A083527 -id:A083527 - cp's OEIS frontend

A025591 Maximal coefficient of Product_{k<=n} (1 + x^k). Number of solutions to +- 1 +- 2 +- 3 +- ... +- n = 0 or 1.

1, 1, 1, 2, 2, 3, 5, 8, 14, 23, 40, 70, 124, 221, 397, 722, 1314, 2410, 4441, 8220, 15272, 28460, 53222, 99820, 187692, 353743, 668273, 1265204, 2399784, 4559828, 8679280, 16547220, 31592878, 60400688, 115633260, 221653776, 425363952, 817175698

Offset: 0

David W. Wilson

If k is allowed to approach infinity, this gives the partition numbers A000009.

a(n) is the maximal number of subsets of {1,2,...,n} that share the same sum.

T. D. Noe, Alois P. Heinz and Ray Chandler, Table of n, a(n) for n = 0..3339 (terms < 10^1000, first 201 terms from T. D. Noe, next 200 terms from Alois P. Heinz)
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance , Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.4.
Vlad-Florin Dragoi and Valeriu Beiu, Fast Reliability Ranking of Matchstick Minimal Networks, arXiv:1911.01153 [cs.DM], 2019.
Steven R. Finch, Signum equations and extremal coefficients, February 7, 2009. [Cached copy, with permission of the author]
Erich Friedman and Mike Keith, Magic Carpets, J. Int Sequences, 3 (2000), Article 00.2.5.
Susumu Kubo, Partially Ordered Sets Corresponding to the Partition Problem, arXiv:2405.05544 [cs.DM], 2024. See pp. 2, 16.
Marco Mondelli, S. Hamed Hassani, and Rüdiger Urbanke, Construction of Polar Codes with Sublinear Complexity, arXiv preprint arXiv:1612.05295 [cs.IT], 2016-2017. See Sect. I.
Robert A. Proctor, Solution of two difficult combinatorial problems with linear algebra, American Mathematical Monthly 89, 721-734.
Blair D. Sullivan, On a conjecture of Adrica and Tomescu, J. Int. Sequences 16 (2013), Article 13.3.1.

Cf. A039828, A063865, A069918, A063866, A063867, A083309, A083527, A086376.

Cf. A053632, A160235, A359319, A359320.

b:= proc(n, i) option remember; `if`(n>i*(i+1)/2, 0,
      `if`(i=0, 1, b(n+i, i-1)+b(abs(n-i), i-1)))
    end:
a:=n-> b(0, n)+b(1, n):
seq(a(n), n=0..40);  # Alois P. Heinz, Mar 10 2014

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]]; Table[Which[Mod[n, 4]==0||Mod[n, 4]==3, f[n, 0], Mod[n, 4]==1||Mod[n, 4]==2, f[n, 1]], {n, 0, 40}]
(* Second program: *)
p = 1; Flatten[{1, Table[p = Expand[p*(1 + x^n)]; Max[CoefficientList[p, x]], {n, 1, 50}]}] (* Vaclav Kotesovec, May 04 2018 *)
b[n_, i_] := b[n, i] = If[n > i(i+1)/2, 0, If[i == 0, 1, b[n+i, i-1] + b[Abs[n-i], i-1]]];
a[n_] := b[0, n] + b[1, n]; a /@ Range[0, 40] (* Jean-François Alcover, Feb 17 2020, after Alois P. Heinz *)

a(n)=if(n<0,0,polcoeff(prod(k=1,n,1+x^k),n*(n+1)\4))

from collections import Counter
def A025591(n):
    c = {0:1,1:1}
    for i in range(2,n+1):
        d = Counter(c)
        for k in c:
            d[k+i] += c[k]
        c = d
    return max(c.values()) # Chai Wah Wu, Jan 31 2024

a(n) = A063865(n) + A063866(n).
a(n) ~ sqrt(6/Pi) * 2^n / n^(3/2) [conjectured by Andrica and Tomescu (2002) and proved by Sullivan (2013)]. - Vaclav Kotesovec, Mar 17 2020
More precise asymptotics: a(n) ~ sqrt(6/Pi) * 2^n / n^(3/2) * (1 - 6/(5*n) + 589/(560*n^2) - 39/(50*n^3) + ...). - Vaclav Kotesovec, Dec 30 2022
a(n) = max_{k>=0} A053632(n,k). - Alois P. Heinz, Jan 20 2023

A083309 a(n) is the number of times that sums 3 +- 5 +- 7 +- 11 +- ... +- prime(2n+1) of the first 2n odd primes is zero. There are 2^(2n-1) choices for the sign patterns.

