A394435 Primes p such that A001414(p-1)^2 + A001414(p+1)^2 is a perfect square, where A001414(n) is the sum of prime factors of n with multiplicity (sopfr).

2, 4271, 4591, 7151, 10973, 13183, 16111, 19469, 31151, 32749, 41887, 43051, 64033, 65729, 73471, 74933, 91541, 122609, 128591, 142993, 144451, 181039, 185071, 187373, 195593, 263759, 321199, 328357, 332477, 363439, 388961, 413129, 453311, 489539, 524087, 528391, 534311, 536717

Offset

1,1

Comments

Primes p for which (A001414(p-1), A001414(p+1)) form the legs of a right triangle with integer hypotenuse.

The reduced Pythagorean triples arising include (3,4,5), (9,40,41), (12,5,13), (33,56,65), and others. For example, a(6)=16111 and a(7)=19469 both yield the triple 16*(12,5,13).

Conjecture: this sequence is infinite (empirical observation; no proof is known).

References

D. E. Knuth, The Art of Computer Programming, Vol. 2, Addison-Wesley, 1998, Section 4.5.4 (on completely additive arithmetic functions).

Links

Table of n, a(n) for n=1..38.

Formula

a(n) is prime and there exists a positive integer h such that A001414(a(n)-1)^2 + A001414(a(n)+1)^2 = h^2.

Equivalently: a(n) is prime and A001414(a(n)-1)/A001414(a(n)+1) is a rational number whose square plus 1 is a perfect rational square (i.e., the ratio is a rational leg-to-leg ratio of a Pythagorean triple).

Example

a(1) = 4271: 4270 = 2 * 5 * 7 * 61, so A001414(4270) = 75; 4272 = 2^4 * 3 * 89, so A001414(4272) = 100; 75^2 + 100^2 = 15625 = 125^2. Triple: 25*(3,4,5).
a(2) = 4591: 4590 = 2 * 3^3 * 5 * 17, so A001414(4590) = 33; 4592 = 2^4 * 7 * 41, so A001414(4592) = 56; 33^2 + 56^2 = 4225 = 65^2. Primitive triple (33,56,65).
a(3) = 7151: A001414(7150) = 36, A001414(7152) = 160; 36^2 + 160^2 = 26896 = 164^2. Triple: 4*(9,40,41).

Maple

sopfr:= proc(n) local t; add(t[1]*t[2], t = ifactors(n)[2]) end proc:
filter:= n -> isprime(n) and issqr(sopfr(n-1)^2 + sopfr(n+1)^2):
select(filter, [2, seq(i, i=3..10^6, 2)]); # Robert Israel, May 21 2026

Mathematica

sopfr[n_] := Total[Times @@@ FactorInteger[n]]; Select[Prime[Range[5, 50000]], IntegerQ[Sqrt[sopfr[#-1]^2 + sopfr[#+1]^2]]&]

Prog

(Python)
from sympy import factorint, nextprime
import math
def sopfr(n):
    if n <= 1:
        return 0
    return sum(p * e for p, e in factorint(n).items())
def is_sq(n):
    r = math.isqrt(n)
    return r * r == n
p = 2
while p < 10**6:
    if is_sq(sopfr(p-1)**2 + sopfr(p+1)**2):
        print(p)
    p = nextprime(p)
(PARI) sopfr(n) = my(f=factor(n)); sum(k=1, matsize(f)[1], f[k, 1]*f[k, 2]);
isok(p) = isprime(p) && issquare(sopfr(p-1)^2+sopfr(p+1)^2); \\ Michel Marcus, May 21 2026

Crossrefs

Cf. A001414 (sopfr), A086299 (Ruth-Aaron pairs: sopfr(n)=sopfr(n+1)), A074981 (Ruth-Aaron numbers), A001222 (Omega, number of prime factors with multiplicity).

Sequence in context: A114722 A135959 A105667 * A178410 A067668 A292391

Adjacent sequences: A394432 A394433 A394434 * A394436 A394437 A394438

Keyword

nonn

Author

Rudra Jadhav, May 21 2026

Status

approved