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 1-4 of 4 results.

A086764 Triangle T(n, k), read by row, related to Euler's difference table A068106 (divide column k of A068106 by k!).

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 9, 11, 7, 3, 1, 44, 53, 32, 13, 4, 1, 265, 309, 181, 71, 21, 5, 1, 1854, 2119, 1214, 465, 134, 31, 6, 1, 14833, 16687, 9403, 3539, 1001, 227, 43, 7, 1, 133496, 148329, 82508, 30637, 8544, 1909, 356, 57, 8, 1
Offset: 0

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Author

Philippe Deléham, Aug 02 2003

Keywords

Comments

The k-th column sequence, k >= 0, without leading zeros, enumerates the ways to distribute n beads, n >= 1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k+1 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords each contribute a factor 1, hence for n=0 one has 1. See A000255 for the description of a fixed cord with beads. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010). - Wolfdieter Lang, Jun 02 2010

Examples

			Formatted as a square array:
      1      3     7    13   21   31  43 57 ... A002061;
      2     11    32    71  134  227 356    ... A094792;
      9     53   181   465 1001 1909        ... A094793;
     44    309  1214  3539 8544             ... A094794;
    265   2119  9403 30637                  ... A023043;
   1854  16687 82508                        ... A023044;
  14833 148329                              ... A023045;
Formatted as a triangular array (mirror of A076731):
       1;
       0      1;
       1      1     1;
       2      3     2     1;
       9     11     7     3    1;
      44     53    32    13    4    1;
     265    309   181    71   21    5    1;
    1854   2119  1214   465  134   31    6   1;
   14833  16687  9403  3539 1001  227   43   7   1;
  133496 148329 82508 30637 8544 1909  356  57   8   1;
		

Crossrefs

Programs

  • Magma
    A086764:= func< n,k | (&+[(-1)^j*Binomial(n-k,j)*Factorial(n-j): j in [0..n]])/Factorial(k) >;
    [A086764(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Mathematica
    T[n_,k_]:=(1/k!)*Sum[(-1)^j*Binomial[n-k,j]*(n-j)!,{j,0,n}];Flatten[Table[T[n,k],{n,0,11},{k,0,n}]] (* Indranil Ghosh, Feb 20 2017 *)
    T[n_, k_] := (n!/k!) HypergeometricPFQ[{k-n},{-n},-1];
    Table[T[n,k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)
  • SageMath
    def A086764(n,k): return sum((-1)^j*binomial(n-k,j)*factorial(n-j) for j in range(n+1))//factorial(k)
    flatten([[A086764(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n, n) = 1; T(n+1, n) = n.
T(n+2, n) = A002061(n+1) = n^2 + n + 1; T(n+3, n) = n^3 + 3*n^2 + 5*n + 2.
T(n, k) = (k + 1)*T(n, k + 1) - T(n-1, k); T(n, n) = 1; T(n, k) = 0, if k > n.
T(n, k) = (n-1)*T(n-1, k) + (n-k-1)*T(n-2, k).
k!*T(n, k) = A068106(n, k). [corrected by Georg Fischer, Aug 13 2022]
Sum_{k>=0} T(n, k) = A003470(n+1).
T(n, k) = (1/k!) * Sum_{j>=0} (-1)^j*binomial(n-k, j)*(n-j)!. - Philippe Deléham, Jun 13 2005
From Peter Bala, Aug 14 2008: (Start)
The following remarks all relate to the array read as a square array: e.g.f for column k: exp(-y)/(1-y)^(k+1); e.g.f. for array: exp(-y)/(1-x-y) = (1 + x + x^2 + x^3 + ...) + (x + 2*x^2 + 3*x^3 + 4*x^4 + ...)*y + (1 + 3*x + 7*x^2 + 13*x^3 + ...)*y^2/2! + ... .
This table is closely connected to the constant e. The row, column and diagonal entries of this table occur in series formulas for e.
Row n for n >= 2: e = n!*(1/T(n,0) + (-1)^n*[1/(1!*T(n,0)*T(n,1)) + 1/(2!*T(n,1)*T(n,2)) + 1/(3!*T(n,2)*T(n,3)) + ...]). For example, row 3 gives e = 6*(1/2 - 1/(1!*2*11) - 1/(2!*11*32) - 1/(3!*32*71) - ...). See A095000.
Column 0: e = 2 + Sum_{n>=2} (-1)^n*n!/(T(n,0)*T(n+1,0)) = 2 + 2!/(1*2) - 3 !/(2*9) + 4!/(9*44) - ... .
Column k, k >= 1: e = (1 + 1/1! + 1/2! + ... + 1/k!) + 1/k!*Sum_{n >= 0} (-1)^n*n!/(T(n,k)*T(n+1,k)). For example, column 3 gives e = 8/3 + 1/6*(1/(1*3) - 1/(3*13) + 2/(13*71) - 6/(71*465) + ...).
Main diagonal: e = 1 + 2*(1/(1*1) - 1/(1*7) + 1/(7*71) - 1/(71*1001) + ...).
First subdiagonal: e = 8/3 + 5/(3*32) - 7/(32*465) + 9/(465*8544) - ... .
Second subdiagonal: e = 2*(1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...). See A143413.
Third subdiagonal: e = 3 - (2*3*5)/(2*53) + (3*4*7)/(53*1214) - (4*5*9)/(1214*30637) + ... .
For the corresponding results for the constants 1/e, sqrt(e) and 1/sqrt(e) see A143409, A143410 and A143411 respectively. For other arrays similarly related to constants see A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)). (End)
G.f. for column k is hypergeom([1,k+1],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
T(n, k) = (n!/k!)*hypergeom([k-n], [-n], -1). - Peter Luschny, Oct 05 2017

