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.

A307746 Triangle read by rows, obtained by omitting all the 1's from the triangle in A307641 (except for the first row).

Original entry on oeis.org

1, 2, 3, 2, 2, 5, 2, 3, 7, 2, 2, 2, 3, 3, 2, 5, 11, 2, 3, 2, 13, 2, 7, 3, 5, 2, 2, 2, 2, 17, 2, 3, 3, 19, 2, 2, 5, 3, 7, 2, 11, 23, 2, 3, 2, 2, 5, 5, 2, 13, 3, 3, 3, 2, 2, 7, 29, 2, 3, 5, 31, 2, 2, 2, 2, 2, 3, 11, 2, 17, 5, 7, 2, 3, 2, 3, 37, 2, 19, 3, 13
Offset: 1

Views

Author

I. V. Serov, Apr 26 2019

Keywords

Comments

Has same shape as the triangle in A027746. The first difference occurs at row 12.

Examples

			Triangle begins:
1;
2;
3;
2, 2;
5;
2, 3;
7;
2, 2, 2;
3, 3;
2, 5;
11;
2, 3, 2;
...
		

Crossrefs

Programs

  • PARI
    f(n)=ispower(n, , &n); if(isprime(n), n, 1); \\ A014963
    row(n) = if (n==1, [1], my(d=divisors(n)); select(x->x!=1, vector(#d, k, f(d[k]))));
    tabl(nn) = for (n=1, nn, print(row(n))); \\ Michel Marcus, Apr 27 2019

Formula

Row(i) = {d|i, A014963(d) > 1} A014963(d).
For i > 1, T(i, A001222(i)) = A088387(i). This is the last term of the i-th row.

A088387 Prime corresponding to largest prime power factor of n, a(1)=1.

Original entry on oeis.org

1, 2, 3, 2, 5, 3, 7, 2, 3, 5, 11, 2, 13, 7, 5, 2, 17, 3, 19, 5, 7, 11, 23, 2, 5, 13, 3, 7, 29, 5, 31, 2, 11, 17, 7, 3, 37, 19, 13, 2, 41, 7, 43, 11, 3, 23, 47, 2, 7, 5, 17, 13, 53, 3, 11, 2, 19, 29, 59, 5, 61, 31, 3, 2, 13, 11, 67, 17, 23, 7, 71, 3, 73, 37, 5, 19, 11, 13, 79, 2, 3, 41, 83, 7, 17
Offset: 1

Views

Author

Reinhard Zumkeller, Sep 28 2003

Keywords

Comments

Most significant prime factor of n: If n = (p_1^e_1)(p_2^e_2)(p_3^e_3)... and max(p_1^e_1,p_2^e_2,...) = p_k^e_k then a(n) = p_k.

Examples

			a(6) = a(2*3) = 3 because 3^1 > 2^1;
a(36) = a((2^2)(3^2)) = 3 because 3^2 > 2^2;
a(12) = a((2^2)*3) = 2 because 2^2 > 3^1.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Sort[ {#[[1]]^#[[2]], #[[1]]} & /@ FactorInteger@ n][[ -1, 2]]; Array[f, 85] (* Robert G. Wilson v, Nov 05 2007 *)
    a[n_] := MaximalBy[FactorInteger[n], Power @@ # &][[1, 1]];
    Array[a, 85] (* Jean-François Alcover, Jun 27 2019 *)
  • PARI
    A088387(n) = if(1==n,1,my(f=factor(n),p=0); isprimepower(vecmax(vector(#f[, 1], i, f[i, 1]^f[i, 2])),&p); (p)); \\ Antti Karttunen, Jul 22 2018
    
  • Python
    from sympy import factorint
    def A088387(n): return max(((p**e,p) for p, e in factorint(n).items()), default=(0,1))[1] # Chai Wah Wu, Apr 17 2023

Formula

A034699(n) = a(n)^A088388(n).
a(n*a(n)) = a(n). - Sam Alexander, Dec 15 2003

Extensions

More terms from Ray Chandler, Dec 20 2003
Edited by N. J. A. Sloane at the suggestion of Stefan Steinerberger, Nov 04 2007

A307662 Triangle T(i,j=1..i) read by rows which contain the naturally ordered divisors-or-ones of the row number i.

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 2, 1, 4, 1, 1, 1, 1, 5, 1, 2, 3, 1, 1, 6, 1, 1, 1, 1, 1, 1, 7, 1, 2, 1, 4, 1, 1, 1, 8, 1, 1, 3, 1, 1, 1, 1, 1, 9, 1, 2, 1, 1, 5, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 2, 3, 4, 1, 6, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13
Offset: 1

Views

Author

I. V. Serov, Apr 20 2019

Keywords

Comments

Replace 0 with 1 in A127093. - Omar E. Pol, Apr 21 2019

Examples

			Triangle begins:
  1,
  1, 2,
  1, 1, 3,
  1, 2, 1, 4,
  1, 1, 1, 1, 5,
  1, 2, 3, 1, 1, 6,
  1, 1, 1, 1, 1, 1, 7,
  1, 2, 1, 4, 1, 1, 1, 8,
  1, 1, 3, 1, 1, 1, 1, 1, 9,
  1, 2, 1, 1, 5, 1, 1, 1, 1,10,
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1,11,
  1, 2, 3, 4, 1, 6, 1, 1, 1, 1, 1,12,
  ...
		

Crossrefs

Programs

  • Mathematica
    Table[Map[If[Mod[n, #] == 0, #, 1] &, Range@ n], {n, 13}] // Flatten (* Michael De Vlieger, Apr 23 2019 *)
  • PARI
    row(n) = vector(n, k, if ((n % k) == 0, k, 1)); \\ Michel Marcus, Apr 21 2019

Formula

For j=1..floor(i/2), T(i,j)=T(i-j,j).
For j=floor(i/2)+1..i-1, T(i,j)=1.
T(i,i) = i.
Let i = A002024(n) and j = A002260(n):
a(n) = a((i-j-1)*(i-j)/2+j) if j=1..floor(i/2).
a(n) = 1 if j=floor(i/2)+1..i-1.
a(n) = i if j=i.
T(i,j) = a(n).

A307723 Naturally ordered prime factorization of n as a quasi-logarithmic word over the binary alphabet {1,0}.

Original entry on oeis.org

10, 1100, 1010, 110100, 101100, 11011000, 101010, 11001100, 10110100, 1101101000, 10110010, 1101100100, 1011011000, 1100110100, 10101010, 1101010100, 1011001100, 110110011000, 1010110100, 110011011000
Offset: 2

Views

Author

I. V. Serov, Apr 24 2019

Keywords

Comments

Let m(n) be the number of digits (letters) in a(n).
m(n) = 2*A064097(n) = 2*(A073933(n)-1).
Split the word a(n) into two parts of equal length. The number of 1's in the left part equals the number of 0's in the right part and vice versa.

Examples

			The sequence begins:
   n a(n)
  -- -----------
   1
   2 10
   3 1100
   4 1010
   5 110100
   6 101100
   7 11011000
   8 101010
   9 11001100
  10 10110100
  11 1101101000
  12 10110010
  ...
		

Crossrefs

Formula

a(1) is empty.
a(n) = concatenation(1, a(n-1), 0) if n is prime.
a(n) = concatenation_{k=1..A001222(n)} a(A307746(n,k)) if n is composite.
a(n) = concatenation(a(n/A088387(n)), a(A088387(n))) if n is composite.
Showing 1-4 of 4 results.