A158092 -id:A158092 - cp's OEIS frontend

A063865 Number of solutions to +- 1 +- 2 +- 3 +- ... +- n = 0.

1, 0, 0, 2, 2, 0, 0, 8, 14, 0, 0, 70, 124, 0, 0, 722, 1314, 0, 0, 8220, 15272, 0, 0, 99820, 187692, 0, 0, 1265204, 2399784, 0, 0, 16547220, 31592878, 0, 0, 221653776, 425363952, 0, 0, 3025553180, 5830034720, 0, 0, 41931984034, 81072032060, 0, 0

Offset: 0

N. J. A. Sloane, suggested by J. H. Conway, Aug 27 2001

Number of sum partitions of the half of the n-th-triangular number by distinct numbers in the range 1 to n. Example: a(7)=8 since triangular(7)=28 and 14 = 2+3+4+5 = 1+3+4+6 = 1+2+5+6 = 3+5+6 = 7+1+2+4 = 7+3+4 = 7+2+5 = 7+1+6. - Hieronymus Fischer, Oct 20 2010
The asymptotic formula below was stated as a conjecture by Andrica & Tomescu in 2002 and proved by B. D. Sullivan in 2013. See his paper and H.-K. Hwang's review MR 2003j:05005 of the JIS paper. - Jonathan Sondow, Nov 11 2013
a(n) is the number of subsets of {1..n} whose sum is equal to the sum of their complement. See example below. - Gus Wiseman, Jul 04 2019

			From _Gus Wiseman_, Jul 04 2019: (Start)
For example, the a(0) = 1 through a(8) = 14 subsets (empty columns not shown) are:
  {}  {3}    {1,4}  {1,6,7}    {3,7,8}
      {1,2}  {2,3}  {2,5,7}    {4,6,8}
                    {3,4,7}    {5,6,7}
                    {3,5,6}    {1,2,7,8}
                    {1,2,4,7}  {1,3,6,8}
                    {1,2,5,6}  {1,4,5,8}
                    {1,3,4,6}  {1,4,6,7}
                    {2,3,4,5}  {2,3,5,8}
                               {2,3,6,7}
                               {2,4,5,7}
                               {3,4,5,6}
                               {1,2,3,4,8}
                               {1,2,3,5,7}
                               {1,2,4,5,6}
(End)

T. D. Noe, N. J. A. Sloane and Ray Chandler, Table of n, a(n) for n = 0..3339 (terms < 10^1000, first 101 terms from T. D. Noe, next 300 terms from N. J. A. Sloane)
Dorin Andrica and Ovidiu Bagdasar, On k-partitions of multisets with equal sums, The Ramanujan J. (2021) Vol. 55, 421-435.
Dorin Andrica, Ovidiu Bagdasar, and George Cătălin Ţurcaş, The Number of Partitions of a Set and Superelliptic Diophantine Equations, Disc. Math. and Applications, Springer, Cham (2020), 35-55.
D. Andrica and E. J. Ionascu, Variations on a result of Erdős and Surányi, INTEGERS 2013 slides.
D. Andrica and I. Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Seq., 5 (2002), Article 02.2.4
Ovidiu Bagdasar and Dorin Andrica, New results and conjectures on 2-partitions of multisets, 2017 7th International Conference on Modeling, Simulation, and Applied Optimization (ICMSAO).
Steven R. Finch, Signum equations and extremal coefficients, February 7, 2009. [Cached copy, with permission of the author]
B. D. Sullivan, On a Conjecture of Andrica and Tomescu, J. Int. Sequences, 16 (2013), Article 13.3.1.
zbMATH, Review of Andrica and Tomescu

Cf. A000980, A025591, A058377, A063866, A063867, A113036, A113037, A141000.
"Decimations": A060468 = 2*A060005, A123117 = 2*A104456.
Analogous sequences for sums of squares and cubes are A158092, A158118, see also A019568. - Pietro Majer, Mar 15 2009
Cf. A053632, A059529, A326173, A326174, A326175.

