A328330 Let f(n) be the number of segments shown on a digital calculator to display n. Then a(n) is the number of terms in the sequence formed by iteration n -> f(n) until n = f(n).

3, 2, 2, 1, 1, 1, 3, 4, 2, 5, 2, 4, 4, 2, 4, 5, 2, 3, 5, 3, 4, 6, 6, 3, 6, 3, 5, 5, 3, 3, 4, 6, 6, 3, 6, 3, 5, 5, 3, 6, 2, 3, 3, 5, 3, 6, 4, 3, 6, 3, 4, 6, 6, 3, 6, 3, 5, 5, 3, 5, 5, 3, 3, 6, 3, 5, 3, 5, 5, 3, 2, 5, 5, 4, 5, 3, 2, 6, 3, 5, 3, 5, 5, 3, 5, 5, 6, 3

Offset

1,1

Comments

Type n on a calculator and count the segments on a calculator display that forms the number. Iterate until you reach a fixed point: 4, 5 or 6. a(n) is the length of the chain.

Links

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Example

The 12th term is 4 as 12 -> 7 -> 3 -> 5 is a chain of 4.
a(8) = 4 because 8 -> 7 -> 3 -> 5 is a chain of length 4.

Prog

(PARI) a(n) = {my(res = 0, on = n, nn = n, cn); while(nn != cn, nn = f(on); cn = on; on = nn; res++); res}
f(n) = {my(d = digits(n), x = [6, 2, 5, 5, 4, 5, 6, 3, 7, 6]); sum(i = 1, #d, x[d[i]+1])} \\ David A. Corneth, Oct 12 2019
(Python)
def f(n):
    return sum((6, 2, 5, 5, 4, 5, 6, 3, 7, 6)[int(d)] for d in str(n))
def A328330(n):
    c, m = 1, f(n)
    while m != n:
        c += 1
        n, m = m, f(m)
    return c # Chai Wah Wu, Oct 27 2020

Crossrefs

Cf. A006942, A338255.

Sequence in context: A010269 A366911 A077450 * A214878 A086138 A170822

Adjacent sequences: A328327 A328328 A328329 * A328331 A328332 A328333

Keyword

nonn,base

Author

Karl Aughton, Oct 12 2019

Status

approved