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 1-10 of 11 results. Next

A094794 a(n) = (1/n!)*A001689(n).

Original entry on oeis.org

44, 309, 1214, 3539, 8544, 18089, 34754, 61959, 104084, 166589, 256134, 380699, 549704, 774129, 1066634, 1441679, 1915644, 2506949, 3236174, 4126179, 5202224, 6492089, 8026194, 9837719, 11962724, 14440269, 17312534, 20624939, 24426264
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Crossrefs

Programs

  • Mathematica
    Table[n^5+10n^4+45n^3+100n^2+109n+44,{n,0,30}] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{44,309,1214,3539,8544,18089},30]
  • PARI
    a(n)=n^5+10*n^4+45*n^3+100*n^2+109*n+44 \\ Charles R Greathouse IV, Oct 16 2015

Formula

a(n) = n^5 + 10*n^4 + 45*n^3 + 100*n^2 + 109*n + 44.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), with a(0)=44, a(1)=309, a(2)=1214, a(3)=3539, a(4)=8544, a(5)=18089. - Harvey P. Dale, Jul 25 2012
G.f.: (x^5 + 10*x^3 + 20*x^2 + 45*x + 44) / (x-1)^6. - Colin Barker, Jun 15 2013
P-recursive: n*a(n) = (n+6)*a(n-1) - a(n-2) with a(0) = 44 and a(1) = 309. Cf. A094791 and A096307. - Peter Bala, Jul 25 2021

A068106 Euler's difference table: triangle read by rows, formed by starting with factorial numbers (A000142) and repeatedly taking differences. T(n,n) = n!, T(n,k) = T(n,k+1) - T(n-1,k).

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 2, 3, 4, 6, 9, 11, 14, 18, 24, 44, 53, 64, 78, 96, 120, 265, 309, 362, 426, 504, 600, 720, 1854, 2119, 2428, 2790, 3216, 3720, 4320, 5040, 14833, 16687, 18806, 21234, 24024, 27240, 30960, 35280, 40320, 133496, 148329, 165016, 183822, 205056, 229080, 256320, 287280, 322560, 362880
Offset: 0

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Author

N. J. A. Sloane, Apr 12 2002

Keywords

Comments

Triangle T(n,k) (n >= 1, 1 <= k <= n) giving number of ways of winning with (n-k+1)st card in the generalized "Game of Thirteen" with n cards.
From Emeric Deutsch, Apr 21 2009: (Start)
T(n-1,k-1) is the number of non-derangements of {1,2,...,n} having largest fixed point equal to k. Example: T(3,1)=3 because we have 1243, 4213, and 3241.
Mirror image of A047920.
(End)

Examples

			Triangle begins:
[0]    1;
[1]    0,    1;
[2]    1,    1,    2;
[3]    2,    3,    4,    6;
[4]    9,   11,   14,   18,   24;
[5]   44,   53,   64,   78,   96,  120;
[6]  265,  309,  362,  426,  504,  600,  720;
[7] 1854, 2119, 2428, 2790, 3216, 3720, 4320, 5040.
		

Crossrefs

Row sums give A002467.
Diagonals give A000142, A001563, A001564, A001565, A001688, A001689, A023043, A023044, A023045, A023046, A023047 (factorials and k-th differences, k=1..10).
See A047920 and A086764 for other versions.
T(2*n, n) is A033815.

Programs

  • Haskell
    a068106 n k = a068106_tabl !! n !! k
    a068106_row n = a068106_tabl !! n
    a068106_tabl = map reverse a047920_tabl
    -- Reinhard Zumkeller, Mar 05 2012
  • Maple
    d[0] := 1: for n to 15 do d[n] := n*d[n-1]+(-1)^n end do: T := proc (n, k) if k <= n then sum(binomial(k, j)*d[n-j], j = 0 .. k) else 0 end if end proc: for n from 0 to 9 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form; Emeric Deutsch, Jul 18 2009
  • Mathematica
    t[n_, k_] := Sum[(-1)^j*Binomial[n-k, j]*(n-j)!, {j, 0, n}]; Flatten[ Table[ t[n, k], {n, 0, 9}, {k, 0, n}]] (* Jean-François Alcover, Feb 21 2012, after Philippe Deléham *)
    T[n_, k_] := n! HypergeometricPFQ[{k-n}, {-n}, -1];
    Table[T[n, k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)

Formula

T(n, k) = Sum_{j>= 0} (-1)^j*binomial(n-k, j)*(n-j)!. - Philippe Deléham, May 29 2005
From Emeric Deutsch, Jul 18 2009: (Start)
T(n,k) = Sum_{j=0..k} d(n-j)*binomial(k, j), where d(i) = A000166(i) are the derangement numbers.
Sum_{k=0..n} (k+1)*T(n,k) = A000166(n+2) (the derangement numbers). (End)
T(n, k) = n!*hypergeom([k-n], [-n], -1). - Peter Luschny, Oct 05 2017
D-finite recurrence for columns: T(n,k) = n*T(n-1,k) + (n-k)*T(n-2,k). - Georg Fischer, Aug 13 2022

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 01 2003
Edited by N. J. A. Sloane, Sep 24 2011

A001688 4th forward differences of factorial numbers A000142.

