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 1-5 of 5 results.

A228562 Composite numbers k that are not prime powers such that binomial(2k-1, k-1) is congruent to 1 (mod k).

Original entry on oeis.org

27173, 2001341, 16024189487
Offset: 1

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Author

Felix Fröhlich, Aug 25 2013

Keywords

Crossrefs

The odd terms of A328497.

Programs

  • Mathematica
    Select[Range[30000], PrimeNu[#] > 1 && Mod[Binomial[2# - 1, # - 1], #] == 1 &] (* Alonso del Arte, May 11 2014 *)
  • PARI
    N=10^9; for(n=2, N, if(Mod(binomial(2*n-1, n-1), n)==1 && !ispower(n) && !isprime(n), print1(n, ", "))); \\ Felix Fröhlich, May 11 2014
    
  • PARI
    vp(n,p)=my(s); while(n\=p, s+=n); s
    is(n)=my(f=factor(n)[,1],G); if(#f==1, return(0)); for(i=1,#f, if(vp(2*n-1,f[i]) > vp(n,f[i])+vp(n-1,f[i]), return(0))); G=prod(i=1,#f,f[i]^(log(n)\log(f[i]))); prod(i=n+1,2*n-1, i/gcd(i,G), Mod(1,n))/prod(i=2,n-1, i/gcd(i,G), Mod(1,n))==1
    forcomposite(n=4,1e9, if(is(n), print1(n", "))) \\ Charles R Greathouse IV, May 12 2014

Formula

A099905(a(n)) = 1. - Jonathan Sondow, Jan 24 2016

A267824 Composite numbers n such that binomial(2n-1, n-1) == 1 (mod n^2).

Original entry on oeis.org

283686649, 4514260853041
Offset: 1

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Author

Jonathan Sondow, Jan 25 2016

Keywords

Comments

Babbage proved the congruence holds if n > 2 is prime.
See A088164 and A263882 for references, links, and additional comments.
Conjecture: n is a term if and only if n = A088164(i)^2 for some i >= 1 (cf. McIntosh, 1995, p. 385). - Felix Fröhlich, Jan 27 2016
The "if" part of the conjecture is true: see the McIntosh reference. - Jonathan Sondow, Jan 28 2016
The above conjecture implies that this sequence and A228562 are disjoint. - Felix Fröhlich, Jan 27 2016
Composites c such that A281302(c) > 1. - Felix Fröhlich, Feb 21 2018

Examples

			a(1) = 16843^2 and a(2) = 2124679^2 are squares of Wolstenholme primes A088164.
		

Crossrefs

A380692 Numbers k such that the least prime dividing k is larger than the maximum exponent in the prime factorization of k; a(1) = 1 by convention.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 25, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 45, 46, 47, 49, 51, 53, 55, 57, 58, 59, 61, 62, 63, 65, 66, 67, 69, 70, 71, 73, 74, 75, 77, 78, 79, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 99
Offset: 1

Views

Author

Amiram Eldar, Jan 30 2025

Keywords

Comments

First differs from A368110 at n = 68: A368110(68) = 98 is not a term of this sequence.
First differs from its subsequence A380695 at n = 428: a(428) = 625 is not a term of A380695.
Numbers k such that A020639(k) > A051903(k).
All the squarefree numbers (A005117) are terms, and all the odd terms of A067259 are terms of this sequence.
Disjoint union of the sequences S_k, k >= 1, where S_k is the sequence of p-rough numbers (numbers whose prime factors are all greater than or equal to p), with p = nextprime(k) = A151800(k), whose maximum exponent in their prime factorization is k (i.e., numbers that are (k+1)-free but not k-free, where k-free numbers are numbers whose prime factorization exponents do not exceed k).
The asymptotic density of this sequence is Sum_{i>=1} d(i) = 0.68213349032332767778..., where d(i), the density of S_i, equals f(i+1) * Product_{primes p <= i} ((1-1/p)/(1-1/p^(i+1))) - f(i) * Product_{primes p <= i} ((1-1/p)/(1-1/p^i)), f(i) = 1/zeta(i) if i >= 2, and f(1) = 0.