M:=400; t1:=1; lprint(0,1); for n from 1 to M do t1:=expand(t1*(x^n+1/x^n)); lprint(n, coeff(t1,x,0)); od: # N. J. A. Sloane, Jul 07 2008

f[n_, s_] := f[n, s]=Which[n==0, If[s==0, 1, 0], Abs[s]>(n*(n+1))/2, 0, True, f[ n-1, s-n]+f[n-1, s+n]]; a[n_] := f[n, 0]
nmax = 50; d = {1}; a1 = {};
Do[
  i = Ceiling[Length[d]/2];
  AppendTo[a1, If[i > Length[d], 0, d[[i]]]];
  d = PadLeft[d, Length[d] + 2 n] + PadRight[d, Length[d] + 2 n];
  , {n, nmax}];
a1 (* Ray Chandler, Mar 13 2014 *)

a(n)=my(x='x); polcoeff(prod(k=1,n,x^k+x^-k)+O(x),0) \\ Charles R Greathouse IV, May 18 2015

a(n)=0^n+floor(prod(k=1,n,2^(n*k)+2^(-n*k)))%(2^n) \\ Tani Akinari, Mar 09 2016

Asymptotic formula: a(n) ~ sqrt(6/Pi)*n^(-3/2)*2^n for n = 0 or 3 (mod 4) as n approaches infinity.
a(n) = 0 unless n == 0 or 3 (mod 4).
a(n) = constant term in expansion of Product_{ k = 1..n } (x^k + 1/x^k). - N. J. A. Sloane, Jul 07 2008
If n = 0 or 3 (mod 4) then a(n) = coefficient of x^(n(n+1)/4) in Product_{k=1..n} (1+x^k). - D. Andrica and I. Tomescu.
a(n) = 2*A058377(n) for any n > 0. - Rémy Sigrist, Oct 11 2017

More terms from Dean Hickerson, Aug 28 2001

Corrected and edited by Steven Finch, Feb 01 2009

A083527 a(n) is the number of times that sums 1+-4+-9+-16+-...+-n^2 of the first n squares is zero. There are 2^(n-1) choices for the sign patterns.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 5, 0, 0, 43, 57, 0, 0, 239, 430, 0, 0, 2904, 5419, 0, 0, 27813, 50213, 0, 0, 348082, 649300, 0, 0, 3913496, 7287183, 0, 0, 50030553, 93696497, 0, 0, 611793542, 1161079907, 0, 0, 8009933135, 15176652567, 0, 0

Offset: 1

T. D. Noe, Apr 29 2003

nonn

The frequency of each possible sum is computed by the Mathematica program without explicitly computing the individual sums.
a(n) is the maximal number of subsets of the first n squares that share the same sum. Cf. A025591, A083309.
a(n)=0 when n==1 or 2 (mod 4).

			a(7) = 1 because there is only one sign pattern of the first seven squares that yields zero: 1+4-9+16-25-36+49.

Alois P. Heinz and Ray Chandler, Table of n, a(n) for n = 1..500 (first 240 terms from Alois P. Heinz)
T. D. Noe, Extremal Sums of Sequences

Cf. A015818, A058498, A063865, A113263, A158092.

b:= proc(n, i) option remember; local m;
      m:= (1+(3+2*i)*i)*i/6;
      `if`(n>m, 0, `if`(n=m, 1, b(abs(n-i^2), i-1) +b(n+i^2, i-1)))
    end:
a:= n-> `if`(irem(n-1, 4)<2, 0, b(n^2, n-1)):
seq(a(n), n=1..40);  # Alois P. Heinz, Oct 31 2011

