A000118 -id:A000118 - cp's OEIS frontend

A008438 Sum of divisors of 2*n + 1.

1, 4, 6, 8, 13, 12, 14, 24, 18, 20, 32, 24, 31, 40, 30, 32, 48, 48, 38, 56, 42, 44, 78, 48, 57, 72, 54, 72, 80, 60, 62, 104, 84, 68, 96, 72, 74, 124, 96, 80, 121, 84, 108, 120, 90, 112, 128, 120, 98, 156, 102, 104, 192, 108, 110, 152, 114, 144, 182, 144, 133, 168

Offset: 0

N. J. A. Sloane

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
Number of ways of writing n as the sum of 4 triangular numbers.
Bisection of A000203. - Omar E. Pol, Mar 14 2012
a(n) is also the total number of parts in all partitions of 2*n + 1 into equal parts. - Omar E. Pol, Feb 14 2021

			Divisors of 9 are 1,3,9, so a(4)=1+3+9=13.
F_2(z) = eta(4z)^8/eta(2z)^4 = q + 4q^3 + 6q^5 +8q^7 + 13q^9 + ...
G.f. = 1 + 4*x + 6*x^2 + 8*x^3 + 13*x^4 + 12*x^5 + 14*x^6 + 24*x^7 + 18*x^8 + 20*x^9 + ...
B(q) = q + 4*q^3 + 6*q^5 + 8*q^7 + 13*q^9 + 12*q^11 + 14*q^13 + 24*q^15 + 18*q^17 + ...

B. C. Berndt, Ramanujan's Notebooks Part III, Springer-Verlag, see p. 139 Ex. (iii).
J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, p. 102.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 19 eq. (6), and p. 283 eq. (8).
W. Dunham, Euler: The Master of Us All, The Mathematical Association of America Inc., Washington, D.C., 1999, p. 12.
H. M. Farkas, I. Kra, Cosines and triangular numbers, Rev. Roumaine Math. Pures Appl., 46 (2001), 37-43.
N. J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 79, Eq. (32.31).
N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer-Verlag, 1984, see p. 184, Prop. 4, F(z).
G. Polya, Induction and Analogy in Mathematics, vol. 1 of Mathematics and Plausible Reasoning, Princeton Univ. Press, 1954, page 92 ff.

N. J. A. Sloane, Table of n, a(n) for n = 0..20000 (first 10000 terms from T. D. Noe)
H. Cohen, Sums involving the values at negative integers of L-functions of quadratic characters, Math. Ann. 217 (1975), no. 3, 271-285. MR0382192 (52 #3080)
M. D. Hirschhorn, The number of representations of a number by various forms, Discrete Mathematics 298 (2005), 205-211
Masao Koike, Modular forms on non-compact arithmetic triangle groups, Unpublished manuscript [Extensively annotated with OEIS A-numbers by N. J. A. Sloane, Feb 14 2021. I wrote 2005 on the first page but the internal evidence suggests 1997.]
K. Ono, S. Robins and P. T. Wahl, On the representation of integers as sums of triangular numbers, Aequationes mathematicae, August 1995, Volume 50, Issue 1-2, pp 73-94. Theorem 3 [Legendre].
H. Rosengren, Sums of triangular numbers from the Frobenius determinant, arXiv:math/0504272 [math.NT], 2005.
Michael Somos, Introduction to Ramanujan theta functions
Min Wang and Zhi-Hong Sun, On the number of representations of n as a linear combination of four triangular numbers II, arXiv:1511.00478 [math.NT], 2015.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions
K. S. Williams, The parents of Jacobi's four squares theorem are unique, Amer. Math. Monthly, 120 (2013), 329-345.

Cf. A000118, A000593, A005879, A096727, A115607, A129588, A225699/A225700.
Number of ways of writing n as a sum of k triangular numbers, for k=1,...: A010054, A008441, A008443, A008438, A008439, A008440, A226252, A007331, A226253, A226254, A226255, A014787, A014809.
Cf. A000203, A002131, A005408, A062731, A099774, A125061, A138741.

a008438 = a000203 . a005408  -- Reinhard Zumkeller, Sep 22 2014

Basis( ModularForms( Gamma0(4), 2), 124) [2]; /* Michael Somos, Jun 12 2014 */

[DivisorSigma(1, 2*n+1): n in [0..70]]; // Vincenzo Librandi, Aug 01 2017

A008438 := proc(n) numtheory[sigma](2*n+1) ; end proc: # R. J. Mathar, Mar 23 2011

DivisorSigma[1, 2 # + 1] & /@ Range[0, 61] (* Ant King, Dec 02 2010 *)
a[ n_] := SeriesCoefficient[ D[ Series[ Log[ QPochhammer[ -x] / QPochhammer[ x]], {x, 0, 2 n + 1}], x], {x, 0 , 2n}]; (* Michael Somos, Oct 15 2019 *)

PARI
```
{a(n) = if( n<0, 0, sigma( 2*n + 1))};
```

{a(n) = if( n<0, 0, n = 2*n; polcoeff( sum( k=1, (sqrtint( 4*n + 1) + 1)\2, x^(k^2 - k), x * O(x^n))^4, n))}; /* Michael Somos, Sep 17 2004 */

{a(n) = my(A); if( n<0, 0, n = 2*n; A = x * O(x^n); polcoeff( (eta(x^4 + A)^2 / eta(x^2 + A))^4, n))}; /* Michael Somos, Sep 17 2004 */

