A289320 a(n) = A289310(n)^2 + A289311(n)^2.

1, 5, 10, 25, 26, 50, 50, 125, 100, 130, 122, 250, 170, 250, 260, 625, 290, 500, 362, 650, 500, 610, 530, 1250, 676, 850, 1000, 1250, 842, 1300, 962, 3125, 1220, 1450, 1300, 2500, 1370, 1810, 1700, 3250, 1682, 2500, 1850, 3050, 2600, 2650, 2210, 6250, 2500

Offset

1,2

Comments

This sequence is totally multiplicative.

a(n) > n^2 for any n > 1.

If n is a square, then a(n) is a square.

If a(n) and a(m) are squares, then a(n*m) is a square.

a(n) is also a square for nonsquares n = 42, 168, 246, 287, 378, 672, 984, 1050, 1148, 1434, 1512, 1673, 2058, 2214, 2583, 2688, ...

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Formula

Totally multiplicative, with a(p^k) = (1 + p^2)^k for any prime p and k > 0.

Sum_{k=1..n} a(k) ~ c * n^3, where c = 2/(Pi^2 * Product_{p prime} (1 - 1/p^2 - 1/p^3 - 1/p^4)) = 0.4778963213... . - Amiram Eldar, Nov 13 2022

Sum_{n>=1} 1/a(n) = 15/Pi^2 (A082020). - Amiram Eldar, Dec 15 2022

Mathematica

f[p_, e_] := (p^2 + 1)^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 13 2022 *)

Prog

(PARI) a(n) = my (f=factor(n)); return (prod(i=1, #f~, (1 + f[i, 1]^2) ^ f[i, 2]))
(Python)
from sympy import factorint
from operator import mul
from functools import reduce
def a(n): return 1 if n==1 else reduce(mul, [(1 + p**2)**k for p, k in factorint(n).items()])
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Aug 03 2017

Crossrefs

Cf. A066872, A082020, A289310, A289311.

Sequence in context: A177432 A269740 A166635 * A078308 A083493 A338046

Adjacent sequences: A289317 A289318 A289319 * A289321 A289322 A289323

Keyword

nonn,mult

Author

Rémy Sigrist, Jul 02 2017

Status

approved