d={1, 1}; nMax=60; zeroLst={0}; Do[p=n^2; d=PadLeft[d, Length[d]+p]+PadRight[d, Length[d]+p]; If[1==Mod[Length[d], 2], AppendTo[zeroLst, d[[(Length[d]+1)/2]]], AppendTo[zeroLst, 0]], {n, 2, nMax}]; zeroLst/2
p = 1; t = {}; Do[p = Expand[p(x^(n^2) + x^(-n^2))]; AppendTo[t, Select[p, NumberQ[ # ] &]/2], {n, 51}]; t (* Robert G. Wilson v, Oct 31 2005 *)

a(n)=sum(i=0,2^(n-1)-1,sum(j=1,n-1,(-1)^bittest(i,j-1)*j^2)==n^2) \\ Charles R Greathouse IV, Nov 05 2012

a(n) is half the coefficient of x^0 in the product_{k=1..n} x^(k^2)+x^(k^-2).
a(n) = A158092(n)/2.
a(n) = [x^(n^2)] Product_{k=1..n-1} (x^(k^2) + 1/x^(k^2)). - Ilya Gutkovskiy, Feb 01 2024

A158118 Number of solutions of +-1+-2^3+-3^3..+-n^3=0.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 4, 2, 0, 0, 4, 124, 0, 0, 536, 712, 0, 0, 4574, 2260, 0, 0, 10634, 73758, 0, 0, 406032, 638830, 0, 0, 4249160, 3263500, 0, 0, 21907736, 82561050, 0, 0, 485798436, 945916970, 0, 0, 5968541478, 6839493576, 0, 0

Offset: 1

Pietro Majer, Mar 12 2009

nonn

Constant term in the expansion of (x + 1/x)(x^8 + 1/x^8)..(x^n^3 + 1/x^n^3).
a(n) = 0 for any n=1 (mod 4) or n=2 (mod 4).
The expansion above and the integral representation formula below are due to Andrica & Tomescu. The asymptotic formula is a conjecture; see Andrica & Ionascu. - Jonathan Sondow, Nov 06 2013

			Example: For n=12 the a(12) = 2 solutions are:
+1+8-27+64-125-216-343+512+729-1000-1331+1728=0,
-1-8+27-64+125+216+343-512-729+1000+1331-1728=0.

Ray Chandler, Table of n, a(n) for n = 1..130
D. Andrica and E. J. Ionascu, Variations on a result of Erdős and Surányi, INTEGERS 2013 slides.
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Sequences, 5 (2002), Article 02.2.4.

Equals twice A113263.

Cf. A063865, A158092, A019568. - Pietro Majer, Mar 15 2009

N:=60: p:=1: a:=[]: for n from 1 to N do p:=expand(p*( x^(n^3) + x^(-n^3) )): a:=[op(a), coeff(p,x,0)]: od:a;

a(n) = 2 * A113263(n).
Integral representation: a(n)=((2^n)/Pi)*int_0^Pi prod_{k=1}^n cos(x*k^3) dx.
Asymptotic formula: a(n)=(2^n)*sqrt(14/(Pi*n^7))*(1+o(1)) as n-->infty; n=-1 or 0 (mod 4).

A158380 Number of solutions to +-1 +- 3 +- 6 +- ... +- n(n+1)/2 = 0.

Original entry on oeis.org

1, 0, 0, 0, 2, 0, 2, 2, 4, 0, 12, 16, 26, 0, 66, 104, 210, 0, 620, 970, 1748, 0, 5948, 10480, 18976, 0, 60836, 111430, 209460, 0, 704934, 1284836, 2387758, 0, 8331820, 15525814, 28987902, 0, 101242982, 190267598, 358969426, 0, 1275032260, 2404124188, 4547419694

Offset: 0

Pietro Majer, Mar 17 2009

Equivalently, number of partitions of the set of the first n triangular numbers {t(1),...,t(n)} into two classes with equal sums.
Constant term in the expansion of (x + 1/x)(x^3 + 1/x^3)...(x^t(n) + 1/x^t(n)).
a(n) = 0 for all n == 1 (mod 4).
Andrica & Tomescu give a more general integral formula than the one below. - Jonathan Sondow, Nov 11 2013

			For n=6 the 2 solutions are +1-3+6-10-15+21 = 0 and -1+3-6+10+15-21 = 0.

