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.
n A262627(n) base-10 representation 1 0 0 2 101 5 3 11001010011 1619
s = {0}; base = 2; z = 20; Do[NestWhile[# + 1 &, 1, ! PrimeQ[tmp = FromDigits[Join[#, IntegerDigits[Last[s]], Reverse[#]] &[IntegerDigits[#, base]], base]] &];
AppendTo[s, FromDigits[IntegerDigits[tmp, base]]], {z}]; s (* A262627 *)
Map[FromDigits[ToString[#], base] &, s] (* A262628 *)
(* Peter J. C. Moses, Sep 01 2015 *)
3 is a term since it is a prime number and its representation in primorial base is 11 (1 * 2# + 1) which is a palindrome.
max = 8; bases = Prime @ Range[max, 1, -1]; nmax = Times @@ bases - 1; Select[Range[nmax], PrimeQ[#] && PalindromeQ @ IntegerDigits[#, MixedRadix[bases]] &]
15 is a term since the binary representation of its prime factors, 3 and 5, are both palindromes: 11 and 101. 1 is a term because it has no prime factors, and "the empty set has every property". - _N. J. A. Sloane_, Jan 16 2022
seq[max_] := Module[{ps = Select[Range[max], PalindromeQ @ IntegerDigits[#, 2] && PrimeQ[#] &], s = {1}, s1, s2}, Do[p = ps[[k]]; emax = Floor@Log[p, max]; s1 = Join[{1}, p^Range[emax]]; s2 = Select[Union[Flatten[Outer[Times, s, s1]]], # <= max &]; s = Union[s, s2], {k, 1, Length[ps]}]; s]; seq[1000]
Join[{1},Module[{bps=Select[Prime[Range[200]],IntegerDigits[#,2] == Reverse[ IntegerDigits[ #,2]]&]},Select[ Range[Max[ bps]],SubsetQ[ bps,FactorInteger[#][[All,1]]]&]]] (* Harvey P. Dale, Jan 16 2022 *)
from sympy import factorint def ispal(s): return s == s[::-1] def ok(n): return n > 0 and all(ispal(bin(f)[2:]) for f in factorint(n)) print([k for k in range(528) if ok(k)]) # Michael S. Branicky, Jan 17 2022
11 is not in this sequence as its representation in base 2 is 1011, in base 3 is 102, in base 5 is 21, in base 7 is 14, none of which are palindromic. 1483 is in this sequence as its representation in base 37 is 131, which is palindromic.
a={}; For[i=1, i<=155, i++, flag=0; For[j=1, Prime[j] < Prime[i] && flag==0, j++, If[PalindromeQ[IntegerDigits[Prime[i], Prime[j]]], flag=1; AppendTo[a, Prime[i]]]]]; a (* Stefano Spezia, Apr 22 2024 *)
from sympy import sieve from sympy.ntheory import digits from itertools import islice def ispal(v): return v == v[::-1] def agen(): yield from (p for p in sieve if any(ispal(digits(p, q)[1:]) for q in sieve.primerange(1, p))) print(list(islice(agen(), 60))) # Michael S. Branicky, Apr 20 2024
Terms 1,2,and 3 are 3,7,31, with respective base-2 valuations of 11, 111, 11111. The fourth term, 73, is the smallest term containing zeros in the base-2 representation: 1001001. Note the middle bit is shared by both halves; the nested palindrome is "1001".
import sympy
def ispal(n):
return str(n) == str(n)[::-1]
def isodd(n): return n%2
outVec = []
for n in range(9999999):
if not sympy.isprime(n): continue
binStr = (bin(n))[2:]
if not ispal(binStr): continue
lenB = len(binStr)
halfB = int(lenB/2)
if isodd(lenB): halfB += 1
if not ispal(binStr[:halfB]): continue
print(n,binStr)
outVec.append(n)
print(outVec)
from sympy import isprime
from itertools import count, islice, product
def pals(d, base=10): # returns a string
digits = "".join(str(i) for i in range(base))
for p in product(digits, repeat=d//2):
if d//2 > 0 and p[0] == "0": continue
left = "".join(p); right = left[::-1]
for mid in [[""], digits][d%2]:
yield left + mid + right
def nbp_gen(): # generator of nested binary palindromes (as strings)
yield '0'
for d in count(1):
yield from (p+p[-1-d%2::-1] for p in pals((d+1)//2, base=2))
def agen(): # generator of terms
yield from filter(isprime, (int(nbp, 2) for nbp in nbp_gen()))
print(list(islice(agen(), 36))) # Michael S. Branicky, Jun 12 2024
(Common Lisp) ; See Links section.
rev := proc (n) local nn: nn := convert(n, base, 10): add(nn[j]*10^(nops(nn)-j), j = 1 .. nops(nn)) end proc: a := proc (n) local n2: n2 := convert(n, binary): if 1 < n and `mod`(n, 2) = 1 and isprime(n) = false and rev(n2) = n2 then n else end if end proc: seq(a(n), n = 1 .. 1500); # Emeric Deutsch, Jul 24 2009
Select[Range[1,1201,2],CompositeQ[#]&&IntegerDigits[#,2]==Reverse[IntegerDigits[#,2]]&] (* Harvey P. Dale, Dec 28 2024 *)
15 = 3*5 = {1,1,1,1}.
21 = 3*7 = {1,0,1,0,1}.
33 = 3*11 = {1,0,0,0,0,1}.
f1[n_]:=Reverse[IntegerDigits[n,2]]==IntegerDigits[n,2]; f2[n_]:=Last/@FactorInteger[n]=={1,1}; lst={};Do[If[f1[n]&&f2[n],AppendTo[lst,n]],{n,8!}];lst
165=3*5*11={1,0,1,0,0,1,0,1} 195=3*5*13={1,1,0,0,0,0,1,1}
f1[n_]:=Reverse[IntegerDigits[n,2]]==IntegerDigits[n,2]; f2[n_]:=Last/@FactorInteger[n]=={1,1,1} lst={};Do[If[f1[n]&&f2[n],AppendTo[lst,n]],{n,8!}];lst
341 is a term since 341 = 101010101_2 is palindromic in base 2, it is composite (= 11 * 31) and 2^340 == 1 (mod 341).
pspQ[n_] := CompositeQ[n] && PowerMod[2, n-1, n] == 1; Select[Range[10^6], PalindromeQ[IntegerDigits[#, 2]] && pspQ[#] &]
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