A059758 -id:A059758 - cp's OEIS frontend

A062209 Numbers k such that the smoothly undulating palindromic number (4*10^k-7)/33 = 121...21 is a prime (or PRP).

7, 11, 43, 139, 627, 1399, 1597, 1979, 7809, 14059, 46499

Offset: 1

Patrick De Geest and Hans Rosenthal (Hans.Rosenthal(AT)t-online.de), Jun 15 2001

Prime versus probable prime status and proofs are given in the author's table.
No further terms < 100000. - Ray Chandler, Aug 17 2011
The corresponding primes, called smoothly undulating palindromic primes (cf. links, A032758 and A059758), are listed in A092696. The number of '12's is given in A056803(n) = (a(n)-1)/2. - M. F. Hasler, Jul 30 2015

			k=11 --> (12*10^11 - 21)/99 = 12121212121.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 139, p. 48, Ellipses, Paris 2008.

Cf. A062210-A062232, A059758, A032758, A092696, A056803.

d[n_]:=IntegerDigits[n]; Length/@d[Select[NestList[FromDigits[Join[d[#],{2,1}]]&,1,1000],PrimeQ]] (* Jayanta Basu, May 25 2013 *)

for(n=1,1e5,ispseudoprime(5^n<<(n+2)\33)&&print1(n",")) \\ M. F. Hasler, Jul 30 2015

a(11) = 46499 from Ray Chandler, Nov 11 2010
Edited by Ray Chandler, Aug 17 2011
Name and other items edited by M. F. Hasler, Jul 30 2015

A062232 Numbers k such that the smoothly undulating palindromic number (98*10^k - 89)/99 is a prime.

Original entry on oeis.org

9, 161, 219, 4859, 21989, 52931, 88595

Offset: 1

Patrick De Geest and Hans Rosenthal (Hans.Rosenthal(AT)t-online.de), Jun 15 2001

Prime versus probable prime status and proofs are given in the author's table.

No further terms < 100000. - Ray Chandler, Aug 17 2011

			k=9 -> (98*10^9 - 89)/99 = 989898989.

Patrick De Geest, SUPP Reference Table - 989
Index entries for sequences related to smoothly undulating palindromic primes

Cf. A062209-A062231, A059758, A032758.

a(6)=52931 from Ray Chandler, Nov 11 2010
a(7)=88595 from Ray Chandler, Jul 23 2011
Edited by Ray Chandler, Aug 17 2011

A059170 Strictly undulating primes (digits alternate and differ by 1).

Original entry on oeis.org

2, 3, 5, 7, 23, 43, 67, 89, 101, 787, 32323, 78787, 1212121, 323232323, 989898989, 12121212121, 32323232323, 787878787878787878787, 787878787878787878787878787, 1212121212121212121212121212121212121212121

Offset: 1

N. J. A. Sloane, Feb 14 2001

Of form ababa... with |a-b| = 1.

The next two terms have 95 and 139 digits respectively. - Jayanta Basu, May 09 2013

C. A. Pickover, "Keys to Infinity", Wiley 1995, pp. 159-160.
C. A. Pickover, "Wonders of Numbers", Oxford New York 2001, Chapter 52, pp. 123-124, 316-317.

Harvey P. Dale, Table of n, a(n) for n = 1..25 (All terms with less than 1000 digits.)
Patrick De Geest, More undulating primes
C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review

Cf. A032758, A059168, A048398, A033619, A046075, A059758.

a[n_]:=DeleteDuplicates[Take[IntegerDigits[n],{1,-1,2}]]; b[n_]:=DeleteDuplicates[Take[IntegerDigits[n],{2,-1,2}]]; t={}; Do[p=Prime[n]; If[p<10, AppendTo[t,p], If[Length[a[p]] == Length[b[p]] == 1 && Abs[a[p][[1]]-b[p][[1]]] == 1, AppendTo[t,p]]], {n,10^5}]; t (* Jayanta Basu, May 08 2013 *)
t1=Join[{2,3,5,7},Select[Range[10,100],PrimeQ[#]&&Abs[Differences[IntegerDigits[#]]]=={1}&]]; Do[a=n*10+(n-1);b=(n-1)*10+n; t1=Join[t1,Select[Table[(a*10^(2*n+1)-b)/99,{n,25}],PrimeQ]]; If[n<=7,c=n*10+(n+1);d=(n+1)*10+n;t1=Join[t1,Select[Table[(c*10^(2*n+1)-d)/99,{n,25}],PrimeQ]]],{n,1,9,2}]; Sort[t1] (* Jayanta Basu, May 09 2013 *)
 With[{c=Flatten[{#,Reverse[#]}&/@Table[{a,a+1},{a,0,8}],1]},Flatten[ Select[ Table[ FromDigits[PadRight[{},n,#]],{n,50}],PrimeQ]&/@c]]//Union (* Harvey P. Dale, Aug 20 2022 *)

Extended by Patrick De Geest, Feb 25 2001

Offset corrected by Arkadiusz Wesolowski, Sep 13 2011

A077799 Numbers m such that a smoothly undulating palindromic prime of the form (rs*10^m-sr)/99 exists, where r and s are two distinct digits and rs and sr denote concatenations of those digits.

