A034178 Number of solutions to n = a^2 - b^2, a > b >= 0.

1, 0, 1, 1, 1, 0, 1, 1, 2, 0, 1, 1, 1, 0, 2, 2, 1, 0, 1, 1, 2, 0, 1, 2, 2, 0, 2, 1, 1, 0, 1, 2, 2, 0, 2, 2, 1, 0, 2, 2, 1, 0, 1, 1, 3, 0, 1, 3, 2, 0, 2, 1, 1, 0, 2, 2, 2, 0, 1, 2, 1, 0, 3, 3, 2, 0, 1, 1, 2, 0, 1, 3, 1, 0, 3, 1, 2, 0, 1, 3, 3, 0, 1, 2, 2, 0, 2, 2, 1, 0, 2, 1, 2, 0, 2, 4, 1, 0, 3

Offset: 1

Author

Erich Friedman

Keywords

• easy
• nonn
• nice

Comments

Also, number of ways n can be expressed as the sum of one or more consecutive odd numbers. (E.g., 45 = 45 = 13+15+17 = 5+7+9+11+13, so a(45)=3.) - Naohiro Nomoto, Feb 26 2002

a(A042965(n))>0, a(A016825(n))=0; also number of occurrences of n in A094728. - Reinhard Zumkeller, May 24 2004

It appears a(n) can be found by adding together the divisor pairs of n and finding the number of even results. For example: n=9 has the divisor pairs (1,9) and (3,3); adding the pairs: 1+9=10 is even and 3+3=6 is even, so a(9)=2. Another example: n=96 has the divisor pairs (1,96) (2,48) (3,32) (4,24) (6,16) (8,12); when each pair is added there are 4 even results, so a(96)=4. - Gregory Bryant, Dec 06 2016

It appears a(n) is the number of nonnegative integers k for which sqrt(k) + sqrt(k + n) is an integer. For example: a(2015) = 4 since there are only four nonnegative integers k for which sqrt(k) + sqrt(k + 2015) is an integer, namely k = 289, 5041, 39601, 1014049. - Joseph Barrera, Nov 29 2020

Examples

			G.f. = x + x^3 + x^4 + x^5 + x^7 + x^8 + 2*x^9 + x^11 + x^12 + x^13 + 2*x^15 + ...
From _Bernard Schott_, Apr 19 2019: (Start)
a(8) = floor((A000005(2) + 1)/2) = floor(3/2) = 1 and 8 = 3^2 - 1^2.
a(9) = floor((A000005(9) + 1)/2) = floor(4/2) = 2 and 9 = 3^2 - 0^2 = 5^2 - 4^2.
a(10) = 0 and a^2 - b^2 = 10 has no solution.
a(11) = floor(A000005(11) + 1)/2 = floor(3/2) = 1 and 11 = 6^2 - 5^2.  (End)
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..2000
• M. A. Nyblom, On the Representation of the Integers as a Difference of Squares, Fibonacci Quart., vol. 40 (2002), no. 3, 243-246.
• Edward T. H. Wang, Problem 1717, Crux Mathematicorum, page 30, Vol. 19, Jan. 93.

Crossrefs

Cf. A000005, A058957, A016825.

Programs

Mathematica

nn = 100; t = Table[0, {nn}]; Do[n = a^2 - b^2; If[n <= nn, t[[n]]++], {a, nn}, {b, 0, a - 1}];t (* T. D. Noe, May 04 2011 *)
Table[Length[FindInstance[a^2-b^2==n&&a>b>=0,{a,b},Integers,10]],{n,100}] (* Harvey P. Dale, Jul 28 2021 *)

PARI

a(n)=sum(k=1, sqrtint(n), (n-k^2)%(2*k)==0) \\ Charles R Greathouse IV, Sep 27 2012

PARI

a(n)=sumdiv(n, d, n>=d^2 && (n-d^2)%(2*d)==0) \\ Charles R Greathouse IV, Sep 27 2012

Python

from sympy import divisor_count as d
def a(n): return (d(n)+1)//2 if n%2==1 else ((d(n//4)+1)//2 if n%4==0 else 0)
# Ely Golden, Jan 26 2025

Formula

From Naohiro Nomoto, Feb 26 2002: (Start)

a(2k) = A038548(2k) - A001227(k).

a(2k+1) = A038548(2k+1). (End)

From Bernard Schott, Apr 11 2019: (Start) (see Crux link)

a(n) = 0 if n == 2 (mod 4)

a(n) = floor((A000005(n) + 1)/2) if n == 1 or n == 3 (mod 4)

a(n) = floor((A000005(n/4) + 1)/2) if n == 0 (mod 4). (End)

G.f.: Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^(2*k-1). - Ilya Gutkovskiy, Apr 18 2019

G.f.: Sum_{n>=1} x^(n^2)/(1-x^(2*n)) (conjecture). - Joerg Arndt, Jan 04 2024