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 1-3 of 3 results.

A096057 a(1) = 1, a(n) = least prime divisor of b(n), where b(1) = 1, b(n) = n*b(n-1) + 1 = A002627(n).

Original entry on oeis.org

1, 3, 2, 41, 2, 1237, 2, 29, 2, 43, 2, 5, 2, 5, 2, 35951249665217, 2, 28001, 2, 1409, 2, 5, 2, 5, 2, 47, 2, 661, 2, 13, 2, 5, 2, 5, 2, 13, 2, 13, 2, 73, 2, 5, 2, 5, 2, 71, 2, 2437159, 2, 31, 2, 5, 2, 5, 2, 13, 2, 3020497643, 2, 23, 2, 5, 2, 5, 2, 5672529813439, 2, 15336863
Offset: 1

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Author

Amarnath Murthy, Jun 17 2004

Keywords

Examples

			a(4) = 41 because b(3) = 3*b(2)+1 = 3*3+1 = 10 and 4*10+1 = 41, which is prime.
b(n) = 1, 3, 10, 41, ... with least prime divisors a(n) = 1, 3, 2, 41, ....
		

Crossrefs

Programs

  • Mathematica
    nxt[{n_,a_}]:={n+1,a(n+1)+1}; FactorInteger[#][[1,1]]&/@NestList[nxt,{1,1},40][[All,2]] (* The program generates the first 41 terms of the sequence. *) (* Harvey P. Dale, Jun 21 2022 *)

Formula

a(n) = A020639(A002627(n)).

Extensions

Corrected and extended by Ray G. Opao, Aug 02 2004
Edited by Jonathan Sondow, Jan 09 2005
More terms from Michel Marcus, Mar 28 2020
More terms from Giovanni Resta, Mar 28 2020

A102469 Largest prime factor of numerator of Sum_{k=0...n} 1/k!, with a(0) = 1.

Original entry on oeis.org

1, 2, 5, 2, 13, 163, 103, 137, 863, 98641, 10687, 31469, 1540901, 522787, 5441, 226871807, 13619, 1276861, 414026539, 2124467, 12670743557, 838025081381, 44659157, 323895443, 337310723185584470837549, 54352957, 11301647941785046703319941, 102505951982728548829
Offset: 0

Views

Author

Jonathan Sondow, Jan 09 2005

Keywords

Comments

It appears that a(n) = A102468(n) (Smarandache number of the same numerator) except when n = 3. The largest prime factor of the corresponding denominator is A007917(n) for n > 1. Omitting the 0th term in the sum, it appears that the largest prime factor and the Kempner number A002034, of the numerator of Sum_{k=1...n} 1/k! are both equal to A096058(n).

Examples

			Sum_{k=0...3} 1/k! = 8/3 and 2 is the largest prime factor 8, so a(3) = 2.
		

Crossrefs

Programs

  • Mathematica
    FactorInteger[#][[-1,1]]&/@Numerator[Accumulate[1/Range[0,30]!]] (* Harvey P. Dale, Nov 14 2012 *)
  • PARI
    a(n) = if(n==0, return(1)); vecmax(factor(numerator(sum(k=0, n, 1/k!)))[,1]); \\ Daniel Suteu, Jun 09 2022

Formula

a(n) = A006530(A061354(n)).

A102468 a(n)! is the smallest factorial divisible by the numerator of Sum_{k=0...n} 1/k!, with a(0) = 1.

Original entry on oeis.org

1, 2, 5, 4, 13, 163, 103, 137, 863, 98641, 10687, 31469, 1540901, 522787, 5441, 226871807, 13619, 1276861, 414026539, 2124467, 12670743557, 838025081381, 44659157, 323895443, 337310723185584470837549, 54352957
Offset: 0

Views

Author

Jonathan Sondow, Jan 09 2005

Keywords

Comments

It appears that a(n) = A102469(n) (largest prime factor of the same numerator) except when n = 3. The smallest factorial divisible by the corresponding denominator is n!. Omitting the 0th term in the sum, it appears that the Kempner number (A002034) and the largest prime factor, of the numerator of Sum_{k=1...n} 1/k! are both equal to A096058(n).
The Mathematica program given below was used to generate the sequence. If the numerator of Sum_{k=0...n}(1/k!) is squarefree, the program prints the value of the numerator's largest prime factor, which must equal a(n). Otherwise, the program prints the complete factorization of the numerator so a(n) can be determined by inspection. - Ryan Propper, Jul 31 2005

Examples

			Sum_{k=0...3} 1/k! = 8/3 and 4! is the smallest factorial divisible by 8, so a(3) = 4.
		

Crossrefs

Programs

  • Mathematica
    Do[l = FactorInteger[Numerator[Sum[1/k!, {k, 0, n}]]]; If[Length[l] == Plus @@ Last /@ l, Print[Max[First /@ l]], Print[l]], {n, 1, 30}] (* Ryan Propper, Jul 31 2005 *)
    nmax = 30; Clear[a]; Do[f = FactorInteger[ Numerator[ Sum[1/k!, {k, 0, n}] ] ]; a[n] = If[Length[f] == Total[f[[All, 2]] ], Max[f[[All, 1]] ], f[[-1, 1]] ], {n, 0, nmax}]; a[3] = 4; Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Sep 16 2015, adapted from Ryan Propper's script *)
  • PARI
    a(n) = {j = 1; s = numerator(sum(k=0, n, 1/k!)); while (j! % s, j++); j;} \\ Michel Marcus, Sep 16 2015

Formula

a(n) = A002034(A061354(n)).

Extensions

More terms from Ryan Propper, Jul 31 2005
Showing 1-3 of 3 results.