A025591 -id:A025591 - cp's OEIS frontend

A350861 Number of solutions to +-1^3 +- 2^3 +- 3^3 +- ... +- n^3 = 0 or 1.

1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 4, 1, 4, 2, 6, 1, 4, 124, 12, 344, 536, 712, 1140, 713, 4574, 2260, 4384, 5956, 10634, 73758, 48774, 197767, 406032, 638830, 1147500, 1097442, 4249160, 3263500, 6499466, 11844316, 21907736, 82561050, 85185855, 261696060

Offset: 0

Ilya Gutkovskiy, Jan 19 2022

nonn

			a(12) = 2: +1^3 + 2^3 - 3^3 + 4^3 - 5^3 - 6^3 - 7^3 + 8^3 + 9^3 - 10^3 - 11^3 + 12^3 = -1^3 - 2^3 + 3^3 - 4^3 + 5^3 + 6^3 + 7^3 - 8^3 - 9^3 + 10^3 + 11^3 - 12^3 = 0.

Robert Israel, Table of n, a(n) for n = 0..99

Cf. A000578, A025591, A113263, A158118.

f:= proc(n) local S,k,x,s;
  S:= mul(1 + x^(2*k^3),k=1..n);
  s:= sum(k^3,k=1..n);
  coeff(S,x,s) + coeff(S,x,s+1)
end proc:
map(f, [$0..50]); # Robert Israel, Mar 15 2023

from functools import lru_cache
@lru_cache(maxsize=None)
def b(n, i):
    if n > (i*(i+1)//2)**2: return 0
    if i == 0: return 1
    return b(n+i**3, i-1) + b(abs(n-i**3), i-1)
def a(n): return b(0, n) + b(1, n)
print([a(n) for n in range(46)]) # Michael S. Branicky, Jan 19 2022

A359348 Maximal coefficient of (1 + x) * (1 + x^3) * (1 + x^6) * ... * (1 + x^(n*(n+1)/2)).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 4, 5, 7, 12, 18, 27, 44, 73, 122, 210, 362, 620, 1050, 1857, 3290, 5949, 10665, 19086, 34330, 62252, 113643, 209460, 383888, 706457, 1300198, 2407535, 4468367, 8331820, 15525814, 28987902, 54180854, 101560631, 190708871, 358969426

Offset: 0

Seiichi Manyama, Dec 27 2022

nonn

			(1 + x) * (1 + x^3) * (1 + x^6) * (1 + x^10) = 1 + x + x^3 + x^4 + x^6 + x^7 + x^9 + 2 * x^10 + x^11 + x^13 + x^14 + x^16 + x^17 + x^19 + x^20. So a(4) = 2.

Seiichi Manyama, Table of n, a(n) for n = 0..200

Cf. A000217, A024940, A025591, A158380, A160235.

a(n) = vecmax(Vec(prod(k=1, n, 1+x^(k*(k+1)/2))));

a(n) ~ sqrt(5) * 2^(n + 3/2) / (sqrt(Pi) * n^(5/2)). - Vaclav Kotesovec, Dec 29 2022

A369629 Number of solutions to +- 1^4 +- 2^4 +- 3^4 +- ... +- n^4 = 0 or 1.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 5, 0, 16, 18, 0, 21, 32, 100, 126, 0, 424, 510, 0, 1428, 2792, 5988, 9786, 7, 29058, 45106, 22, 150437, 276828, 473854, 836737, 1838, 2455340, 4777436, 15847, 14696425, 27466324, 46429640, 83010230, 738627

Offset: 0

Ilya Gutkovskiy, Jan 28 2024

nonn

Chai Wah Wu, Table of n, a(n) for n = 0..73

Cf. A000583, A025591, A111253, A158465, A350403, A350861.

from itertools import count, islice
from collections import Counter
def A369629_gen(): # generator of terms
    ccount = Counter({0:1})
    yield 1
    for i in count(1):
        bcount = Counter()
        for a in ccount:
            bcount[a+(j:=i**4)] += ccount[a]
            bcount[a-j] += ccount[a]
        ccount = bcount
        yield(ccount[0]+ccount[1])
A369629_list = list(islice(A369629_gen(),20)) # Chai Wah Wu, Jan 29 2024

a(46)-a(50) from Chai Wah Wu, Jan 29 2024

A342602 Number of solutions to 1 +-* 2 +-* 3 +-* ... +-* n = 0.

