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

1,3

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.

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

Prog

(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

Crossrefs

Cf. A000032, A355439,

Cf. A020344, A020345.

Sequence in context: A077248 A282342 A137248 * A358427 A136260 A296932

Adjacent sequences: A355435 A355436 A355437 * A355439 A355440 A355441

Keyword

nonn,base,look

Author

Michel Marcus, Jul 02 2022

Status

approved