Extensions

More terms from David Wasserman, Mar 28 2005
Additional comments from Zerinvary Lajos, Mar 30 2006
Edited by N. J. A. Sloane, Sep 24 2011

A247490 Square array read by antidiagonals: A(k, n) = (-1)^(n+1)* hypergeom([k, -n+1], [], 1) for n>0 and A(k,0) = 0 (n>=0, k>=1).

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 2, 3, 2, 0, 1, 3, 7, 11, 9, 0, 1, 4, 13, 32, 53, 44, 0, 1, 5, 21, 71, 181, 309, 265, 0, 1, 6, 31, 134, 465, 1214, 2119, 1854, 0, 1, 7, 43, 227, 1001, 3539, 9403, 16687, 14833, 0, 1, 8, 57, 356, 1909, 8544, 30637, 82508, 148329, 133496
Offset: 0

Views

Author

Peter Luschny, Sep 20 2014

Keywords

Examples

			k\n
[1], 0, 1, 0,  1,   2,    9,   44,    265,      1854, ...  A000166
[2], 0, 1, 1,  3,  11,   53,   309,  2119,     16687, ...  A000255
[3], 0, 1, 2,  7,  32,  181,  1214,  9403,     82508, ...  A000153
[4], 0, 1, 3, 13,  71,  465,  3539,  30637,   296967, ...  A000261
[5], 0, 1, 4, 21, 134, 1001,  8544,  81901,   870274, ...  A001909
[6], 0, 1, 5, 31, 227, 1909, 18089, 190435,  2203319, ...  A001910
[7], 0, 1, 6, 43, 356, 3333, 34754, 398959,  4996032, ...  A176732
[8], 0, 1, 7, 57, 527, 5441, 61959, 770713, 10391023, ...  A176733
The referenced sequences may have a different offset or other small deviations.
		

Crossrefs

Programs

  • Maple
    A := (k,n) -> `if`(n<2,n,hypergeom([k,-n+1],[],1)*(-1)^(n+1));
    seq(print(seq(round(evalf(A(k,n),100)), n=0..8)), k=1..8);
  • Sage
    from mpmath import mp, hyp2f0
    mp.dps = 25; mp.pretty = True
    def A247490(k, n):
        if n < 2: return n
        if k == 1 and n == 2: return 0  # (failed to converge)
        return int((-1)^(n+1)*hyp2f0(k, -n+1, 1))
    for k in (1..8): print([k], [A247490(k, n) for n in (0..8)])

A284207 Eleventh column of Euler's difference table in A068106.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 3628800, 36288000, 402796800, 4906137600, 64988179200, 929459059200, 14266826784000, 233845982899200, 4075249496774400, 75225258805132800, 1465957162768492800, 30071395843421184000, 647624841502298284800
Offset: 1

Views

Author

Enrique Navarrete, Mar 22 2017

Keywords

Comments

For n >= 11, this is the number of permutations of [n] that avoid substrings j(j+10), 1 <= j <= n-10.