Original entry on oeis.org

0, 0, 1, 1, 1, 4, 6, 14, 29, 63, 129, 300, 756, 1677, 4134, 9525, 22841, 57175, 141819, 354992, 882420, 2218078, 5588989, 14173217, 35918542

Offset: 1

Scott R. Shannon, Mar 27 2021

Normal operator precedence is followed, so multiplication is performed before addition or subtraction. Unlike A058377, which uses only addition and subtraction, this sequence has solutions for all values of n >= 3.

The author thanks Ursula Ponting for useful discussions.

			a(3) = 1 as 1 + 2 - 3 = 0 is the only solution.
a(4) = 1 as 1 - 2 - 3 + 4 = 0 is the only solution.
a(5) = 1 as 1 * 2 - 3 - 4 + 5 = 0 is the only solution. This is the first term where a solution exists while no corresponding solution exists in A058377.
a(6) = 4. The solutions, all of which use multiplication, are
         1 + 2 * 3 + 4 - 5 - 6 = 0,
         1 - 2 + 3 * 4 - 5 - 6 = 0,
         1 - 2 * 3 + 4 - 5 + 6 = 0,
         1 * 2 + 3 - 4 + 5 - 6 = 0.
a(10) = 63. An example solution is
         1 - 2 * 3 * 4 - 5 - 6 - 7 * 8 + 9 * 10 = 0.
a(20) = 354992. An example solution is
         1 * 2 * 3 * 4 * 5 * 6 * 7 + 8 * 9 + 10 * 11 - 12 * 13 + 14 * 15
             - 16 * 17 * 18 - 19 * 20 = 0
  which includes thirteen multiplications.

Cf. A342804 (using +-*/), A342995 (using +-/), A058377 (using +-), A063865, A000217, A025591, A161943.

Table[Length@Select[Tuples[{"+","-","*"},k-1],ToExpression[""<>Riffle[ToString/@Range@k,#]]==0&],{k,11}] (* Giorgos Kalogeropoulos, Apr 02 2021 *)

from itertools import product
def a(n):
  nn = [str(i) for i in range(1, n+1)]
  return sum(eval("".join([*sum(zip(nn, ops+("",)), ())])) == 0 for ops in product("+-*", repeat=n-1))
print([a(n) for n in range(1, 14)]) # Michael S. Branicky, Apr 02 2021

A342995 Number of solutions to 1 +-/ 2 +-/ 3 +-/ ... +-/ n = 0.

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 4, 8, 0, 3, 37, 80, 6, 17, 461, 868, 190, 364, 5570, 11342, 3993, 7307, 78644

Offset: 1

Scott R. Shannon, Apr 01 2021

Normal operator precedence is followed, so division is performed before addition or subtraction. Unlike A058377, which uses only addition and subtraction, this sequence has solutions for all values of n >= 10.