ModularForms( Gamma0(4), 2, prec=124).1;  # Michael Somos, Jun 12 2014

Expansion of q^(-1/2) * (eta(q^2)^2 / eta(q))^4 = psi(q)^4 in powers of q where psi() is a Ramanujan theta function. - Michael Somos, Apr 11 2004
Expansion of Jacobi theta_2(q)^4 / (16*q) in powers of q^2. - Michael Somos, Apr 11 2004
Euler transform of period 2 sequence [4, -4, 4, -4, ...]. - Michael Somos, Apr 11 2004
a(n) = b(2*n + 1) where b() is multiplicative and b(2^e) = 0^n, b(p^e) =(p^(e+1) - 1) / (p - 1) if p>2. - Michael Somos, Jul 07 2004
Given g.f. A(x), then B(q) = q * A(q^2) satisfies 0 = f(B(q), B(q^2), B(q^4)) where f(u, v, w) = v^3 + 8*w*v^2 + 16*w^2*v - u^2*w - Michael Somos, Apr 08 2005
Given g.f. A(x), then B(q) = q * A(q^2) satisfies 0 = f(B(q), B(q^3), B(q^9)) where f(u, v, w) = v^4 - 30*u*v^2*w + 12*u*v*w*(u + 9*w) - u*w*(u^2 + 9*w*u + 81*w^2).
Given g.f. A(x), then B(q) = q * A(q^2) satisfies 0 = f(B(q), B(q^2), B(q^3), B(q^6)) where f(u1, u2, u3, u6) = u2^3 + u1^2*u6 + 3*u2*u3^2 + 27*u6^3 - u1*u2*u3 - 3*u1*u3*u6 - 7*u2^2*u6 - 21*u2*u6^2. - Michael Somos, May 30 2005
G.f.: Sum_{k>=0} (2k + 1) * x^k / (1 - x^(2k + 1)).
G.f.: (Product_{k>0} (1 - x^k) * (1 + x^k)^2)^4. - Michael Somos, Apr 11 2004
G.f. Sum_{k>=0} a(k) * x^(2*k + 1) = x * (Product_{k>0} (1 - x^(4*k))^2 / (1 - x^(2*k)))^4 = x * (Sum_{k>0} x^(k^2 - k))^4 = Sum_{k>0} k * (x^k / (1 - x^k) - 3 * x^(2*k) / (1 - x^(2*k)) + 2 * x^(4*k) / (1 - x^(4*k))). - Michael Somos, Jul 07 2004
Number of solutions of 2*n + 1 = (x^2 + y^2 + z^2 + w^2) / 4 in positive odd integers. - Michael Somos, Apr 11 2004
8 * a(n) = A005879(n) = A000118(2*n + 1). 16 * a(n) = A129588(n). a(n) = A000593(2*n + 1) = A115607(2*n + 1).
a(n) = A000203(2*n+1). - Omar E. Pol, Mar 14 2012
G.f. is a period 1 Fourier series which satisfies f(-1 / (4 t)) = (1/4) (t/i)^2 g(t) where q = exp(2 Pi i t) and g() is the g.f. for A096727. Michael Somos, Jun 12 2014
a(0) = 1, a(n) = (4/n)*Sum_{k=1..n} A002129(k)*a(n-k) for n > 0. - Seiichi Manyama, May 06 2017
G.f.: exp(Sum_{k>=1} 4*(x^k/k)/(1 + x^k)). - Ilya Gutkovskiy, Jul 31 2017
From Peter Bala, Jan 10 2021: (Start)
a(n) = A002131(2*n+1).
G.f.: Sum_{n >= 0} x^n*(1 + x^(2*n+1))/(1 - x^(2*n+1))^2. (End)
Sum_{k=1..n} a(k) ~ Pi^2 * n^2 / 8. - Vaclav Kotesovec, Aug 07 2022
Convolution of A125061 and A138741. - Michael Somos, Mar 04 2023

Comments from Len Smiley, Enoch Haga

A064987 a(n) = n*sigma(n).

Original entry on oeis.org

1, 6, 12, 28, 30, 72, 56, 120, 117, 180, 132, 336, 182, 336, 360, 496, 306, 702, 380, 840, 672, 792, 552, 1440, 775, 1092, 1080, 1568, 870, 2160, 992, 2016, 1584, 1836, 1680, 3276, 1406, 2280, 2184, 3600, 1722, 4032, 1892, 3696, 3510, 3312, 2256, 5952

Offset: 1

Vladeta Jovovic, Oct 30 2001

Dirichlet convolution of sigma_2(n)=A001157(n) with phi(n)=A000010(n). - Vladeta Jovovic, Oct 27 2002
Equals row sums of triangle A143311 and of triangle A143308. - Gary W. Adamson, Aug 06 2008
a(n) is also the sum of all n's present in A244580, or in other words, a(n) is also the volume (or number of cubes) below the terraces of the n-th level of the staircase described in A244580 (see also A237593). - Omar E. Pol, Oct 11 2018
If n is a superperfect number then sigma(n) is a Mersenne prime and a(n) is a perfect number, a(A019279(k)) = A000396(k), k >= 1, assuming there are no odd perfect numbers. - Omar E. Pol, Apr 15 2020

B. C. Berndt, Ramanujan's theory of theta-functions, Theta functions: from the classical to the modern, Amer. Math. Soc., Providence, RI, 1993, pp. 1-63. MR 94m:11054. see page 43.
G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, AMS Chelsea Publishing, Providence, Rhode Island, 2002, pp. 166-167.

Harry J. Smith, Table of n, a(n) for n = 1..1000
Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
Passawan Noppakaew and Prapanpong Pongsriiam, Product of Some Polynomials and Arithmetic Functions, J. Int. Seq. (2023) Vol. 26, Art. 23.9.1.
Michel Planat, Twelve-dimensional Pauli group contextuality with eleven rays, arXiv:1201.5455 [quant-ph], 2012.

Main diagonal of A319073.

Cf. A000203, A038040, A002618, A000010, A001157, A143308, A143311, A004009, A006352, A000594, A126832, A069097 (Mobius transform), A001001 (inverse Mobius transform), A237593, A244580.

a:=List([1..50],n->n*Sigma(n));; Print(a); # Muniru A Asiru, Jan 01 2019

a064987 n = a000203 n * n  -- Reinhard Zumkeller, Jan 21 2014

[n*SumOfDivisors(n): n in [1..70]]; // Vincenzo Librandi, Jan 01 2019

with(numtheory): [n*sigma(n)$n=1..50]; # Muniru A Asiru, Jan 01 2019

# DivisorSigma[1,#]&/@Range[80]  (* Harvey P. Dale, Mar 12 2011 *)

numlib::sigma(n)*n$ n=1..81 // Zerinvary Lajos, May 13 2008

PARI
```
{a(n) = if ( n==0, 0, n * sigma(n))}
```

{ for (n=1, 1000, write("b064987.txt", n, " ", n*sigma(n)) ) } \\ Harry J. Smith, Oct 02 2009