Ray Chandler, Table of n, a(n) for n = 0..633
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Sequences, 5 (2002), Article 02.2.4.

Cf. A058498, A063865, A158092, A158118.

N:=70: p:=1: a:=[]: for n from 0 to N do
p:=expand(p*(x^(n*(n+1)/2)+x^(-n*(n+1)/2))):
a:=[op(a), coeff(p, x, 0)]: od:a;
# second Maple program:
b:= proc(n, i) option remember; (m-> `if`(n>m, 0,
      `if`(n=m, 1, b(abs(n-i*(i+1)/2), i-1)+
      b(n+i*(i+1)/2, i-1))))((2+(3+i)*i)*i/6)
    end:
a:= n-> `if`(irem(n, 4)=1, 0, b(0, n)):
seq(a(n), n=0..50);  # Alois P. Heinz, Sep 17 2017

a[n_] := With[{t = Table[k(k+1)/2, {k, 1, n}]}, Coefficient[Times @@ (x^t + 1/x^t), x, 0]];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 50}] (* Jean-François Alcover, Jun 16 2018 *)

t(k) = k*(k+1)/2;
a(n) = polcoeff(prod(k=1, n, (x^t(k)+ 1/x^t(k))), 0); \\ Michel Marcus, May 19 2015

a(n) = (2^n/Pi) * Integral_{x=0..Pi} cos(x)*cos(3x)*...*cos(n(n+1)x/2) dx.
a(n) ~ 2^(n+1)*sqrt(10/Pi)*n^(-5/2)*(1+o(1)) as n --> infinity, n !== 1 (mod 4).
a(n) = 2 * A058498(n) for n > 0. - Alois P. Heinz, Nov 01 2011

a(0) = 1 prepended by Joerg Arndt, Sep 17 2017

Example corrected by Ilya Gutkovskiy, Feb 02 2022

A350249 a(n) is the constant term in expansion of Product_{k=1..n} (x^(k^2) + 1 + 1/x^(k^2)).

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 7, 19, 43, 95, 189, 429, 1003, 2457, 6319, 16165, 41601, 107969, 280253, 737065, 1950865, 5201941, 13954313, 37593679, 101695957, 276296549, 753191093, 2061201397, 5658850121, 15583938539, 43040609115, 119182143639, 330841253283, 920550527585

Offset: 0

Ilya Gutkovskiy, Jan 28 2022

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..200

Cf. A007576, A158092, A350495, A350880.

b:= proc(n) option remember; `if`(n=0, 1,
      expand((x^(n^2)+1+1/x^(n^2))*b(n-1)))
    end:
a:= n-> coeff(b(n),x,0):
seq(a(n), n=0..33);  # Alois P. Heinz, Jan 28 2022

Table[Coefficient[Product[x^(k^2) + 1 + 1/x^(k^2), {k, 1, n}], x, 0], {n, 0, 30}] (* Vaclav Kotesovec, Feb 05 2022 *)

Conjecture: a(n) ~ sqrt(5) * 3^(n + 1/2) / (2*sqrt(Pi)*n^(5/2)). - Vaclav Kotesovec, Feb 04 2022

A348165 Number of solutions to +-1^2 +- 2^2 +- 3^2 +- ... +- n^2 = n.

