A345409 -id:A345409 - cp's OEIS frontend

A345408 Numbers that are the sum of an emirp and its reversal in more than one way.

1090, 2662, 2992, 3212, 4334, 4994, 5104, 5324, 6776, 7106, 9328, 9548, 10450, 10670, 10780, 11110, 11330, 11440, 11660, 12122, 12452, 12892, 13222, 15004, 16786, 17446, 17666, 29092, 29482, 31912, 36352, 44644, 44834, 45454, 46654, 46664, 47474, 47864, 49094, 49294, 49484, 49684, 49894, 50104

Offset: 1

J. M. Bergot and Robert Israel, Jun 18 2021

Numbers that are in A345409 in more than one way.

Interchanging an emirp and its reversal is not counted as a different way.

			a(3) = 2992 is a member because 2992 = 1091 + 1901 = 1181+1811 where 1091 and 1181 and their reversals 1901 and 1811 are primes.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A006567, A345409.

revdigs:= proc(n) local L,i; L:= convert(n,base,10); add(L[-i]*10^(i-1),i=1..nops(L)) end proc:
isemirp1:= proc(n) local r;
if not isprime(n) then return false fi;
r:= revdigs(n);
r > n and isprime(r)
end proc:
E:= select(isemirp1, [seq(seq(seq(i*10^d+j,j=1..10^d-1,2),i=[1,3,7,9]),d=1..4)]):
V:= sort(map(t -> t+revdigs(t),E)):
M:= select(t -> V[t+1]=V[t], [$1..nops(V)-1]):
sort(convert(convert(V[M],set),list));

from collections import Counter
from sympy import isprime, nextprime
def epgen(start=1, end=float('inf')): # generates unique emirp/prime pairs
    p = nextprime(start-1)
    while p <= end:
        revp = int(str(p)[::-1])
        if p < revp and isprime(revp): yield (p, revp)
        p = nextprime(p)
def aupto(lim):
    c = Counter(sum(ep) for ep in epgen(1, lim) if sum(ep) <= lim)
    return sorted(s for s in c if c[s] > 1)
print(aupto(50105)) # Michael S. Branicky, Jun 18 2021

A345410 a(n) is the least number that is the sum of an emirp and its reversal in exactly n ways.

Original entry on oeis.org

44, 1090, 10450, 5104, 88888, 10780, 289982, 299992, 482174, 478874, 868868, 499994, 1073270, 1087790, 1071070, 1069970, 10904990, 10794980, 1091090, 10892990, 1100000, 29955992, 1101100, 26688662, 31022002, 27599572, 46400354, 44688644, 29821792, 45289244, 30122092, 26988962

Offset: 1

J. M. Bergot and Robert Israel, Jun 18 2021

Interchanging an emirp and its reversal is not counted as a different way.
a(n) is the least number k such that there are exactly n unordered pairs of distinct primes (p,p') such that p' is the digit reversal of p and p+p' = k.
Are terms not divisible by 3? Amiram Eldar finds proof they are; A056964(n) = n + reverse(n) is divisible by 3 if and only if n is divisible by 3. But emirps are primes (other than 3) so they are not divisible by 3. - David A. Corneth, Jun 19 2021

			a(3) = 10450 because 10450 = 1229+9221 = 1409+9041 = 3407+7043.

David A. Corneth, Table of n, a(n) for n = 1..423
David A. Corneth, A few examples

Cf. A006567, A345408, A345409.

revdigs:= proc(n) local L,i; L:= convert(n,base,10); add(L[-i]*10^(i-1),i=1..nops(L)) end proc:
isemirp1:= proc(n) local r;
if not isprime(n) then return false fi;
r:= revdigs(n);
r > n and isprime(r)
end proc:
E:= select(isemirp1, [seq(seq(seq(i*10^d+j,j=1..10^d-1,2),i=[1,3,7,9]),d=1..5)]):
V:= sort(map(t -> t+revdigs(t),E)):
N:= nops(V):
W:= Vector(16):
i:= 1:
while i < N do
for j from 1 to N-i while V[i+j]=V[i] do od:
if j <= 16 and W[j] = 0 then W[j]:= V[i] fi;
  i:= i+j;
od:
convert(W,list);

from itertools import product
from collections import Counter
from sympy import isprime, nextprime
def epgen(start=1, end=float('inf')): # generates unique emirp/prime pairs
    digits = 2
    while True:
      for first in "1379":
        for last in "1379":
          if last < first: continue
          for mid in product("0123456789", repeat=digits-2):
            strp = first + "".join(mid) + last
            revstrp = strp[::-1]
            if strp >= revstrp: continue
            p = int(strp)
            if p > end: return
            revp = int(strp[::-1])
            if isprime(p) and isprime(revp): yield (p, revp)
      digits += 1
def aupto(lim):
    alst = []
    c = Counter(sum(ep) for ep in epgen(1, lim) if sum(ep) <= lim)
    r = set(c.values())
    for i in range(1, max(r)+1):
        if i in r: alst.append(min(s for s in c if c[s] == i))
        else: break
    return alst
print(aupto(11*10**5)) # Michael S. Branicky, Jun 19 2021

More terms from David A. Corneth, Jun 18 2021

cp's OEIS Frontend

A345408 Numbers that are the sum of an emirp and its reversal in more than one way.

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