			a(3) = 1 as 1 + 2 - 3 = 0 is the only solution.
a(4) = 1 as 1 - 2 - 3 + 4 = 0 is the only solution.
a(5) = 0, as in A058377.
a(6) = 1 as 1 - 2 / 3 / 4 - 5 / 6 = 0 is the only solution. This is the first term where a solution exists while no corresponding solution exists in A058377.
a(8) = 8. Seven of the solutions involve just addition and subtraction, matching those in A058377, but one additional solution exists using division:
        1 / 2 / 3 / 4 + 5 / 6 - 7 / 8 = 0.
a(10) = 3. All three solutions require division:
        1 + 2 / 3 / 4 + 5 / 6 + 7 - 8 + 9 - 10 = 0,
        1 - 2 / 3 / 4 - 5 / 6 + 7 - 8 - 9 + 10 = 0,
        1 - 2 / 3 / 4 - 5 / 6 - 7 + 8 + 9 - 10 = 0.
a(15) = 461. Of these, 361 use only addition and subtraction, the other 100 also require division. One example of the latter is
        1 / 2 / 3 / 4 - 5 - 6 - 7 / 8 + 9 / 10 + 11 + 12 - 13 + 14 / 15 = 0.
a(20) = 11342. An example solution is
        1 / 2 / 3 - 4 / 5 / 6 + 7 / 8 / 9 + 10 + 11 / 12 - 13 + 14 / 15 / 16
             + 17 / 18 + 19 / 20 = 0
  which sums seven fractions that include eleven divisions.

Cf. A342804 (using +-*/), A342602 (using +-*), A058377 (using +-), A063865, A000217, A025591, A161943.

Table[Length@Select[Tuples[{"+","-","/"},k-1],ToExpression[""<>Riffle[ToString/@Range@k,#]]==0&],{k,11}] (* Giorgos Kalogeropoulos, Apr 02 2021 *)

from itertools import product
from fractions import Fraction
def a(n):
  nn = ["Fraction("+str(i)+", 1)" for i in range(1, n+1)]
  return sum(eval("".join([*sum(zip(nn, ops+("",)), ())])) == 0 for ops in product("+-/", repeat=n-1))
print([a(n) for n in range(1, 11)]) # Michael S. Branicky, Apr 02 2021

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

A369708 Maximal coefficient of (1 + x^3) * (1 + x^5) * (1 + x^7) * ... * (1 + x^prime(n)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 4, 5, 8, 14, 23, 40, 70, 126, 221, 394, 711, 1290, 2354, 4344, 8015, 14868, 27585, 51094, 95160, 178436, 335645, 634568, 1202236, 2261052, 4267640, 8067296, 15318171, 29031484, 55248527, 105251904, 200711160, 383580180, 733704990

Offset: 0

Ilya Gutkovskiy, Jan 29 2024

nonn

Cf. A024939, A025591, A350457.

b:= proc(n) option remember; `if`(n<2, 1, expand(b(n-1)*(1+x^ithprime(n)))) end:
a:= n-> max(coeffs(b(n))):
seq(a(n), n=0..40);  # Alois P. Heinz, Jan 29 2024

Table[Max[CoefficientList[Product[(1 + x^Prime[k]), {k, 2, n}], x]], {n, 0, 40}]

a(n) = vecmax(Vec(prod(i=2, n, 1+x^prime(i)))); \\ Michel Marcus, Jan 29 2024

from collections import Counter
from sympy import prime
def A369708(n):
    c = {0:1}
    for i in range(2,n+1):
        p, d = prime(i), Counter(c)
        for k in c:
            d[k+p] += c[k]
        c = d
    return max(c.values()) # Chai Wah Wu, Jan 31 2024

A369712 Maximal coefficient of (1 + x) * (1 + x^2)^2 * (1 + x^3)^3 * ... * (1 + x^n)^n.

Original entry on oeis.org

1, 1, 2, 9, 79, 1702, 78353, 7559080, 1509040932, 619097417818, 519429629728698, 887531129680197018, 3078434842626707386602, 21627792113204714623569767, 307257554772242590850211062866, 8813577747274880345454470354985336, 509819403352972623999938010230619997952

Offset: 0

Ilya Gutkovskiy, Jan 29 2024

nonn

Cf. A025591, A026007, A047653.

b:= proc(n) option remember; `if`(n=0, 1, expand(b(n-1)*(1+x^n)^n)) end:
a:= n-> max(coeffs(b(n))):
seq(a(n), n=0..16);  # Alois P. Heinz, Jan 29 2024

Table[Max[CoefficientList[Product[(1 + x^k)^k, {k, 1, n}], x]], {n, 0, 16}]

a(n) = vecmax(Vec(prod(k=1, n, (1+x^k)^k))); \\ Michel Marcus, Jan 30 2024

A350287 Number of solutions to +-1 +- 3 +- 6 +- 10 +- ... +- n*(n + 1)/2 = 0 or 1.

