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 1-8 of 8 results.

A375274 Decimal expansion of the asymptotic density of the exponentially Fibonacci numbers (A115063).

Original entry on oeis.org

9, 4, 4, 3, 3, 5, 9, 0, 5, 0, 6, 4, 0, 6, 3, 3, 2, 4, 4, 8, 0, 5, 7, 3, 1, 3, 7, 7, 5, 6, 6, 6, 8, 8, 0, 5, 6, 1, 4, 6, 3, 4, 5, 8, 3, 2, 2, 2, 0, 2, 3, 5, 5, 5, 9, 2, 3, 6, 8, 3, 7, 7, 0, 4, 5, 5, 9, 3, 9, 5, 3, 8, 4, 6, 5, 4, 4, 6, 8, 5, 8, 7, 1, 9, 4, 1, 4, 2, 8, 0, 5, 2, 0, 3, 3, 7, 9, 2, 7, 4, 7, 9, 7, 2, 4
Offset: 0

Views

Author

Amiram Eldar, Aug 09 2024

Keywords

Comments

This constant was apparently first calculated by Juan Arias-de-Reyna and Peter J. C. Moses in 2015 (see A115063).

Examples

			0.94433590506406332448057313775666880561463458322202...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = m = 500; em = 16; f[x_] := Log[(1 - x) * (1 + Sum[x^Fibonacci[e], {e, 2, em}])]; c = Rest[CoefficientList[Series[f[x], {x, 0, m}], x] * Range[0, m]]; RealDigits[Exp[NSum[Indexed[c, k]*PrimeZetaP[k]/k, {k, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 105][[1]]
  • PARI
    c(imax) = prodeulerrat((1-1/p)*(1 + sum(i = 2, imax, 1/p^fibonacci(i))));
    f(prec) = {default(realprecision, prec); my(k = 2, c1 = 0, c2 = c(k)); while(c1 != c2, k++; c1 = c2; c2 = c(k)); c1;}
    f(120)

Formula

Equals Product_{p prime} (1 + Sum_{i>=2} (u(i) - u(i-1))/p^i), where u(i) = A010056(i) is the characteristic function of the Fibonacci numbers (A000045) (first formula at A115063).
Equals Product_{p prime} (1 + Sum_{i>=4} (-1)^(i+1)/p^A259623(i)).
Equals Product_{p prime} ((1 - 1/p) * (1 + Sum_{i>=2} 1/p^Fibonacci(i))).

A209061 Exponentially squarefree numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69
Offset: 1

Views

Author

Reinhard Zumkeller, Mar 13 2012

Keywords

Comments

Numbers having only squarefree exponents in their canonical prime factorization.
According to the formula of Theorem 3 [Toth], the density of the exponentially squarefree numbers is 0.9559230158619... (A262276). - Peter J. C. Moses and Vladimir Shevelev, Sep 10 2015
From Vladimir Shevelev, Sep 24 2015: (Start)
A generalization. Let S be a finite or infinite increasing integer sequence s=s(n), s(0)=0.
Let us call a positive number N an exponentially S-number, if all exponents in its prime power factorization are in the sequence S.
Let {u(n)} be the characteristic function of S. Then, for the density h=h(S) of the exponentially S-numbers, we have the representations
h(S) = Product_{prime p} Sum_{j in S} (p-1)/p^(j+1) = Product_{p} (1 + Sum_{j>=1} (u(j) - u(j-1))/p^j). In particular, if S = {0,1}, then the exponentially S-numbers are squarefree numbers; if S consists of 0 and {2^k}A138302%20(see%20%5BShevelev%5D,%202007);%20if%20S%20consists%20of%200%20and%20squarefree%20numbers,%20then%20u(n)=%7Cmu(n)%7C,%20where%20mu(n)%20is%20the%20M%C3%B6bius%20function%20(A008683),%20we%20obtain%20the%20density%20h%20of%20the%20exponentially%20squarefree%20numbers%20(cf.%20Toth's%20link,%20Theorem%203);%20the%20calculation%20of%20h%20with%20a%20very%20high%20degree%20of%20accuracy%20belongs%20to%20_Juan%20Arias-de-Reyna">{k>=0}, then the exponentially S-numbers form A138302 (see [Shevelev], 2007); if S consists of 0 and squarefree numbers, then u(n)=|mu(n)|, where mu(n) is the Möbius function (A008683), we obtain the density h of the exponentially squarefree numbers (cf. Toth's link, Theorem 3); the calculation of h with a very high degree of accuracy belongs to _Juan Arias-de-Reyna (A262276). Note that if S contains 1, then h(S) >= 1/zeta(2) = 6/Pi^2; otherwise h(S) = 0. Indeed, in the latter case, the density of the sequence of exponentially S-numbers does not exceed the density of A001694, which equals 0. (End)
The term "exponentially squarefree number" was apparently coined by Subbarao (1972). - Amiram Eldar, May 28 2025

Crossrefs

Programs

  • Haskell
    a209061 n = a209061_list !! (n-1)
    a209061_list = filter
       (all (== 1) . map (a008966 . fromIntegral) . a124010_row) [1..]
    