Examples

			a(14)=64988179200 since this is the number of permutations in S14 that avoid substrings {1(11),2(12),3(13),4(14)}.
		

Crossrefs

Also 3628800 times A176736.
Cf. A068106.

Programs

  • Mathematica
    Table[Sum[(-1)^j*Binomial[n - 10, j]*(n - j)!, {j, 0, n - 10}], {n, 22}] (* Michael De Vlieger, Apr 03 2017 *)

Formula

For n>=11: a(n) = Sum_{j=0..n-10} (-1)^j*binomial(n-10,j)*(n-j)!.
Note a(n)/n! ~ 1/e.

A336246 Array read by upwards antidiagonals: T(n,k) is the number of ways to place n persons on different seats such that each person number p, 1 <= p <= n, differs from the seat number s(p), 1 <= s(p) <= n+k, k >= 0.

Original entry on oeis.org

0, 1, 1, 2, 3, 2, 9, 11, 7, 3, 44, 53, 32, 13, 4, 265, 309, 181, 71, 21, 5, 1854, 2119, 1214, 465, 134, 31, 6, 14833, 16687, 9403, 3539, 1001, 227, 43, 7, 133496, 148329, 82508, 30637, 8544, 1909, 356, 57, 8, 1334961, 1468457, 808393, 296967, 81901, 18089, 3333, 527, 73, 9
Offset: 1

Views

Author

Gerhard Kirchner, Jul 19 2020

Keywords

Comments

T(n,0) = !n (subfactorial) is the number of derangements or fixed-point-free permutations, see A000166(n) below: n persons are placed on n seats such that no person sits on a seat with the same number. The generalization of a permutation is a variation (n persons and n+k seats such that k seats remain free). In this sense, T(n,k) is the number of fixed-point-free variations. I am rather sure that such variations have been examined, but I cannot find a reference.
Some subsequences T(n,k) with k=const:
T(n,0) = A000166(n); T(n,1) = A000255(n); T(n,2) = A000153(n-1);
T(n,3) = A000261(n-1); T(n,4) = A001909(n-3); T(n,5) = A001910(n-4);
T(n,6) = A176732(n); T(n,7) = A176733(n); T(n,8) = A176734(n);
T(n,9) = A176735(n); T(n,10) = A176736(n).

Examples

			For k=1, the n-tuples of seat numbers are:
- for n=1: 2 => T(1,1) = 1.
- for n=2: 21, 23, 31 => T(2,1) = 3,
     21: person 1 sits on seat 2 and vice versa.
     A counterexample is 13 because person 1 would sit on seat 1.
- for n=3: 214,231,234,241,312,314,341,342,412,431,432 => T(3,1) = 11.
Array begins:
   0   1    2    3    4 ...
   1   3    7   13   21 ...
   2  11   32   71  134 ...
   9  53  181  465 1001 ...
  44 309 1214 3539 8544 ...
  .. ... .... .... ....
		

Crossrefs

Programs

  • Maxima
    block(nr: 0, k: -1,  mmax: 55,
        /*First mmax terms are returned, recurrence used*/
       a: makelist(0, n, 1, mmax),
       while nr
    				
  • Maxima
    block(n: 1, k: 0,  mmax: 55,
        /*First mmax terms are returned, explicit formula used*/
       a: makelist(0, n, 1, mmax),
       for m from 1 thru mmax do (su: 0,
         for r from 0 thru n do su: su+(-1)^r*binomial(n,r)*(n+k-r)!/k!,
         a[m]: su, if n=1 then (n: k+2, k: 0) else (n: n-1, k: k+1)),
      return(a));

Formula

T(n,k) = (n+k-1)*T(n-1,k) + (n-1)*T(n-2,k) for n >= 2, k >= 0 with T(0,k)=1 and T(1,k)=k.
For n=0, there is one empty variation. T(0,k) is used for the recurrence only, not in the table. For n=1, the person can be placed on seat number 2..k+1 (if k > 0).
You also find the recurrence in the formula section of A000166 (k=0) and in the name section of the other sequences listed above (1 <= k <= 10). Some sequences have a different offset.
T(n,k) = Sum_{r=0..n} (-1)^r*binomial(n,r)*(n+k-r)!/k!.
Proofs see link.
Showing 1-4 of 4 results.