A355438 Lucas(a(n)) is least Lucas number beginning with n.

1, 0, 2, 3, 13, 23, 4, 14, 19, 24, 5, 10, 15, 39, 20, 25, 49, 6, 11, 35, 59, 16, 64, 21, 45, 69, 26, 50, 7, 31, 55, 12, 36, 60, 17, 151, 41, 65, 22, 156, 46, 70, 27, 94, 51, 252, 8, 32, 166, 56, 190, 13, 281, 37, 305, 61, 18, 85, 42, 109, 310, 66, 267, 23, 224, 47, 181, 71, 138, 339

Offset: 1

Author

Michel Marcus, Jul 02 2022

Keywords

• nonn
• base
• look

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..4000 from Michel Marcus)
• Ron Knott, Every number starts some Fibonacci Number, The Mathematical Magic of the Fibonacci Numbers.

Crossrefs

Cf. A000032, A355439,

Cf. A020344, A020345.

Programs

PARI

L(n) = real((2 + quadgen(5)) * quadgen(5)^n); \\ A000032
isok(k, dn) = my(dk=digits(L(k))); if (#dk >= #dn, Vec(dk, #dn) == dn);
a(n) = my(k=0, dn=digits(n)); while (!isok(k, dn), k++); k;

Python

def aupton(nn):
    ans, f, g, k = dict(), 2, 1, 0
    while len(ans) < nn:
        sf = str(f)
        for i in range(1, len(sf)+1):
            if int(sf[:i]) > nn:
                break
            if sf[:i] not in ans:
                ans[sf[:i]] = k
        f, g, k = g, f+g, k+1
    return [int(ans[str(i)]) for i in range(1, nn+1)]
print(aupton(70)) # Michael S. Branicky, Jul 08 2022

Formula

Trivially a(n) >= log_phi(n-1) for n > 1. Probably upper bounds are obtainable using the theory of linear forms in logarithms. - Charles R Greathouse IV, Jul 08 2022