A001563 -id:A001563 - cp's OEIS frontend

A240796 Total number of occurrences of the pattern 1<2 in all preferential arrangements (or ordered partitions) of n elements.

0, 1, 15, 186, 2330, 31065, 447405, 6979588, 117745668, 2141106795, 41810587775, 873474855726, 19451904450654, 460209050303821, 11531197020389025, 305122289460210120, 8503747639606509128, 249020038061419770783, 7645072502094118876755, 245564189847880300238290

Offset: 1

N. J. A. Sloane, Apr 13 2014

nonn

There are A000670(n) preferential arrangements of n elements - see A000670, A240763.
The number that avoid the pattern 1<2  is 2^(n-1).
The total number of occurrences of the pattern 1<2 in all permutations on n elements is (n-1)*(n-1)! (cf. A010027, A001563).

			The 13 preferential arrangements on 3 points and the number of times the pattern 1<2 occurs are:
1<2<3, 3
1<3<2, 2
2<1<3, 2
2<3<1, 1
3<1<2, 1
3<2<1, 0
1=2<3, 2
1=3<2, 1
2=3<1, 0
1<2=3, 2
2<1=3, 1
3<1=2, 0
1=2=3, 0,
for a total of a(3) = 15.

Alois P. Heinz, Table of n, a(n) for n = 1..420

Cf. A000670, A240763, A240796-A240800, A010027, A001563, A264082, A381299.

b:= proc(n, t) option remember; `if`(n=0, [1, 0], add((p-> p+
      [0, p[1]*j*t/2])(b(n-j, t+j))*binomial(n, j), j=1..n))
    end:
a:= n-> b(n, 0)[2]:
seq(a(n), n=1..25);  # Alois P. Heinz, Dec 08 2014

b[n_, t_] := b[n, t] = If[n == 0, {1, 0}, Sum[Function[{p}, p + {0, p[[1]]*j*t/2}][b[n - j, t + j]]*Binomial[n, j], {j, 1, n}]]; a[n_] := b[n, 0][[2]]; Table[a[n], {n, 1, 25}] (* Jean-François Alcover, Jun 08 2015, after Alois P. Heinz *)

a(n) ~ n! * n^2 / (8 * (log(2))^(n+1)). - Vaclav Kotesovec, May 03 2015

a(n) = Sum_{k=0..binomial(n,2)} k * A381299(n,k). - Alois P. Heinz, Feb 22 2025

a(8)-a(20) from Alois P. Heinz, Dec 08 2014

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

Benoit Cloitre, Jun 11 2004

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

Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001563, A001564, A001565, A001688, A001689, A023043.

Cf. A094791, A094793, A094794, A094795.

with(combinat):seq(fibonacci(4, i)-1,i=1..40); # Zerinvary Lajos, Mar 20 2008

Table[n^3+3*n^2+5*n+2,{n,0,70}] (* Vladimir Joseph Stephan Orlovsky, May 04 2011 *)

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

A239069 Decimal expansion of gamma - Ei(-1).

Original entry on oeis.org

7, 9, 6, 5, 9, 9, 5, 9, 9, 2, 9, 7, 0, 5, 3, 1, 3, 4, 2, 8, 3, 6, 7, 5, 8, 6, 5, 5, 4, 2, 5, 2, 4, 0, 8, 0, 0, 7, 3, 2, 0, 6, 6, 2, 9, 3, 4, 6, 8, 3, 1, 8, 0, 6, 3, 8, 3, 7, 4, 5, 8, 4, 7, 9, 5, 8, 4, 3, 6, 4, 2, 5, 3, 3, 6, 8, 0, 6, 2, 1, 5, 6, 5, 9, 1, 5, 7, 3, 1, 4, 3, 2, 6, 8, 8, 3, 9, 9, 9, 4

Offset: 0

Jonathan Sondow, Mar 12 2014

See crossrefs sequences for other comments, references, links, and formulas.

			0.7965995992970531342836758655425240800732066293468...

Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 37, table 37:7:1 at page 355.

G. C. Greubel, Table of n, a(n) for n = 0..10000
J. C. Lagarias, Euler's constant: Euler's work and modern developments, arXiv:1303.1856 [math.NT], 2013-2014; Bull. Amer. Math. Soc., 50 (2013), 527-628; see p. 553.

Cf. A001008, A001113, A001563, A001620, A002805, A073003, A099285.

