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A033620
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Numbers all of whose prime factors are palindromes.
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9
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1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 27, 28, 30, 32, 33, 35, 36, 40, 42, 44, 45, 48, 49, 50, 54, 55, 56, 60, 63, 64, 66, 70, 72, 75, 77, 80, 81, 84, 88, 90, 96, 98, 99, 100, 101, 105, 108, 110, 112, 120, 121, 125, 126, 128, 131
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OFFSET
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1,2
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COMMENTS
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LINKS
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FORMULA
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EXAMPLE
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10 = 2 * 5 is a term since both 2 and 5 are palindromes.
110 = 2 * 5 * 11 is a term since 2, 5 and 11 are palindromes.
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MAPLE
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N:= 5: # to get all terms of up to N digits
digrev:= proc(t) local L; L:= convert(t, base, 10);
add(L[-i-1]*10^i, i=0..nops(L)-1);
end proc:
PPrimes:= [2, 3, 5, 7, 11]:
for d from 3 to N by 2 do
m:= (d-1)/2;
PPrimes:= PPrimes, select(isprime, [seq(seq(n*10^(m+1)+y*10^m+digrev(n), y=0..9), n=10^(m-1)..10^m-1)]);
od:
PPrimes:= map(op, [PPrimes]):
M:= 10^N:
B:= Vector(M);
B[1]:= 1:
for p in PPrimes do
for k from 1 to floor(log[p](M)) do
R:= [$1..floor(M/p^k)];
B[p^k*R] := B[p^k*R] + B[R]
od
od:
# alternative
isA033620:= proc(n)
for d in numtheory[factorset](n) do
if not isA002113(op(1, d)) then
return false;
end if;
end do;
true ;
end proc:
for n from 1 to 300 do
if isA033620(n) then
printf("%d, ", n) ;
end if;
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MATHEMATICA
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palQ[n_]:=Reverse[x=IntegerDigits[n]]==x; Select[Range[131], And@@palQ/@First/@FactorInteger[#]&] (* Jayanta Basu, Jun 05 2013 *)
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PROG
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(Haskell)
a033620 n = a033620_list !! (n-1)
a033620_list = filter chi [1..] where
chi n = a136522 spf == 1 && (n' == 1 || chi n') where
n' = n `div` spf
(PARI) ispal(n)=n=digits(n); for(i=1, #n\2, if(n[i]!=n[#n+1-i], return(0))); 1
(Python)
from sympy import isprime, primefactors
def pal(n): s = str(n); return s == s[::-1]
def ok(n): return all(pal(f) for f in primefactors(n))
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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