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.

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A119029 Numerator of Sum_{k=1..n} n^(k-1)/k!.

Original entry on oeis.org

1, 2, 4, 25, 217, 203, 6743, 69511, 1184417, 728102, 5720654791, 601499, 4670663321629, 42568060798, 888330615353, 15438515749903, 1676770323947695709, 30538296012677, 16858207434636875406943
Offset: 1

Views

Author

Alexander Adamchuk, Jul 22 2006

Keywords

Comments

Apparently, the three sequences T_1(n) = Sum_{k=1..n} n^(k-1)/k!, T_2(n) = Sum_{k=0..n} n^k/k!, and T_3(n) = Sum_{k=1..n} n^k/k!, with numerators in A119029, A120266, and A120267, respectively, have the same denominators, listed in A214401. This, however, is not immediately obvious. - Petros Hadjicostas, May 12 2020

Examples

			The first few fractions are 1, 2, 4, 25/3, 217/12, 203/5, 6743/72, 69511/315, 1184417/2240, 728102/567, ... = A119029/A214401. - _Petros Hadjicostas_, May 12 2020
		

Crossrefs

Programs

  • Mathematica
    Numerator[Table[Sum[n^(k-1)/k!,{k,1,n}],{n,1,30}]]

Formula

a(n) = numerator(Sum_{k=1..n} n^(k-1)/k!).
a(n) = A120267(n)/n.

A120267 Numerator of Sum_{k=1..n} n^k/k!.

Original entry on oeis.org

1, 4, 12, 100, 1085, 1218, 47201, 556088, 10659753, 7281020, 62927202701, 7217988, 60718623181177, 595952851172, 13324959230295, 247016251998448, 28505095507110827053, 549689328228186, 320305941258100632731917
Offset: 1

Views

Author

Alexander Adamchuk, Jun 30 2006

Keywords

Comments

n divides a(n) and a(n)/n = A119029(n). - Alexander Adamchuk, Oct 08 2006
Apparently, the three sequences T_1(n) = Sum_{k=1..n} n^(k-1)/k!, T_2(n) = Sum_{k=0..n} n^k/k!, and T_3(n) = Sum_{k=1..n} n^k/k!, with numerators in A119029, A120266, and A120267, respectively, have the same denominators, listed in A214401. This, however, is not immediately obvious. - Petros Hadjicostas, May 12 2020

Examples

			The first few fractions are 1, 4, 12, 100/3, 1085/12, 1218/5, 47201/72, 556088/315, 10659753/2240, 7281020/567, ... = A120267/A214401. - _Petros Hadjicostas_, May 12 2020
		

Crossrefs

Programs

  • Mathematica
    Numerator[Table[Sum[n^k/k!, {k,1,n}], {n,1,30}]]

Formula

a(n) = numerator(Sum_{k=1..n} n^k/k!).
a(n) = n*A119029(n). - Alexander Adamchuk, Oct 08 2006

Extensions

Various sections edited by Petros Hadjicostas, May 12 2020~

A214402 Cancellation factor in reducing Sum_{k=0...n} n^k/k! to lowest terms.

Original entry on oeis.org

1, 2, 6, 8, 10, 144, 70, 128, 162, 6400, 22, 6220800, 26, 100352, 182250, 425984, 170, 429981696, 38, 163840000, 13502538, 317194240, 46, 247669456896, 31250, 1417674752, 15943230, 80564191232, 9802, 25076532510720000000, 62, 10737418240, 38196790434, 1241245548544
Offset: 1

Views

Author

Jonathan Sondow, Jul 15 2012

Keywords

Crossrefs

Programs

  • Mathematica
    Table[n!/Denominator[Sum[n^k/k!, {k, 0, n}]], {n, 1, 30}]
  • PARI
    a(n) = n!/denominator(sum(k=0, n, n^k/k!)); \\ Michel Marcus, Apr 20 2021

Formula

a(n) = n!/A214401(n).

