A035497 Happy primes: primes that eventually reach 1 under iteration of "x -> sum of squares of digits of x".

7, 13, 19, 23, 31, 79, 97, 103, 109, 139, 167, 193, 239, 263, 293, 313, 331, 367, 379, 383, 397, 409, 487, 563, 617, 653, 673, 683, 709, 739, 761, 863, 881, 907, 937, 1009, 1033, 1039, 1093, 1151, 1277, 1303, 1373, 1427, 1447, 1481, 1487, 1511, 1607, 1663

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• base

Comments

The 2nd and 3rd repunit primes, 1111111111111111111 and 11111111111111111111111 are happy primes. - Thomas M. Green, Oct 23 2009

There are 200 terms up to 10^4, 1465 up to 10^5, 11144 up to 10^6, 91323 up to 10^7, 812371 up to 10^8, 7408754 up to 10^9, and 67982202 up to 10^10. These are consistent with b*prime(n) < a(n) < c*prime(n) with constants 0 < b < c. - Charles R Greathouse IV, Jan 06 2016

References

• R. K. Guy, Unsolved Problems Number Theory, Sect. E34.

Links

• Nathaniel Johnston, Table of n, a(n) for n = 1..10000
• Shyam Sunder Gupta, Happy Numbers, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 7, 209-222.
• Carlos Rivera, Puzzle 21. Happy primes, The Prime Puzzles and Problems Connection.
• Eric Weisstein's World of Mathematics, Happy Number
• Doctor Who, Episode 42
• Wikipedia, Happy number
• Wikipedia, Doctor Who, Episode 42

Crossrefs

Cf. A007770 (happy numbers), A046519.

Programs

Mathematica

g[n_] := Total[ IntegerDigits[n]^2]; fQ[n_] := NestWhileList[g@# &, n, UnsameQ, All][[-1]] == 1; Select[Prime@ Range@ 300, fQ@# &] (* Robert G. Wilson v, Jan 03 2013 *)
hpQ[p_]:=NestWhile[Total[IntegerDigits[#]^2]&,p,#!=1&,1,50]==1; Select[Prime[ Range[ 300]],hpQ] (* Harvey P. Dale, Jun 07 2022 *)

PARI

has(n)=while(n>6, n=norml2(digits(n))); n==1
is(n)=has(n) && isprime(n) \\ Charles R Greathouse IV, Dec 14 2015

Python

from sympy import isprime
def swb(n): return sum(map(lambda x: x*x, map(int, str(n))))
def happy(bd):
    while bd not in [1, 4]: bd = swb(bd) # iterate to fixed point or cycle
    return bd == 1
def ok(n): return isprime(n) and happy(n)
def aupto(n): return [k for k in range(1, n+1) if ok(k)]
print(aupto(2012)) # Michael S. Branicky, Jul 13 2022

Extensions

More terms from Patrick De Geest, Oct 15 1999