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.

A135212 a(n) = A078456(n)/A120271(n).

Original entry on oeis.org

1, 1, 2, 4, 8, 576, 1152, 2304, 4608, 18432, 552960, 59719680, 2388787200, 100329062400, 200658124800, 802632499200, 1605264998400, 288947699712000, 6356849393664000, 444979457556480000, 10679506981355520000
Offset: 1

Views

Author

Alexander Adamchuk, Nov 23 2007

Keywords

Crossrefs

Programs

  • Mathematica
    Table[ Det[ DiagonalMatrix[ Table[ Prime[i+1]-1, {i, 1, n-1} ] ] + 1 ], {n, 1, 50} ] / Numerator[ Table[ Sum[ 1/(Prime[i]-1), {i, 1, n} ], {n, 1, 50}] ]
  • PARI
    a(n) = matdet(matrix(n-1, n-1, j, k, if (j==k, prime(j+1), 1)))/numerator(sum(k=1, n, 1/(prime(k)-1))); \\ Michel Marcus, Oct 02 2016

Formula

a(n) = A078456(n)/A120271(n).

A120271 a(n) = numerator(Sum_{k=1..n} 1/(prime(k)-1)).

Original entry on oeis.org

1, 3, 7, 23, 121, 21, 173, 1597, 17927, 127469, 129317, 43619, 44081, 44521, 1033223, 13538159, 395369371, 132680013, 400467919, 402757063, 1214947859, 1221110939, 50305908619, 50529880549, 101470376303, 509322834499, 8691337402883
Offset: 1

Views

Author

Alexander Adamchuk, Jul 01 2006

Keywords

Comments

a(n) is squarefree except for n = 5, 14, 49, ... where squared prime factors are 11, 211, 479, ...
a(n)/A128646(n) is the asymptotic mean over the positive integers of the number of prime divisors that are not greater than prime(n), counted with multiplicity (cf. A007814, A169611, A356006). - Amiram Eldar, Jul 23 2022

Crossrefs

Cf. A128646 (denominators), A119686, A006093, A000040.

Programs

  • Maple
    R:= [seq(1/(ithprime(k)-1),k=1..40)]:
    S:= ListTools:-PartialSums(R):
    A:= map(numer,S); # Robert Israel, Jan 12 2025
  • Mathematica
    Numerator[Table[Sum[1/(Prime[i]-1),{i,1,n}],{n,1,50}]]
    Accumulate[1/(Prime[Range[30]]-1)]//Numerator (* Harvey P. Dale, May 03 2025 *)
  • PARI
    a(n) = numerator(sum(k=1, n, 1/(prime(k)-1))); \\ Michel Marcus, Oct 02 2016

Formula

a(n) = numerator(Sum_{k=1..n} 1/(prime(k)-1)).
a(n) = A078456(n) * A135212(n). - Alexander Adamchuk, Nov 23 2007

A068499 Numbers m such that m! reduced modulo (m+1) is not zero.

Original entry on oeis.org

1, 2, 3, 4, 6, 10, 12, 16, 18, 22, 28, 30, 36, 40, 42, 46, 52, 58, 60, 66, 70, 72, 78, 82, 88, 96, 100, 102, 106, 108, 112, 126, 130, 136, 138, 148, 150, 156, 162, 166, 172, 178, 180, 190, 192, 196, 198, 210, 222, 226, 228, 232, 238, 240, 250, 256, 262, 268, 270
Offset: 1

Views

Author

Benoit Cloitre, Mar 11 2002

Keywords

Comments

Also n such that tau((n+1)!) = 2* tau(n!)
For n > 2, a(n) is the smallest number such that a(n) !== -1 (mod a(k)+1) for any 1 < k < n. [Franklin T. Adams-Watters, Aug 07 2009]
Also n such that sigma((n+1)!) = (n+2)* sigma(n!), which is the same as A062569(n+1) = (n+2)*A062569(n). - Zak Seidov, Aug 22 2012
This sequence is obtained by the following sieve: keep 1 in the sequence and then, at the k-th step, keep the smallest number, x say, that has not been crossed off before and cross off all the numbers of the form k*(x + 1) - 1 with k > 1. The numbers that are left form the sequence. - Jean-Christophe Hervé, Dec 12 2015
a(n) = A039915(n-1) for 3 < n <= 1000. - Georg Fischer, Oct 19 2018

Examples

			Illustration of the sieve: keep 1 = a(1) and then
1st step: take 2 = a(2) and cross off 5, 8, 11, 14, 17, 20, 23, 26, etc.
2nd step: take 3 = a(3) and cross off 7, 11, 15, 19, 23, 27, etc.
3rd step: take 4 = a(4) and cross off 9, 14, 19, 24, etc.
4th step: take 6 = a(5) and cross off 13, 19, 25 etc.
10 is obtained at next step and the smallest crossed off numbers are then 21 and 28. This gives the beginning of the sequence up to 22 = a(10): 1, 2, 3, 4, 6, 10, 12, 16, 18, 22. - _Jean-Christophe Hervé_, Dec 12 2015
		

Crossrefs

Cf. A000040, A039915, A062569, A166460 (almost complement).