Original entry on oeis.org

1, 1, 0, 0, 2, 1, 2, 2, 4, 7, 12, 16, 26, 42, 66, 104, 210, 318, 620, 970, 1748, 3281, 5948, 10480, 18976, 34233, 60836, 111430, 209460, 378529, 704934, 1284836, 2387758, 4466874, 8331820, 15525814, 28987902, 54162165, 101242982, 190267598, 358969426, 674845081

Offset: 0

Ilya Gutkovskiy, Jan 19 2022

nonn

			a(4) = 2: -1 - 3 - 6 + 10 = +1 + 3 + 6 - 10 = 0.

Cf. A000217, A025591, A058498, A158380.

from functools import lru_cache
@lru_cache(maxsize=None)
def b(n, i):
    if n > i*(i+1)*(i+2)//6: return 0
    if i == 0: return 1
    return b(n+i*(i+1)//2, i-1) + b(abs(n-i*(i+1)//2), i-1)
def a(n): return b(0, n) + b(1, n)
print([a(n) for n in range(41)]) # Michael S. Branicky, Jan 19 2022

A350504 Maximal coefficient of (1 + x) * (1 + x^3) * (1 + x^5) * ... * (1 + x^(2*n-1)).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 5, 8, 13, 22, 38, 68, 118, 211, 380, 692, 1262, 2316, 4277, 7930, 14745, 27517, 51541, 96792, 182182, 343711, 650095, 1231932, 2338706, 4447510, 8472697, 16164914, 30884150, 59086618, 113189168, 217091832, 416839177, 801247614, 1541726967, 2969432270

Offset: 0

Ilya Gutkovskiy, Jan 28 2022

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A025591, A039828, A160235, A350457.

b:= proc(n) option remember; `if`(n=0, 1,
      expand((1+x^(2*n-1))*b(n-1)))
    end:
a:= n-> max(coeffs(b(n))):
seq(a(n), n=0..40);  # Alois P. Heinz, Jan 28 2022

b[n_] := b[n] = If[n == 0, 1, Expand[(1 + x^(2*n - 1))*b[n - 1]]];
a[n_] := Max[CoefficientList[b[n], x]];
Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 01 2022, after Alois P. Heinz *)

a(n) = vecmax(Vec(prod(k=1, n, 1+x^(2*k-1)))); \\ Seiichi Manyama, Jan 28 2021

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

cp's OEIS Frontend

A350861 Number of solutions to +-1^3 +- 2^3 +- 3^3 +- ... +- n^3 = 0 or 1.

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A359348 Maximal coefficient of (1 + x) * (1 + x^3) * (1 + x^6) * ... * (1 + x^(n*(n+1)/2)).

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A369629 Number of solutions to +- 1^4 +- 2^4 +- 3^4 +- ... +- n^4 = 0 or 1.

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Extensions

A342602 Number of solutions to 1 +-* 2 +-* 3 +-* ... +-* n = 0.

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Mathematica

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A342995 Number of solutions to 1 +-/ 2 +-/ 3 +-/ ... +-/ n = 0.

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

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Python

A369708 Maximal coefficient of (1 + x^3) * (1 + x^5) * (1 + x^7) * ... * (1 + x^prime(n)).

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Python

A369712 Maximal coefficient of (1 + x) * (1 + x^2)^2 * (1 + x^3)^3 * ... * (1 + x^n)^n.

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