A046384
Palindromes with exactly 10 palindromic prime factors (counted with multiplicity).
Original entry on oeis.org
8448, 46464, 65856, 69696, 255552, 426624, 639936, 2346432, 21544512, 28898289882, 422868868224, 475119911574, 2323896983232, 2325778775232, 2578986898752, 4362464642634, 5269476749625, 21337599573312, 21354877845312, 23487999978432, 23679788797632
Offset: 1
The palindrome 28898289882 is a term since it has 10 factors 2 3^2 7^3 11^2 101 383, 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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