This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
Select[Range[12000],#==IntegerReverse[#]&&PrimeNu[#]==PrimeOmega[#]==3&] (* Harvey P. Dale, Aug 29 2016 *)
The palindrome 996699 is a term since it has 3 factors 3 11 30203, all palindromic.
Number of divisors of palindrome number 22 with divisors 1, 2, 11 and 22 is 4 (palindrome number).
[n: n in [1..1000] | Intseq(n, 10) eq Reverse(Intseq(n, 10)) and Intseq(NumberOfDivisors(n), 10) eq Reverse(Intseq(NumberOfDivisors(n), 10))]
ispali:= proc(n) local L; L:= convert(n,base,10); L = ListTools:-Reverse(L) end proc: select(t -> ispali(t) and ispali(numtheory:-tau(t)), [$1..10000]); # Robert Israel, Mar 26 2019
Select[Range@ 600, And[PalindromeQ@ #, PalindromeQ@ DivisorSigma[0, #]] &] (* Michael De Vlieger, Mar 24 2019 *)
ispal(n) = my(d=digits(n)); Vecrev(d) == d; isok(n) = ispal(n) && ispal(numdiv(n)); \\ Michel Marcus, Mar 23 2019
a(5) = 117 is a term because 117 = 3^2 * 13 has 3 prime factors, counted with multiplicity, and so does its reversal 711 = 3^2 * 79.
rev:= proc(n) local L,i; L:= convert(n,base,10); add(L[-i]*10^(i-1),i=1..nops(L)) end proc: select(t -> numtheory:-bigomega(t) = 3 and numtheory:-bigomega(rev(t))=3, [$1..10000]);
Select[Range[710], PrimeOmega[#]==3&&PrimeOmega[FromDigits[Reverse[IntegerDigits[#]]]]==3&] (* Stefano Spezia, Nov 07 2023 *)
from sympy import factorint def tp(n): return sum(factorint(n).values()) == 3 def ok(n): return tp(n) and tp(int(str(n)[::-1])) print([k for k in range(10**3) if ok(k)]) # Michael S. Branicky, Nov 21 2023
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 *)
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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