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 1-7 of 7 results.

A093101 Cancellation factor in reducing Sum_{k=0...n} 1/k! to lowest terms.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 20, 1, 10, 1, 8, 5, 2, 5, 4, 1, 130, 1, 4000, 1, 2, 5, 52, 5, 494, 1, 40, 1, 10, 13, 4, 25, 38, 5, 16, 13, 230, 13, 20, 1, 46, 5, 104, 475, 62, 1, 20, 1, 130, 31, 832, 2755, 74, 5, 4, 13, 50, 1, 40, 23, 2, 2795, 76, 34385, 2, 1, 80, 1, 650, 1, 2812, 5, 74, 5
Offset: 0

Views

Author

Jonathan Sondow, May 10 2004, Oct 18 2006

Keywords

Comments

Same as n!/A061355(n) and (1+n+n(n-1)+n(n-1)(n-2)+...+n!)/A061354(n).
a(n) is relatively prime to n.
gcd(a(n),a(n+1)) = 1.

Examples

			E.g. 1/0!+1/1!+1/2!+1/3!=16/6=(2*8)/(2*3) so a(3)=2.
		

Crossrefs

(n+1)!/(a(n)*a(n+1)) = A123899(n).
(n+3)!/(a(n)*a(n+1)*a(n+2)) = A123900(n).
(n+3)/GCD(a(n), a(n+2)) = A123901(n).
Cf. also A000522, A061354, A061355.

Programs

  • Mathematica
    f[n_] := n! / Denominator[ Sum[1/k!, {k, 0, n}]]; Table[ f[n], {n, 0, 74}] (* Robert G. Wilson v *)
    (* Second program: *)
    A[n_] := If[n==0,1,n*A[n-1]+1]; Table[GCD[A[n],n! ], {n, 0, 74}]
  • PARI
    A000522(n) = sum(k=0, n, binomial(n, k)*k!); \\ This function from Joerg Arndt, Dec 14 2014
    A093101(n) = gcd(n!,A000522(n)); \\ Antti Karttunen, Jul 12 2017

Formula

a(n) = gcd(n!, 1+n+n(n-1)+n(n-1)(n-2)+...+n!).
a(n) = gcd(n!, A(n)) where A(0) = 1, A(n) = n*A(n-1)+1.

Extensions

More terms from Robert G. Wilson v, May 14 2004

A124779 a(n) = gcd(A(n), A(n+2))/gcd(d(n), d(n+2)) where A(n) = Sum_{k=0..n} n!/k! and d(n) = gcd(A(n), n!).

Original entry on oeis.org

1, 2, 5, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 37, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Jonathan Sondow, Nov 07 2006

Keywords

Comments

The next term > 1 is a(460) = 463. The primes 2, 5, 13, 37, 463 are the only terms > 1 up to n = 600000. If a(n) > 1 with n > 1, then a(n) = n+3 is prime. This uses A(n+2) = (n+2)(n+1)*A(n) + n+3. The terms > 1 are A064384 = primes p such that p divides 0!-1!+2!-3!+...+(-1)^{p-1}(p-1)!. The proof uses (n-1)!/(n-k-1)! = (n-1)(n-2)...(n-k) == (-1)^k k! (mod n). Cf. Cloitre's comment in A064383.
An integer p > 1 is in the sequence if and only if p is prime and p|A(p-1), where A(0) = 1 and A(n) = n*A(n-1)+1 for n > 0. - Jonathan Sondow, Dec 22 2006
Michael Mossinghoff has calculated that there are only five primes in the sequence up to 150 million. Heuristics suggest it contains infinitely many. - Jonathan Sondow, Jun 12 2007

Examples

			a(2) = gcd(A(2), A(4))/gcd(d(2), d(4)) = gcd(5, 65)/gcd(1, 1) = 5/1 = 5.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, Springer-Verlag, 3rd edition, 2004, B43.

Crossrefs

A(n) = A000522, d(n) = A093101, gcd(A(n), A(n+2)) = A124780, gcd(d(n), d(n+2)) = A124781, (n+3)/gcd(A(n), A(n+2)) = A124782, (n+3)/gcd(d(n), d(n+2)) = A123901. Cf. A061354, A061355, A123899, A123900.
Cf. A129924.

