cp's OEIS Frontend

Denominators in series for sqrt(Pi/4)*erf(x): sqrt(Pi/4)*erf(x)= x/1 - x^3/3 + x^5/10 - x^7/42 + x^9/216 -+ ...

Appears to be the BinomialMean transform of A000354 (after truncating the first term of A000354). (See A075271 for the definition of BinomialMean.) - John W. Layman, Apr 16 2003

Number of permutations p of {1,2,...,n+2} such that max|p(i)-i|=n+1. Example: a(1)=3 since only the permutations 312,231 and 321 of {1,2,3} satisfy the given condition. - Emeric Deutsch, Jun 04 2003

Stirling transform of A000670(n+1) = [3, 13, 75, 541, ...] is a(n) = [3, 10, 42, 216, ...]. - Michael Somos, Mar 04 2004

Stirling transform of a(n) = [2, 10, 42, 216, ...] is A052875(n+1) = [2, 12, 74, ...]. - Michael Somos, Mar 04 2004

A related sequence also arises in evaluating indefinite integrals of sec(x)^(2k+1), k=0, 1, 2, ... Letting u = sec(x) and d = sqrt(u^2-1), one obtains a(0) = log(u+d) 2*k*a(k) = (2*k-1)*u^(2*k-1)*d + a(k-1). Viewing these as polynomials in u gives 2^k*k!*a(k) = a(0) + d*Sum(i=0..k-1){ (2*i+1)*i!*2^i*u^(2*i+1) }, which is easily proved by induction. Apart from the power of 2, which could be incorporated into the definition of u (or by looking at erf(ix/2)/ i (i=sqrt(-1)), the sum's coefficients form our series and are the reciprocals of the power series terms for -sqrt(-Pi/4)*erf(ix/2)). This yields a direct but somewhat mysterious relationship between the power series of erf(x) and integrals involving sec(x). - William A. Huber (whuber(AT)quantdec.com), Mar 14 2002

When written in factoradic ("factorial base"), this sequence from a(1) onwards gives the smallest number containing two adjacent digits, increasing when read from left to right, whose difference is n-1. - Christian Perfect, May 03 2016

a(n-1)^2 is the number of permutations p of [1..2n] such that Sum_{i=1..2n} abs(p(i)-i) = 2n^2-2. - Fang Lixing, Dec 07 2018

A standard series for the calculation of coordinates on a clothoid (also called cornuspiral):

x = s*(a(0) - (tau^2/a(2)) + (tau^4/a(4)) - (tau^6/a(6)) + ...)

y = s*((tau/a(1)) + (tau^3/a(3)) - (tau^5/a(5)) + ...).

s is the arclength from the clothoids origin to the desired point p(x,y). The tangent at the clothoids origin intersects with the tangent at the point p(x,y) with an angle of tau. - Thomas Scheuerle, Oct 13 2021

a(n) = P_n(1) where P_n(x) is the Pidduck polynomials. - Michael Somos, May 27 2023

Solution of the discrete parking problem when infinite lattice randomly filled with 3-length segments.

Solution of the discrete parking problem when infinite lattice randomly filled with 2-length segments is equal to 1-1/e^2 (see A219863).

Also, the limit of a(n) = (3 + 2*(n-3)*a(n-3) + (n-1)*(n-3)*a(n-1))/(n*(n-2)); a(0) = 0; a(1) = 0; a(2) = 0 as n tends to infinity.

If the length of the segments that randomly cover infinite lattice tends to infinity, then the fraction of occupied places is equal to Rényi's parking constant (see A050996).

The alternating sum of reciprocals of the factorials of the positive half-integers.