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 1-3 of 3 results.

A147612 If n is a Jacobsthal number then 1 else 0.

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0

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Author

Reinhard Zumkeller, Nov 08 2008

Keywords

Comments

a(A001045(n)) = 1; a(A147613(n)) = 0.

Crossrefs

Programs

Formula

a(n) = 0^(j(n,1)*j(n,-1)) with j(n,i) = if n mod 2 = 0 then n else j((n+i)/2,-i).
a(n) = A105348(n), for n <> 1. - R. J. Mathar, Nov 19 2008
For n > 0, a(n) = A000035(A281228(A265746(n))), where A000035(A281228(n)) is the characteristic function of powers of 3 (A000244). - Antti Karttunen, Oct 09 2017

A293431 a(n) is the number of Jacobsthal numbers dividing n.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 3, 1, 1, 2, 1, 2, 3, 2, 1, 2, 2, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 2, 1, 1, 2, 2, 1, 3, 2, 2, 3, 1, 1, 2, 1, 2, 2, 1, 1, 2, 3, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 3, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 2, 1, 2, 2, 1, 1, 3, 3, 2, 2, 2, 1, 3, 1, 1, 2, 1, 2, 2, 1, 1, 3, 2, 1, 2, 1, 1, 4
Offset: 1

Views

Author

Antti Karttunen, Oct 09 2017

Keywords

Examples

			For n = 15, whose divisors are [1, 3, 5, 15], the first three, 1, 3 and 5 are all in A001045, thus a(15) = 3.
For n = 21, whose divisors are [1, 3, 7, 21], 1, 3 and 21 are in A001045, thus a(21) = 3.
For n = 105, whose divisors are [1, 3, 5, 7, 15, 21, 35, 105], only the divisors 1, 3, 5 and 21 are in A001045, thus a(105) = 4.
		

Crossrefs

Programs

  • Mathematica
    With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Array[DivisorSum[#, 1 &, MemberQ[s, #] &] &, 105]] (* Michael De Vlieger, Oct 09 2017 *)
  • PARI
    A147612aux(n,i) = if(!(n%2),n,A147612aux((n+i)/2,-i));
    A147612(n) = 0^(A147612aux(n,1)*A147612aux(n,-1));
    A293431(n) = sumdiv(n,d,A147612(d));
    
  • Python
    from sympy import divisors
    def A293431(n): return sum(1 for d in divisors(n,generator=True) if (m:=3*d+1).bit_length()>(m-3).bit_length()) # Chai Wah Wu, Apr 18 2025

Formula

a(n) = Sum_{d|n} A147612(d).
a(n) = A293433(n) + A147612(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{n>=2} 1/A001045(n) = 1.718591611927... . - Amiram Eldar, Jan 01 2024

A293434 a(n) is the sum of the proper divisors of n that are Jacobsthal numbers (A001045).

Original entry on oeis.org

0, 1, 1, 1, 1, 4, 1, 1, 4, 6, 1, 4, 1, 1, 9, 1, 1, 4, 1, 6, 4, 12, 1, 4, 6, 1, 4, 1, 1, 9, 1, 1, 15, 1, 6, 4, 1, 1, 4, 6, 1, 25, 1, 12, 9, 1, 1, 4, 1, 6, 4, 1, 1, 4, 17, 1, 4, 1, 1, 9, 1, 1, 25, 1, 6, 15, 1, 1, 4, 6, 1, 4, 1, 1, 9, 1, 12, 4, 1, 6, 4, 1, 1, 25, 6, 44, 4, 12, 1, 9, 1, 1, 4, 1, 6, 4, 1, 1, 15, 6, 1, 4, 1, 1, 30
Offset: 1

Views

Author

Antti Karttunen, Oct 09 2017

Keywords

Examples

			For n = 15, whose proper divisors are [1, 3, 5], all of them are in A001045, thus a(15) = 1 + 3 + 5 = 9.
For n = 21, whose proper divisors are [1, 3, 7], both 1 and 3 are in A001045, thus a(21) = 1 + 3 = 4.
For n = 21845, whose proper divisors are [1, 5, 17, 85, 257, 1285, 4369], only 1, 5, 85 are in A001045, thus a(21845) = 1 + 5 + 85 = 91.
		

Crossrefs

Programs

  • Mathematica
    With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Table[DivisorSum[n, # &, And[MemberQ[s, #], # != n] &], {n, 105}]] (* Michael De Vlieger, Oct 09 2017 *)
  • PARI
    A147612aux(n,i) = if(!(n%2),n,A147612aux((n+i)/2,-i));
    A147612(n) = 0^(A147612aux(n,1)*A147612aux(n,-1));
    A293434(n) = sumdiv(n,d,(dA147612(d)*d);
    
  • Python
    from sympy import divisors
    def A293434(n): return sum(d for d in divisors(n,generator=True) if d(m-3).bit_length()) # Chai Wah Wu, Apr 18 2025

Formula

a(n) = Sum_{d|n, dA147612(d)*d.
a(n) = A293432(n) - (A147612(n)*n).
Showing 1-3 of 3 results.