cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A071313 a(n) is the smallest number that cannot be obtained from the numbers {1,3,...,2*n-1} using each number at most once and the operators +, -, *, /, where intermediate subexpressions must be integers.

Original entry on oeis.org

2, 5, 11, 41, 92, 733, 4337, 28972, 195098, 1797746
Offset: 1

Views

Author

Koksal Karakus (karakusk(AT)hotmail.com), Jun 11 2002

Keywords

Comments

If noninteger subexpressions are permitted, a(5) = 122 and not 92 since 92 = (3+7)*(9 + 1/5). - Michael S. Branicky, Jul 01 2022

Examples

			a(2)=5 because using {1,3} and the four operations we can obtain 1=1, 3-1=2, 3=3, 3+1=4 but we cannot obtain 5 in the same way.
		

Crossrefs

Programs

  • Python
    def a(n):
        R = dict() # index of each reachable subset is [card(s)-1][s]
        for i in range(n): R[i] = dict()
        for i in range(1, n+1): R[0][(2*i-1,)] = {2*i-1}
        reach = set(range(1, 2*n, 2))
        for j in range(1, n):
            for i in range((j+1)//2):
                for s1 in R[i]:
                    for s2 in R[j-1-i]:
                        if set(s1) & set(s2) == set():
                            s12 = tuple(sorted(set(s1) | set(s2)))
                            if s12 not in R[len(s12)-1]:
                                R[len(s12)-1][s12] = set()
                            for a in R[i][s1]:
                                for b in R[j-1-i][s2]:
                                    allowed = [a+b, a*b, a-b, b-a]
                                    if a!=0 and b%a==0: allowed.append(b//a)
                                    if b!=0 and a%b==0: allowed.append(a//b)
                                    R[len(s12)-1][s12].update(allowed)
                                    reach.update(allowed)
        k = 1
        while k in reach: k += 1
        return k
    print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Jul 01 2022

Extensions

a(10) from Michael S. Branicky, Jul 01 2022

A071603 Number of different positive integers that we can obtain from the integers {1,2,...,n} using each number at most once and the operators +, -, *, /, where intermediate subexpressions must be integers.

Original entry on oeis.org

1, 3, 9, 31, 121, 542, 2868, 16329, 106762, 758155, 6142570
Offset: 1

Views

Author

Koksal Karakus (karakusk(AT)hotmail.com), Jun 02 2002

Keywords

Examples

			a(4)=31 because we can obtain the positive integers 1,2,...,28 and 30,32,36 by using the integers {1, 2, 3, 4} at most once and the four operations. For example 30 = 3*2*(4+1).
		

Crossrefs

Programs

  • Python
    def a(n):
        R = dict() # index of each reachable subset is [card(s)-1][s]
        for i in range(n): R[i] = dict()
        for i in range(1, n+1): R[0][(i,)] = {i}
        reach = set(range(1, n+1))
        for j in range(1, n):
            for i in range((j+1)//2):
                for s1 in R[i]:
                    for s2 in R[j-1-i]:
                        if set(s1) & set(s2) == set():
                            s12 = tuple(sorted(set(s1) | set(s2)))
                            if s12 not in R[len(s12)-1]:
                                R[len(s12)-1][s12] = set()
                            for a in R[i][s1]:
                                for b in R[j-1-i][s2]:
                                    allowed = [a+b, a*b, a-b, b-a]
                                    if a!=0 and b%a==0: allowed.append(b//a)
                                    if b!=0 and a%b==0: allowed.append(a//b)
                                    R[len(s12)-1][s12].update(allowed)
                                    reach.update(allowed)
        return len(set(r for r in reach if r > 0 and r.denominator == 1))
    print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Jul 01 2022

Extensions

a(10)-a(11) from Michael S. Branicky, Jul 01 2022
Showing 1-2 of 2 results.