James S. DeArmon - cp's OEIS frontend

A387199 Numbers which are not themselves palindromes, but a single swap of two digits creates a palindrome.

110, 112, 113, 114, 115, 116, 117, 118, 119, 122, 133, 144, 155, 166, 177, 188, 199, 211, 220, 221, 223, 224, 225, 226, 227, 228, 229, 233, 244, 255, 266, 277, 288, 299, 311, 322, 330, 331, 332, 334, 335, 336, 337, 338, 339, 344, 355, 366, 377, 388, 399, 411, 422

Offset: 1

James S. DeArmon, Aug 21 2025

Might be called "single-transposition palindromes".

Leading zeros are not allowed in either the initial number or the resultant palindrome.

			110 is a term since a swap of the second and third digits yields the palindrome 101.

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Cf. A002113.

def pal(s): return s == s[::-1]
def swaps(s): yield from (t for i in range(len(s)-1) for j in range(i+1, len(s)) if (t:=s[:i]+s[j]+s[i+1:j]+s[i]+s[j+1:])[0]!='0')
def ok(n): return not pal(s:=str(n)) and any(pal(t) for t in swaps(s))
print([k for k in range(425) if ok(k)]) # Michael S. Branicky, Aug 22 2025

More terms from Michael S. Branicky, Aug 22 2025

A383502 Position of the first instance of prime(n), in base 3, in the ternary representation of Pi after the ternary point.

Original entry on oeis.org

2, 4, 6, 2, 12, 20, 6, 10, 27, 16, 85, 3, 35, 78, 95, 6, 38, 96, 19, 27, 9, 66, 157, 171, 81, 191, 12, 127, 52, 3, 36, 275, 88, 589, 283, 40, 290, 952, 47, 10, 1213, 750, 572, 84, 126, 2, 282, 162, 125, 480, 26, 66, 185, 157, 1490, 1310, 832, 1321, 352

Offset: 1

James S. DeArmon, Apr 28 2025

Positions are numbered starting from 1 for the first ternary digit after the ternary point in Pi.

			The ternary digits of Pi and their numbering, after the ternary point, begin
          1 2 3 4 5 6 7 8 9 ...
   0 1  . 0 2 1 1 0 1 2 2 2 2 0 1 0 2 ...
                    \---/
prime(7) is 17, or 122_3, which first appears at position 6.

Cf. A000040, A178707, A007089, A004602, A004601.

import gmpy2
from sympy import isprime
def to_base3(n):
    if n == 0: return '0'
    d = []
    while n: d.append(str(n % 3)); n //= 3
    return ''.join(reversed(d))
gmpy2.get_context().precision = 1200000
pi_ternary = gmpy2.digits(gmpy2.const_pi(),3)[0][4:]  # skip "10."
out = []
for p in range(2, 280):
    if isprime(p):
        pos = pi_ternary.find(to_base3(p)) + 1
        out.append(pos)
print(out)

A381041 Smallest prime p such that 3^n + p + 1 is prime.

Original entry on oeis.org

3, 3, 3, 3, 7, 7, 3, 19, 7, 3, 3, 19, 79, 7, 7, 43, 67, 139, 127, 103, 7, 97, 3, 31, 31, 13, 379, 61, 109, 433, 3, 79, 127, 79, 67, 139, 127, 229, 7, 109, 271, 313, 3, 151, 7, 103, 67, 283, 421, 67, 43, 373, 97, 97, 97, 19, 61, 3, 157, 331, 127, 37, 139, 439, 421

Offset: 0

James S. DeArmon, Apr 20 2025

nonn

			a(0) = 3, since 3 + (3^0+1) = 5 is prime and 2 + (3^0+1) = 4 is not.
a(1) = 3, since 3 + (3^1+1) = 7 is prime and 2 + (3^1+1) = 6 is not.

Robert Israel, Table of n, a(n) for n = 0..2000

Cf. A007051, A020483, A056206, A056208, A057673.

f:= proc(n) local t,p;
 p:= 1: t:= 3^n+1;
 do
   p:= nextprime(p);
   if isprime(p+t) then return p fi
 od;
end proc:
map(f, [$0..100]); # Robert Israel, Jun 19 2025

a[n_]:=Module[{p=2},While[!PrimeQ[p+3^n+1], p=NextPrime[p]]; p]; Array[a,65,0] (* Stefano Spezia, Apr 25 2025 *)

a(n) = my(p=2, x=3^n+1); while (!isprime(p+x), p=nextprime(p+1)); p; \\ Michel Marcus, Apr 24 2025

from sympy import isprime, nextprime
def a(n):
    p, b = 2, 3**n+1
    while not isprime(p+b):
        p = nextprime(p)
    return p
print([a(n) for n in range(65)]) # Michael S. Branicky, Apr 23 2025

from sympy import nextprime, isprime
def A381041(n):
    p = 3**n+1
    q = nextprime(p)
    while not isprime(q-p):
        q = nextprime(q)
    return q-p # Chai Wah Wu, May 01 2025

a(n) = A020483(A007051(n)). - Robert Israel, Jun 19 2025

More terms from Michael S. Branicky, Apr 23 2025

A382667 Position of the first instance of prime(n), in base 2, in the binary representation of Pi after the binary point.