Original entry on oeis.org

1, 1, 0, 0, 1, 2, 0, 0, 2, 4, 0, 0, 19, 29, 0, 0, 127, 208, 0, 0, 1121, 1917, 0, 0, 10479, 19360, 0, 0, 113213, 204121, 0, 0, 1290968, 2363982, 0, 0, 15303057, 28397538, 0, 0, 187446097, 351339307, 0, 0, 2355979330, 4455357992, 0, 0, 30360404500, 57630025172

Offset: 0

Ilya Gutkovskiy, Jan 28 2022

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..200

Cf. A063890, A158092, A348892.

b:= proc(n, i) option remember; (m-> `if`(n>m, 0, `if`(n=m, 1,
      b(abs(n-i^2), i-1)+b(n+i^2, i-1))))((1+(3+2*i)*i)*i/6)
    end:
a:= n-> `if`(irem(n, 4)>1, 0, b(n$2)):
seq(a(n), n=0..50);  # Alois P. Heinz, Jan 28 2022

b[n_, i_] := b[n, i] = Function[m, If[n > m, 0, If[n == m, 1, b[Abs[n - i^2], i - 1] + b[n + i^2, i - 1]]]][(1 + (3 + 2*i)*i)*i/6];
a[n_] := If[Mod[n, 4] > 1, 0, b[n, n]];
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Feb 26 2022, after Alois P. Heinz *)

from functools import lru_cache
@lru_cache(maxsize=None)
def b(n, i):
    if n > i*(i+1)*(2*i+1)//6: return 0
    if i == 0: return 1
    return b(n+i**2, i-1) + b(abs(n-i**2), i-1)
def a(n): return b(n, n)
print([a(n) for n in range(50)]) # Michael S. Branicky, Jan 28 2022

a(n) = [x^n] Product_{k=1..n} (x^(k^2) + 1/x^(k^2)).

A231015 Least k such that n = +- 1^2 +- 2^2 +- 3^2 +- 4^2 +- ... +- k^2 for some choice of +- signs.

Original entry on oeis.org

7, 1, 4, 2, 3, 2, 3, 6, 7, 6, 4, 6, 3, 5, 3, 5, 7, 6, 7, 6, 4, 5, 4, 5, 7, 9, 7, 5, 4, 5, 4, 6, 7, 6, 7, 5, 7, 5, 7, 6, 7, 6, 7, 9, 8, 5, 8, 5, 7, 6, 7, 6, 11, 5, 8, 5, 7, 6, 7, 6, 7, 10, 7, 6, 7, 6, 7, 9, 7, 10, 7, 6, 7, 6, 8, 9, 8, 9, 8, 9, 7, 6, 7, 6, 11, 9

Offset: 0

Jonathan Sondow, Nov 02 2013

Erdős and Surányi proved that for each n there are infinitely many k satisfying the equation.
A158092(k) is the number of solutions to 0 = +-1^2 +- 2^2 +- ... +- k^2. The first nonzero value is A158092(7) = 2, so a(0) = 7.
a(n) is also defined for n < 0, and clearly a(-n) = a(n).
See A158092 and the Andrica-Ionascu links for more comments.
The integral formula (3.6) in Andrica-Vacaretu (see Theorem 3 of the INTEGERS 2013 slides which has a typo) gives in this case the number of representations of n as +- 1^2 +- 2^2 +- ... +- k^2 for some choice of +- signs. This integral formula is (2^n/2*Pi)*Integral_{t=0..2*Pi} cos(n*t) * Product_{j=1..k} cos(j^2*t) dt. Clearly the number of such representations of n is the coefficient of z^n in the expansion (z^(1^2) + z^(-1^2))*(z^(2^2) + z^(-2^2))*...*(z^(k^2) + z^(-k^2)). Andrica-Vacaretu used this generating function to prove the integral formula. Section 4 of Andrica-Vacaretu gives a table of the number of such representations of n for k=1,...,9. - Dorin Andrica, Nov 12 2013

			0 = 1^2 + 2^2 - 3^2 + 4^2 - 5^2 - 6^2 + 7^2.
1 = 1^2.
2 = - 1^2 - 2^2 - 3^2 + 4^2.
3 = - 1^2 + 2^2.
4 = - 1^2 - 2^2 + 3^2.