Extensions

More terms from Michel Marcus, Apr 20 2021

A214401 Denominator of Sum_{k=0..n} n^k/k!.

Original entry on oeis.org

1, 1, 1, 3, 12, 5, 72, 315, 2240, 567, 1814400, 77, 239500800, 868725, 7175168, 49116375, 2092278988800, 14889875, 3201186852864000, 14849255421, 3783802880000, 3543572316375, 562000363888803840000, 2505147019375, 496358721386591551488
Offset: 1

Views

Author

Jonathan Sondow, Jul 15 2012

Keywords

Comments

Apparently, the three sequences T_1(n) = Sum_{k=1..n} n^(k-1)/k!, T_2(n) = Sum_{k=0..n} n^k/k!, and T_3(n) = Sum_{k=1..n} n^k/k!, with numerators in A119029, A120266, and A120267, respectively, have the same denominators, listed in the current sequence. This, however, is not immediately obvious. - Petros Hadjicostas, May 12 2020

Crossrefs

Numerators are A120266.

Programs

  • Mathematica
    Denominator[Table[Sum[n^k/k!, {k, 0, n}], {n, 1, 30}]]
  • PARI
    a(n) = denominator(sum(k=0, n, n^k/k!)); \\ Michel Marcus, Apr 20 2021

Formula

a(n) = n!/A214402(n).

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)).

A102581 Numbers n such that denominator of Sum_{k=0 to 2n+1} 1/k! is (2n+1)!/2.

Original entry on oeis.org

1, 2, 6, 10, 30, 32, 42, 46, 56, 62, 70, 80, 82, 96, 120, 122, 136, 150, 160, 162, 170, 172, 176, 186, 192, 196, 200, 210, 222, 230, 236, 252, 262, 266, 276, 290, 292, 300, 302, 306, 312, 326, 356, 366, 380, 382, 400, 416, 422, 426, 452, 460, 486, 490, 496, 500
Offset: 1

Views

Author

Jonathan Sondow, Jan 21 2005

Keywords

Comments

The denominator of Sum_{k=0 to m} 1/k! is m!/d, where d = A093101(m). If m > 1 is odd, say m = 2n+1, then d is even. n is a member when d = 2. If m > 3 and m = 3 (mod 4), so that n > 1 is odd, then d is divisible by 4. So except for 1 the members are even.

Examples

			1/0! + 1/1! + 1/2! + 1/3! = 8/3 and 3 = (2*1+1)!/2, so 1 is a member.
		

Crossrefs

n is a member <=> A093101(2n+1) = 2 <=> A061355(2n+1) = (2n+1)!/2 <=> n = 1 or n/2 is a member of A102582.

Programs

  • Mathematica
    fQ[n_] := (Denominator[ Sum[1/k!, {k, 0, 2n + 1}]] == (2n + 1)!/2); Select[ Range[ 500], fQ[ # ] &] (* Robert G. Wilson v, Jan 24 2005 *)

Formula

a(n) = 2*A102582(n-1) for n > 1.

Extensions

More terms from Robert G. Wilson v, Jan 24 2005

A102582 Numbers n such that denominator of Sum_{k=0..4n+1} 1/k! is (4n+1)!/2.

Original entry on oeis.org

1, 3, 5, 15, 16, 21, 23, 28, 31, 35, 40, 41, 48, 60, 61, 68, 75, 80, 81, 85, 86, 88, 93, 96, 98, 100, 105, 111, 115, 118, 126, 131, 133, 138, 145, 146, 150, 151, 153, 156, 163, 178, 183, 190, 191, 200, 208, 211, 213, 226, 230, 243, 245, 248, 250, 256, 260, 261, 265
Offset: 1

Views

Author

Jonathan Sondow, Jan 21 2005

Keywords

Comments

The denominator of Sum_{k=0 to m} 1/k! is m!/d, where d = A093101(m). If m = 4n+1 > 1, then d is even. n is a member when d = 2.

