A046383
Palindromes with exactly 9 palindromic prime factors (counted with multiplicity).
Original entry on oeis.org
4224, 6336, 23232, 213312, 276672, 21755712, 23888832, 63577536, 63788736, 65499456, 67566576, 286121682, 2118338112, 2158888512, 2365885632, 2536556352, 2559999552, 2627117262, 2745665472, 5781111875, 6338118336, 42282628224, 52173037125, 80219291208
Offset: 1
The palindrome 6338118336 is a term since it has 9 factors 2^6 3 11 3001003, all palindromic.
A348050
Palindromes setting a new record of their number of prime divisors A001222.
Original entry on oeis.org
1, 2, 4, 8, 88, 252, 2112, 4224, 8448, 44544, 48384, 405504, 4091904, 405909504, 677707776, 4285005824, 21128282112, 29142024192, 4815463645184, 445488555884544, 27874867776847872, 40539458585493504, 63556806860865536, 840261068860162048, 4870324782874230784
Offset: 1
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m=0;lst=Union@Flatten[Table[{FromDigits@Join[s=IntegerDigits@n,Reverse@s],FromDigits@Join[w=IntegerDigits@n,Rest@Reverse@w]},{n,10^5}]];Do[t=PrimeOmega@lst[[n]];If[t>m,Print@lst[[n]];m=t],{n,Length@lst}] (* Giorgos Kalogeropoulos, Oct 25 2021 *)
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from sympy import factorint
from itertools import product
def palsthru(maxdigits):
midrange = [[""], [str(i) for i in range(10)]]
for digits in range(1, maxdigits+1):
for p in product("0123456789", repeat=digits//2):
left = "".join(p)
if len(left) and left[0] == '0': continue
for middle in midrange[digits%2]:
yield int(left+middle+left[::-1])
def afind(maxdigits):
record = -1
for p in palsthru(maxdigits):
f = factorint(p, multiple=True)
if p > 0 and len(f) > record:
record = len(f)
print(p, end=", ")
afind(10) # Michael S. Branicky, Oct 25 2021
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