A132365 Least number k such that the Lucas number A000032(k) contains n.

1, 0, 2, 3, 13, 9, 4, 6, 7, 24, 5, 10, 15, 26, 20, 25, 49, 6, 11, 16, 13, 12, 10, 21, 45, 40, 20, 36, 7, 31, 50, 12, 35, 19, 17, 15, 41, 36, 22, 23, 39, 39, 14, 21, 41, 60, 8, 32, 19, 56, 20, 13, 45, 37, 51, 44, 17, 56, 42, 22, 25, 62, 35, 15, 71, 47, 25, 24, 43, 32, 17, 45, 49, 38

Offset: 1

Author

Jonathan Vos Post, Nov 08 2007

Keywords

• base
• easy
• nonn

Comments

Values such that a(n)=n (fixed points) are 1, 62. I don't know if there are any other fixed points. The first time a(n)=a(n+1) occurs because L(39)=141422324 which includes both 41 and 42 (and later on in the sequence, because it contains 141 and 142). [Sean A. Irvine, Nov 30 2009]

Links

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

Crossrefs

Cf. A000032, A000045, A000350, A050816, A038546, A052000, A023172.

Programs

Python

def A132365(n):
    a, b, m, s = 2, 1, 0, str(n)
    while True:
        if s in str(a):
            return m
        m += 1
        a, b = b, a+b # Chai Wah Wu, Jun 06 2017

Formula

a(n) = Min{k such that A000032(k) contains the decimal digit substring which represents the integer n}.

Extensions

Incorrect comment removed by Sean A. Irvine, Nov 30 2009

Corrected and extended by Sean A. Irvine, Nov 30 2009