A228564 Largest odd divisor of n^2 + 1.

1, 1, 5, 5, 17, 13, 37, 25, 65, 41, 101, 61, 145, 85, 197, 113, 257, 145, 325, 181, 401, 221, 485, 265, 577, 313, 677, 365, 785, 421, 901, 481, 1025, 545, 1157, 613, 1297, 685, 1445, 761, 1601, 841, 1765, 925, 1937, 1013, 2117, 1105, 2305, 1201, 2501, 1301, 2705

Offset: 0

Author

Jeremy Gardiner, Aug 25 2013

Keywords

• nonn
• easy

Comments

From Lamine Ngom, Jan 04 2023: (Start)

For n>2, a(n) = hypotenuse c of the primitive Pythagorean triple (a, b, c) such that n*a = b + c.

Terms that appear twice (1, 5, 145, 4901, ...) are the positive terms of A076218. Equivalently, the products of two consecutive terms of A001653, or one more than the squares of A001542.

These duplicated terms appear at indices i and j (i>j) such that (i^2-1)/2 = j^2 (A001541). In addition, they are hypotenuse in two primitive Pythagorean triples: (i, j^2, a(i)) and (2*j, j^2-1, a(i)). (End)

Examples

			A002522(3) = 3^2+1 = 10 => a(3) = 10/2 = 5.
		

Links

• Jeremy Gardiner, Table of n, a(n) for n = 0..1000
• Wikipedia, Quasi-polynomial.
• Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Crossrefs

Cf. A000265, A002522, A164900, A191871.

Programs

BASIC

for n = 0 to 45 : t=n^2+1
x: if not t mod 2 then t=t/2 : goto x
print str$(t);", "; : next n
print
end

GAP

List([0..60],n->NumeratorRat((n^2+1)/(n+1))); # Muniru A Asiru, Feb 20 2019

Magma

[(n^2+1)*(3+(-1)^n)/4: n in [0..60]]; // Bruno Berselli, Aug 26 2013

Magma

[Denominator(2*n^2/(n^2+1)): n in [0..60]]; // Vincenzo Librandi, Aug 19 2014

Maple

lod:= t -> t/2^padic:-ordp(t,2):
seq(lod(n^2+1),n=0..60); # Robert Israel, Aug 19 2014

Mathematica

Table[(n^2 + 1) (3 + (-1)^n)/4, {n, 0, 60}] (* Bruno Berselli, Aug 26 2013 *)

PARI

a(n)=if(n<2,n>0,m=n\4;[4*a(2*m)-3,2*a(2*m)+4*m-1,4*a(2*m)+16*m+1,2*a(2*m)+12*m+3][(n%4)+1]) \\ Ralf Stephan, Aug 26 2013

PARI

a(n)=(n^2+1)/2^valuation(n^2+1,2) \\ Ralf Stephan, Aug 26 2013

Sage

[(n^2+1)*(3+(-1)^n)/4 for n in (0..60)] # G. C. Greubel, Feb 20 2019

Formula

a(n) = A000265(A002522(n)).

From Ralf Stephan, Aug 26 2013: (Start)

a(4n) = 4*a(2n) - 3.

a(4n+1) = 2*a(2n) + 4*n - 1.

a(4n+2) = 4*a(2n) + 16*n + 1.

a(4n+3) = 2*a(2n) + 12*n + 3. (End)

From Bruno Berselli, Aug 26 2013: (Start)

G.f.: (1 + x + 2*x^2 + 2*x^3 + 5*x^4 + x^5) / (1-x^2)^3.

a(n) = 3*a(n-2) -3*a(n-4) +a(n-6).

a(n) = (n^2+1)*(3+(-1)^n)/4. (End)

From Peter Bala, Feb 14 2019: (Start)

a(n) = numerator((n^2 + 1)/(n + 1)).

a(n) is a quasi-polynomial in n: a(2*n) = 4*n^2 + 1; a(2*n + 1) = 2*n^2 + 2*n + 1. (End)

From Amiram Eldar, Aug 09 2022: (Start)

a(n) = numerator((n^2+1)/2).

Sum_{n>=0} 1/a(n) = (1 + coth(Pi/2)*Pi/2 + tanh(Pi/2)*Pi)/2. (End)

E.g.f.: ((2 + x + 2*x^2)*cosh(x) + (1 + x)^2*sinh(x))/2. - Stefano Spezia, Aug 04 2025