A342804 -id:A342804 - cp's OEIS frontend

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

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

A363802 Numbers whose digits can be interposed with one or more of the arithmetic operators +, -, *, /, with no parentheses or concatenation, to yield 10 as the result.

Original entry on oeis.org

19, 25, 28, 37, 46, 52, 55, 64, 73, 82, 91, 109, 118, 119, 125, 127, 128, 133, 136, 137, 145, 146, 152, 154, 155, 163, 164, 172, 173, 181, 182, 190, 191, 208, 215, 217, 218, 219, 224, 226, 229, 234, 235, 242, 244, 250, 251, 253, 262, 271, 274, 280, 281, 286, 291, 298, 307

Offset: 1

Evan Gillard, Jun 23 2023

This sequence is a variant of a "game" you can play using the numbers on train carriages (usually 4 digits in Australia's case), ignoring prefixed zeros, preventing re-ordering of the digits and allowing only addition, subtraction, multiplication and division.
No parentheses or concatenation are allowed and expressions follow operator precedence (*/) then (+-), and left to right within the same level of precedence: "2 + 3 * 2 / 6" is 2 + ((3*2)/6) = 2 + 1 = 3.
Infinite since A052224 is a subsequence. - Michael S. Branicky, Jun 24 2023

			1 + 9 = 2 + 8 = 1 * 9 + 1 = 2 * 9 - 8 = 10 so 19, 28, 191 and 298 are terms.
110 is not a term even though 1 * 10 = 10 since concatenation is disallowed.

Supersequence of A052224.

Cf. A342804, A357272.

from itertools import product
from fractions import Fraction
def is_A363802(n):
    s = [f"Fraction({d}, 1)" for d in str(n)]
    for ops in product("+-*/", repeat=len(s)-1):
        try: v = eval("".join(sum(zip(ops, s[1:]), (s[0],))))
        except: v = None
        if v == 10: return True
    return False
# Evan Gillard and Michael S. Branicky, Jun 23 2023

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

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

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A363802 Numbers whose digits can be interposed with one or more of the arithmetic operators +, -, *, /, with no parentheses or concatenation, to yield 10 as the result.

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Python