Original entry on oeis.org

3, 11, 16, 11, 16, 15, 25, 60, 91, 14, 11, 126, 58, 393, 207, 18, 14, 13, 6, 180, 141, 169, 58, 243, 47, 326, 168, 475, 15, 291, 451, 108, 64, 87, 327, 421, 358, 41, 356, 468, 343, 16, 618, 107, 80, 179, 57, 206, 291, 325, 361, 205, 427, 12, 95, 108, 436, 6, 996

Offset: 1

James S. DeArmon, Apr 02 2025

Positions are numbered starting from 1 for the first bit after the binary point in Pi.

			For n=19, the bits of Pi and their numbering, after the binary point, begin
          1 2 3 4 5 6 7 8 9 ...
   1 1 .  0 0 1 0 0 1 0 0 0 0 1 1 1 1 ...
                    \-----------/
                    prime(19) = 67
prime(19) = 1000011_2 begins at position a(19) = 6.
prime(58) = 271 = 100001111_2 also starts at 6 => a(58) = 6.

Cf. A000040, A004601, A178707, A233836, A378472, A382307.

p=Drop[RealDigits[Pi,2,1010][[1]],2](* increase for n>73 *);a[n_]:=First[SequencePosition[p,IntegerDigits[Prime[n],2]][[1]]] (* James C. McMahon, Apr 26 2025 *)

import gmpy2
from sympy import isprime
gmpy2.get_context().precision = 12000000
gmpy2.get_context().round = gmpy2.RoundDown
pi = gmpy2.const_pi()
binary_pi = gmpy2.digits(pi, 2)[0][2:] # Get binary digits and remove "11"
print([binary_pi.find(bin(cand)[2:])+1 for cand in range(2, 700) if isprime(cand)])

a(n) = A178707(A000040(n)). - Pontus von Brömssen, Apr 12 2025

A382307 Position of start of first run of alternating bit values in the base-2 representation of Pi, or -1 if no such run exists.

Original entry on oeis.org

1, 2, 2, 19, 19, 19, 19, 19, 1195, 1697, 1890, 1890, 1890, 1890, 15081, 63795, 206825, 206825, 206825, 470577, 470577, 557265, 557265, 557265, 557265, 557265, 447666572, 447666572, 699793337, 699793337, 2049646803, 2250772991

Offset: 1

James S. DeArmon, Mar 21 2025

In base-2, Pi is: 11.00100100001111110110101010001... For this sequence, the integer part of Pi is ignored, and the first fractional bit is numbered one.
The bit substring may begin with either 0 or 1.
a(27) > 4*10^6. - Michael S. Branicky, Mar 23 2025
Conjecture: no term is -1 (would follow from a proof that Pi is normal in base 2). - Sean A. Irvine, Mar 30 2025
a(33) > 4*10^9. - Pontus von Brömssen, Apr 06 2025

			The first alternating bit run is "0", at position 1, so a(1) = 1. The second and third alternating bit runs are "01" and "010", starting at position 2, so a(2) and a(3) are both 2.
a(4)-a(8) = 19 since the binary digits of Pi are "10101010" starting at position 19.

Cf. A004601, A175945, A178708, A233836, A378472.

def binary_not(binary_string):
    return ''.join('1' if bit == '0' else '0' for bit in binary_string)
# !pip install gmpy2   # may be necessary
import gmpy2
gmpy2.get_context().precision = 12000000
pi = gmpy2.const_pi()
# Convert Pi to binary representation
binary_pi = gmpy2.digits(pi, 2)[0] # zero-th element is the string of bits
binary_pi = binary_pi[2:] # remove leading "11" left of decimal point
outVec = []
strSearch0 = "" # this substring starts with "0"
for lenRun in range(1,30):
  strSearch0 += "0" if lenRun%2==1 else "1"
  strSearch1 = binary_not(strSearch0)
  l0 = binary_pi.find(strSearch0)+1 # incr origin-0 result
  l1 = binary_pi.find(strSearch1)+1
  outVec.append(min(l0,l1))
print(outVec)

a(26)-a(32) from Pontus von Brömssen, Apr 06 2025

A381158 Prime numbers where digit values decrease while alternating parity.