Multiplicative with a(p^e) = p^e * (p^(e+1) - 1) / (p - 1).
G.f.: Sum_{n>0} n^2*x^n/(1-x^n)^2. - Vladeta Jovovic, Oct 27 2002
G.f.: phi_{2, 1}(x) where phi_{r, s}(x) = Sum_{n, m>0} m^r * n^s * x^{m*n}. - Michael Somos, Apr 02 2003
G.f. is also (Q - P^2) / 288 where P, Q are Ramanujan Lambert series. - Michael Somos, Apr 02 2003. See the Hardy reference, p. 136, eq. (10.5.4) (with a proof). For Q and P, (10.5.6) and (10.5.5), see E_4 A004009 and E_2 A006352, respectively. - Wolfdieter Lang, Jan 30 2017
Convolution of A000118 and A186690. Dirichlet convolution of A000027 and A000290. - Michael Somos, Mar 25 2012
Dirichlet g.f.: zeta(s-1)*zeta(s-2). - R. J. Mathar, Feb 16 2011
a(n) = A009194(n)*A009242(n). - Michel Marcus, Oct 23 2013
a(n) (mod 5) = A126832(n) = A000594(n) (mod 5). See A126832 for references. - Wolfdieter Lang, Feb 03 2017
L.g.f.: Sum_{k>=1} k*x^k/(1 - x^k) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, May 13 2017
Sum_{k>=1} 1/a(k) = 1.4383899259334187832765458631783591251241657856627653748389234270650138768... - Vaclav Kotesovec, Sep 20 2020
From Peter Bala, Jan 21 2021: (Start)
G.f.: Sum_{n >= 1} n*q^n*(1 + q^n)/(1 - q^n)^3 (use the expansion x*(1 + x)/(1 - x)^3 = x + 2^2*x^2 + 3^2*x^3 + 4^2*x^4 + ...).
A faster converging g.f.: Sum_{n >= 1} q^(n^2)*( n^3*q^(3*n) - (n^3 + 3*n^2 - n)*q^(2*n) - (n^3 - 3*n^2 - n)*q^n + n^3 )/(1 - q^n)^3 - differentiate equation 5 in Arndt w.r.t. both x and q and then set x = 1. (End)
From Richard L. Ollerton, May 07 2021: (Start)
a(n) = Sum_{k=1..n} sigma_2(gcd(n,k)).
a(n) = Sum_{k=1..n} sigma_2(n/gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)). (End)
From Peter Bala, Jan 22 2024: (Start)
a(n) = Sum_{1 <= j, k <= n} sigma_1( gcd(j, k, n) ).
a(n) = Sum_{d divides n} sigma_1(d)*J_2(n/d) = Sum_{d divides n} sigma_2(d)* phi(n/d), where the Jordan totient function J_2(n) = A007434(n). (End)

A281976 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that both x and x + 24*y are squares.

Original entry on oeis.org

1, 2, 3, 2, 2, 3, 3, 2, 1, 3, 4, 2, 1, 2, 2, 2, 2, 3, 5, 2, 3, 3, 2, 1, 1, 4, 5, 4, 2, 2, 4, 3, 3, 3, 6, 2, 6, 5, 3, 3, 3, 7, 6, 2, 2, 5, 4, 1, 2, 3, 7, 6, 8, 4, 5, 5, 2, 4, 5, 2, 3, 5, 3, 4, 2, 5, 9, 4, 5, 4, 5, 1, 3, 5, 4, 5, 5, 4, 2, 3, 3

Offset: 0

Zhi-Wei Sun, Feb 04 2017

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 8, 12, 23, 24, 47, 71, 168, 344, 632, 1724).
By the linked JNT paper, any nonnegative integer can be written as the sum of a fourth power and three squares.
We have verified a(n) > 0 for all n = 0..10^7.
See also A281977, A282013 and A282014 for similar conjectures.
a(n) <= A273404(n). Starts to differ from A273404 at n=145. - R. J. Mathar, Feb 12 2017
Qing-Hu Hou at Tianjin Univ. has verified a(n) > 0 for all n = 0..10^10.
I would like to offer 2400 US dollars for the first proof of my conjecture that a(n) > 0 for any nonnegative integer n. - Zhi-Wei Sun, Feb 14 2017

			a(8) = 1 since 8 = 0^2 + 0^2 + 2^2 + 2^2 with 0 = 0^2 and 0 + 24*0 = 0^2.
a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + 3^2 with 1 = 1^2 and 1 + 24*1 = 5^2.
a(23) = 1 since 23 = 1^2 + 2^2 + 3^2 + 3^2 with 1 = 1^2 and 1 + 24*2 = 7^2.
a(24) = 1 since 24 = 4^2 + 0^2 + 2^2 + 2^2 with 4 = 2^2 and 4 + 24*0 = 2^2.
a(47) = 1 since 47 = 1^2 + 1^2 + 3^2 + 6^2 with 1 = 1^2 and 1 + 24*1 = 5^2.
a(71) = 1 since 71 = 1^2 + 5^2 + 3^2 + 6^2 with 1 = 1^2 and 1 + 24*5 = 11^2.
a(168) = 1 since 168 = 4^2 + 4^2 + 6^2 + 10^2 with 4 = 2^2 and 4 + 24*4 = 10^2.
a(344) = 1 since 344 = 4^2 + 0^2 + 2^2 + 18^2 with 4 = 2^2 and 4 + 24*0 = 2^2.
a(632) = 1 since 632 = 0^2 + 6^2 + 14^2 + 20^2 with 0 = 0^2 and 0 + 24*6 = 12^2.
a(1724) = 1 since 1724 = 25^2 + 1^2 + 3^2 + 33^2 with 25 = 5^2 and 25 + 24*1 = 7^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.
Zhi-Wei Sun, The 24-conjecture with $2400 prize, a message to Number Theory List, Feb. 14, 2017.

Cf. A000118, A000290, A000583, A270969, A273404, A281939, A281941, A281975, A281977, A281980, A282013, A282014.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[x^2+24y],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,Sqrt[(n-x^4-y^2)/2]}];Print[n," ",r];Continue,{n,0,80}]

A271510 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0, z >= 0 and w >= 0 such that x^2 + 8y^2 + 16z^2 is a square.

Original entry on oeis.org

1, 3, 3, 2, 4, 4, 1, 1, 3, 4, 5, 2, 3, 5, 2, 1, 4, 5, 5, 3, 4, 2, 2, 1, 1, 8, 5, 4, 4, 4, 2, 2, 3, 3, 7, 2, 6, 7, 3, 3, 5, 6, 4, 6, 2, 4, 4, 1, 3, 6, 9, 4, 8, 5, 6, 2, 2, 6, 10, 4, 1, 5, 3, 7, 4, 10, 3, 5, 5, 2, 4, 1, 5, 6, 7, 2, 6, 1, 7, 4, 4

