A365125 Put a positive charge at 0 and a negative charge at 1, then keep adding alternating charges at points of zero potential; this is the decimal expansion of the limit.

6, 8, 7, 8, 4, 1, 8, 1, 0, 3, 2, 8, 3, 8, 9, 2, 6, 3, 2, 7, 1, 3, 4, 4, 0, 4, 4, 0, 9, 8, 8, 3, 3, 4, 8, 6, 1, 1, 5, 8, 3, 9, 7, 9, 4, 8, 7, 6, 6, 8, 9, 5, 4, 1, 1, 7, 4, 7, 5, 8, 6, 6, 9, 4, 4, 1, 0, 7, 8, 5, 2, 8, 1, 7, 2, 1, 2, 4, 7, 5, 3, 8, 9, 1, 0, 8, 7, 9, 1, 2, 6, 5, 7, 8, 9, 7, 8, 5, 3, 6

Offset

0,1

Comments

The potential is inversely proportional to the distance: a positive charge at p has potential 1/abs(x-p). Starting from a positive charge at 0 and a negative charge at 1, the potential is zero at the midpoint 1/2, so another positive charge is added there. Then the potential is zero at 0.7886... and a negative charge is added there, and so on. The first few zero potential points are: 0.5, 0.7886..., 0.6325..., 0.7179..., 0.6714..., ... The limit is the constant of this sequence.

After initial points p_0 = 0 and p_1 = 1, each new zero point fits between previous two points and is the solution to Sum_{n>=0} 1/(x-p_n) = 0.

Links

Table of n, a(n) for n=0..99.

Rok Cestnik, Visualization

Example

0.68784181032838926327134404409883348611583979...

Mathematica

p={0, 1};
For[r=1, r<39, ++r, p=Append[p, x/.NSolve[{Sum[1/(x-p[[k]]), {k, 1, Length[p]}]==0, (x-p[[-1]])*(x-p[[-2]])<0}, x, WorkingPrecision -> 30][[1, 1]]]; ];
A365125 = RealDigits[Floor[10^10*p[[-1]]]][[1]]

Prog

(Python)
from decimal import Decimal, getcontext, ROUND_UP, ROUND_DOWN
getcontext().prec = 100
def nm(f, df, x):
    for i in range(10):
        x -= f(x)/df(x)
    return x
def flip_rounding():
    if getcontext().rounding == ROUND_UP: getcontext().rounding = ROUND_DOWN
    else: getcontext().rounding = ROUND_UP
def get_zero(vs, rounding):
    getcontext().rounding = rounding
    def p(x, v):
        flip_rounding(); t = x-v
        flip_rounding(); return 1/t
    def dp(x, v):
        flip_rounding(); t = x-v; t = t**2
        flip_rounding(); return -1/t
    def f(x): return sum(p(x, vs[n]) for n in range(len(vs)))
    def df(x): return sum(dp(x, vs[n]) for n in range(len(vs)))
    sign = -1 if rounding == ROUND_DOWN else 1
    return nm(f, df, (vs[-1]+vs[-2])/2+sign*abs(vs[-1]-vs[-2])/3)
v_lo = [Decimal(0), Decimal(1)]
v_up = [Decimal(0), Decimal(1)]
for r in range(150):
    v_lo.append(get_zero(v_lo, ROUND_DOWN))
    v_up.append(get_zero(v_up, ROUND_UP))
lower_bounds = [v_lo[i] for i in range(0, len(v_lo), 2)]
upper_bounds = [v_up[i] for i in range(1, len(v_up), 2)]
right = True
A365125 = [int(l) for l, u in zip(str(lower_bounds[-1])[2:], str(upper_bounds[-1])[2:]) if right and (right := (l == u))]

Crossrefs

Sequence in context: A225037 A157852 A327839 * A088608 A323984 A375067

Adjacent sequences: A365122 A365123 A365124 * A365126 A365127 A365128

Keyword

nonn,cons

Author

Rok Cestnik, Aug 22 2023

Status

approved