A320122 Numbers that are not Keith numbers in any base.

12, 30, 390, 1170, 1200, 1560, 2340, 2760, 3120, 3900, 4680, 6120, 6240, 7680, 7800, 8460, 10020, 10140, 10950, 11580, 15090, 15480, 17160, 17580, 18360, 19140, 20280, 20700, 20940, 21480, 23040, 23280, 24060, 24210, 24960, 26550, 28740, 29250, 29520, 29670, 30060, 31080, 32400

Offset: 1

Author

Robert FERREOL, Oct 06 2018

Keywords

• nonn
• base

Comments

A number N >= 2 is a Keith number in a base b <= N if the Fibonacci sequence u(i) whose initial terms are the t digits of N in the base b, and later terms are given by rule that u(i) = sum of t previous terms, contains N itself. Here a(n) is the n-th number N that is not a Keith number in any base b <= N.

Examples

			a(1) = 12 because 12 is not a Keith number in any base from 2 to 12, while all previous numbers are in some base.
For example, with b = 2, the sequence is : 1, 1, 0, 0, 2, 3, 5, 10, 20, ...; it doesn't contain 12. See A251703.
		

Links

• Arie Bos, Inventory of n-step Fibonacci sequences, 2015.

Crossrefs

Cf. A007629 (Keith numbers in base 10).

Programs

Maple

fibo:=proc(n, b) local L,m,M,k:
L:=convert(n,base,b):m:=nops(L):M:=seq(L[m+1-k],k=1..m):
while M[m]
				

PARI

iskb(n, b) = if(nA007629
isok(n) = if (n<=2, 0, for(b=2, n-1, if (iskb(n, b), return(0))); return (1)); \\ Michel Marcus, Oct 08 2018

Python

def digits(n, b):
    r = []
    m = n
    while m > 0:
        r = [m % b] + r
        m = m // b
    return r
def fibo(n, b):
    L = digits(n, b)
    m = len(L) - 1
    while L[m] < n:
        L.append(sum(k for k in L))
        L.pop(0)
    return L[m] == n
def test(n):
    for b in range(2, n + 1):
        if fibo(n, b):
            return True
    return False
print([n for n in range(2, 2001) if not test(n)])

Extensions

More terms from Michel Marcus, Oct 08 2018