  • Mathematica
    Select[Range@ 69, Times @@ Boole@ Map[SquareFreeQ, Last /@ FactorInteger@ #] > 0 &] (* Michael De Vlieger, Sep 07 2015 *)
  • PARI
    is(n)=my(f=factor(n)[,2]); for(i=1,#f,if(!issquarefree(f[i]), return(0))); 1 \\ Charles R Greathouse IV, Sep 02 2015

Formula

A166234(a(n)) <> 0.
Product_{k=1..A001221(n)} A008966(A124010(n,k)) = 1.
One can prove that the principal term of Toth's asymptotics for the density of this sequence (cf. Toth's link, Theorem 3) equals also Product_{prime p}(Sum_{j in S}(p-1)/p^{j+1})*x, where S is the set of 0 and squarefree numbers. The remainder term O(x^(0.2+t)), where t>0 is arbitrarily small, was obtained by L. Toth while assuming the Riemann Hypothesis. - Vladimir Shevelev, Sep 12 2015

A197680 Numbers whose exponents in their prime power factorization are squares.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 48, 51, 53, 55, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 101
Offset: 1

Views

Author

A. Neves, Oct 17 2011

Keywords

Comments

Numbers whose prime factorization has the form Product_i p_i^e_i where the e_i are all squares.
All squarefree numbers (A005117) are in the sequence. - Vladimir Shevelev, Nov 16 2015
Let h_k be the density of the subsequence of A197680 of numbers whose prime power factorization (PPF) has the form Product_i p_i^e_i where the e_i all squares <= k^2. Then for every k>1 there exists eps_k>0 such that for any x from the interval (h_k-eps_k, h_k) there is no sequence S of positive integers such that x is the density of numbers whose PPF has the form Product_i p_i^e_i where the e_i are all in S. - For a proof, see [Shevelev], second link. - Vladimir Shevelev, Nov 17 2015
Numbers with an odd number of exponential divisors (A049419). - Amiram Eldar, Nov 05 2021

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; local k; for k from 1+
          `if`(n=1, 0, a(n-1)) while 0=mul(`if`(issqr(
           i[2]), 1, 0), i=ifactors(k)[2]) do od; k
        end:
    seq(a(n), n=1..80);  # Alois P. Heinz, Jun 30 2016
  • Mathematica
    Select[Range[100], Union[IntegerQ /@ Sqrt[Transpose[FactorInteger[#]][[2]]]][[1]] &] (* T. D. Noe, Oct 18 2011 *)
  • PARI
    isok(n) = {my(f = factor(n)[,2]); #select(x->issquare(x), f) == #f; } \\ Michel Marcus, Oct 23 2015

Formula

Sum_{i<=x, i is in A197680} 1 = h*x + O(sqrt(x)*log x*e^(c*sqrt(log x)/(log(log x))), where c=4*sqrt(2.4/log 2)=7.44308... and h=Product_{prime p} (1+Sum_{i>=2} (u(i)-u(i-1))/p^i)=0.641115... where u(n) is the characteristic function of sequence A000290. The calculations of h in the formula were done independently by Juan Arias-de-Reyna and Peter J. C. Moses. For a proof of the formula, see the first Shevelev link. - Vladimir Shevelev, Nov 17 2015

Extensions

Reformulation of the name by Vladimir Shevelev, Oct 14 2015

A369939 Numbers whose maximal exponent in their prime factorization is a Fibonacci number.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71
Offset: 1

Views

Author

Amiram Eldar, Feb 06 2024

Keywords

Comments

First differs from its subsequence A115063 at n = 2448. a(2448) = 2592 = 2^5 * 3^4 is not a term of A115063.
First differs from A209061 at n = 62.
Numbers k such that A051903(k) is a Fibonacci number.
The asymptotic density of this sequence is 1/zeta(4) + Sum_{k>=5} (1/zeta(Fibonacci(k)+1) - 1/zeta(Fibonacci(k))) = 0.94462177878047854647... .

Crossrefs

Similar sequences: A368714, A369937, A369938.

Programs

  • Mathematica
    fibQ[n_] := Or @@ IntegerQ /@ Sqrt[5*n^2 + {-4, 4}];
    Select[Range[100], fibQ[Max[FactorInteger[#][[;; , 2]]]] &]
  • PARI
    isfib(n) = issquare(5*n^2 - 4) || issquare(5*n^2 + 4);
    is(n) = n == 1 || isfib(vecmax(factor(n)[, 2]));

A386803 Numbers without an exponent 4 in their prime factorization.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70
Offset: 1

Views

Author

Amiram Eldar, Aug 03 2025

Keywords

Comments

First differs from its subsequence A209061 at n = 246: a(246) = 256 = 2^8 is not a term of A209061.
First differs from its subsequences A115063 and A369939 at n = 62: a(62) = 64 = 2^6 is not a term of A115063.
The complement of this sequence is a subsequence of A336595.
These numbers were named semi-4-free integers by Suryanarayana (1971).
The asymptotic density of this sequence is Product_{p prime} (1 - 1/p^4 + 1/p^5) = 0.95908865419555719109... (Suryanarayana, 1971).

