[go: up one dir, main page]

login
A355438
Lucas(a(n)) is least Lucas number beginning with n.
2
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
KEYWORD
nonn,base,look
AUTHOR
Michel Marcus, Jul 02 2022
STATUS
approved