors[2 n + 1], {n, 0, 22}] // Flatten (* Michael De Vlieger, Jul 18 2016 *)
  • PARI
    row(n) = divisors(2*n+1); \\ Amiram Eldar, May 02 2025

Formula

T(n, k) = k-th divisor of 2*n+1 in increasing order.

A274661 Triangle read by rows: T(n, m) gives the m-th contribution T(n, m)*cos((2*m+1)*v) to the coefficient of q^n in the Fourier expansion of Jacobi's elliptic cn(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, -17, 23, 6, -18, -1, 11, 0, -6, 0, 3, 0, -2, 0, 1
Offset: 0

Views

Author

Wolfdieter Lang, Jul 27 2016

Keywords

Comments

If one takes the row polynomials as P(n, x) = Sum_{m=0..n} T(n, m)*x^m, n >= 0, Jacobi's elliptic function cn(u|k) in terms of the new variables v and q becomes cn(u|k) = Sum_{n>=0} P(n, x)*q^n, if in P(n, x) one replaces x^j by cos((2*j+1)*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 cn(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 0.5959766014 to be compared with cn(1|sqrt(1/2)) approximately 0.5959765676.
For the derivation of the Fourier series formula of cn 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 sn see A274659 (differently signed triangle).
The sum of entries in row n is P(n, 1) = A000007(n): 1, repeat 0. Proof: due to the g.f. identity (from the convolution)
Sum_{n >= 0} x^n/(1 + x^(2*n+1)) = (Sum_{n >= 0} x^(n*(n+1)))^2.
This is proved by bisecting the g.f. on the l.h.s. which generates c(n, 1) = (-1)^n*Sum_{2*r+1 | 2*n+1} (-1)^n. The part with n = 2*k+1 vanishes due to r_2(4*k+1)/4 = 0, where r_2(n) is the number of solutions of n as a sum of two squares. See the Grosswald reference. The part with n = 2*k becomes Sum_{k >= 0} x^(2*k) r_2(4*k+1)/4 which is the r.h.s. See A008441, the Broadhurst Oct 20 2002 comment.
For another version of this expansion of cn see A275791.
See also the W. Lang link, eqs. (43) and (44). - Wolfdieter Lang, Aug 26 2016

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
...
n = 4: c(0, x)*a(4) + c(2, x)*a(2) + c(4, x)*a(0) = (+x^1)*3 +  (+x^1 + x^5)*(-2) + (+x^1 - x^3 + x^9)*1 = +2*x^1 - x^3 - 2*x^5 + 0*x^7 + x^9. Hence row n=4 is 2, -1, -2, 0, 1.
From A274660, row n = 4: c(4, x) = +x^1 - x^3 +x^9.
n = 4: P(4, x) = 2 - 1*x^1 - 2*x^2 + 1*x^4, that is the contribution of order q^4 to cn in the new variables is (2*cos(v)  - 1*cos(3*v) - 2*cos(5*v) + 1*cos(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. Grosswald, Representations of Integers as Sums of Squares. Springer-Verlag, NY, 1985, p. 15, Theorem 3.
  • E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, fourth edition, reprinted, 1958, Cambridge at the University Press.

Links

Crossrefs

Cf. A000007, A099774, A274621, A274659, A274660, A275791.

Formula

T(n, m) = [x^(2*m+1)]Sum_{j=0..n} c(j, x)*a(n-j), with a(k) = A274621(k/2) if k is even and a(k) = 0 if k is odd, and c(j, x) = (-1)^j*Sum_{2*r+1 | 2*j+1} (-1)^r*x^(2*r+1) = Sum_{k=1..A099774(j+1)} sign(A274660(j, k))*x^(abs(A274660(j, k))), for j >= 0.

A274662 Triangle T(n, m) appearing in the expansion of Jacobi's elliptic function sn(u, k) divided by sin(v) in terms of the Jacobi nome q and even powers of 2*cos(v), with v = u/((2/Pi)*K(k)).

