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.

A375429 The indices of the terms of A375428 in the Fibonacci sequence.

Original entry on oeis.org

0, 2, 2, 3, 2, 2, 2, 4, 3, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 3, 2, 2, 2, 4, 3, 2, 4, 3, 2, 2, 2, 5, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 3, 3, 2, 2, 4, 3, 3, 2, 3, 2, 4, 2, 4, 2, 2, 2, 3, 2, 2, 3, 5, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 3, 3, 2, 2, 2, 4, 4, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 3, 2, 2, 2, 5, 2, 3, 3, 3, 2, 2, 2, 4, 2
Offset: 1

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Author

Amiram Eldar, Aug 15 2024

Keywords

Comments

Since 1 appears twice in the Fibonacci sequence (1 = Fibonacci(1) = Fibonacci(2)), its index here is chosen to be 2.

Examples

			For n = 16 = 2^4, the Zeckendorf representation of 4 is 101, i.e., 4 = Fibonacci(2) + Fibonacci(4). Therefore 16 = 2^(Fibonacci(2) + Fibonacci(4)) = 2^Fibonacci(2) * 2^Fibonacci(4), and a(16) = 4.
		

Crossrefs

Programs

  • Mathematica
    A130233[n_] := Module[{k = 2}, While[Fibonacci[k] <= n, k++]; k-1]; a[n_] := A130233[Max[FactorInteger[n][[;;, 2]]]]; a[1] = 0; Array[a, 100]
  • PARI
    A130233(n) = {my(k = 2); while(fibonacci(k) <= n, k++); k-1;}
    a(n) = if(n == 1, 0, A130233(vecmax(factor(n)[,2])));

Formula

a(n) = A130233(A375428(n)).
a(n) = A130233(A051903(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3 - 1/zeta(2) + Sum_{k>=4} (1 - 1/zeta(Fibonacci(k))) = 2.59996215929231584366... .
If the chosen index for 1 is 1 instead of 2, then the asymptotic mean is 3 - 2/zeta(2) + Sum_{k>=4} (1 - 1/zeta(Fibonacci(k))) = 1.99203505743828921499... .

A375432 Numbers k such that A375428(k) > A375430(k).

Original entry on oeis.org

8, 24, 27, 32, 40, 54, 56, 64, 72, 88, 96, 104, 108, 120, 125, 135, 136, 152, 160, 168, 184, 189, 192, 200, 216, 224, 232, 243, 248, 250, 256, 264, 270, 280, 288, 296, 297, 312, 320, 328, 343, 344, 351, 352, 360, 375, 376, 378, 392, 408, 416, 424, 440, 448, 456
Offset: 1

Views

Author

Amiram Eldar, Aug 15 2024

Keywords

Comments

First differs from A374590 at n = 31.
For numbers k that are not in this sequence A375428(k) = A375430(k).
Numbers k such that A051903(k)+1 is not of the form Fibonacci(m)-1, m >= 3.
The asymptotic density of this sequence is 1 - 1/zeta(2) - Sum_{k>=4} (1/zeta(Fibonacci(k)) - 1/zeta(Fibonacci(k)-1)) = 0.12330053981922224451... .

Examples

			8 is a term since A375428(8) = 3 > 2 = A375430(8).
		

Crossrefs

Programs

  • Mathematica
    fibQ[n_] := n >= 2 && Or @@ IntegerQ /@ Sqrt[5*n^2 + {-4, 4}]; Select[Range[300], !fibQ[Max[FactorInteger[#][[;;, 2]]] + 1] &]
  • PARI
    isfib(n) = n >= 2 && (issquare(5*n^2-4) || issquare(5*n^2+4));
    is(n) = n > 1 && !isfib(vecmax(factor(n)[,2]) + 1);

A375430 The maximum exponent in the unique factorization of n in terms of distinct terms of A115975 using the dual Zeckendorf representation of the exponents in the prime factorization of n; a(1) = 0.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Aug 15 2024

Keywords

Comments

First differs from A299090 at n = 128. Differs from A046951 and A159631 at n = 1, 36, 64, 72, ... .
When the exponents in the prime factorization of n are expanded as sums of distinct Fibonacci numbers using the dual Zeckendorf representation (A104326), we get a unique factorization of n in terms of distinct terms of A115975, i.e., n is represented as a product of prime powers (A246655) whose exponents are Fibonacci numbers. a(n) is the maximum exponent of these prime powers. Thus all the terms are Fibonacci numbers.

Examples

			For n = 8 = 2^3, the dual Zeckendorf representation of 3 is 11, i.e., 3 = Fibonacci(2) + Fibonacci(3) = 1 + 2. Therefore 8 = 2^(1+2) = 2^1 * 2^2, and a(8) = 2.
		

Crossrefs

Programs

  • Mathematica
    A130312[n_] := Module[{k = 0}, While[Fibonacci[k] <= n, k++]; Fibonacci[k-2]]; a[n_] := A130312[1 + Max[FactorInteger[n][[;;, 2]]]]; a[1] = 0; Array[a, 100]
  • PARI
    A130312(n) = {my(k = 0); while(fibonacci(k) <= n, k++); fibonacci(k-2);}
    a(n) = if(n == 1, 0, A130312(1 + vecmax(factor(n)[,2])));

Formula

a(n) = A130312(1 + A051903(n)).
a(n) = A000045(A375431(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + Sum_{k>=4} Fibonacci(k) * (1 - 1/zeta(Fibonacci(k)-1)) = 1.48543763231328442311... .

A386468 The maximum exponent in the prime factorization of the largest exponentially squarefree divisor of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 3, 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, 3, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 2, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jul 22 2025

Keywords

Comments

First differs from A375428 at n = 64.
Differs from A368105 at n = 1, 36, 64, 72, 100, ... .
Except for a(1), all the terms are by definition squarefree numbers.

Crossrefs

Programs

  • Mathematica
    a[n_] := Module[{k = Max[FactorInteger[n][[;; , 2]]]}, While[! SquareFreeQ[k], k--]; k]; a[1] = 0; Array[a, 100]
  • PARI
    a(n) = if(n == 1, 0, my(k = vecmax(factor(n)[,2])); while(!issquarefree(k), k--); k);

Formula

a(n) = A051903(A365683(n)).
a(n) = A070321(A051903(n)) for n >= 2.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + Sum_{k>=2} A378085(k-1)*(1-1/zeta(k)) = 1.66055078443790141429... .
Showing 1-4 of 4 results.