RealDigits[EulerGamma - ExpIntegralEi[-1], 10, 100][[1]]

Euler + eint1(1,1)[1] \\ Michel Marcus, Aug 01 2020

Equals (the Euler-Mascheroni constant) - (the exponential integral at -1) = A001620 + A099285.
Equals (the Euler-Mascheroni constant) + (Gompertz's constant / e) = A001620 + (A073003 / A001113).
Equals Sum_{n>=1} (-1)^(n-1) / A001563(n) = Sum_{n>=1} (-1)^(n-1) / (n*n!).
Equals -Integral_{x=0..1} log(x)/exp(x) dx. - Amiram Eldar, Aug 01 2020
Equals (1/e) * Sum_{k>=1} H(k)/k!, where H(k) = A001008(k)/A002805(k) is the k-th harmonic number. - Amiram Eldar, Jun 25 2021

A255566 a(0) = 0; after which, a(2n) = A255411(a(n)), a(2n+1) = A256450(a(n)).

Original entry on oeis.org

0, 1, 4, 2, 18, 6, 12, 3, 96, 24, 48, 8, 72, 15, 16, 5, 600, 120, 240, 30, 360, 56, 60, 10, 480, 87, 88, 20, 90, 21, 22, 7, 4320, 720, 1440, 144, 2160, 270, 288, 36, 2880, 416, 420, 67, 432, 73, 66, 13, 3600, 567, 568, 107, 570, 109, 108, 26, 576, 111, 112, 27, 114, 28, 52, 9, 35280, 5040, 10080, 840, 15120, 1584, 1680, 168

Offset: 0

Antti Karttunen, May 05 2015

This sequence can be represented as a binary tree. Each left hand child is produced as A255411(n), and each right hand child as A256450(n), when parent contains n >= 1:
                                      0
                                      |
                   ...................1...................
                  4                                       2
       18......../ \........6                  12......../ \........3
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
   96       24         48       8          72       15         16       5
600  120 240  30    360  56   60 10     480  87   88  20     90  21   22 7
etc.
Because all terms of A255411 are even it means that odd terms can occur only in odd positions (together with some even terms, for each one of which there is a separate infinite cycle), while terms in even positions are all even.
After its initial 1, A255567 seems to give all the terms like 2, 3, 12, ... where the left hand child of the right hand child is one more than the right hand child of the left hand child (as for 2: 16 = 15+1, as for 3: 22 = 21+1, as for 12: 88 = 87+1).

Inverse: A255565.
Cf. A000142, A001511, A001563, A083318, A255411, A256450, A257679.
Cf. also A255567 and arrays A257503, A257505.
Related or similar permutations: A273666, A273667.

a(0) = 0; after which, a(2n) = A255411(a(n)), a(2n+1) = A256450(a(n)).
Other identities:
For all n >= 0, a(2^n) = A001563(n+1). [The leftmost branch of the binary tree is given by n*n!]
For all n >= 0, a(A083318(n)) = A000142(n+1). [And the next innermost vertices by (n+1)! This follows because A256450(n*n! - 1) = (n+1)! - 1.]
For all n >= 1, A257679(a(n)) = A001511(n).

Formula changed because of the changed starting offset of A256450 - Antti Karttunen, May 30 2016

A285793 Sum T(n,k) of the k-th entries in all cycles of all permutations of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 4, 2, 18, 13, 5, 96, 83, 43, 18, 600, 582, 342, 192, 84, 4320, 4554, 2874, 1824, 1068, 480, 35280, 39672, 26232, 17832, 11784, 7080, 3240, 322560, 382248, 261288, 185688, 131256, 88920, 54360, 25200, 3265920, 4044240, 2834640, 2078640, 1534320, 1110960, 765360, 473760, 221760

Offset: 1

Alois P. Heinz, Apr 26 2017

Each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements.

			T(3,2) = 13 because the sum of the second entries in all cycles of all permutations of [3] ((123), (132), (12)(3), (13)(2), (1)(23), (1)(2)(3)) is 2+3+2+3+3+0 = 13.
Triangle T(n,k) begins:
:      1;
:      4,      2;
:     18,     13,      5;
:     96,     83,     43,     18;
:    600,    582,    342,    192,     84;
:   4320,   4554,   2874,   1824,   1068,   480;
:  35280,  39672,  26232,  17832,  11784,  7080,  3240;
: 322560, 382248, 261288, 185688, 131256, 88920, 54360, 25200;

Wikipedia, Permutation

Columns k=1-2 give: A001563, A285795.
Main diagonal and first lower diagonal give: A038720(n-1) (for n>1), A286175.
Row sums give A000142 * A000217 = A180119.
Cf. A285439, A285595, A286231.

