A343591 Smoothly undulating alternating primes.

2, 3, 5, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89, 101, 181, 383, 727, 787, 929, 18181, 32323, 72727, 74747, 78787, 94949, 1212121, 1616161, 323232323, 383838383, 727272727, 929292929, 989898989, 12121212121, 14141414141, 32323232323, 383838383838383, 38383838383838383, 72727272727272727

Offset: 1

Author

Bernard Schott, Apr 21 2021

Keywords

• nonn
• base

Comments

Equivalently, numbers that are primes, smoothly undulating = in which the digits alternate: ababab... with a <> b (A032758) and alternating = in which parity of the digits alternates (A030144).

Charles W. Trigg was the first to use the word 'smoothly' for these integers.

If we note a(ba) the terms where the substring (ba) is repeated in their decimal expansion, there exist only 16 possibilities with a odd <> 5 and b even <> 0 to get such primes. Indeed, there exist primes of the form 1(21), 1(41), 1(61), 1(81), 3(23), 3(83), 7(27), 7(47), 7(87), 9(29), 9(49), 9(89). There do not exist terms of the form 3(63), 7(67), 9(69), as they are always composite.

Now, what about possible terms of the form 3(43)? If (43) is repeated 3k times, 3(43) is divisible by 3; if (43) is repeated 3k+1 times, 3(43) is divisible by 7; so if such a prime exists, then the substring (43) must be repeated 3k+2 times, but it is not known if such smoothly undulating prime 3(43) exists and if it exists, (43) must be repeated at least 9302 times, so k >= 3100 (link).

Some properties:

-> Every term has two digits or an odd number of digits.

-> All terms with an odd number of digits are palindromic (A059758).

-> Only 2 and the nine 2-digit terms begin with an even digit.

Examples

			1616161 is a term as it is prime and the digits 1 and 6 have odd and even parity and alternate.
		

References

• Charles W. Trigg, Special Palindromic Primes, Journal of Recreational Mathematics, 4 (July 1971) 169-170.

Links

• Robert Israel, Table of n, a(n) for n = 1..69
• Mathematics Stack Exchange, (Unsolved) In this infinite sequence, no term is a prime: prove/disprove.

Crossrefs

Intersection of A030144 and A032758.

Subsequence of A343590.

Cf. A059758, A242541.

Programs

Maple

f:= proc(n) local i,a,b,c,d;
  c:= add(10^i,i=1..n-1,2);
  d:= add(10^i,i=0..n-1,2);
  if n = 2 then op(select(isprime,[seq(seq(a*c+b*d, b=[1,3,7,9]),a=[2,4,6,8])]))
    else op(select(isprime, [seq(seq(a*c+b*d, a=[0,2,4,6,8]),b=[1,3,7,9])]))
  fi
end proc:
f(1):= (2,3,5,7):
map(f, [1,2,seq(i,i=3..17,2)]); # Robert Israel, Nov 09 2023

Mathematica

f[o_,e_,n_,m_] := FromDigits @ Riffle[ConstantArray[o,n], ConstantArray[e,n-m]]; seq[n_,m_] := Module[{o = Range[1,9,2], e = Range[0,8,2]}, Select[Union[f @@@ Join[Tuples[{o, e, {n}, {m}}], Tuples[{Rest @ e, o, {n}, {m}}]]], PrimeQ]]; s = seq[1, 1]; Do[s = Join[s, seq[m, Boole[m > 1]]], {m, 1, 10}]; s (* Amiram Eldar, Apr 21 2021 *)

Python

from sympy import isprime
def agenthru(maxdigits):
  if maxdigits >= 1: yield from [2, 3, 5, 7]
  for digits in [2]*(maxdigits >= 2) + list(range(3, maxdigits+1, 2)):
    hlf, odd = (digits+1)//2, digits%2
    d1range = "1379" if digits%2 == 1 else "123456789"
    d2range = "1379" if digits%2 == 0 else "0123456789"
    for d1 in d1range:
      for d2 in d2range:
        if int(d1)%2 == int(d2)%2: continue
        t = int("".join([*sum(zip(d1*hlf, d2*(digits-hlf)), ())]+[d1*odd]))
        if isprime(t): yield t
print([p for p in agenthru(17)]) # Michael S. Branicky, Apr 21 2021