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.

Showing 1-4 of 4 results.

A359411 a(n) is the number of divisors of n that are both infinitary and exponential.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

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Author

Amiram Eldar, Dec 30 2022

Keywords

Comments

First differs from A318672 and A325989 at n = 32.
If e > 0 is the exponent of the highest power of p dividing n (where p is a prime), then for each divisor d of n that is both an infinitary and an exponential divisor, the exponent of the highest power of p dividing d is a number k such that k | e and the bitwise AND of e and k is equal to k.
The least term that is higher than 2 is a(216) = 4.
The position of the first appearance of a prime p in this sequence is 2^A359081(p), if A359081(p) > -1. E.g., 2^39 = 549755813888 for p = 3, 2^175 = 4.789...*10^52 for p = 5, and 2^1275 = 6.504...*10^383 for p = 7.
This sequence is unbounded since A246600 is unbounded (see A359082).

Examples

			a(8) = 2 since 8 has 2 divisors that are both infinitary and exponential: 2 and 8.
		

Crossrefs

Programs

  • Mathematica
    s[n_] := DivisorSum[n, 1 &, BitAnd[n, #] == # &]; f[p_, e_] := s[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    s(n) = sumdiv(n, d, bitand(d, n)==d);
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, s(f[i,2]));}
    
  • Python
    from math import prod
    from sympy import divisors, factorint
    def A359411(n): return prod(sum(1 for d in divisors(e,generator=True) if e|d == e) for e in factorint(n).values()) # Chai Wah Wu, Sep 01 2023

Formula

Multiplicative with a(p^e) = A246600(e).
a(n) = 1 if and only if n is in A138302.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + Sum_{k>=1} A246600(k)/p^k) = 1.135514937... .

A359082 Indices of records in A246600.

Original entry on oeis.org

1, 3, 15, 63, 255, 495, 4095, 96255, 98175, 130815, 203775, 1048575, 5810175, 6455295, 16777215, 67096575, 88062975, 389656575, 553517055, 850917375, 1157349375, 9141354495, 12826279935, 22828220415, 26818379775, 31684427775, 68719476735, 242870910975, 1168231038975
Offset: 1

Views

Author

Amiram Eldar, Dec 15 2022

Keywords

Comments

Numbers k with a record number of divisors d such that the bitwise OR of k and d is equal to k (or equivalently, the bitwise AND of k and d is equal to d).
All the terms are odd since A246600(2*k) = A246600(k).
This sequence is infinite since A246600(2^m-1) = A000005(2^m-1) = A046801(m), and A046801 is unbounded (A046801(2^(m+1)) > A046801(2^m) for all m >= 0).
The corresponding record values are 1, 2, 4, 6, 8, 11, 24, 25, 28, 32, 35, 48, 56, 89, 96, 105, 121, 127, 148, 162, 216, 243, 245, 256, 319, 358, 512, 633, 768, ... .
2*10^11 < a(28) <= 2^48 - 1.

Crossrefs

Programs

  • Mathematica
    s[n_] := DivisorSum[n, 1 &, BitAnd[n, #] == # &]; seq={}; sm = 0; Do[If[(sn = s[n]) > sm, sm = sn; AppendTo[seq, n]], {n, 1, 10^6}]; seq
  • PARI
    lista(nmax) = {my(list = List(), ndmax = 0, d, s); for(n = 1, nmax, nd = sumdiv(n, d, bitand(d, n)==d); if(nd > ndmax, ndmax = nd; listput(list, n))); Vec(list)};

Extensions

a(28)-a(29) from Martin Ehrenstein, Dec 19 2022

A359080 Numbers k such that A246600(k) = A000005(k).

Original entry on oeis.org

1, 3, 5, 7, 11, 13, 15, 17, 19, 23, 27, 29, 31, 37, 41, 43, 47, 51, 53, 59, 61, 63, 67, 71, 73, 79, 83, 85, 89, 95, 97, 101, 103, 107, 109, 111, 113, 119, 123, 125, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 173, 179, 181, 187, 191, 193, 197, 199, 211, 219
Offset: 1

Views

Author

Amiram Eldar, Dec 15 2022

Keywords

Comments

Numbers k such that for all the divisors d of k the bitwise OR of k and d is equal to k (or equivalently, the bitwise AND of k and d is equal to d).
Subsequence of A102553. Terms of A102553 that are not in this sequence: 2, 135, 175, 243, 343, ... .
All the terms are odd since if k is even and d = 1 then bitor(k, 1) > k and thus A246600(k) < A000005(k).
All the odd primes are terms.
All the numbers of the form 2^k-1 (A000225) are terms.
Numbers k such that the bitwise OR(k, d_1, d_2, ..., d_m) = k, where d_1, ..., d_m are the divisors of k. - Chai Wah Wu, Dec 18 2022

Crossrefs

Subsequence of A102553.
Subsequences: A000225, A065091.

Programs

  • Mathematica
    s[n_] := DivisorSum[n, 1 &, BitAnd[n, #] == # &]; Select[Range[250], s[#] == DivisorSigma[0, #] &]
  • PARI
    is(n) = sumdiv(n, d, bitor(d, n) == n) == numdiv(n);
    
  • Python
    from itertools import count, islice
    from operator import ior
    from functools import reduce
    from sympy import divisors
    def A359080_gen(startvalue=1):  # generator of terms >= startvalue
        return filter(
            lambda n: n | reduce(ior, divisors(n, generator=True)) == n,
            count(max(startvalue, 1)),
        )
    A359080_list = list(islice(A359080_gen(), 20))  # Chai Wah Wu, Dec 18 2022
    print(A359080_list)

A359083 Numbers k such that A246600(k) = A000005(k) and A000005(k) sets a new record.

Original entry on oeis.org

1, 3, 15, 63, 255, 891, 4095, 262143, 1048575, 16777215, 68719476735
Offset: 1

Views

Author

Amiram Eldar, Dec 15 2022

Keywords

Comments

Numbers k with a record number of divisors, such that for all the divisors d of k the bitwise OR of k and d is equal to k (or equivalently, the bitwise AND of k and d is equal to d).
All the terms are odd since all the terms of A359080 are odd.
The corresponding numbers of divisors are 1, 2, 4, 6, 8, 10, 24, 32, 48, 96, 512, ... .
a(12) > 3*10^11, if it exists.

Crossrefs

Subsequence of A359080.

Programs

  • Mathematica
    s[n_] := DivisorSum[n, 1 &, BitAnd[n, #] == # &]; seq={}; dm = 0; Do[d = DivisorSigma[0, n]; If[d > dm && d == s[n], dm = d; AppendTo[seq, n]], {n, 1, 2*10^7}]; seq
  • PARI
    lista(nmax) = {my(list = List(), ndmax = 0, d, s); for(n = 1, nmax, nd = numdiv(n); if(nd > ndmax && sumdiv(n, d, bitand(d, n)==d) == nd, ndmax = nd; listput(list, n))); Vec(list)};
Showing 1-4 of 4 results.