Programs

  • Mathematica
    (A[n_] := Sum[n!/k!, {k, 0, n}]; d[n_] := GCD[A[n],n! ]; Table[GCD[A[n],A[n+2]]/GCD[d[n],d[n+2]], {n,0,100}])
  • PARI
    A124779(n)={my(An=A000522(n),A2=A000522(n+2));gcd(An, A2)/gcd([An,n!,A2,(n+2)!])} \\ M. F. Hasler, Jun 04 2019

Formula

a(n) = A124780(n)/A124781(n) = A124782(n)/A123901(n).
a(n) = gcd(A(n), A(n+2))/gcd(A(n), A(n+2), n!) where A(n)=1+n+n(n-1)+...+n!. - Jonathan Sondow, Nov 10 2006
a(n) = gcd(N(n), N(n+2)), where N(n) = A061354(n) = numerator of Sum[1/k!,{k,0,n}]. - Jonathan Sondow, Jun 12 2007

A123901 a(n) = (n+3)/gcd(d(n), d(n+2)) where d(n) = cancellation factor in reducing Sum_{k=0..n} 1/k! to lowest terms.

Original entry on oeis.org

3, 4, 5, 3, 7, 4, 9, 1, 11, 6, 13, 7, 3, 8, 17, 9, 19, 2, 21, 11, 23, 12, 5, 1, 27, 14, 29, 3, 31, 16, 33, 17, 7, 18, 37, 19, 3, 4, 41, 21, 43, 22, 9, 23, 47, 24, 49, 5, 51, 2, 53, 27, 11, 28, 57, 29, 59, 6, 61, 31, 63, 32, 1, 33, 67, 34, 69, 7, 71, 36, 73, 1, 15, 38, 77, 3, 79, 8, 81
Offset: 0

Views

Author

Jonathan Sondow, Oct 18 2006

Keywords

Examples

			a(5) = 4 because (5+3)/gcd(d(5),d(7)) = 8/gcd(2,20) = 8/2 = 4.
		

Crossrefs

Programs

  • Mathematica
    (A[n_] := If[n==0,1,n*A[n-1]+1]; d[n_] := GCD[A[n],n! ]; Table[(n+3)/GCD[d[n],d[n+2]], {n,0,79}])
    (* Second program, faster: *)
    Table[(n + 3)/Apply[GCD, Map[GCD[#!, Floor[E*#!] - Boole[# == 0]] &, n + {0, 2}]], {n, 0, 78}] (* Michael De Vlieger, Jul 12 2017 *)
  • PARI
    A000522(n) = sum(k=0, n, binomial(n, k)*k!); \\ This function from Joerg Arndt, Dec 14 2014
    A093101(n) = gcd(n!,A000522(n));
    m1=m2=1; for(n=0,4096,m=m1; m1=m2; m2 = A093101(n+2); m124781 = gcd(m,m2); write("b093101.txt", n, " ", m); write("b124781.txt", n, " ", m124781); write("b123901.txt", n, " ", (n+3)/m124781)); \\ Antti Karttunen, Jul 12 2017

Formula

a(n) = (n+3)/A124781(n) = (n+3)/gcd(A093101(n),A093101(n+2)) where A093101(n) = gcd(n!,1+n+n(n-1)+...+n!).

A123900 a(n) = (n+3)!/(d(n)*d(n+1)*d(n+2)) where d(n) = cancellation factor in reducing Sum_{k=0...n} 1/k! to lowest terms.

Original entry on oeis.org

6, 12, 60, 180, 2520, 1008, 18144, 18144, 3991680, 5987520, 155675520, 1089728640, 26153487360, 523069747200, 17784371404800, 12312257126400, 935731541606400, 4678657708032, 12772735542927360, 140500090972200960
Offset: 0

Views

Author

Jonathan Sondow, Oct 18 2006

Keywords

Examples

			a(2) = 60 because (2+3)!/(d(2)*d(3)*d(4)) = 5!/(GCD(2,5)*GCD(6,16)*GCD(24,65)) = 120/2 = 60.
		

Crossrefs

Programs

  • Mathematica
    (A[n_] := If[n==0,1,n*A[n-1]+1]; d[n_] := GCD[A[n],n! ]; Table[(n+3)!/(d[n]*d[n+1]*d[n+2]), {n,0,21}])

Formula

a(n) = (n+3)!/(A093101(n)*A093101(n+1)*A093101(n+2)) where A093101(n) = gcd(n!,1+n+n(n-1)+...+n!).

A124781 a(n) = gcd(A093101(n), A093101(n+2)) where A093101(n) = gcd(n!, A(n)) and A(n) = A000522(n) = Sum_{k=0..n} n!/k!.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 5, 26, 1, 2, 1, 10, 1, 2, 1, 2, 5, 2, 1, 2, 13, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 26, 1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 65, 2, 1, 2, 1, 10, 1, 2, 1, 74, 5, 2, 1, 26, 1, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 13, 2, 1, 2, 5, 2, 1, 2, 1
Offset: 0

Views

Author

Jonathan Sondow, Nov 07 2006

Keywords

Comments

a(n) divides n+3 because A(n+2) = (n+2)(n+1)*A(n) + n+3.