Original entry on oeis.org

3, 5, 7, 9, 11, 15, 17, 21, 23, 25, 27, 31, 33, 37, 39, 43, 45, 51, 55, 57, 63, 65, 71, 77, 81, 83, 89, 95, 99, 109, 133, 139, 143, 145, 149, 161, 163, 195, 209, 219, 225, 229, 237, 243, 245, 277, 315, 357, 479, 513, 515, 537, 551, 561, 567, 583, 627, 849, 857

Offset: 1

Patrick De Geest, Nov 16 2002

Ray Chandler, Table of n, a(n) for n = 1..146
P. De Geest, SUPP's Sorted By Length
Index entries for sequences related to smoothly undulating palindromic primes

Cf. A062209-A062232, A059758, A032758.

Edited by Ray Chandler, Aug 17 2011

Name clarified by Sean A. Irvine, Jun 14 2025

A343591 Smoothly undulating alternating primes.

Original entry on oeis.org

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

Bernard Schott, Apr 21 2021

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.

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

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

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.

Intersection of A030144 and A032758.
Subsequence of A343590.
Cf. A059758, A242541.

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

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 *)

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

A062210 Numbers k such that the smoothly undulating palindromic number (14*10^k - 41)/99 is a prime.

Original entry on oeis.org

11, 277, 479, 583, 1631, 6343, 14689

Offset: 1

Patrick De Geest and Hans Rosenthal (Hans.Rosenthal(AT)t-online.de), Jun 15 2001

Prime versus probable prime status and proofs are given in the author's table.

No further terms < 100000. - Ray Chandler, Aug 17 2011

			11 is in the sequence because (14*10^11 - 41)/99 = 14141414141 is prime.

Cf. A062209-A062232, A059758, A032758.

Edited by Ray Chandler, Aug 17 2011

Name and Example edited by Jon E. Schoenfield, Jun 25 2017

A062216 Numbers k such that the smoothly undulating palindromic number (31*10^k - 13)/99 is a prime.

Original entry on oeis.org

3, 51, 83, 225, 561, 10419, 18255, 43869

Offset: 1

Patrick De Geest and Hans Rosenthal (Hans.Rosenthal(AT)t-online.de), Jun 15 2001

Prime versus probable prime status and proofs are given in the author's table.

No further terms < 100000. - Ray Chandler, Aug 17 2011

			k=51 -> (31*10^51 - 13)/99 = 313131313131313131313131313131313131313131313131313.

Cf. A062209-A062232, A059758, A032758.

43869 from Ray Chandler, Sep 30 2010

Edited by Ray Chandler, Aug 17 2011

A062220 Numbers k such that the smoothly undulating palindromic number (38*10^k - 83)/99 is a prime.

Original entry on oeis.org

3, 9, 15, 17, 21, 57, 4233, 4335, 13221, 26447, 29897, 91997

Offset: 1

Patrick De Geest and Hans Rosenthal (Hans.Rosenthal(AT)t-online.de), Jun 15 2001

Prime versus probable prime status and proofs are given in the author's table.

No further terms < 100000. - Ray Chandler, Aug 17 2011

			k=21 -> (38*10^21 - 83)/99 = 383838383838383838383.

Cf. A062209-A062232, A059758, A032758.

a(12)=91997 from Ray Chandler, Jul 29 2011

Edited by Ray Chandler, Aug 17 2011

A062231 Numbers k such that the smoothly undulating palindromic number (97*10^k - 79)/99 is a prime.

Original entry on oeis.org

9, 27, 45, 237

Offset: 1

Patrick De Geest and Hans Rosenthal (Hans.Rosenthal(AT)t-online.de), Jun 15 2001

Prime versus probable prime status and proofs are given in the author's table.

No further terms < 100000. - Ray Chandler, Aug 17 2011

			k=27 -> (97*10^27 - 79)/99 = 979797979797979797979797979.

Patrick De Geest, SUPP Reference Table - 979
Index entries for sequences related to smoothly undulating palindromic primes

Cf. A062209-A062232, A059758, A032758.

Select[Range[1,239,2],PrimeQ[FromDigits[PadRight[{},#,{9,7}]]]&] (* Harvey P. Dale, Mar 24 2021 *)

Edited by Ray Chandler, Aug 17 2011

A062211 Numbers k such that the smoothly undulating palindromic number (15*10^k - 51)/99 is a prime.

Original entry on oeis.org

3, 15, 63, 89, 245, 583, 1791, 2123, 7233, 24787, 44653

Offset: 1

Patrick De Geest and Hans Rosenthal (Hans.Rosenthal(AT)t-online.de), Jun 15 2001

Prime versus probable prime status and proofs are given in the author's table.

No further terms < 100000. - Ray Chandler, Aug 17 2011

			k=15 -> (15*10^15 - 51)/99 = 151515151515151.

Cf. A062209-A062232, A059758, A032758.

a(11)=44653 from Ray Chandler, Nov 11 2010

Edited by Ray Chandler, Aug 17 2011

cp's OEIS Frontend

A062209 Numbers k such that the smoothly undulating palindromic number (4*10^k-7)/33 = 121...21 is a prime (or PRP).

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A062232 Numbers k such that the smoothly undulating palindromic number (98*10^k - 89)/99 is a prime.

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A059170 Strictly undulating primes (digits alternate and differ by 1).

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A077799 Numbers m such that a smoothly undulating palindromic prime of the form (rs*10^m-sr)/99 exists, where r and s are two distinct digits and rs and sr denote concatenations of those digits.

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A343591 Smoothly undulating alternating primes.

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A062210 Numbers k such that the smoothly undulating palindromic number (14*10^k - 41)/99 is a prime.

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A062216 Numbers k such that the smoothly undulating palindromic number (31*10^k - 13)/99 is a prime.

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A062220 Numbers k such that the smoothly undulating palindromic number (38*10^k - 83)/99 is a prime.

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