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A035134
Squarefree composite palindromes.
2
6, 22, 33, 55, 66, 77, 111, 141, 161, 202, 222, 262, 282, 303, 323, 393, 434, 454, 474, 494, 505, 515, 535, 545, 555, 565, 595, 606, 626, 646, 707, 717, 737, 767, 777, 818, 838, 858, 878, 898, 939, 949, 959, 969, 979, 989, 1001, 1111, 1221, 1441, 1551, 1661
OFFSET
1,1
COMMENTS
Palindromes with at least two and all distinct prime factors.
LINKS
Eric Weisstein's World of Mathematics, Squarefree
MAPLE
N:= 4: # to get all terms with <= N digits
revdigs:= proc(n) local L, j, nL;
L:= convert(n, base, 10); nL:= nops(L);
add(L[j]*10^(nL-j), j=1..nL);
end proc:
palis:= $0..9:
for d from 2 to N do
if d::even then
palis:= palis, seq(x*10^(d/2)+revdigs(x), x=10^(d/2-1)..10^(d/2)-1)
else
palis:= palis, seq(seq(x*10^((d+1)/2)+y*10^((d-1)/2)+revdigs(x), y=0..9), x=10^((d-3)/2)..10^((d-1)/2)-1);
fi
od:
select(t -> not(isprime(t)) and numtheory:-issqrfree(t), [palis][3..-1]): # Robert Israel, Sep 18 2016
MATHEMATICA
sqfQ[n_]:=Max[Transpose[FactorInteger[n]][[2]]]<=1; palQ[n_]:=FromDigits[Reverse[IntegerDigits[n]]]==n; Select[Range[2, 1662], !PrimeQ[#] && sqfQ[#] && palQ[#] &] (* Jayanta Basu, May 12 2013 *)
Select[Range[2000], PalindromeQ[#]&&SquareFreeQ[#]&&CompositeQ[#]&] (* Harvey P. Dale, Apr 10 2022 *)
PROG
(PARI) isA002113(n)=n=digits(n); for(i=1, #n\2, if(n[i]!=n[#n+1-i], return(0))); 1;
is(n) = n>1 && isA002113(n) && issquarefree(n) && !isprime(n) \\ Altug Alkan, Sep 19 2016
\\ in and output digits as a vector.
(PARI) nxtA002113(n)={my(d=n); i=(#d+1)\2; while(i&&d[i]==9, d[i]=0; d[#d+1-i]=0; i--); if(i, d[i]++; d[#d+1-i]=d[i], d=vector(#d+1); d[1]=d[#d]=1); d}\\sum(i=1, #d, 10^(#d-i)*d[i])}
\\ all terms up to n digits
lista(n) = {my(p = [6], l=List(), sp, i); while(#p <= n, sp = sum(i=1, #p, p[i]*10^(#p-i)); if(issquarefree(sp)&&!isprime(sp), listput(l, sp)); p=nxtA002113(p)); l} \\ David A. Corneth, Sep 19 2016
(Python)
from itertools import product
from sympy import factorint, isprime
def pals(d, base=10): # all d-digit palindromes
digits = "".join(str(i) for i in range(base))
for p in product(digits, repeat=d//2):
if d > 1 and p[0] == "0": continue
left = "".join(p); right = left[::-1]
for mid in [[""], digits][d%2]: yield int(left + mid + right)
def ok(pal): f = factorint(pal); return len(f)>1 and all(f[p]<2 for p in f)
print(list(filter(ok, (p for d in range(1, 5) for p in pals(d) if ok(p))))) # Michael S. Branicky, Jun 22 2021
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Nov 15 1998
STATUS
approved