Examples

			1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! = 163/60 and 60 = 5!/2 = (4*1+1)!/2, so 1 is a member.
		

Crossrefs

n is a member <=> 2n is a member of A102581 <=> A093101(4n+1) = 2 <=> A061355(4n+1) = (4n+1)!/2.

Programs

  • Mathematica
    fQ[n_] := (Denominator[ Sum[1/k!, {k, 0, 4n + 1}]] == (4n + 1)!/2); Select[ Range[0, 274], fQ[ # ] &] (* Robert G. Wilson v, Jan 24 2005 *)

Formula

a(n) = A102581(n+1)/2.

Extensions

More terms from Robert G. Wilson v, Jan 24 2005

A102470 Numbers n such that denominator of Sum_{k=0 to n} 1/k! is n!.

Original entry on oeis.org

0, 1, 2, 4, 6, 8, 10, 16, 18, 20, 26, 28, 40, 46, 48, 58, 66, 68, 70, 80, 86, 96, 98, 118, 126, 130, 136, 146, 150, 170, 176, 178, 180, 188, 190, 206, 208, 210, 216, 230, 260, 266, 268, 278, 286, 288, 300, 306, 308, 326, 328, 338, 346, 358, 366, 370, 378, 380, 388
Offset: 1

Views

Author

Jonathan Sondow, Jan 14 2005

Keywords

Comments

a(n) is even for n > 1, as Sum_{k=0 to n} 1/k! reduces to lower terms when n > 1 is odd.

Examples

			1/0! + 1/1! + 1/2! + 1/3! +1/4! = 65/24 and 24 = 4!, so 4 is a member. But 1/0! + 1/1! + 1/2! + 1/3! = 8/3 and 3 < 3!, so 3 is not a member.
		

Crossrefs

For n > 0, n is a member <=> A093101(n) = 1 <=> A061355(n) = n! <=> A061355(n) = A002034(A061355(n))! <=> A061354(n) = 1+n+n(n-1)+n(n-1)(n-2)+...+n!. See also A102471.

Programs

  • Mathematica
    fQ[n_] := (Denominator[Sum[1/k!, {k, 0, n}]] == n!); Select[ Range[0, 389], fQ[ # ] &] (* Robert G. Wilson v, Jan 15 2005 *)

Formula

a(n) = 2*A102471(n-1) for n > 1.

Extensions

More terms from Robert G. Wilson v, Jan 15 2005

A102471 Numbers n such that the denominator of Sum_{k=0 to 2n} 1/k! is (2n)!.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 9, 10, 13, 14, 20, 23, 24, 29, 33, 34, 35, 40, 43, 48, 49, 59, 63, 65, 68, 73, 75, 85, 88, 89, 90, 94, 95, 103, 104, 105, 108, 115, 130, 133, 134, 139, 143, 144, 150, 153, 154, 163, 164, 169, 173, 179, 183, 185, 189, 190, 194, 195, 198, 199, 204
Offset: 1

Views

Author

Jonathan Sondow, Jan 14 2005

Keywords

Comments

n is a member <=> A093101(2n) = 1 <=> A061355(2n) = (2n)! <=> A061355(2n) = A002034(A061355(2n))!.

Examples

			Sum_{k=0 to 6} 1/k! = 1957/720 and 720 = 6! = (2*3)!, so 3 is a member. But Sum_{k=0 to 12} 1/k! = 260412269/95800320 and 95800320 < 12! = (2*6)!, so 6 is not a member.
		

Crossrefs

Programs

  • Mathematica
    fQ[n_] := (Denominator[Sum[1/k!, {k, 0, 2n}]] == (2n)!); Select[ Range[0, 204], fQ[ # ] &] (* Robert G. Wilson v, Jan 15 2005 *)

Formula

a(n) = A102470(n+1)/2 for n > 0.

Extensions

More terms from Robert G. Wilson v, Jan 15 2005
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