A035134 Squarefree composite palindromes.

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

Robert Israel, Table of n, a(n) for n = 1..10000

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

Cf. A002113, A005117, A035132, A035135.

Sequence in context: A069824 A015709 A020195 * A075799 A046392 A046408

Adjacent sequences: A035131 A035132 A035133 * A035135 A035136 A035137

Keyword

nonn,base

Author

Patrick De Geest, Nov 15 1998

Status

approved