Original entry on oeis.org

2, 3, 5, 7, 41, 43, 61, 83, 521, 541, 743, 761, 941, 983, 6521, 8521, 8543, 8741, 8761, 76541, 76543, 94321, 98321, 98543

Offset: 1

James S. DeArmon, Feb 15 2025

			41 is a term, since the digits decrease in value and alternate even and odd.

Cf. A052014, A000040, A028867, A028864, A052015, A382027.

b:= proc(n) `if`(isprime(n), n, [][]), seq(b(10*n+j), j=irem(n, 10)-1..1, -2) end:
{seq(b(n), n=1..9)}[];  # Alois P. Heinz, Mar 12 2025

ad=Select[Prime[Range[10^6]],Reverse[Union[IntegerDigits[#]]]==IntegerDigits[#]&];fQ[n_] := Block[{m = Mod[ IntegerDigits@ n, 2]}, m == Split[m, UnsameQ][[1]]];Select[ad,fQ] (* James C. McMahon, Mar 20 2025 *)

from sympy import isprime
from itertools import combinations
def ap(s): return all((s[i] in "13579") == (s[i+1] in "02468") for i in range(len(s)-1))
def afull(): return sorted(t for d in range(1, 10) for c in combinations("987654321", d) if ap(c) and isprime(t:=int("".join(c))))
print(afull()) # Michael S. Branicky, Mar 10 2025

A380596 Numbers with embedded palindromes as proper substrings of the term.

Original entry on oeis.org

100, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 122, 133, 144, 155, 166, 177, 188, 199, 200, 211, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 233, 244, 255, 266, 277, 288, 299, 300, 311, 322, 330, 331, 332, 333, 334, 335, 336, 337, 338, 339, 344, 355, 366, 377, 388, 399

Offset: 1

James S. DeArmon, Jan 27 2025

An embedded palindrome is a substring of at least two contiguous digits (since a single digit is trivially a palindrome). E.g., 121 is a palindrome, but has no embedded palindromes; 110 has the embedded palindrome "11".

Alternatively, k contains a proper substring of the form dd or ded, where d and e are single decimal digits (i.e., a length-2 or -3 palindrome). - Michael S. Branicky, Feb 08 2025

			100 is a term, since "00" is a palindrome; 1001 is a term for the same reason.
1020 is a term, since "020" is a palindrome; 10201 is a term for the same reason.

Cf. A002113.

Significant overlap with A044821 for terms below 1000.

from itertools import combinations
def get_all_substrings(string):
    length = len(string) + 1
    return [string[x:y] for x, y in combinations(range(length), r=2)]
def is_palindrome(n):
    return str(n) == str(n)[::-1]
def ok(n):
    subsets = get_all_substrings(str(n))
    subsets = [subset for subset in subsets if is_palindrome(subset) and len(subset)>1 and len(subset)0
print([k for k in range (100,400) if ok(k)])

def ok(n):
    s = str(n)
    return any(p == p[::-1] and len(p) < len(s) for p in (s[i:i+j] for j in (2, 3) for i in range(len(s)-j+1)))
print([k for k in range(400) if ok(k)]) # Michael S. Branicky, Feb 08 2025

A379750 First prime of cousin prime pairs which differ, in their binary representation, by a single bit.

Original entry on oeis.org

3, 19, 43, 67, 97, 163, 193, 307, 313, 379, 457, 499, 643, 673, 739, 769, 859, 883, 907, 937, 1009, 1297, 1483, 1489, 1579, 1609, 1867, 1873, 1993, 2083, 2137, 2203, 2347, 2377, 2473, 2539, 2617, 2659, 2683, 2689, 2707, 2833, 2857, 2953, 3019, 3163, 3187, 3217

Offset: 1

James S. DeArmon, Jan 01 2025

The first prime of a cousin prime pair is a prime p for which p+4 is also prime.

The only way for p and p+4 to differ at a single bit position is when p has a 0 bit at its "4" position, so p == {0,1,2,3} (mod 8), and so this sequence is the intersection of A023200 and A047471.

			3 is a term since it's a cousin prime with 7 and their respective binary representations 011 and 111 differ at a single bit position.
13 is not a term since, although it's a cousin prime with 17, their respective binary representations 1101 and 10001 differ at more than a single bit position.

Cf. A023200 (cousin primes), A047471, A071695.

Select[Prime[Range[480]], PrimeQ[#+4]&&Mod[#,8]<4&] (* James C. McMahon, Mar 01 2025 *)

import sympy
def ok(n): return (n&5)==1 and sympy.isprime(n) and sympy.isprime(n+4)

a(45)-a(48) from James C. McMahon, Mar 01 2025

A377370 Smallest prime ending in n alternating decimal digits 0 and 1.

