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A343145
a(n) is the least positive number k such that applying x->prime(x) n times results in a number that ends in k.
1
1, 7, 6447, 7, 1, 1, 69, 9, 1, 1, 1, 7, 1, 1
OFFSET
0,2
EXAMPLE
a(1) = 7 since prime(7) = 17 which ends in 7.
a(2) = 6447 since prime(prime(6447)) = 806447 which ends in 6447.
a(3) = 7 since prime(prime(prime(7)))=277 which ends in 7.
a(4) = 1 since prime(prime(prime(prime(1)))) = 11 which ends in 1.
a(5) = 1 since prime(prime(prime(prime(prime(1))))) = 31 which ends in 1.
a(6) = 69 since prime(prime(prime(prime(prime(prime(69)))))) = 54615469 which ends in 69.
a(7) = 9 since prime(prime(prime(prime(prime(prime(prime(9))))))) = 4535189 which ends in 9.
PROG
(Python)
def A343145(n):
k = 1
while True:
m = k
for _ in range(n):
m = prime(m)
if m % 10**(len(str(k))) == k:
return k
k += 1
while not (k % 2 and k % 5):
k += 1
(PARI) primemap(n, tms) = my(x=n); for(i=1, tms, x=prime(x)); x
enddigits(n, len) = ((n/10^len - floor(n/10^len)) * 10^len)
a(n) = for(k=1, oo, my(x=k); if(enddigits(primemap(k, n), #Str(k))==k, return(k))) \\ Felix Fröhlich, Apr 14 2021
CROSSREFS
KEYWORD
nonn,hard,base,more
AUTHOR
Chai Wah Wu, Apr 06 2021
STATUS
approved