Offset: 0

Zhi-Wei Sun, Apr 09 2016

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 7, 23, 71, 77, 105, 191, 215, 311, 335, 2903, 4^k*q (k = 0,1,2,... and q = 6, 15, 47, 138).
(ii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with x >= y >= 0, z >=0 and w >= 0 such that 4*x^2 + 21*y^2 + 24*z^2 (or 5*x^2 + 40*y^2 + 4*z^2, 20*x^2 + 85*y^2 +16*z^2, 25*x^2 + 480*y^2 + 96*z^2, 36*x^2 + 45*y^2 + 40*z^2, 40*x^2 + 72*y^2 + 9*z^2) is a square.
(iii) For any ordered pair (b, c) = (48, 112), (63, 7), (112, 1008), (136, 24), (136, 216), (360, 40), (840, 280), (1008, 112), each natural number can be written as x^2 + y^2 + z^2 + w^2 with x >= y >= 0, z >=0 and w >= 0 such that 9*x^2 + b*y^2 + c*z^2 is a square.
(iv) For any ordered pair (b, c) = (80, 25), (81, 48), (144, 9), (144, 153), (177, 48), each natural number can be written as x^2 + y^2 + z^2 + w^2 with x >= y >= 0, z >=0 and w >= 0 such that 16*x^2 + b*y^2 + c*z^2 is a square.
This conjecture is much stronger than Lagrange's four-square theorem. It is apparent that a(m^2*n) >= a(n) for all m,n = 1,2,3,....
See also A271513 and A271518 for related conjectures.
Conjectures (i), including the "a(n) = 1" part, (ii), (iii), and (iv) have been verified for n <= 10^9. - Mauro Fiorentini, Jun 19 2024

			a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 1 = 1 and 1^2 + 8*1^2 + 16*0^2 = 3^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 = 1 and 1^2 + 8*1^2 + 16*1^2 = 5^2.
a(15) = 1 since 15 = 3^2 + 1^2 + 2^2 + 1^2 with 3 > 1 and 3^2 + 8*1^2 + 16*2^2 = 9^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 3 > 1 and 3^2 + 8*1^2 + 16*2^2 = 9^2.
a(47) = 1 since 47 = 3^2 + 2^2 + 5^2 + 3^2 with 3 > 2 and 3^2 + 8*2^2 + 16*5^2 = 21^2.
a(71) = 1 since 71 = 7^2 + 2^2 + 3^2 + 3^2 with 7 > 2 and 7^2 + 8*2^2 + 16*3^2 = 15^2.
a(77) = 1 since 77 = 5^2 + 4^2 + 6^2 + 0^2 with 5 > 4 and 5^2 + 8*4^2 + 16*6^2 = 27^2.
a(105) = 1 since 105 = 6^2 + 2^2 + 4^2 + 7^2 with 6 > 2 and 6^2 + 8*2^2 + 16*4^2 = 18^2.
a(138) = 1 since 138 = 3^2 + 2^2 + 5^2 + 10^2 with 3 > 2 and 3^2 + 8*2^2 + 16*5^2 = 21^2.
a(191) = 1 since 191 = 9^2 + 3^2 + 1^2 + 10^2 with 9 > 3 and 9^2 + 8*3^2 + 16*1^2 = 13^2.
a(215) = 1 since 215 = 11^2 + 7^2 + 6^2 + 3^2 with 11 > 7 and 11^2 + 8*7^2 + 16*6^2 = 33^2.
a(311) = 1 since 311 = 15^2 + 6^2 + 1^2 + 7^2 with 15 > 6 and 15^2 + 8*6^2 + 16*1^2 = 23^2.
a(335) = 1 since 335 = 17^2 + 1^2 + 3^2 + 6^2 with 17 > 1 and 17^2 + 8*1^2 + 16*3^2 = 21^2.
a(2903) = 1 since 2903 = 49^2 + 14^2 + 15^2 + 9^2 with 49 > 14 and 49^2 + 8*14^2 + 16*15^2 = 87^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. Also available from arXiv:1604.06723 [math.NT], 2016-2017.

Cf. A000118, A000290, A270969, A271513, A271518.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x^2+8y^2+16z^2],r=r+1],{y,0,Sqrt[n/2]},{x,y,Sqrt[n-y^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,0,80}]

A271513 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with 3x^2 + 4y^2 + 9*z^2 a square, where w, x, y and z are nonnegative integers.

Original entry on oeis.org

1, 3, 2, 1, 4, 6, 3, 2, 2, 5, 6, 1, 2, 5, 4, 2, 4, 4, 3, 2, 6, 5, 1, 1, 3, 8, 6, 2, 4, 6, 6, 4, 2, 3, 8, 3, 7, 7, 1, 6, 6, 8, 6, 1, 2, 11, 7, 1, 2, 12, 8, 2, 7, 5, 9, 4, 4, 4, 7, 2, 4, 9, 4, 7, 4, 11, 6, 1, 5, 8, 7

Offset: 0

Zhi-Wei Sun, Apr 09 2016

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 3, 11, 23, 43, 47, 67, 83, 107, 155, 323, 683, 803, 4^k*m (k = 0,1,2,... and m = 22, 38). [Conjecture verified for all natural numbers up to 10^9. - Mauro Fiorentini, Jul 04 2024]
(ii) Any natural number can be written as w^2 + x^2 + y^2 + z^2 with x, y, z integers and a*x^2 + b*y^2 + c*z^2 a square, whenever (a,b,c) is among the following triples: (1,3,12), (1,3,18), (1,3,21), (1,3,60), (1,5,15), (1,8,24), (1,12,15), (1,24,56), (1,24,72), (1,48,72), (1,48,168), (1,120,180), (1,192,288), (1,280,560), (3,9,13), (4,5,12), (4,5,60), (4,9,60), (4,12,21), (4,12,45), (4,12,69), (4,12,93), (4,12,237), (4,21,24), (4,21,36), (4,21,504), (4,24,93), (4,28,77), (4,45,120), (4,45,540), (4,45,600), (5,36,40), (7,9,126), (7,9,588), (8,16,73), (8,16,97), (8,49,112), (9,13,27), (9,16,24), (9,19,36), (9,21,91), (9,24,232), (9,28,63), (9,40,45), (9,40,56), (9,40,120), (9,45,115),(9,45,235), (12,13,24), (12,13,36), (12,36,37), (12,36,133), (13,36,72), (13,36,108), (15,24,25), (15,49,105), (16,17,48), (16,20,45), (16,21,84), (16,33,72), (16,33,176), (16,45,180), (16,48,57), (16,48,105), (16,48,233), (16,48,249), (19,45,57), (19,45,180), (21,25,35), (21,25,75), (21,28,36), (21,28,60), (21,43,105), (21,100,105),(24,25,72), (24,25,120), (24,48,97), (24,81,184), (24,120,145), (25,36,75), (25,40,56), (25,45,51), (25,45,99), (25,48,96), (25,48,144), (25,54,90), (25,75,81), (25,80,184), (25,96,120), (25,200,216), (28,33,36), (28,36,77), (28,72,189), (32,64,73), (33,36,220), (33,48,144), (33,72,256), (33,88,144), (36,45,100), (36,45,172), (37,81,243), (40,81,120), (40,81,240), (41,64,256), (45,48,76), (48,144,177), (49,56,64), (49,63,72), (55,141,165), (57,64,192), (60,105,196), (64,65,160), (72,73,144), (81,160,240), (85,140,196), (105,112,144), (112,144,153), (136,144,153), (144,145,240), (144,160,225),(148,189,252), (175,189,225). [Conjecture verified for all triples and all natural numbers up to 10^9. - Mauro Fiorentini, Jul 04 2024]
(iii) If a, b and c are positive integers such that any natural number can be written as w^2 + x^2 + y^2 + z^2 with x, y, z integers and a*x^2 + b*y^2 + c*z^2 a square, then a, b and c cannot be pairwise coprime.
This conjecture is stronger than Lagrange's four-square theorem. Moreover, there are many other suitable triples (a,b,c) for our purpose not listed in part (ii) of the conjecture. If a, b and c are positive integers such that any natural number can be written as w^2 + x^2 + y^2 + z^2 with x, y, z integers and a*x^2 + b*y^2 + c*z^2 a square, then one of a+b+c, 4*a+b+c, a+4*b+c and a+b+4*c must be a square since 2^2 + 1^2 + 1^2 + 1^2 is the unique way to express 7 as a sum of four squares.
Obviously, a(m^2*n) >= a(n) for all m,n = 1,2,3,....
See also A271510 and A271518 for related conjectures.