Original entry on oeis.org

9, 53, 362, 2790, 24024, 229080, 2399760, 27422640, 339696000, 4536362880, 64988179200, 994447238400, 16190733081600, 279499828608000, 5100017213491200, 98087346669312000, 1983334021853184000, 42063950934061056000, 933754193111900160000
Offset: 0

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Author

Keywords

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Mathematica
    Table[(n^4 + 6*n^3 + 17*n^2 + 20*n + 9) n!, {n, 0, 20}] (* T. D. Noe, Aug 09 2012 *)
    Differences[Range[0,30]!,4] (* Harvey P. Dale, Jun 06 2017 *)
  • PARI
    a(n)=if(n<0,0,n!*(n^4 + 6*n^3 + 17*n^2 + 20*n + 9))

Formula

For n>=0 a(n) = n!*(n^4 + 6*n^3 + 17*n^2 + 20*n + 9). - Benoit Cloitre, Jun 10 2004
G.f.: -log(-x+1)+1+2/(x-1)^4*x*(4-3*x+2*x^2). - Simon Plouffe, Master's Thesis, Uqam 1992
E.g.f.: (9 + 8*x + 6*x^2 + x^4)/(1 - x)^5. - Ilya Gutkovskiy, Jan 20 2017
a(n) = (n+5)*a(n-1) - (n-1)*a(n-2) with a(0) = 9 and a(1) = 53. Cf. A095177. - Peter Bala, Jul 22 2021

A096307 E.g.f.: exp(x)/(1-x)^6.

Original entry on oeis.org

1, 7, 55, 481, 4645, 49171, 566827, 7073725, 95064361, 1369375615, 21054430591, 344231563897, 5964569413645, 109196040092491, 2106381399472435, 42705264827626261, 907920105215691217, 20198878182718877815
Offset: 0

Views

Author

Philippe Deléham, Jun 26 2004

Keywords

Comments

Sum_{k = 0..n} A094816(n,k)*x^k give A000522(n), A001339(n), A082030(n), A095000(n), A095177(n) for x = 1, 2, 3, 4, 5 respectively.

Crossrefs

Programs

  • Mathematica
    Table[HypergeometricPFQ[{6, -n}, {}, -1], {n, 0, 20}] (* Benedict W. J. Irwin, May 27 2016 *)
    With[{nn = 250}, CoefficientList[Series[Exp[x]/(1 - x)^6, {x, 0, nn}], x] Range[0, nn]!] (* G. C. Greubel, May 27 2016 *)

Formula

a(n) = Sum_{k = 0..n} A094916(n, k)*6^k.
a(n) = Sum_{k = 0..n} binomial(n, k)*(k+5)!/5!.
a(n) = 2F0(6,-n;;-1). - Benedict W. J. Irwin, May 27 2016
From Peter Bala, Jul 25 2021: (Start)
a(n) = (n+6)*a(n-1) - (n-1)*a(n-2) with a(0) = 1 and a(1) = 7. Cf. A001689.
First-order recurrence: P(n-1)*a(n) = n*P(n)*a(n-1) - 1 with a(0) = 1, where P(n) = n^5 + 10*n^4 + 45*n^3 + 100*n^2 + 109*n + 44 = A094794(n).
(End)

A094793 a(n) = (1/n!)*A001688(n).

Original entry on oeis.org

9, 53, 181, 465, 1001, 1909, 3333, 5441, 8425, 12501, 17909, 24913, 33801, 44885, 58501, 75009, 94793, 118261, 145845, 178001, 215209, 257973, 306821, 362305, 425001, 495509, 574453, 662481, 760265, 868501, 987909, 1119233, 1263241
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Comments

Number of injections from {1,2,3,4} to {1,2,...,n} with no fixed points. - Fiona T. Brunk (fbrunk(AT)mcs.st-and.ac.uk), May 23 2006
In general (cf. A094792, A094794, A094795, etc.), the number of injections [k] -> [n] with no fixed points is given by Sum_{i=0..k} (-1)^i*binomial(k,i)*(n-i)!/(n-k)!, which is equal to (1/n!)*f_k(n) where f_k(n) gives the k-th differences of factorial numbers. - Fiona T. Brunk (fbrunk(AT)mcs.st-and.ac.uk), May 23 2006

