A085750 -id:A085750 - cp's OEIS frontend

A263646 Coefficients for an expansion of the Schwarzian derivative of a power series.

1, 1, 2, 1, 3, 1, 1, 4, 1, 1, 5, 1, 1, 1, 6, 1, 1, 1, 7, 1, 1, 1, 1, 8, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 14, 1, 1, 1, 1, 1, 1, 1, 15, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Tom Copeland, Oct 31 2015

Coefficients for an expansion of the Schwarzian derivative of a power series f(x), with f'(0) = 1, expressed in terms of an expansion of the natural logarithm of the derivative of the function G(x) = log(D f(x)) = Sum_{n >= 1} -F(n) x^n/n.

			Partitions by powers of x^n:
n=0: -(F2 + F1^2/2)
n=1: -(2 F3 + F1 F2)
n=2: -(3 F4 + F1 F3 + F2^2/2) =  -[3 F4 + (F1 F3 + F2 F2 + F3 F1) / 2]
n=3: -(4 F5 + F1 F4 + F2 F3)  =  -[4 F5 + (F1 F4 + F2 F3 + F3 F2 + F4 F1) / 2]
n=4: -(5 F6 + F1 F5 + F2 F4 + F3^2/2)
--------------------
Example series:
f(x)= (1/2) / (1-x)^2 = 1/2 + x + (3/2) x^2 + 2x^3 + (5/2)x^4 + ... .
log(f'(x)) = log(1 + 3x + 6x^2 + 10x^3 + ...) = 3x + 3 x^2/2 + 3 x^3/3  + ... .
Then F(n) = -3 for n>=1, and the Schwarzian derivative series is
S{f(x)} = - [(-3 + 3^2/2) + (-2*3 + 3^2) x + (-3*3 + 3^2 + 3^2/2) x^2 + ...] = -3/2 - 3x - (9/2)x^2 - 6x^3 - (15/2)x^4 - ... .
--------------------
The Schwarzian vanishes if and only if acting on a Moebius, or linear fractional, transformation. This corresponds to F(n) = (-1)^(n+1) 2 * d^n, where d is an arbitrary constant.
--------------------
Example polynomial:
f(x) = (x-x1)(x-x2)/-(x1+x2)
log(f'(x)) = log[1 - 2x/(x1+x2)] = Sum_{n>= 1} -(2/(x1+x2))^n x^n/n.
Then F(n) = (2/(x1+x2))^n, and the Schwarzian derivative series is
S{f(x)} = (Sum_{n>= 0} -6(n+1) 2^n (x/(x1+x2))^n) / (x1+x2)^2 = -6 / (x1+x2-2*x)^2 (cf. A001787 and A085750).

Cf. A263916, A001787, A085750, A051340.

print(sum(([n]+[1]*((n+1)//2) for n in range(1, 18)), [])) # Andrey Zabolotskiy, Mar 07 2024

Schwarzian{f(x)} = S{f(x)} = (D^3 f(x)) / (D f(x)) - (3/2) [(D^2 f(x)) / D f(x)]^2 = D [(D^2 f(x)) / D f(x)] - (1/2) [(D^2 f(x)) / D f(x)]^2 = D^2 log[D f(x)] - (1/2) [D log[D f(x)]]^2.
Then, with G(x) = log[D f(x)], S{f(x)} = D^2 G(x) - (1/2) [D G(x)]^2.
With f'(0) = 1, G(x) = log[D f(x)] = sum[n >= 1, -F(n) * x^n/n], and F(n) as Fn,
S{f(x)} = -[(F2 + F1^2/2) + (2 F3 + F1 F2) x + (3 F4 + F1 F3 + F2^2/2) x^2 + (4 F5 + F1 F4 + F2 F3) x^3 + (5 F6 + F1 F5 + F2 F4 + F3^2/2) x^4 + (6 F7 + F1 F6 + F2 F5 + F3 F4) x^5 + (7 F8 + F1 F7 + F2 F6 + F3 F5 + F4^2/2) x^6 + ...] .
This entry's a(m) are the numerators of the coefficients of the binary partitions in the brackets. For the singular partition of the integer n, the coefficient is (n-1); for the symmetric partition, 1/2; and for the rest, 1.
More symmetrically, x^2 S{f(x)}= - sum{n>=2, x^n [(n-1)F(n) + (1/2) sum(k=1 to n-1, F(n-k) F(k))]}.
With f(x)= c(0) + x + c(2) x^2 + ... , F(n) are given by the Faber polynomials of A263916: F(n) = Faber(n,2c(2),3c(3),..,(n+1)c(n+1)).