			a(3) = 1 since 3 = 0^2 + 1^2 + 1^2 + 1^2 with 3*1^2 + 4*1^2 + 9*1^2 = 4^2.
a(11) = 1 since 11 = 1^2 + 3^2 + 0^2 + 1^2 with 3*3^2 + 4*0^2 + 9*1^2 = 6^2.
a(22) = 1 since 22 = 4^2 + 2^2 + 1^2 + 1^2 with 3*2^2 + 4*1^2 + 9*1^2 = 5^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 3*1^2 + 4*2^2 + 9*3^2 = 10^2.
a(38) = 1 since 38 = 0^2 + 6^2 + 1^2 + 1^2 with 3*6^2 + 4*1^2 + 9*1^2 = 11^2.
a(43) = 1 since 43 = 4^2 + 3^2 + 3^2 + 3^2 with 3*3^2 + 4*3^2 + 9*3^2 = 12^2.
a(47) = 1 since 47 = 3^2 + 6^2 + 1^2 + 1^2 with 3*6^2 + 4*1^2 + 9*1^2 = 11^2.
a(67) = 1 since 67 = 8^2 + 1^2 + 1^2 + 1^2 with 3*1^2 + 4*1^2 + 9*1^2 = 4^2.
a(83) = 1 since 83 = 0^2 + 9^2 + 1^2 + 1^2 with 3*9^2 + 4*1^2 + 9*1^2 = 16^2.
a(107) = 1 since 107 = 9^2 + 3^2 + 4^2 + 1^2 with 3*3^2 + 4*4^2 + 9*1^2 = 10^2.
a(155) = 1 since 155 = 0^2 + 9^2 + 5^2 + 7^2 with 3*9^2 + 4*5^2 + 9*7^2 = 28^2.
a(323) = 1 since 323 = 3^2 + 15^2 + 8^2 + 5^2 with 3*15^2 + 4*8^2 + 9*5^2 = 34^2.
a(683) = 1 since 683 = 15^2 + 11^2 + 16^2 + 9^2 with 3*11^2 + 4*16^2 + 9*9^2 = 46^2.
a(803) = 1 since 803 = 24^2 + 13^2 + 7^2 + 3^2 with 3*13^2 + 4*7^2 + 9*3^2 = 28^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. Also available from arXiv:1604.06723 [math.NT], 2016-2017.

Cf. A000118, A000290, A270969, A271510, A271518.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[3x^2+4y^2+9z^2],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,0,70}]

A025428 Number of partitions of n into 4 nonzero squares.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 3, 0, 1, 2, 0, 1, 2, 1, 2, 2, 1, 2, 1, 0, 3, 2, 1, 2, 1, 2, 1, 2, 2, 1, 4, 1, 2, 3, 0, 2, 4, 1, 3, 2, 1, 4, 1, 1, 3, 3, 2, 2, 4, 2, 1, 3, 2, 3, 4, 2, 3, 3, 1, 2, 5, 2, 4, 3, 2, 4, 1, 1, 6, 4, 3, 4, 2, 3, 0, 4, 4, 3, 5, 1, 5, 5, 1, 4, 5, 2

Offset: 0

David W. Wilson

Records occur at n= 4, 28, 52, 82, 90, 130, 162, 198, 202, 210,.... - R. J. Mathar, Sep 15 2015

Cf. A000414, A000534, A025357-A025375, A216374, A025416 (greedy inverse).

Column k=4 of A243148.

A025428 := proc(n)
    local a,i,j,k,lsq ;
    a := 0 ;
    for i from 1 do
        if 4*i^2 > n then
            return a;
        end if;
        for j from i do
            if i^2+3*j^2 > n then
                break;
            end if;
            for k from j do
                if i^2+j^2+2*k^2 > n then
                    break;
                end if;
                lsq := n-i^2-j^2-k^2 ;
                if lsq >= k^2 and issqr(lsq) then
                    a := a+1 ;
                end if;
            end do:
        end do:
    end do:
end proc:
seq(A025428(n),n=1..40) ; # R. J. Mathar, Jun 15 2018
# second Maple program:
b:= proc(n, i, t) option remember; `if`(n=0, `if`(t=0, 1, 0),
     `if`(i<1 or t<1, 0, b(n, i-1, t)+`if`(i^2>n, 0, b(n-i^2, i, t-1))))
    end:
a:= n-> b(n, isqrt(n), 4):
seq(a(n), n=0..100);  # Alois P. Heinz, Apr 14 2019

nn = 100; lim = Sqrt[nn]; t = Table[0, {nn}]; Do[n = a^2 + b^2 + c^2 + d^2; If[n <= nn, t[[n]]++], {a, lim}, {b, a, lim}, {c, b, lim}, {d, c, lim}]; t (* T. D. Noe, Sep 28 2012 *)
f[n_] := Length@ IntegerPartitions[n, {4}, Range[ Floor[ Sqrt[n - 1]]]^2]; Array[f, 105] (* Robert G. Wilson v, Sep 28 2012 *)

