A343128
Numbers k such that prime(prime(prime(k))) ends in k.
Original entry on oeis.org
7, 229, 417, 657, 26203, 32553, 50971, 93487, 231221, 17064941, 54784601, 93007099, 981668491, 16040988367
Offset: 1
417 is a term since prime(prime(prime(417))) = 302417.
26203 is a term since prime(prime(prime(26203))) = 73226203.
17064941 is a term since prime(prime(prime(17064941))) = 169217064941.
prime(prime(prime(54784601))) = 647454784601, prime(prime(prime(93007099))) = 1185993007099, prime(prime(prime(981668491))) = 17148981668491, prime(prime(prime(16040988367))) = 390416040988367. - _Chai Wah Wu_, Apr 14 2021
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.
Original entry on oeis.org
1, 7, 6447, 7, 1, 1, 69, 9, 1, 1, 1, 7, 1, 1
Offset: 0
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.
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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
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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
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