More terms from Tom Copeland, Oct 01 2016

A297477 Triangle read by rows: T(n, k) gives the coefficients of x^k of the characteristic polynomial P(n, x) of the n X n matrix M with entries M(i, j) = 1 if i = 1 or j = 1, -1 if i = j > 1, and 0 otherwise. T(0, 0) := 0.

Original entry on oeis.org

0, 1, -1, -2, 0, 1, 3, 3, -1, -1, -4, -8, -3, 2, 1, 5, 15, 14, 2, -3, -1, -6, -24, -35, -20, 0, 4, 1, 7, 35, 69, 65, 25, -3, -5, -1, -8, -48, -119, -154, -105, -28, 7, 6, 1, 9, 63, 188, 308, 294, 154, 28, -12, -7, -1, -10, -80, -279, -552, -672, -504, -210, -24, 18, 8, 1

Offset: 0

Mats Granvik, Dec 30 2017

The norm of the matrix M appears to be sqrt(n), where with norm is meant the eigenvalue of the largest magnitude, negative or positive. Row sums appear to be A085750 [see below for the proof].
Also the coefficients of the characteristic polynomial of the matrix defined by the recurrence: A(n, k) =  if n < k then if and(n > 1, k > 1) then Sum_{i=1..k-1} -A(k-i, n) else 0 else if and(n > 1, k > 1) then Sum_{i=1..n-1} -A(n-i, k) else 0.
By letting the upper summation indexes "k-1" and "n-1" in the recurrence above, change place with each other one gets the number theoretic matrix A191898, and it appears that the eigenvalue norm sqrt(n) of this matrix is a lower bound for the eigenvalue norm of matrix A191898 which in turn for n>10 appears to be close to A007917, the previous prime sequence. If the eigenvalue norm of matrix A191898 also can be proven to be less than n+1, then one could say that there is always a prime gap between sqrt(n) and n+1.
From Wolfdieter Lang, Feb 02 2018: (Start)
The characteristic polynomial P(n, x) = Det(M_n - x*1_n), with the n X n matrix M_n defined in the name and 1_n the n dimensional unit matrix, satisfies, after expanding the last row, the recurrence: P(n, x) = -z*P(n-1, x) + (-1)^(n-1)*z^(n-2), for n >= 2, and input P(1, x) = y, where y = 1-x and z = 1+x. The solution is P(n, x) = y*(-z)^(n-1) - (n-1)*(-z)^(n-2) = (-1)^n*(1 + x)^(n-2)*(x^2 - n), for n >= 1. After picking the coefficient of x^k this becomes the formula for T(n, k) given in the formula section.
The Determinant of M_n is P(n, 0) = T(n, 0) = (-1)^n*n = A181983(n).
The eigenvalues of M_n are +1 for n = 1 and for n >= 2 they are +sqrt(n), -sqrt(n), and n-2 times -1.
Therefore the spectral radius (absolute value of the maximal eigenvalue) is rho_n = sqrt(n), and the spectral norm of M_n (square root of the maximal eigenvalue of (M_n)^+ M_n) is also sqrt(n), for n >= 1. See the conjecture in the first comment above.
The square of the Frobenius norm (aka Hilbert-Schmidt norm) of M_n is max_{i,j=1..n} |M_n(i,j)|^2 = 3*n - 2 = A016777(n-1), for n >= 1.
The row sums are P(n, 1) = (-1)^(n-1)*(n-1)*2^(n-2) = A085750(n), for n >= 1, and for n=0 the row sum is 0. The alternating row sums are P(n, -1) = 2 for n=1, -1 for n = 2, and zero otherwise.
The column sequence (without leading zero) for k = 1 is (-1)^(n+1)*n*(n-2), for n >= 1, which is -A131386(n).  For k = 2 it is (-1)^n*(1 - n*binomial(n-2, 2)) for n >= 2 which is (-1)^n*A110427(n-1). Other columns follow from the formula for T(n, k). (End)

			The matrix for these characteristic polynomials starts:
  {
  {1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
  {1, -1, 0, 0, 0, 0, 0, 0, 0, 0},
  {1, 0, -1, 0, 0, 0, 0, 0, 0, 0},
  {1, 0, 0, -1, 0, 0, 0, 0, 0, 0},
  {1, 0, 0, 0, -1, 0, 0, 0, 0, 0},
  {1, 0, 0, 0, 0, -1, 0, 0, 0, 0},
  {1, 0, 0, 0, 0, 0, -1, 0, 0, 0},
  {1, 0, 0, 0, 0, 0, 0, -1, 0, 0},
  {1, 0, 0, 0, 0, 0, 0, 0, -1, 0},
  {1, 0, 0, 0, 0, 0, 0, 0, 0, -1}
  }
----------------------------------------------------------------------
The table T(n, k) begins:
  n\k   0   1    2    3    4    5    6   7   8   9  10 ...
  0:    0
  1:    1  -1
  2:   -2   0    1
  3:    3   3   -1   -1
  4:   -4  -8   -3    2    1
  5;    5  15   14    2   -3   -1
  6:   -6 -24  -35  -20    0    4    1
  7:    7  35   69   65   25   -3   -5  -1
  8:   -8 -48 -119 -154 -105  -28    7   6   1
  9:    9  63  188  308  294  154   28 -12  -7  -1
  10: -10 -80 -279 -552 -672 -504 -210 -24  18   8   1
  ... reformatted by _Wolfdieter Lang_, Feb 02 2018.