Original entry on oeis.org

0, 0, 1, 2, 7, 19, 63, 197, 645, 2172, 7423, 25534, 89218, 317284, 1130526, 4033648, 14515742, 52625952, 191790090, 702333340, 2585539586, 9570549372, 35562602950, 131774529663, 491713178890, 1842214901398, 6909091641548

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. Let S = 3 + 5 + 7 + ... + prime(2n+1). Because the primes do not grow very fast, it is easy to show that, for n > 2, all even numbers between -S+20 and S-20 occur at least once as a sum.
a(n) is the maximal number of subsets of {prime(2), prime(3), ..., prime(n+1)} that share the same sum. Cf. A025591, A083527.
See A238894 for a more general sequence that looks at all sums formed. - T. D. Noe, Mar 07 2014

			a(3) = 1 because there is only one sign pattern of the first six odd primes that yields zero: 3 + 5 + 7 - 11 + 13 - 17.

Ray Chandler, Table of n, a(n) for n = 1..1000 (first 100 terms from T. D. Noe)
T. D. Noe, Extremal Sums of Sequences.

Cf. A015818, A063865, A238894.

Cf. A022894 (use all primes in the sum), A022895 (r.h.s. = 1), A022896 (r.h.s. = 2), A022897 (interleaved 0 for odd number of terms), ..., A022903 (using primes >= 7), A022904, A022920; A261061 - A261063 and A261044 (r.h.s. = -1); A261057, A261059, A261060, A261045 (r.h.s. = -2).

d={1, 0, 0, 1}; nMax=32; zeroLst={}; Do[p=Prime[n+1]; d=PadLeft[d, Length[d]+p]+PadRight[d, Length[d]+p]; If[0==Mod[n, 2], AppendTo[zeroLst, d[[(Length[d]+1)/2]]]], {n, 2, nMax}]; zeroLst/2

A083309(n, rhs=0, firstprime=2)={rhs-=prime(firstprime); my(p=vector(2*n-2+bittest(rhs, 0), i, prime(i+firstprime))); sum(i=1, 2^#p-1, sum(j=1, #p, (-1)^bittest(i, j-1)*p[j])==rhs)} \\ For illustrative purpose, too slow for n >> 10. - M. F. Hasler, Aug 08 2015

a(n) = A022897(2n). - M. F. Hasler, Aug 08 2015

A158092 Number of solutions to +- 1 +- 2^2 +- 3^2 +- 4^2 +- ... +- n^2 = 0.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 2, 10, 0, 0, 86, 114, 0, 0, 478, 860, 0, 0, 5808, 10838, 0, 0, 55626, 100426, 0, 0, 696164, 1298600, 0, 0, 7826992, 14574366, 0, 0, 100061106, 187392994, 0, 0, 1223587084, 2322159814, 0, 0, 16019866270, 30353305134, 0, 0

Offset: 1

Pietro Majer, Mar 12 2009

nonn

Twice A083527.
Number of partitions of the half of the n-th-square-pyramidal number into parts that are distinct square numbers in the range 1 to n^2. Example: a(7)=2 since, squarePyramidal(7)=140 and 70=1+4+16+49=9+25+36. - Hieronymus Fischer, Oct 20 2010
Erdős & Surányi prove that this sequence is unbounded. More generally, there are infinitely many ways to write a given number k as such a sum. - Charles R Greathouse IV, Nov 05 2012
The expansion and integral representation formulas below are due to Andrica & Tomescu. The asymptotic formula is a conjecture; see Andrica & Ionascu. - Jonathan Sondow, Nov 11 2013

			For n=8 the a(8)=2 solutions are: +1-4-9+16-25+36+49-64=0 and -1+4+9-16+25-36-49+64=0.

Alois P. Heinz and Ray Chandler, Table of n, a(n) for n = 1..500 (first 240 terms from Alois P. Heinz)
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.
P. Erdős and J. Surányi, Egy additív számelméleti probléma (in Hungarian; Russian and German summaries), Mat. Lapok 10 (1959), pp. 284-290.

Cf. A063865, A158118, A019568, A083527, A231015.