Original entry on oeis.org

11, 101, 101, 20101, 210101, 4010101, 61010101, 1601010101, 8101010101, 260101010101, 3110101010101, 11010101010101, 11010101010101, 1201010101010101, 6101010101010101, 180101010101010101, 1710101010101010101, 5010101010101010101, 41010101010101010101

Offset: 1

James S. DeArmon, Dec 27 2024

Leading zeros are not allowed, so that prime p = 101 does not end 0101 and is not a candidate for a(4).
For n>=5, the ending is not itself prime, so that the prefix must be a positive integer.
Some terms are repeated, e.g., 11010101010101 ends with "010101010101" and also ends with the next element in alternating series "1010101010101".

			For n=4, the required ending is the 4 digits 0101, and the smallest prime ending that way is a(4) = 20101.

Robert Israel, Table of n, a(n) for n = 1..992

Cf. A036952, A065720.

f:= proc(n) local t0,t,k;
  t0:= add(10^k,k=0..n-1,2);
  if n::even then t0:=t0 + 10^n fi;
  for t from t0 by 10^n do if isprime(t) then return t fi od
end proc:
map(f, [$1..20]); # Robert Israel, Feb 26 2025

s={};d={};Do[If[OddQ[n],PrependTo[d,1],PrependTo[d,0]];If[PrimeQ[FromDigits[d]&&OddQ[n]],p=FromDigits[d],i=0;p=FromDigits[Prepend[d,i]];Until[PrimeQ[p],i++;p=FromDigits[Prepend[d,i]]]];AppendTo[s,p],{n,19}];s (* James C. McMahon, Feb 08 2025 *)

from sympy import isprime
from itertools import count
def a(n):
    ending = int(("01"*n)[-n:])
    if n&1 and isprime(ending): return ending
    return next(i for i in count(10**n+ending, 10**n) if isprime(i))
print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Jan 26 2025

A379150 Smallest prime ending in "3", with n preceding "0" digits.

Original entry on oeis.org

103, 2003, 70003, 100003, 1000003, 20000003, 500000003, 40000000003, 40000000003, 100000000003, 2000000000003, 230000000000003, 3100000000000003, 11000000000000003, 20000000000000003, 100000000000000003, 1000000000000000003, 310000000000000000003, 500000000000000000003

Offset: 1

James S. DeArmon, Dec 16 2024

Leading zeros are not allowed, e.g., "03".

a(997) has 1001 digits. - Michael S. Branicky, Dec 16 2024

			a(1) = 103, is the smallest prime ending in "03";
a(2) = 2003, is the smallest prime ending in "003".

Michael S. Branicky, Table of n, a(n) for n = 1..996

Cf. A070854, A070847.

Table[i=1;While[!PrimeQ[m=FromDigits[Join[IntegerDigits[i],Table[0,n],{3}]]],i++];m,{n,19}] (* James C. McMahon, Dec 23 2024 *)

a(n)=for(i=1, oo, if(isprime(i*10^(n+1)+3), return(i*10^(n+1)+3))) \\ Johann Peters, Dec 27 2024

import sympy
def prime3_finder():
  outVec = []
  power = 2
  for n in range(100,999999999):
      if not n & 3 == 3: continue # speed-up over simple MOD operation
      if not n % 10**power == 3: continue
      if not sympy.isprime(n): continue
      outVec.append(n)
      power += 1
  return outVec
outvec = prime3_finder()
print(outvec)

from sympy import isprime
from itertools import count
def a(n): return next(i for i in count(10**(n+1)+3, 10**(n+1)) if isprime(i))
print([a(n) for n in range(1, 20)]) # Michael S. Branicky, Dec 16 2024

More terms from Michael S. Branicky, Dec 16 2024

cp's OEIS Frontend

User: James S. DeArmon

A387199 Numbers which are not themselves palindromes, but a single swap of two digits creates a palindrome.

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A383502 Position of the first instance of prime(n), in base 3, in the ternary representation of Pi after the ternary point.

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A381041 Smallest prime p such that 3^n + p + 1 is prime.

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A382667 Position of the first instance of prime(n), in base 2, in the binary representation of Pi after the binary point.

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A382307 Position of start of first run of alternating bit values in the base-2 representation of Pi, or -1 if no such run exists.

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A381158 Prime numbers where digit values decrease while alternating parity.

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Maple

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A380596 Numbers with embedded palindromes as proper substrings of the term.

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A379750 First prime of cousin prime pairs which differ, in their binary representation, by a single bit.

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