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
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Squarefree
Programs
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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 *)
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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.
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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
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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
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