Programs

  • Mathematica
    Select[Range[300],Mod[#!,#+1]!=0&] (* Harvey P. Dale, Apr 11 2012 *)
  • PARI
    {plnt=1 ; nfa=1; mxind=60 ;  for(k=1, 10^7, nfa=nfa*k;
    if(nfa % (k+1) != 0 , print1(k, ", "); plnt++ ;
    if(mxind <  plnt, break() )))} \\ Douglas Latimer, Apr 25 2012
    
  • PARI
    a(n)=if(n<5,n,prime(n-1)-1) \\ Charles R Greathouse IV, Apr 25 2012
    
  • Python
    from sympy import prime
    def A068499(n): return prime(n-1)-1 if n>3 else n # Chai Wah Wu, Aug 27 2024

Formula

For n >= 4, a(n) = prime(n-1) - 1 = A006093(n-1).
For n <> 3, all terms are one less prime. - Zak Seidov, Aug 22 2012
a(n) = Integer part of A078456(n+1)/A078456(n). - Eric Desbiaux, May 07 2013

A070918 Triangle of T(n,k) coefficients of polynomials with first n prime numbers as roots.

Original entry on oeis.org

1, -2, 1, 6, -5, 1, -30, 31, -10, 1, 210, -247, 101, -17, 1, -2310, 2927, -1358, 288, -28, 1, 30030, -40361, 20581, -5102, 652, -41, 1, -510510, 716167, -390238, 107315, -16186, 1349, -58, 1, 9699690, -14117683, 8130689, -2429223, 414849, -41817, 2451, -77, 1
Offset: 0

Views

Author

Rick L. Shepherd, May 20 2002

Keywords

Comments

Analog of the Stirling numbers of the first kind (A008275): The Stirling numbers (beginning with the 2nd row) are the coefficients of the polynomials having exactly the first n natural numbers as roots. This sequence (beginning with first row) consists of the coefficients of the polynomials having exactly the first n prime numbers as roots.

Examples

			Row 4 of this sequence is 210, -247, 101, -17, 1 because (x-prime(1))(x-prime(2))(x-prime(3))(x-prime(4)) = (x-2)(x-3)(x-5)(x-7) = x^4 - 17*x^3 + 101*x^2 - 247*x + 210.
Triangle begins:
        1;
       -2,      1;
        6,     -5,       1;
      -30,     31,     -10,      1;
      210,   -247,     101,    -17,      1;
    -2310,   2927,   -1358,    288,    -28,    1;
    30030, -40361,   20581,  -5102,    652,  -41,   1;
  -510510, 716167, -390238, 107315, -16186, 1349, -58, 1;
  ...
		

Crossrefs

Cf. A008275 (Stirling numbers of first kind).
Cf. A005867 (absolute values of row sums).
Cf. A054640 (sum of absolute values of terms in rows).

Programs

  • Maple
    T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(mul(x-ithprime(i), i=1..n)):
    seq(T(n), n=0..10);  # Alois P. Heinz, Aug 18 2019
  • Mathematica
    Table[CoefficientList[Expand[Times@@(x-Prime[Range[n]])],x],{n,0,10}]// Flatten (* Harvey P. Dale, Feb 12 2020 *)
  • PARI
    p=1; for(k=1,10,p=p*(x-prime(k)); for(n=0,k,print1(polcoeff(p,n),",")))

Formula

From Alois P. Heinz, Aug 18 2019: (Start)
T(n,k) = [x^k] Product_{i=1..n} (x-prime(i)).
Sum_{k=0..n} |T(n,k)| = A054640(n).
|Sum_{k=0..n} T(n,k)| = A005867(n).
|Sum_{k=0..n} k * T(n,k)| = A078456(n). (End)

Extensions

First term T(0,0)=1 prepended by Alois P. Heinz, Aug 18 2019

A096294 Triangle T(n,k) read by rows: for n >=0 and n >= k >=0, the fraction of positive integers with exactly k of the first n primes as divisors is T(n,k)/A002110(n).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 8, 14, 7, 1, 48, 92, 56, 13, 1, 480, 968, 652, 186, 23, 1, 5760, 12096, 8792, 2884, 462, 35, 1, 92160, 199296, 152768, 54936, 10276, 1022, 51, 1, 1658880, 3679488, 2949120, 1141616, 239904, 28672, 1940, 69, 1, 36495360, 82607616
Offset: 0

Views

Author

Matthew Vandermast, Jun 24 2004

Keywords

Comments

Sum of entries in n-th row is A002110(n), the product of the first n primes (primorial numbers, first definition).
From Peter Munn, Apr 10 2017: (Start)
T(n,k) is a count of those integers in any interval of A002110(n) integers that have exactly k of the first n primes as divisors. The count is the same for each such interval because each of the first n primes is a factor of an integer m if and only if it is a factor of m + A002110(n).
A284411(m) is least p=prime(n) such that 2*Sum_{k=0..m-1} T(n,k) < A002110(n).
(End)

Examples

			Triangle begins:
1
1 1
2 3 1
8 14 7 1
48 92 56 13 1
480 968 652 186 23 1
		

Crossrefs

First column is A005867; second column is A078456. See also A096180.

Programs

  • PARI
    primo(n) = prod(k=1, n, prime(k));
    row(n) = {v = vector(n+1); for (k=1, primo(n), f = factor(k)[,1]; v[1+sum(j=1, #f, primepi(f[j])<=n)]++;); v;} \\ Michel Marcus, Apr 29 2017
Showing 1-5 of 5 results.