Alois P. Heinz, Table of n, a(n) for n = 0..20000
D. Andrica and E. J. Ionascu, Variations on a result of Erdős and Surányi, Integers Conference 2013 Abstract.
D. Andrica and E. J. Ionascu, Variations on a result of Erdős and Surányi, INTEGERS 2013 slides.
D. Andrica and D. Vacaretu, Representation theorems and almost unimodal sequences, Studia Univ. Babes-Bolyai, Mathematica, Vol. LI, 4 (2006), 23-33.
P. Erdős and J. Surányi, Egy additív számelméleti probléma (in Hungarian; Russian and German summaries), Mat. Lapok 10 (1959), pp. 284-290.

Cf. A000330, A158092, A231071, A231272.

b:= proc(n, i) option remember; local m; m:=i*(i+1)*(2*i+1)/6;
      n<=m and (n=m or b(n+i^2, i-1) or b(abs(n-i^2), i-1))
    end:
a:= proc(n) local k; for k while not b(n, k) do od; k end:
seq(a(n), n=0..100);  # Alois P. Heinz, Nov 03 2013

b[n_, i_] := b[n, i] = Module[{m}, m = i*(i+1)*(2*i+1)/6; n <= m && (n == m || b[n+i^2, i-1] || b[Abs[n-i^2], i-1])]; a[n_] := Module[{k}, For[k = 1, !b[n, k] , k++]; k]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jan 28 2014, after Alois P. Heinz *)

a(n(n+1)(2n+1)/6) = a(A000330(n)) = n for n > 0.

a((n(n+1)(2n+1)/6)-2) = a(A000330(n)-2) = n for n > 0.

a(4) corrected and a(5)-a(85) from Donovan Johnson, Nov 03 2013

A158465 Number of solutions to +-1+-2^4+-3^4+-4^4...+-n^4=0.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 16, 18, 0, 0, 32, 100, 0, 0, 424, 510, 0, 0, 2792, 5988, 0, 0, 29058, 45106, 0, 0, 276828, 473854, 0, 0, 2455340, 4777436, 0, 0, 27466324, 46429640, 0, 0, 280395282, 526489336, 0, 0, 3193589950, 5661226928, 0, 0

Offset: 1

Pietro Majer, Mar 19 2009

nonn

Constant term in the expansion of (x + 1/x)(x^16 + 1/x^16)..(x^n^4 + 1/x^n^4).
a(n)=0 for any n=1 (mod 4) or n=2 (mod 4).
Andrica & Tomescu give a more general integral formula than the one below. The asymptotic formula below is a conjecture by Andrica & Ionascu; it remains unproven. - Jonathan Sondow, Nov 11 2013

			For n=16 the a(16) = 2 solutions are +1 +16 +81 +256 -625 -1296 -2401 +4096 +6561 +10000 +14641 +20736 -28561 -38416 -50625 +65536 = 0 and the opposite.

D. Andrica and E. J. Ionascu, Variations on a result of Erdős and SurányiINTEGERS 2013 slides.
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Sequences, 5 (2002), Article 02.2.4.

Cf. A063865, A158092, A158118, A158380, A019568.

A111253(n) = a(n)/2. - Alois P. Heinz, Oct 31 2011

N:=32: p:=1 a:=[]: for n from 32 to N do p:=expand
(p*(x^(n^4)+x^(-n^4))): a:=[op(a), coeff(p,x,0)]: od:a;

Integral representation: a(n) = ((2^n)/Pi)*int_0^pi prod_{k=1}^n cos(x*k^4) dx.

Asymptotic formula: a(n) = (2^n)*sqrt(18/(Pi*n^9))*(1+o(1)) as n->infinity; n=-1 or 0 (mod 4).

a(35)-a(58) from Alois P. Heinz, Oct 31 2011

A292496 Number of solutions to +- 1^2 +- 3^2 +- 5^2 +- 7^2 +- ... +- (4*n-1)^2 = 0.