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5,-10,10,-5,1},{9,53,181,465,1001},40] (* Harvey P. Dale, May 23 2016 *)

Formula

a(n) = n^4 + 6*n^3 + 17*n^2 + 20*n + 9.
a(n) = Sum_{i=0..4} (-1)^i*binomial(4,i)*(n-i)!/(n-4)!. - Fiona T. Brunk (fbrunk(AT)mcs.st-and.ac.uk), May 23 2006
G.f.: -(x^4+6*x^2+8*x+9) / (x-1)^5. - Colin Barker, Jun 16 2013
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Fung Lam, Apr 17 2014
P-recursive: n*a(n) = (n+5)*a(n-1) - a(n-2) with a(0) = 9 and a(1) = 53. Cf. A094791. - Peter Bala, Jul 25 2021

A094792 a(n) = (1/n!)*A001565(n).

Original entry on oeis.org

2, 11, 32, 71, 134, 227, 356, 527, 746, 1019, 1352, 1751, 2222, 2771, 3404, 4127, 4946, 5867, 6896, 8039, 9302, 10691, 12212, 13871, 15674, 17627, 19736, 22007, 24446, 27059, 29852, 32831, 36002, 39371, 42944, 46727, 50726, 54947, 59396, 64079
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Comments

Number of injections from {1,2,3} to {1,2,...,n} with no fixed points. - Fiona T. Brunk (fbrunk(AT)mcs.st-and.ac.uk), May 23 2006

Crossrefs

Programs

Formula

a(n) = n^3 + 3*n^2 + 5*n + 2.
a(n) = Sum_{i=0..3} (-1)^i*binomial(3,i)*(n-i)!/(n-3)!. - Fiona T. Brunk (fbrunk(AT)mcs.st-and.ac.uk), May 23 2006
G.f.: (x^3+3*x+2) / (x-1)^4. - Colin Barker, Jun 15 2013
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Fung Lam, Apr 17 2014
P-recursive: n*a(n) = (n+4)*a(n-1) - a(n-2) with a(0) = 2 and a(1) = 11. Cf. A094791. - Peter Bala, Jul 25 2021

A094795 a(n) = (1/n!)*A023043(n).

Original entry on oeis.org

265, 2119, 9403, 30637, 81901, 190435, 398959, 770713, 1395217, 2394751, 3931555, 6215749, 9513973, 14158747, 20558551, 29208625, 40702489, 55744183, 75161227, 99918301, 131131645, 170084179, 218241343, 277267657, 349044001, 435685615
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{7,-21,35,-35,21,-7,1},{265,2119,9403,30637,81901,190435,398959},30] (* Harvey P. Dale, Aug 29 2023 *)

Formula

a(n) = n^6 + 15*n^5 + 100*n^4 + 355*n^3 + 694*n^2 + 689*n + 265.
G.f.: -(265 + 264*x + 135*x^2 + 40*x^3 + 15*x^4 + x^6)/(x-1)^7. - R. J. Mathar, Nov 15 2019
P-recursive: n*a(n) = (n+7)*a(n-1) - a(n-2) with a(0) = 265 and a(1) = 2119. Cf. A094791. - Peter Bala, Jul 25 2021

A023044 7th differences of factorial numbers.

Original entry on oeis.org

1854, 16687, 165016, 1781802, 20886576, 264398280, 3597143040, 52370755920, 812752093440, 13397819541120, 233845982899200, 4309095479673600, 83609603781580800, 1704092533657113600, 36403110891295948800
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    Differences[Range[0, 25]!, 7] (* Paolo Xausa, May 26 2025 *)

A061312 Triangle T[n,m]: T[n,-1] = 0; T[0,0] = 0; T[n,0] = n*n!; T[n,m] = T[n,m-1] - T[n-1,m-1].

