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Solving the nome q for the parameter m gives m(q) = (theta_2^4(q))/(theta_3^4(q)) (1) = (16eta^8(1/2tau)eta^(16)(2tau))/(eta^(24)(tau)), (2) where theta_i(q)=theta_i(0,q) is ...

A polynomial is said to be irreducible if it cannot be factored into nontrivial polynomials over the same field. For example, in the field of rational polynomials Q[x] (i.e., ...

An isolated point of a graph is a node of degree 0 (Hartsfield and Ringel 1990, p. 8; Harary 1994, p. 15; D'Angelo and West 2000, p. 212; West 2000, p. 22). The number of ...

The sequence of numbers {j_n} giving the number of digits in the three-fold power tower n^(n^n). The values of n^(n^n) for n=1, 2, ... are 1, 16, 7625597484987, ... (OEIS ...

Define the juggler sequence for a positive integer a_1=n as the sequence of numbers produced by the iteration a_(k+1)={|_a_k^(1/2)_| for even a_k; |_a_k^(3/2)_| for odd a_k, ...

d sum OEIS 0 23.10344 A082839 1 16.17696 A082830 2 19.25735 A082831 3 20.56987 A082832 4 21.32746 A082833 5 21.83460 A082834 6 22.20559 A082835 7 22.49347 A082836 8 22.72636 ...

Klee's identity is the binomial sum sum_(k=0)^n(-1)^k(n; k)(n+k; m)=(-1)^n(n; m-n), where (n; k) is a binomial coefficient. For m=0, 1, ... and n=0, 1,..., the following ...

The problem of determining how many nonattacking knights K(n) can be placed on an n×n chessboard. For n=8, the solution is 32 (illustrated above). In general, the solutions ...

A tree with its nodes labeled. The number of labeled trees on n nodes is n^(n-2), the first few values of which are 1, 1, 3, 16, 125, 1296, ... (OEIS A000272). Cayley (1889) ...

Arrange copies of the n digits 1, ..., n such that there is one digit between the 1s, two digits between the 2s, etc. For example, the unique (modulo reversal) n=3 solution ...