Original entry on oeis.org

1, 0, 0, 0, 2, 0, 12, 0, 40, 10, 516, 124, 5020, 1828, 48570, 32806, 527890, 444480, 6137942, 6482314, 70573856, 93276044, 853480374, 1300190254, 10660384742, 18371629260, 134129890382, 259804151324, 1728886287134, 3667061002286, 22672130669968

Offset: 0

Seiichi Manyama, Sep 17 2017

nonn

			For n=4 the 2 solutions are +1^2-3^2-5^2+7^2-9^2+11^2+13^2-15^2 = 0 and -1^2+3^2+5^2-7^2+9^2-11^2-13^2+15^2 = 0.

Seiichi Manyama, Table of n, a(n) for n = 0..100

Cf. A158092, A292476, A292497.

a(n) = polcoeff(prod(k=1, 2*n, x^(2*k-1)^2+1/x^(2*k-1)^2), 0); \\ Michel Marcus, Sep 18 2017

Constant term in the expansion of Product_{k=1..2*n} (x^(2*k-1)^2+1/x^(2*k-1)^2).

A231071 Number of solutions to n = +- 1^2 +- 2^2 +- 3^2 +- 4^2 +- ... +- k^2 for minimal k giving at least one solution.

Original entry on oeis.org

2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 3, 2, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 9, 1, 3, 1, 1, 1, 2, 1, 1, 6, 1, 1, 1, 1, 1, 2, 1, 5, 1, 1, 1, 1, 4, 3, 1, 2, 1, 2, 2, 1, 2, 1, 14, 2, 1, 3, 2, 1, 2, 1, 1, 7, 1, 3, 2, 5, 1, 2, 1

Offset: 0

Alois P. Heinz, Nov 03 2013

This type of sequence was first studied by Andrica and Vacaretu. - Jonathan Sondow, Nov 06 2013

			a(8) = 3: 8 = -1-4-9-16+25-36+49 = -1-4+9+16-25-36+49 = -1+4+9-16+25+36-49.
a(9) = 2: 9 = -1-4+9+16+25-36 = 1+4+9-16-25+36.
a(10) = 1: 10 = -1+4-9+16.

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Dorin Andrica and Daniel Vacaretu, Representation theorems and almost unimodal sequences, Studia Univ. Babes-Bolyai, Mathematica, Vol. LI, 4 (2006), 23-33.

Cf. A000330, A231015, A231272.

Cf. A083527, A158092 (extremal sums).

b:= proc(n, i) option remember; (m->`if`(n>m, 0, `if`(n=m, 1,
      b(n+i^2, i-1) +b(abs(n-i^2), i-1))))((1+(3+2*i)*i)*i/6)
    end:
a:= proc(n) local k; for k while b(n, k)=0 do od; b(n, k) end:
seq(a(n), n=0..100);

b[n_, i_] := b[n, i] = Function[m, If[n > m, 0, If[n == m, 1,
   b[n+i^2, i-1] + b[Abs[n-i^2], i-1]]]][(1+(3+2*i)*i)*i/6];
a[n_] := Module[{k}, For[k = 1, b[n, k] == 0, k++]; b[n, k]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Sep 01 2022, after Alois P. Heinz *)

From Jonathan Sondow, Nov 03 2013: (Start)
a(n(n+1)(2n+1)/6) = 1 for n > 0: n(n+1)(2n+1)/6 = 1+4+9+...+n^2. See A000330.
a(n(n+1)(2n+1)/6 - 2) = 1 for n > 1: n(n+1)(2n+1)/6 - 2 = -1+4+9+...+n^2. (End)

cp's OEIS Frontend

A063865 Number of solutions to +- 1 +- 2 +- 3 +- ... +- n = 0.

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A158380 Number of solutions to +-1 +- 3 +- 6 +- ... +- n(n+1)/2 = 0.

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A348165 Number of solutions to +-1^2 +- 2^2 +- 3^2 +- ... +- n^2 = n.

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A231015 Least k such that n = +- 1^2 +- 2^2 +- 3^2 +- 4^2 +- ... +- k^2 for some choice of +- signs.

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