From Pietro Majer, Mar 15 2009: (Start)
N:=60: p:=1: a:=[]: for n from 1 to N do p:=expand(p*(x^(n^2)+x^(-n^2))):
a:=[op(a), coeff(p, x, 0)]: od:a; (End)
# second Maple program:
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, 2*b(n^2, n-1)):
seq(a(n), n=1..60);  # Alois P. Heinz, Nov 05 2012

b[n_, i_] := b[n, i] = With[{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]]]]; a[n_] := If[Mod[n-1, 4]<2, 0, 2*b[n^2, n-1]]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Mar 13 2015, after Alois P. Heinz *)

a(n)=2*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

from itertools import count, islice
from collections import Counter
def A158092_gen(): # generator of terms
    ccount = Counter({0:1})
    for i in count(1):
        bcount = Counter()
        for a in ccount:
            bcount[a+(j:=i**2)] += ccount[a]
            bcount[a-j] += ccount[a]
        ccount = bcount
        yield(ccount[0])
A158092_list = list(islice(A158092_gen(),20)) # Chai Wah Wu, Jan 29 2024

Constant term in the expansion of (x + 1/x)(x^4 + 1/x^4)..(x^n^2 + 1/x^n^2).
a(n)=0 for any n == 1 or 2 (mod 4).
Integral representation: a(n)=((2^n)/pi)*int_0^pi prod_{k=1}^n cos(x*k^2) dx
Asymptotic formula: a(n) = (2^n)*sqrt(10/(pi*n^5))*(1+o(1)) as n-->infty; n == -1 or 0 (mod 4).
a(n) = 2 * A083527(n). - T. D. Noe, Mar 12 2009
min{n : a(n) > 0} = A231015(0) = 7. - Jonathan Sondow, Nov 06 2013

a(51)-a(56) from R. H. Hardin, Mar 12 2009

Edited by N. J. A. Sloane, Sep 15 2009

A113263 a(n) is the number of ways the set {1^3, 2^3, ..., n^3} can be partitioned into two sets of equal sums.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 2, 62, 0, 0, 268, 356, 0, 0, 2287, 1130, 0, 0, 5317, 36879, 0, 0, 203016, 319415, 0, 0, 2124580, 1631750, 0, 0, 10953868, 41280525, 0, 0, 242899218, 472958485, 0, 0, 2984270739, 3419746788, 0, 0

Offset: 1

Floor van Lamoen, Oct 21 2005

nonn

a(n)=0 when n == 1 or 2 mod 4.

Alois P. Heinz and Ray Chandler, Table of n, a(n) for n = 1..130 (first 100 terms from Alois P. Heinz)

Cf. A058498, A083527.

A113263:=proc(n) local i,p,t; t:= NULL; p:=1; for i to n do p:=p*(x^(i^3)+x^(-i^3)); t:=t,coeff(p,x,0)/2; od; t; end;

p = 1; t = {}; Do[p = Expand[p(x^(n^3) + x^(-n^3))]; AppendTo[t, Select[ p, NumberQ[ # ] &]/2], {n, 56}]; t (* Robert G. Wilson v *)

a(n) is half the coefficient of x^0 in product(x^(k^3)+x^(k^-3), k=1..n).

a(n) = [x^(n^3)] Product_{k=1..n-1} (x^(k^3) + 1/x^(k^3)). - Ilya Gutkovskiy, Feb 01 2024

More terms from Robert G. Wilson v and Tony Noe, Oct 27 2005

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)

A111253 a(n) is the number of ways the set {1^4, 2^4, ..., n^4} can be partitioned into two sets of equal sums.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 8, 9, 0, 0, 16, 50, 0, 0, 212, 255, 0, 0, 1396, 2994, 0, 0, 14529, 22553, 0, 0, 138414, 236927, 0, 0, 1227670, 2388718, 0, 0, 13733162, 23214820, 0, 0, 140197641, 263244668, 0, 0, 1596794975, 2830613464, 0, 0

Offset: 1

Robert G. Wilson v, Oct 31 2005

nonn

a(n)=0 when n == 1 or 2 (mod 4).

Cf. A058377, A083527, A113263.

b:= proc(n, i) option remember; local m;
      m:= (-1+(10+(15+6*i)*i)*i^2)*i/30;
      `if`(n>m, 0, `if`(n=m, 1, b(abs(n-i^4), i-1) +b(n+i^4, i-1)))
    end:
a:= n-> `if`(irem(n-1, 4)<2, 0, b(n^4, n-1)):
seq(a(n), n=1..38);  # Alois P. Heinz, Oct 30 2011

d = {1, 1}; nMax=50; zeroLst = {0}; Do[p = n^4; 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 (* T. D. Noe, Oct 31 2005 *)
p = 1; t = {}; Do[p = Expand[p(x^(n^4) + x^(-n^4))]; AppendTo[t, Select[p, NumberQ[ # ] &]/2], {n, 30}]; t

a(n) is half the coefficient of x^0 in product_{k=1..n} x^(k^4)+x^(k^-4).

a(n) = [x^(n^4)] Product_{k=1..n-1} (x^(k^4) + 1/x^(k^4)). - Ilya Gutkovskiy, Feb 01 2024

a(51)-a(54) from T. D. Noe, Nov 01 2005

Corrected a(51)-a(52) and extended up to a(58) by Alois P. Heinz, Oct 31 2011

A300268 Number of solutions to 1 +- 4 +- 9 +- ... +- n^2 == 0 (mod n).