Examples

			a(3) = gcd(d(3),d(5)) = gcd(gcd(3!,16), gcd(5!,326)) = gcd(2,2) = 2.
		

Crossrefs

Programs

  • Mathematica
    (A[n_] := Sum[n!/k!, {k,0,n}]; d[n_] := GCD[n!,A[n]]; Table[GCD[d[n],d[n+2]], {n,0,100}])
    (* Second program, faster: *)
    Table[GCD @@ Map[GCD[#!, Floor[E*#!] - Boole[# == 0]] &, n + {0, 2}], {n, 0, 96}] (* Michael De Vlieger, Jul 12 2017 *)
  • PARI
    A000522(n) = sum(k=0, n, binomial(n, k)*k!); \\ This function from Joerg Arndt, Dec 14 2014
    A093101(n) = gcd(n!,A000522(n));
    m1=m2=1; for(n=0,4096,m=m1; m1=m2; m2 = A093101(n+2); m124781 = gcd(m,m2); write("b093101.txt", n, " ", m); write("b124781.txt", n, " ", m124781); write("b123901.txt", n, " ", (n+3)/m124781)); \\ Antti Karttunen, Jul 12 2017

Formula

a(n) = gcd(A093101(n), A093101(n+2)) = (n+3)/A123901(n).
a(n) = gcd(A(n), A(n+2), n!) where A(n)=1+n+n(n-1)+...+n!. - Jonathan Sondow, Nov 13 2006

Extensions

Replaced d(n) in the name with A093101(n). - Antti Karttunen, Jul 12 2017

A124780 a(n) = gcd(A(n), A(n+2)) where A(n) = A000522(n) = Sum_{k=0..n} n!/k!.

Original entry on oeis.org

1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 13, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 5, 26, 1, 2, 1, 10, 1, 2, 1, 2, 5, 2, 37, 2, 13, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 26, 1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 65, 2, 1, 2, 1, 10, 1, 2, 1, 74, 5, 2, 1, 26, 1, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 13, 2, 1, 2, 5, 2, 1, 2, 1
Offset: 0

Views

Author

Jonathan Sondow, Nov 07 2006

Keywords

Comments

a(n) divides n+3 because A(n+2) = (n+2)(n+1)*A(n) + n+3.

Examples

			a(2) = gcd(A(2), A(4)) = gcd(5, 65) = 5.
		

Crossrefs

Programs

  • Mathematica
    (A[n_] := Sum[n!/k!, {k, 0, n}]; Table[GCD[A[n],A[n+2]], {n,0,100}])
    GCD[#[[1]],#[[3]]]&/@Partition[Table[Sum[n!/k!,{k,0,n}],{n,0,100}],3,1] (* Harvey P. Dale, Jun 14 2022 *)
  • PARI
    A000522(n) = sum(k=0, n, binomial(n, k)*k!); \\ This function from Joerg Arndt, Dec 14 2014
    A124780(n) = gcd(A000522(n),A000522(n+2)); \\ Antti Karttunen, Jul 07 2017

Formula

a(n) = gcd(A000522(n), A000522(n+2)) = (n+3)/A124782(n)

A124782 a(n) = (n+3)/gcd(A(n), A(n+2)) where A(n) = A000522(n) = Sum_{k=0..n} n!/k!.

Original entry on oeis.org

3, 2, 1, 3, 7, 4, 9, 1, 11, 6, 1, 7, 3, 8, 17, 9, 19, 2, 21, 11, 23, 12, 5, 1, 27, 14, 29, 3, 31, 16, 33, 17, 7, 18, 1, 19, 3, 4, 41, 21, 43, 22, 9, 23, 47, 24, 49, 5, 51, 2, 53, 27, 11, 28, 57, 29, 59, 6, 61, 31, 63, 32, 1, 33, 67, 34, 69, 7, 71, 36, 73, 1, 15, 38, 77, 3, 79, 8, 81, 41
Offset: 0

Views

Author

Jonathan Sondow, Nov 07 2006

Keywords

Comments

a(n) is an integer since A(n+2) = (n+2)(n+1)*A(n) + n+3.

Examples

			a(3) = (3+3)/gcd(A(3), A(5)) = 6/gcd(16, 326) = 6/2 = 3.
		

Crossrefs

Programs

Formula

a(n) = (n+3)/A124780(n) = (n+3)/gcd(A000522(n), A000522(n+2)).
Showing 1-7 of 7 results.