

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



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
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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: A325524 A010269 A077450 * A214878 A086138 A170822
Adjacent sequences: A328327 A328328 A328329 * A328331 A328332 A328333


KEYWORD

nonn,base


AUTHOR

Karl Aughton, Oct 12 2019


STATUS

approved