Original entry on oeis.org

0, 1, 1, 4, 3, 2, 18, 14, 11, 9, 96, 78, 64, 53, 44, 600, 504, 426, 362, 309, 265, 4320, 3720, 3216, 2790, 2428, 2119, 1854, 35280, 30960, 27240, 24024, 21234, 18806, 16687, 14833, 322560, 287280, 256320, 229080, 205056, 183822, 165016, 148329
Offset: 0

Views

Author

Wouter Meeussen, Jun 06 2001

Keywords

Comments

Appears in the (n,k)-matching problem A076731. [Johannes W. Meijer, Jul 27 2011]

Examples

			0,
1, 1,
4, 3, 2,
18, 14, 11, 9,
96, 78, 64, 53, 44,
600, 504, 426, 362, 309, 265,
4320, 3720, 3216, 2790, 2428, 2119, 1854,
35280, 30960, 27240, 24024, 21234, 18806, 16687, 14833,
		

Crossrefs

Cf. A061018.
From Johannes W. Meijer, Jul 27 2011: (Start)
The row sums equal A193465. (End)

Programs

  • Magma
    [[(&+[(-1)^j*Binomial(k+1,j)*Factorial(n-j+1): j in [0..k+1]]): k in [0..n]]: n in [0..20]]; // G. C. Greubel, Aug 13 2018
  • Maple
    A061312 := proc(n,m): add(((-1)^j)*binomial(m+1,j)*(n+1-j)!, j=0..m+1) end: seq(seq(A061312(n,m), m=0..n), n=0..7); # Johannes W. Meijer, Jul 27 2011
  • Mathematica
    T[n_, k_]:= Sum[(-1)^j*Binomial[k + 1, j]*(n + 1 - j)!, {j, 0, k + 1}]; Table[T[n, k], {n, 0, 100}, {k, 0, n}] // Flatten  (* G. C. Greubel, Aug 13 2018 *)
  • PARI
    for(n=0,20, for(k=0,n, print1(sum(j=0,k+1, (-1)^j*binomial(k+1,j) *(n-j+1)!), ", "))) \\ G. C. Greubel, Aug 13 2018
    

Formula

T[n,m] = T[n,m-1]-T[n-1,m-1] with T[n,-1] = 0 and T[n,0] = A001563(n) = n*n!
T(n,m) = sum(((-1)^j)*binomial(m+1,j)*(n+1-j)!, j=0..m+1) [Johannes W. Meijer, Jul 27 2011]

A094791 Triangle read by rows giving coefficients of polynomials arising in successive differences of (n!)_{n>=0}.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 3, 5, 2, 1, 6, 17, 20, 9, 1, 10, 45, 100, 109, 44, 1, 15, 100, 355, 694, 689, 265, 1, 21, 196, 1015, 3094, 5453, 5053, 1854, 1, 28, 350, 2492, 10899, 29596, 48082, 42048, 14833, 1, 36, 582, 5460, 32403, 124908, 309602, 470328, 391641, 133496
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Comments

Let D_0(n)=n! and D_{k+1}(n)=D_{k}(n+1)-D_{k}(n), then D_{k}(n)=n!*P_{k}(n) where P_{k} is a polynomial with integer coefficients of degree k.
The horizontal reversal of this triangle arises as a binomial convolution of the derangements coefficients der(n,i) (numbers of permutations of size n with i derangements = A098825(n,i) = number of permutations of size n with n-i rencontres = A008290(n,n-i), see formula section). - Olivier Gérard, Jul 31 2011

Examples

			D_3(n) = n!*(n^3 + 3*n^2 + 5*n + 2).
D_4(n) = n!*(n^4 + 6*n^3 + 17*n^2 + 20*n + 9).
Table begins:
  1
  1  0
  1  1   1
  1  3   5   2
  1  6  17  20    9
  1 10  45 100  109   44
  1 15 100 355  694  689  265
  ...
		

Crossrefs

Successive differences of factorial numbers: A001563, A001564, A001565, A001688, A001689, A023043.
Rencontres numbers A008290. Partial derangements A098825.
Row sum is A000255. Signed version in A126353.

Programs

  • Maple
    with(LREtools): A094791_row := proc(n)
    delta(x!,x,n); simplify(%/x!); seq(coeff(%,x,n-j),j=0..n) end:
    seq(print(A094791_row(n)),n=0..9); # Peter Luschny, Jan 09 2015
  • Mathematica
    d[0][n_] := n!; d[k_][n_] := d[k][n] = d[k - 1][n + 1] - d[k - 1][n] // FullSimplify;
    row[k_] := d[k][n]/n! // FullSimplify // CoefficientList[#, n]& // Reverse;
    Array[row, 10, 0] // Flatten (* Jean-François Alcover, Aug 02 2019 *)

Formula

T(n, n) = A000166(n).
T(2, k) = A000217(k).
Sum_{k=0..n} T(n,n-k)*x^k = Sum_{i=0..n} der(n,i)*binomial( n+x, i) (an analog of Worpitzky's identity). - Olivier Gérard, Jul 31 2011

Extensions

Edited and T(0,0) corrected according to the author's definition by Olivier Gérard, Jul 31 2011
Showing 1-10 of 11 results. Next