Original entry on oeis.org

1, 0, 1, 0, -3, 1, 0, 4, -5, 1, 0, -3, 13, -7, 1, 0, 6, -25, 26, -9, 1, 0, -12, 43, -70, 43, -11, 1, 0, 8, -70, 157, -147, 64, -13, 1, 0, -3, 109, -315, 408, -264, 89, -15, 1, 0, 13, -155, 582, -984, 872, -429, 118, -17, 1, 0, -18, 201, -1001, 2142, -2464, 1641, -650, 151, -19, 1
Offset: 0

Views

Author

Wolfdieter Lang, Aug 08 2016

Keywords

Comments

The representation of Jacobi's elliptic sn(u, k) function in terms of quotients of theta functions of the variables q (Jacobi nome) and v = u/((2/Pi)*K(k)) with the real quarter period K is
sn(u, k) = (theta_3(0, q)/theta_2(0, q)) * (theta_1(v, q)/theta_4(v, q)).
This can be written either in terms of infinite sums or products. (see e.g., Tricomi, p. 176, eq. (3.87), p. 156, eq. (3.51), p. 167, eq. (3.71) with (3.71'), p. 173, eq. (3.81)).
The sums representation involves sin((2*n+1)*v) and cos(2*n*v) functions. Using Chebyshev S and T polynomial (A049310 and A053120) one can write sn(u, k)/sin(v) = Sum_{n >= 0} q^n*Sum_{m = 0..n} T(n, m) * (2*cos(v))^(2*m).
The product representation involves directly (2*cos(v))^2 powers in the q expansion:
sn(u, k)/sin(v) = Product_{n >= 1} (1 - (q^(2*n)/(1 + q^(2*n))^2)*(2*cos(v))^2) / (1 - (q^(2*n-1)/(1 + q^(2*n-1))^2)*(2*cos(v))^2) = Sum_{n >=0} q^n * Sum_{m = 1..n} T(n, m)*(2*cos(v))^(2*m).
This sn expansion in the v and q variables is used in the scaled phase space coordinate qhat(v, q) of the plane pendulum. See A275790.
An alternative expansion of sn in the variables v and q is given in A274659.
See also the W. Lang link, equations (52) and (53).

Examples

			The triangle T(n, m) begins:
n\m 0   1    2    3    4    5    6   7   8 9
0:  1
1:  0   1
2:  0  -3    1
3:  0   4   -5    1
4:  0  -3   13   -7    1
5:  0   6  -25   26   -9    1
6:  0 -12   43  -70   43  -11    1
7:  0   8  -70  157 -147   64  -13   1
8:  0  -3  109 -315  408 -264   89 -15   1
9:  0  13 -155  582 -984  872 -429 118 -17 1
...
row n=10: 0 -18 201 -1001 2142 -2464 1641 -650 151 -19 1
...
n=4: the q^4 term of sn(u, k)/sin(v) is -3*(2*cos(v))^2 + 13*(2*cos(v))^4 - 7*(2*cos(v))^6 + (2*cos(v))^8.
One can check the identity for example for u = 1 and k = sqrt(1/2), belonging to v = 0.8472130848  and q = 0.04321391815 (Maple 10 digits), with the result from Maple's sn function sn(1, sqrt(1/2)) = 0.8030018249 (10 digits). If one takes the expansion up to q^4 inclusive one obtains .8030012888 (10 digits).

References

  • F. Tricomi, Elliptische Funktionen (German translation by M. Krafft of: Funzioni ellittiche), Akademische Verlagsgesellschaft Geest & Portig K.-G., Leipzig, 1948.

Links

Crossrefs

Cf. A002103, A005797, A038534/A056982, A049310, A053120, A274659, A275790.

Formula

sn(u, k) = sin(v)*Sum_{n >= 0} q^n*Sum_{m = 0..n} T(n, m)*(2*cos(v))^(2*m), becoming an identity when q, the Jacobi nome, is replaced by exp(-Pi*K'(k)/K(k)) and v by u/((2/Pi)*K(k)) with the real and imaginary quarter periods K' and K, respectively. For the expansions of q = q(k) see A005797 or better A002103 for q = q((1-k^2)^(1/4)), and for (2/Pi)*K(k) see A038534 / A056982.
Showing 1-4 of 4 results.

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