A025428(n)=sum(a=1,n,sum(b=1,a,sum(c=1,b,sum(d=1,c,a^2+b^2+c^2+d^2==n))))

A025428(n)=sum(a=1,sqrtint(max(n-3,0)), sum(b=1,min(sqrtint(n-a^2-2),a), sum(c=1,min(sqrtint(n-a^2-b^2-1),b),issquare(n-a^2-b^2-c^2,&d) & d <= c )))

A025428(n)=sum(a=sqrtint(max(n,4)\4),sqrtint(max(n-3,0)), sum(b=sqrtint((n-a^2)\3-1)+1,min(sqrtint(n-a^2-2),a), sum(c=sqrtint((t=n-a^2-b^2)\2-1)+1, min(sqrtint(t-1),b), issquare(t-c^2) ))) \\ - M. F. Hasler, Sep 17 2012
for(n=1,100,print1(A025428(n),","))

T(n)={a=matrix(n,4,i,j,0);for(d=1,sqrtint(n),forstep(i=n,d*d+1,-1,for(j=2,4,a[i,j]+=sum(k=1,j,if(k0,a[i-k*d*d,j-k],if(k==j&&i-k*d*d==0,1)))));a[d*d,1]=1);for(i=1,n,print(i" "a[i,4]))} /* Robert Gerbicz, Sep 28 2012 */

For n>0, a(n) = ( A063730(n) + 6*A213024(n) + 3*A063725(n/2) + 8*A092573(n) + 6*A010052(n/4) ) / 24. - Max Alekseyev, Sep 30 2012
a(n) = ( A000118(n) - 4*A005875(n) - 6*A004018(n) - 12*A000122(n) - 15*A000007(n) + 12*A014455(n) - 24*A033715(n) - 12*A000122(n/2) + 12*A004018(n/2) + 32*A033716(n) - 32*A000122(n/3) + 48*A000122(n/4) ) / 384. - Max Alekseyev, Sep 30 2012
a(n) = [x^n y^4] Product_{k>=1} 1/(1 - y*x^(k^2)). - Ilya Gutkovskiy, Apr 19 2019
a(n) = Sum_{k=1..floor(n/4)} Sum_{j=k..floor((n-k)/3)} Sum_{i=j..floor((n-j-k)/2)} A010052(i) * A010052(j) * A010052(k) * A010052(n-i-j-k). - Wesley Ivan Hurt, Apr 19 2019

Values of a(0..10^4) double-checked by M. F. Hasler, Sep 17 2012

A046897 Sum of divisors of n that are not divisible by 4.

Original entry on oeis.org

1, 3, 4, 3, 6, 12, 8, 3, 13, 18, 12, 12, 14, 24, 24, 3, 18, 39, 20, 18, 32, 36, 24, 12, 31, 42, 40, 24, 30, 72, 32, 3, 48, 54, 48, 39, 38, 60, 56, 18, 42, 96, 44, 36, 78, 72, 48, 12, 57, 93, 72, 42, 54, 120, 72, 24, 80, 90, 60, 72, 62, 96, 104, 3, 84, 144, 68, 54, 96, 144, 72

Offset: 1

N. J. A. Sloane

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

The o.g.f. is (theta_3(0,x)^4 - 1)/8, see the Hardy reference, eqs. 9.2.1, 9.2.3 and 9.2.4 on p. 133 for Sum' m*u_m. Also Hardy-Wright, p. 314. See also the Somos, Jan 25 2008 formula below. - Wolfdieter Lang, Dec 11 2016

			G.f. = q + 3*q^2 + 4*q^3 + 3*q^4 + 6*q^5 + 12*q^6 + 8*q^7 + 3*q^8 + 13*q^9 + ...

J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 194.
G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, AMS Chelsea Publishing, Providence, Rhode Island 2002, p. 133.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Clarendon Press, Oxford, Fifth edition, 1979, p. 314.
P. A. MacMahon, Combinatory Analysis, Cambridge Univ. Press, London and New York, Vol. 1, 1915 and Vol. 2, 1916; see vol. 2, p 31, Article 273.
C. J. Moreno and S. S. Wagstaff, Jr., Sums of Squares of Integers, Chapman & Hall, 2006.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Michael Somos, Introduction to Ramanujan theta functions.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.

Cf. A000203, A000118, A051731, A069733, A027748, A124010, A190621, A000593 (not divis. by 2), A046913 (not divis. by 3), A116073 (not divis. by 5).

a046897 1 = 1
a046897 n = product $ zipWith
            (\p e -> if p == 2 then 3 else div (p ^ (e + 1) - 1) (p - 1))
            (a027748_row n) (a124010_row n)
-- Reinhard Zumkeller, Aug 12 2015

A := Basis( ModularForms( Gamma0(4), 2), 72); B := (A[1] - 1)/8 + A[2]; B; /* Michael Somos, Dec 30 2014 */

A046897 := proc(n) if n mod 4 = 0 then numtheory[sigma](n)-4*numtheory[sigma](n/4) ; else numtheory[sigma](n) ; end if; end proc: # R. J. Mathar, Mar 23 2011

a[n_] := Sum[ Boole[ !Divisible[d, 4]]*d, {d, Divisors[n]}]; Table[ a[n], {n, 1, 71}] (* Jean-François Alcover, Dec 12 2011 *)
DivisorSum[#1, # &, Mod[#, 4] != 0 &] & /@ Range[71] (* Jayanta Basu, Jun 30 2013 *)
a[ n_] := SeriesCoefficient[ (EllipticTheta[ 3, 0, q]^4 - 1) / 8, {q, 0, n}]; (* Michael Somos, Dec 30 2014 *)
f[2, e_] := 3; f[p_, e_] := (p^(e+1)-1)/(p-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 15 2020 *)