Cf. A016777, A085750 (row sums), A067998, A110427 (column k=2), -A131386 (column k=1), A181983 (Det M_n), A191898.

f:= proc(n) local M,P,lambda,k;
  M:= Matrix(n,n, proc(i,j) if i=1 or j=1 then 1 elif i=j then -1 else 0 fi end proc);
  P:= (-1)^n*LinearAlgebra:-CharacteristicPolynomial(M,lambda);
  seq(coeff(P,lambda,k),k=0..n)
end proc:
f(0):= 0:
for n from 0 to 10 do f(n) od; # Robert Israel, Feb 02 2018

Clear[A, x, t];
Table[t[n_, 1] = 1;
  t[1, k_] = 1;
  t[n_, k_] :=
   t[n, k] =
    If[n < k,
     If[And[n > 1, k > 1], Sum[-t[k - i, n], {i, 1, k - 1}], 0],
     If[And[n > 1, k > 1], Sum[-t[n - i, k], {i, 1, n - 1}], 0]];
  A = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}];
  CoefficientList[CharacteristicPolynomial[A, x], x], {nn, 1, 10}];
Flatten[%]

From Wolfdieter Lang, Feb 02 2018: (Start)
T(n, k) = [x*k] P(n, x), for n >= 1, with P(n, x) = Det(M_n - x*1_n), and the matrix M_n defined in the name (1_n is the n dimensional unit matrix). T(0, 0):= 0.
T(n, k) = (-1)^(n+1)*n for k = 0, (-1)^(n+1)*n*(n-2) for k = 1, and (-1)^n*(binomial(n-2, k-2) - n*binomial(n-2, k)) for k >= 2, with n >= 0 and 0 <= k <= n. T(n, k) = 0 for k > n. (End)

Edited by Wolfdieter Lang, Feb 02 2018

A357503 a(n) is the hafnian of the 2n X 2n symmetric matrix whose element (i,j) equals abs(i-j).

Original entry on oeis.org

1, 1, 8, 174, 7360, 512720, 53245824, 7713320944, 1486382446592, 367691598791424, 113570289012090880

Offset: 0

Stefano Spezia, Oct 01 2022

			a(2) = M_{1,2}*M_{3,4} + M_{1,3}*M_{2,4} + M_{1,4}*M_{2,3} = 8 is the hafnian of
    0, 1, 2, 3;
    1, 0, 1, 2;
    2, 1, 0, 1;
    3, 2, 1, 0.

Wikipedia, Hafnian
Wikipedia, Symmetric matrix

Cf. A049581, A085750 (determinant of M(n)), A085807 (permanent of M(n)), A094053 (super- and subdiagonal sums of M(n) in reversed order), A144216 (row- and column sums of M(n)), A338456.

M[i_, j_, n_]:=Part[Part[Table[Abs[r-c], {r, n}, {c, n}], i], j]; a[n_]:=Sum[Product[M[Part[PermutationList[s, 2n], 2i-1], Part[PermutationList[s, 2n], 2i], 2n], {i, n}], {s, SymmetricGroup[2n]//GroupElements}]/(n!*2^n); Array[a, 6, 0]

tm(n) = matrix(n, n, i, j, abs(i-j));
a(n) = my(m = tm(2*n), s=0); forperm([1..2*n], p, s += prod(j=1, n, m[p[2*j-1], p[2*j]]); ); s/(n!*2^n); \\ Michel Marcus, May 02 2023

a(6) from Michel Marcus, May 02 2023

a(7)-a(10) from Pontus von Brömssen, Oct 15 2023

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A263646 Coefficients for an expansion of the Schwarzian derivative of a power series.

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A297477 Triangle read by rows: T(n, k) gives the coefficients of x^k of the characteristic polynomial P(n, x) of the n X n matrix M with entries M(i, j) = 1 if i = 1 or j = 1, -1 if i = j > 1, and 0 otherwise. T(0, 0) := 0.

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A357503 a(n) is the hafnian of the 2n X 2n symmetric matrix whose element (i,j) equals abs(i-j).

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