Examples

			6 = 2^1 * 3^1 is a term since 2 > 1.
8 = 2^3 is not a term since 2 < 3.
		

Crossrefs

Subsequence of A380693.
Subsequences: A005117, A136327, the intersection of A005408 and A067259, A380694, A380695.

Programs

  • Mathematica
    q[k_] := k == 1 || Module[{f = FactorInteger[k]}, f[[1, 1]] > Max[f[[;; , 2]]]]; Select[Range[100], q]
  • PARI
    isok(k) = if(k == 1, 0, my(f = factor(k), e = f[, 2]); f[1, 1] > vecmax(e));

A335024 Ratios of consecutive terms of A056612.

Original entry on oeis.org

1, 1, 2, 1, 18, 1, 4, 1, 10, 1, 12, 1, 14, 15, 8, 1, 54, 1, 100, 63, 22, 1, 8, 1, 26, 3, 28, 1, 30, 1, 16, 363, 34, 35, 36, 1, 38, 39, 40, 1, 294, 1, 4, 45, 46, 1, 48, 1, 50, 51, 52, 1, 162, 55, 56, 57, 58, 1, 60, 1, 62, 189, 32, 65, 198, 1, 68, 23, 70, 1, 24, 1, 74, 75, 76, 847, 78, 1, 80
Offset: 1

Views

Author

Petros Hadjicostas, May 19 2020

Keywords

Comments

Conjecture 1: a(n) = 1 if and only if n + 1 = p^k for some prime p and some positive integer k < p.
Conjecture 2 (due to Alois P. Heinz): a(n) = 1 <=> n+1 in A136327.

Crossrefs

Programs

  • Maple
    b:= proc(n) b(n):= (1/n +`if`(n=1, 0, b(n-1))) end:
    g:= proc(n) g(n):= (f-> igcd(b(n)*f, f))(n!) end:
    a:= n-> g(n+1)/g(n):
    seq(a(n), n=1..80);  # Alois P. Heinz, May 20 2020
  • Mathematica
    g[n_] := GCD[n!, n! Sum[1/k, {k, 1, n}]];
    a[n_] := g[n + 1]/g[n];
    Array[a, 80] (* Jean-François Alcover, Dec 01 2020, after PARI *)
  • PARI
    g(n) = gcd(n!, n!*sum(k=1, n, 1/k)); \\ A056612
    a(n) = g(n+1)/g(n); \\ Michel Marcus, May 20 2020

Formula

a(n) = A056612(n+1)/A056612(n).

A343533 a(n) is the largest value of k such that binomial(2*m-1, m-1) == 1 (mod m^k) for m = 2*n + 1.

Original entry on oeis.org

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

Views

Author

Felix Fröhlich, Apr 18 2021

Keywords

Comments

If 2*n + 1 is a prime >= 5, then a(n) >= 3 by Wolstenholme's theorem.
If 2*n + 1 is a Wolstenholme prime (A088164), then a(n) >= 4.
If 2*n + 1 is a term of A267824, then a(n) >= 2.
If 2*n + 1 is the square of an odd prime, the cube of a prime >= 5 or a term of A228562, then a(n) >= 1.

Crossrefs

Programs

  • Maple
    a := proc(n) local x, x0, y, k, bound; bound := 1000;
    x := 2*n + 1; x0 := x;
    y := binomial(4*n + 1, 2*n);
    for k from 0 to bound while y mod x = 1 do
        x := x * x0 od;
    if k < bound then k else print("No k below ", bound) fi end:
    seq(a(n), n = 1..100); # Peter Luschny, Apr 22 2021
  • PARI
    a(n) = my(x=2*n+1, b=binomial(2*x-1, x-1)); for(k=1, oo, if(Mod(b, x^k)!=1, return(k-1)))
Showing 1-5 of 5 results.