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A364479
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Happy palindromic primes.
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1
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7, 313, 383, 11311, 15451, 30103, 30803, 35053, 36263, 71317, 74047, 94349, 94649, 95959, 98689, 1221221, 1257521, 1262621, 1281821, 1311131, 1444441, 1551551, 1594951, 1597951, 1658561, 1703071, 1737371, 1764671, 1829281, 1924291, 1957591, 1970791, 1981891, 1988891, 3001003
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internal format)
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OFFSET
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1,1
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COMMENTS
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All terms in the sequence are prime numbers that read the same backward and forward. Each term is a happy number, meaning it converges to 1 under the process of repeatedly summing the squares of its digits. The sequence is a subset of both the palindromic prime numbers (A002385) and the happy numbers (A007770).
There are no happy palindromic primes with an even number of digits: every palindromic number with an even number of digits is divisible by 11, so 11 itself is the only palindromic prime, and it is not a happy number.
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LINKS
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EXAMPLE
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313 is a term as it is palindromic (can be reversed), is a prime and is happy: 3^2 + 1^2 + 3^2 = 19, 1^2 + 9^2 = 82, 8^2 + 2^2 = 68, 6^2 + 8^2 = 100, 1^2 + 0^2 + 0^2 = 1.
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MATHEMATICA
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happyQ[n_] := NestWhile[Plus @@ (IntegerDigits[#]^2) &, n, UnsameQ, All] == 1; Select[Prime[Range[220000]], PalindromeQ[#] && happyQ[#] &] (* Amiram Eldar, Jul 28 2023 *)
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PROG
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(Python)
def is_prime(num):
return num > 1 and all(num % i != 0 for i in range(2, int(num ** 0.5) + 1))
def is_palindrome(num):
return str(num) == str(num)[::-1]
def is_happy(num):
while num != 1 and num != 4:
num = sum(int(digit) ** 2 for digit in str(num))
return num == 1
happy_palindromic_primes = [num for num in range(1, 10000000) if is_prime(num) and is_palindrome(num) and is_happy(num)]
print(happy_palindromic_primes)
(Python)
from itertools import islice
from sympy import isprime
def A364479_gen(): # generator of terms
n = 1
while True:
for z in (1, 3, 5, 7, 9):
for y in range(z*n, (z+1)*n):
k, m = y//10, 0
while k >= 10:
k, r = divmod(k, 10)
m = 10*m + r
if isprime(a:=y*n + 10*m + k):
b = a
while b not in {1, 37, 58, 89, 145, 42, 20, 4, 16}:
b = sum((0, 1, 4, 9, 16, 25, 36, 49, 64, 81)[ord(d)-48] for d in str(b))
if b == 1:
yield a
n *= 10
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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