{a(n) = if( n<1, 0, sumdiv(n, d, if(d%4, d)))};

a(n) = (-1)^(n+1)*Sum_{d divides n} (-1)^(n/d+d)*d. Multiplicative with a(2^e) = 3, a(p^e) = (p^(e+1)-1)/(p-1) for an odd prime p. - Vladeta Jovovic, Sep 10 2002 [For a proof of the multiplicative property, see for example Moreno and Wagstaff, p. 33. - N. J. A. Sloane, Nov 09 2016]
G.f.: Sum_{k>0} x^k/(1+(-x)^k)^2, or Sum_{k>0} k*x^k/(1+(-x)^k). - Vladeta Jovovic, Dec 16 2002
Expansion of (1 - phi(q)^4) / 8 in powers of q where phi() is a Ramanujan theta function. - Michael Somos, Jan 25 2008
Equals inverse Mobius transform of A190621. - Gary W. Adamson, Jul 03 2008
A000118(n) = 8*a(n) for all n>0.
Dirichlet g.f.: (1 - 4^(1-s)) * zeta(s) * zeta(s-1). - Michael Somos, Oct 21 2015
L.g.f.: log(Product_{k>=1} (1 - x^(4*k))/(1 - x^k)) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, Mar 14 2018
From Peter Bala, Dec 19 2021: (Start)
Logarithmic g.f.: Sum_{n >= 1} a(n)*x^n/n = Sum_{n >= 1} x^n*(1 + x^n + x^(2*n))/( n*(1 - x^(4*n)) )
G.f.: Sum_{n >= 1} x^n*(x^(6*n) + 2*x^(5*n) + 3*x^(4*n) + 3*x^(2*n) + 2*x^n + 1)/(1 - x^(4*n))^2. (End)
Sum_{k=1..n} a(k) ~ (Pi^2/16) * n^2. - Amiram Eldar, Oct 04 2022

A271714 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 such that (10w+5x)^2 + (12y+36z)^2 is a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 2, 1, 3, 1, 2, 1, 2, 3, 1, 4, 4, 2, 2, 1, 3, 3, 5, 2, 2, 5, 2, 1, 2, 3, 3, 3, 2, 3, 2, 3, 4, 4, 2, 3, 9, 2, 3, 1, 1, 6, 2, 3, 4, 6, 4, 1, 2, 5, 3, 3, 4, 3, 5, 1, 4, 5, 1, 3, 6, 6, 1, 3, 4, 5, 12, 2, 4, 6, 2, 4

Offset: 1

Zhi-Wei Sun, Apr 12 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 7, 9, 19, 49, 133, 589, 2^k, 2^k*3, 4^k*q (k = 0,1,2,... and q = 14, 67, 71, 199).
(ii) If P(y,z) is one of 2y-3z, 2y-8z and 4y-6z, then any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (w-x)^2 + P(y,z)^2 is a square.
(iii) For each triple (a,b,c) = (1,4,4), (1,12,12), (2,4,8), (2,6,6), (2,12,12), (3,4,4), (3,4,8), (3,8,8), (3,12,12), (3,12,36), (5,4,4), (5,4,8), (5,8,16), (5,36,36), (6,4,4), (7,12,12), (7,20,20), (7,24,24), (9,4,4), (9,12,12),(9,36,36), (11,12,12), (13,4,4), (15,12,12), (16,12,12), (21,20,20), (21,24,24), (23,12,12), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (w+a*x)^2 + (b*y-c*z)^2 is a square.
See also A271510, A271513, A271518, A271644, A271665, A271721 and A271724 for other conjectures refining Lagrange's four-square theorem.

			a(2) = 1 since 2 = 1^2 + 1^2 + 0^2 + 0^2 with (10*1+5*1)^2 + (12*0+36*0)^2 = 15^2 + 0^2 = 15^2.
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with (10*1+5*1)^2 + (12*0+36*1)^2 = 15^2 + 36^2 = 39^2.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with (10*2+5*0)^2 + (12*0+36*0)^2 = 20^2 + 0^2 = 20^2.
a(6) = 1 since 6 = 2^2 + 0^2 + 1^2 + 1^2 with (10*2+5*0)^2 + (12*1+36*1)^2 = 20^2 + 48^2 = 52^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with (10*1+5*2)^2 + (12*1+36*1)^2 = 20^2 + 48^2 = 52^2.
a(9) = 1 since 9 = 3^2 + 0^2 + 0^2 + 0^2 with (10*3+5*0)^2 + (12*0+36*0)^2 = 30^2 + 0^2 = 30^2.
a(19) = 1 since 19 = 3^2 + 0^2 + 3^2 + 1^2 with (10*3+5*0)^2 + (12*3+36*1)^2 = 30^2 + 72^2 = 78^2.
a(49) = 1 since 49 = 7^2 + 0^2 + 0^2 + 0^2 with (10*7+5*0)^2 + (12*0+36*0)^2 = 70^2 + 0^2 = 70^2.
a(133) = 1 since 133 = 9^2 + 0^2 + 6^2 + 4^2 with (10*9+5*0)^2 + (12*6+36*4)^2 = 90^2 + 216^2 = 234^2.
a(589) = 1 since 589 = 17^2 + 10^2 + 2^2 + 14^2 with (10*17+5*10)^2 + (12*2+36*14)^2 = 220^2 + 528^2 = 572^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723, 2016.

Cf. A000118, A000290, A271510, A271513, A271518, A271608, A271644, A271665, A271719, A271721, A271724.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(10*Sqrt[n-x^2-y^2-z^2]+5x)^2+(12y+36z)^2],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]

A271665 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 such that w^2 + 4xy + 8yz + 32zx is a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 1, 1, 6, 3, 1, 3, 1, 6, 2, 1, 7, 10, 1, 1, 9, 3, 2, 6, 2, 2, 3, 3, 8, 10, 1, 1, 10, 2, 2, 3, 5, 8, 11, 1, 7, 13, 2, 6, 16, 6, 1, 2, 6, 2, 3, 1, 3, 16, 4, 7, 9, 3, 2, 10, 4, 9, 4, 1, 8, 15, 1, 1, 15, 5, 2, 9, 6, 8, 2, 3, 10, 13, 4, 2, 17, 7, 1, 6

Offset: 1

Zhi-Wei Sun, Apr 12 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*3^m, 4^k*3^m*43, 4^k*9^m*q (k,m = 0, 1, 2, ... and q = 7, 15, 79, 95, 141, 159, 183).
(ii) Any positive integer n can be written as w^2 + x^2 + y^2 + z^2 with w*x + x*y + 2*y*z + 3*z*x (or w*x + 3*x*y + 8*y*z + 5*z*x) twice a square, where w is a positive integer and x,y,z are nonnegative integers.
(iii) For each k = 1, 2, 8, any positive integer can be written as w^2 + x^2 + y^2 + z^2 with w^2 + k*(x*y+y*z) a square, where w is a positive integer and x,y,z are nonnegative integers.
(iv) For each ordered pair (b,c) = (16,4), (24,4), (32,16), any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x^2 + b*y^2 + c*x*z + c*y*z + c*z*w is a square.
We also guess that for each triple (b,c,d) = (1,3,4), (1,6,8), (1,7,24), (1,8,15), (1,10,24), (1,12,35), (1,14,48), (1,20,48), (2,1,2), (2,4,2), (2,4,7), (2,6,7), (2,8,4), (2,8,14), (2,8,31), (2,10,23), (2,12,14), (2,12,34), (2,14,47), (3,1,1), (3,1,4), (3,1,16), (3,2,14), (3,2,17), (3,2,38), (3,3,3), (3,3,13), (3,4,1), (3,4,4), (3,5,11), (3,6,3), (3,6,6), (3,6,26), (3,8,2), (3,8,13), (3,8,22), (3,9,39), (3,12,12), (3,12,33), (3,15,1), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and x^2 + b*y^2 + c*x*z + d*y*z a square.
See also A271510, A271513, A271518 and A271644 for other conjectures refining Lagrange's four-square theorem.