Original entry on oeis.org

1, 0, 2, 4, 6, 0, 10, 48, 32, 0, 94, 344, 370, 0, 1268, 4608, 3856, 0, 13798, 55960, 50090, 0, 182362, 721952, 690496, 0, 2485592, 9586984, 9256746, 0, 34636834, 135335936, 130150588, 0, 493452348, 1908875264, 1857293524, 0, 7049188508, 27603824928

Offset: 1

Seiichi Manyama, Mar 01 2018

nonn

			Solutions for n = 7:
-------------------------------
1 +4 +9 +16 +25 +36 +49 =  140.
1 +4 +9 +16 +25 +36 -49 =   42.
1 +4 +9 -16 -25 -36 +49 =  -14.
1 +4 +9 -16 -25 -36 -49 = -112.
1 +4 -9 +16 -25 -36 +49 =    0.
1 +4 -9 +16 -25 -36 -49 =  -98.
1 -4 +9 -16 +25 -36 +49 =   28.
1 -4 +9 -16 +25 -36 -49 =  -70.
1 -4 -9 +16 +25 -36 +49 =   42.
1 -4 -9 +16 +25 -36 -49 =  -56.

Seiichi Manyama, Table of n, a(n) for n = 1..3334 (terms 1..1000 from Alois P. Heinz)

Number of solutions to 1 +- 2^k +- 3^k +- ... +- n^k == 0 (mod n): A300190 (k=1), this sequence (k=2), A300269 (k=3).

Cf. A083527, A215573.

b:= proc(n, i, m) option remember; `if`(i=0, `if`(n=0, 1, 0),
      add(b(irem(n+j, m), i-1, m), j=[i^2, m-i^2]))
    end:
a:= n-> b(0, n-1, n):
seq(a(n), n=1..60);  # Alois P. Heinz, Mar 01 2018

b[n_, i_, m_] := b[n, i, m] = If[i == 0, If[n == 0, 1, 0],
     Sum[b[Mod[n + j, m], i - 1, m], {j, {i^2, m - i^2}}]];
a[n_] := b[0, n - 1, n];
Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Mar 19 2022, after Alois P. Heinz *)

a(n) = my (v=vector(n,k,k==1)); for (i=2, n, v = vector(n, k, v[1 + (k-i^2)%n] + v[1 + (k+i^2)%n])); v[1] \\ Rémy Sigrist, Mar 01 2018

def A(n)
  ary = [1] + Array.new(n - 1, 0)
  (1..n).each{|i|
    i2 = 2 * i * i
    a = ary.clone
    (0..n - 1).each{|j| a[(j + i2) % n] += ary[j]}
    ary = a
  }
  ary[(n * (n + 1) * (2 * n + 1) / 6) % n] / 2
end
def A300268(n)
  (1..n).map{|i| A(i)}
end
p A300268(100)

A307877 Number of ways of partitioning the set of the first n positive squares into two subsets whose sums differ at most by 1.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 2, 1, 1, 5, 2, 1, 5, 13, 43, 43, 57, 193, 274, 239, 430, 1552, 3245, 2904, 5419, 18628, 31048, 27813, 50213, 188536, 372710, 348082, 649300, 2376996, 4197425, 3913496, 7287183, 27465147, 53072709, 50030553, 93696497, 351329160, 650125358

Offset: 0

Alois P. Heinz, Jun 04 2019

nonn

			a(6) = 2: 1,9,36/4,16,25; 1,4,16,25/9,36.
a(7) = 1: 1,4,16,49/9,25,36.