Crossrefs

Subsequences: A115063, A209061, A369939.
Numbers without an exponent k in their prime factorization: A001694 (k=1), A337050 (k=2), A386799 (k=3), this sequence (k=4), A386807 (k=5).
Numbers that have exactly m exponents in their prime factorization that are equal to 4: this sequence (m=0), A386804 (m=1), A386805 (m=2), A386806 (m=3).

Programs

  • Mathematica
    Select[Range[100], !MemberQ[FactorInteger[#][[;; , 2]], 4] &]
  • PARI
    isok(k) = vecsum(apply(x -> if(x == 4, 1, 0), factor(k)[, 2])) == 0;

A384913 The number of unordered factorizations of n into exponentially Fibonacci powers of primes (A115975).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 6, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 8, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jun 12 2025

Keywords

Comments

First differs from A384912 at n = 64.

Examples

			a(4) = 2 since 4 has 2 factorizations: 2^1 * 2^1 and 2^2, with exponents 1 and 2 that are Fibonacci numbers.
		

Crossrefs

Programs

  • Mathematica
    fib[n_] := Boole[Or @@ IntegerQ /@ Sqrt[5*n^2 + {-4, 4}]];
    s[n_] := s[n] = If[n == 0, 1, Sum[Sum[d * fib[d], {d, Divisors[j]}] * s[n-j], {j, 1, n}] / n];
    f[p_, e_] := s[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    isfib(n) = issquare(5*n^2 - 4) || issquare(5*n^2 + 4);
    s(n) = if(n < 1, 1, sum(j = 1, n, sumdiv(j, d, d*isfib(d)) * s(n-j))/n);
    a(n) = vecprod(apply(s, factor(n)[, 2]));

Formula

Multiplicative with a(p^e) = A003107(e).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} f(1/p) = 2.05893526314055968638..., where f(x) = (1-x) / Product_{k>=2} (1-x^A000045(k)).

A375766 The maximum exponent in the prime factorization of the numbers whose exponents in their prime factorization are all Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 5, 1, 2, 2, 2, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Aug 27 2024

Keywords

Comments

First differs from A375768 at n = 2448.
All the terms are Fibonacci numbers by definition.

Crossrefs

Programs

  • Mathematica
    fibQ[n_] := Or @@ IntegerQ /@ Sqrt[5*n^2 + {-4, 4}]; s[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, If[AllTrue[e, fibQ], Max[e], Nothing]]; s[1] = 0; Array[s, 100]
  • PARI
    isfib(n) = issquare(5*n^2 - 4) || issquare(5*n^2 + 4);
    lista(kmax) = {my(e, ans); print1(0, ", "); for(k = 2, kmax, e = factor(k)[,2]; ans = 1; for(i = 1, #e, if(!isfib(e[i]), ans = 0; break)); if(ans, print1(vecmax(e), ", ")));}

Formula

a(n) = A051903(A115063(n)).
a(n) = A000045(A375767(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = (1/zeta(2) + Sum_{k>=3} (Fibonacci(k) * (d(k) - d(k-1)))) / A375274 = 1.52546070254904121983..., where d(k) = Product_{p prime} ((1-1/p)*(1 + Sum_{i=2..k} 1/p^Fibonacci(i))) for k >= 3, and d(2) = 1/zeta(2).

A115105 Numbers of the form p^F(i)*q^F(j), where p and q are distinct primes; F(i) and F(j) are Fibonacci numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 65, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 79, 82, 83
Offset: 1

Views

Author

Giovanni Teofilatto, Mar 03 2006; corrected Mar 04 2006

Keywords

Comments

All the primes are in the sequence, since they are of the form p^F(0)*q^F(1) with p^F(0) = p^0 = 1.

Examples

			a(1) = 1 because 2^0*3^0 = 1;
a(3) = 3 because 3^1*5^0 = 3;
a(4) = 4 because 2^2*3^0 = 4;
a(6) = 6 because 2^1*3^1 = 6.
		

Crossrefs

Cf. A115063.

Programs

  • Mathematica
    fibQ[n_] := IntegerQ @ Sqrt[5 n^2 - 4] || IntegerQ @ Sqrt[5 n^2 + 4]; aQ[n_] :=Length[ (e=FactorInteger[n][[;;, 2]])]<3 && AllTrue[e, fibQ]; Select[Range[100], aQ] (* Amiram Eldar, Oct 06 2019 *)

Extensions

2 missing terms inserted by Amiram Eldar, Oct 06 2019
Showing 1-8 of 8 results.