			a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 1^2 + 4*0*1 + 8*1*1 + 32*1*0 = 3^2.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 + 4*2*0 + 8*0*0 + 32*0*0 = 2^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1^2 + 4*2*1 + 8*1*1 + 32*1*2 = 9^2.
a(15) = 1 since 15 = 1^2 + 2^2 + 1^2 + 3^2 with 1^2 + 4*2*1 + 8*1*3 + 32*3*2 = 15^2.
a(43) = 1 since 43 = 3^2 + 3^2 + 4^2 + 3^2 with 3^2 + 4*3*4 + 8*4*3 + 32*3*3 = 21^2.
a(79) = 1 since 79 = 5^2 + 3^2 + 6^2 + 3^2 with 5^2 + 4*3*6 + 8*6*3 + 32*3*3 = 23^2.
a(95) = 1 since 95 = 5^2 + 6^2 + 5^2 + 3^2 with 5^2 + 4*6*5 + 8*5*3 + 32*3*6 = 29^2.
a(129) = 1 since 129 = 5^2 + 6^2 + 8^2 + 2^2 with 5^2 + 4*6*8 + 8*8*2 + 32*2*6 = 27^2.
a(141) = 1 since 141 = 8^2 + 5^2 + 4^2 + 6^2 with 8^2 + 4*5*4 + 8*4*6 + 32*6*5 = 36^2.
a(159) = 1 since 159 = 11^2 + 1^2 + 6^2 + 1^2 with 11^2 + 4*1*6 + 8*6*1 + 32*1*1 = 15^2.
a(183) = 1 since 183 = 1^2 + 9^2 + 10^2 + 1^2 with 1^2 + 4*9*10 + 8*10*1 + 32*1*9 = 27^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723, 2016.

Cf. A000118, A000290, A271510, A271513, A271518, A271608, A271644.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[4x*y+8*y*z+32*z*x+(n-x^2-y^2-z^2)],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]

A271724 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w(x+2y+3*z) a square, where w,x,y,z are nonnegative integers with x > 0.

Original entry on oeis.org

1, 3, 2, 1, 4, 4, 1, 3, 4, 6, 4, 2, 4, 7, 1, 1, 10, 8, 5, 6, 8, 5, 1, 4, 7, 10, 7, 2, 11, 13, 2, 3, 8, 9, 8, 6, 7, 13, 3, 6, 15, 8, 4, 4, 13, 8, 1, 2, 8, 15, 11, 4, 14, 18, 5, 7, 6, 6, 12, 5, 12, 17, 5, 1, 16, 21, 3, 11, 16, 12, 1, 8, 8, 18, 16, 5, 16, 12, 4, 6

Offset: 1

Zhi-Wei Sun, Apr 13 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 7, 15, 47, 151, 4^k*q (k = 0,1,2,... and q = 1, 23, 71).
(ii) For positive integers a,b,c with gcd(a,b,c) squarefree, any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and w*(a*x+b*y+c*z) a square, if and only if {a,b,c} is among {1,2,3}, {1,3,6}, {1,6,9}, {5,6,9}, {18,30,114}.
(iii) For each quadruple (a,b,c,d) = (1,1,2,12), (1,2,7,60), (1,3,9,48), (1,4,11,48), (1,5,8,24), (1,8,11,24), (2,6,8,15), (3,5,6,24), (3,6,15,40), (3,6,18,40), (3,12,15,20), (4,4,8,15), (4,8,12,21), (4,8,12,45), (4,8,20,15), (4,8,36,45), (5,10,15,24), (6,9,15,20), (7,14,28,60), (7,21,28,60), (7,21,42,60), (12,36,48,55), (14,21,28,60), (3,9,18,112), (3,21,33,80), (4,5,9,120), (4,12,16,105), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (a*x+b*y+c*z)^2 + (d*w)^2 is a square.
See also A271510, A271513, A271518, A271644, A271665, A271714 and A271721 for other conjectures refining Lagrange's four-square theorem.

			a(1) = 1 since 1 = 0^2 + 1^2 + 0^2 + 0^2 with 1 > 0 and 0*(1+2*0+3*0) = 0^2.
a(3) = 2 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 1*(1+2*0+3*1) = 2^2, and 3 = 0^2 + 1^2 + 1^2 + 1^2 with 0*(1+2*1+3*1) = 0^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1*(1+2*1+3*2) = 3^2.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^2 + 1^2 with 2*(3+2*1+3*1) = 4^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with 1*(3+2*2+3*3) = 4^2.
a(31) = 2 since 31 = 2^2 + 1^2 + 1^2 + 5^2 with 2*(1+2*1+3*5) = 6^2, and also 31 = 2^2 + 3^2 + 3^2 + 3^2 with 2*(3+2*3+3*3) = 6^2.
a(47) = 1 since 47 = 1^2 + 1^2 + 3^2 + 6^2 with 1*(1+2*3+3*6) = 5^2.
a(71) = 1 since 71 = 1^2 + 6^2 + 5^2 + 3^2 with 1*(6+2*5+3*3) = 5^2.
a(151) = 1 since 151 = 9^2 + 6^2 + 5^2 + 3^2 with 9*(6+2*5+3*3) = 15^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723, 2016.

Cf. A000118, A000290, A259789, A271510, A271513, A271518, A271608, A271644, A271665, A271714, A271721.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[Sqrt[n-x^2-y^2-z^2](x+2y+3z)],r=r+1],{x,1,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Label[aa];Continue,{n,1,80}]

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A271513 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with 3*x^2 + 4*y^2 + 9*z^2 a square, where w, x, y and z are nonnegative integers.

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A025428 Number of partitions of n into 4 nonzero squares.

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A271510 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0, z >= 0 and w >= 0 such that x^2 + 8y^2 + 16z^2 is a square.

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