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

Cf. A000290, A069918, A083527, A158092, A306443, A308682.

s:= proc(n) s(n):= `if`(n=0, 1, n^2+s(n-1)) end:
b:= proc(n, i) option remember; `if`(i=0, `if`(n<=1, 1, 0),
      `if`(n>s(i), 0, (p-> b(n+p, i-1)+b(abs(n-p), i-1))(i^2)))
    end:
a:= n-> ceil(b(0, n)/2):
seq(a(n), n=0..45);

s[n_] := s[n] = If[n == 0, 1, n^2 + s[n - 1]];
b[n_, i_] := b[n, i] = If[i == 0, If[n <= 1, 1, 0], If[n > s[i], 0, Function[p, b[n + p, i - 1] + b[Abs[n - p], i - 1]][i^2]]];
a[n_] := Ceiling[b[0, n]/2];
a /@ Range[0, 45] (* Jean-François Alcover, Dec 07 2020, after Alois P. Heinz *)

a(n) = A083527(n) if n == 0 or 3 (mod 4).

A292497 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, 1, 0, 6, 0, 20, 5, 258, 62, 2510, 914, 24285, 16403, 263945, 222240, 3068971, 3241157, 35286928, 46638022, 426740187, 650095127, 5330192371, 9185814630, 67064945191, 129902075662, 864443143567, 1833530501143, 11336065334984, 25990268638322

Offset: 0

Seiichi Manyama, Sep 17 2017

nonn

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

Cf. A083527, A156700, A292496.

For n>0 constant term in the expansion of 1/2 * Product_{k=1..2*n} (x^(2*k-1)^2+1/x^(2*k-1)^2).

a(n) = A292496(n)/2 for n>0.

A350403 Number of solutions to +-1^2 +- 2^2 +- 3^2 +- ... +- n^2 = 0 or 1.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 2, 2, 2, 5, 2, 2, 10, 13, 43, 86, 114, 193, 274, 478, 860, 1552, 3245, 5808, 10838, 18628, 31048, 55626, 100426, 188536, 372710, 696164, 1298600, 2376996, 4197425, 7826992, 14574366, 27465147, 53072709, 100061106, 187392994, 351329160

Offset: 0

Ilya Gutkovskiy, Dec 29 2021

nonn

			a(8) = 2: +1^2 - 2^2 - 3^2 + 4^2 - 5^2 + 6^2 + 7^2 - 8^2 =
          -1^2 + 2^2 + 3^2 - 4^2 + 5^2 - 6^2 - 7^2 + 8^2 = 0.

Cf. A000290, A025591, A083527, A158092.

b:= proc(n, i) option remember; `if`(n>i*(i+1)*(2*i+1)/6, 0,
      `if`(i=0, 1, b(n+i^2, i-1)+b(abs(n-i^2), i-1)))
    end:
a:=n-> b(0, n)+b(1, n):
seq(a(n), n=0..42);  # Alois P. Heinz, Jan 16 2022

b[n_, i_] := b[n, i] = If[n > i*(i + 1)*(2*i + 1)/6, 0,
     If[i == 0, 1, b[n + i^2, i - 1] + b[Abs[n - i^2], i - 1]]];
a[n_] := b[0, n] + b[1, n];
Table[a[n], {n, 0, 42}] (* Jean-François Alcover, Mar 01 2022, after Alois P. Heinz *)

from itertools import product
def a(n):
    if n == 0: return 1
    nn = ["0"] + [str(i)+"**2" for i in range(1, n+1)]
    return sum(eval("".join([*sum(zip(nn, ops+("", )), ())])) in {0, 1} for ops in product("+-", repeat=n))
print([a(n) for n in range(18)]) # Michael S. Branicky, Jan 16 2022

from functools import cache
@cache
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(0, n) + b(1, n)
print([a(n) for n in range(43)]) # Michael S. Branicky, Jan 16 2022 after Alois P. Heinz

cp's OEIS Frontend

A025591 Maximal coefficient of Product_{k<=n} (1 + x^k). Number of solutions to +- 1 +- 2 +- 3 +- ... +- n = 0 or 1.

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A083309 a(n) is the number of times that sums 3 +- 5 +- 7 +- 11 +- ... +- prime(2n+1) of the first 2n odd primes is zero. There are 2^(2n-1) choices for the sign patterns.

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A158092 Number of solutions to +- 1 +- 2^2 +- 3^2 +- 4^2 +- ... +- n^2 = 0.

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A113263 a(n) is the number of ways the set {1^3, 2^3, ..., n^3} can be partitioned into two sets of equal sums.

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

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A111253 a(n) is the number of ways the set {1^4, 2^4, ..., n^4} can be partitioned into two sets of equal sums.

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A300268 Number of solutions to 1 +- 4 +- 9 +- ... +- n^2 == 0 (mod n).

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