T(n,1) = n * n!.

T(n,n) = floor((n-1)!*(n+2)/2).

A318215 Expansion of e.g.f. exp(x/(1 + x)^2).

Original entry on oeis.org

1, 1, -3, 7, 1, -219, 2581, -22973, 162177, -554039, -10506419, 343049631, -6846400703, 113528248237, -1609627861659, 17371462450651, -36066494745599, -5681921495461743, 243263898097515037, -7398126521141652809, 193119003246643917441, -4476119490014676723659, 89171014860669488040757

Offset: 0

Ilya Gutkovskiy, Aug 21 2018

sign

Cf. A001563, A082579, A111884, A181983.

A318215 := proc(n)
    add((-1)^(n-k)*binomial(n+k-1,2*k-1)*n!/k!,k=0..n) ;
end proc:
seq(A318215(n),n=0..42) ; # R. J. Mathar, Aug 20 2021

nmax = 22; CoefficientList[Series[Exp[x/(1 + x)^2], {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[(-1)^(n - k) Binomial[n + k - 1, 2 k - 1] n!/k!, {k, 0, n}], {n, 0, 22}]
a[n_] := a[n] = Sum[(-1)^(k + 1) k k! Binomial[n - 1, k - 1] a[n - k], {k, 1, n}]; a[0] = 1; Table[a[n], {n, 0, 22}]
Join[{1}, Table[(-1)^(n + 1) n n! HypergeometricPFQ[{1 - n, 1 + n}, {3/2, 2}, 1/4], {n, 22}]]

E.g.f.: Product_{k>=1} exp((-1)^(k+1)*k*x^k).
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n+k-1,2*k-1)*n!/k!.
a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k+1)*k*k!*binomial(n-1,k-1)*a(n-k).
D-finite with recurrence a(n) +(3*n-4)*a(n-1) +(n-1)*(3*n-5)*a(n-2) +(n-1)*(n-2)*(n-3)*a(n-3)=0. - R. J. Mathar, Aug 20 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

Benoit Cloitre, Jun 11 2004

Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A001563, A001564, A001565, A001688, A001689, A023043.

Cf. A094791, A094792, A094793, A094794.

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

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

A143946 Triangle read by rows: T(n,k) is the number of permutations of [n] for which the sum of the positions of the left-to-right maxima is k (1 <= k <= n(n+1)/2).

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 2, 1, 0, 1, 6, 0, 6, 3, 2, 3, 2, 1, 0, 1, 24, 0, 24, 12, 8, 18, 8, 10, 3, 6, 3, 2, 1, 0, 1, 120, 0, 120, 60, 40, 90, 64, 50, 39, 42, 23, 28, 13, 10, 8, 6, 3, 2, 1, 0, 1, 720, 0, 720, 360, 240, 540, 384, 420, 234, 372, 198, 208, 168, 124, 98, 75, 60, 35, 34, 13, 16, 8, 6, 3

Offset: 1

Emeric Deutsch, Sep 21 2008

Row n contains n*(n+1)/2 = A000217(n) entries.

Sum of entries in row n = n! = A000142(n).

			T(4,6)=3 because we have 1243, 1342 and 2341 with left-to-right maxima at positions 1,2,3.
Triangle starts:
   1;
   1,  0,  1;
   2,  0,  2,  1,  0,  1;
   6,  0,  6,  3,  2,  3,  2,  1,  0,  1;
  24,  0, 24, 12,  8, 18,  8, 10,  3,  6,  3,  2,  1,  0,  1;
  ...

Alois P. Heinz, Rows n = 1..50, flattened
I. Kortchemski, Asymptotic behavior of permutation records, arXiv: 0804.0446 [math.CO], 2008-2009.

Cf. A000142, A000217, A001563, A001804, A143947.

T(n,n) gives A368246.

P:=proc(n) options operator, arrow: sort(expand(product(t^j+j-1,j=1..n))) end proc: for n to 7 do seq(coeff(P(n),t,i),i=1..(1/2)*n*(n+1)) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n) option remember; `if`(n=0, 1,
      expand(b(n-1)*(x^n+n-1)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n)):
seq(T(n), n=1..7);  # Alois P. Heinz, Aug 05 2020

row[n_] := CoefficientList[Product[t^k + k - 1, {k, 1, n}], t] // Rest;
Array[row, 7] // Flatten (* Jean-François Alcover, Nov 28 2017 *)

T(n,1) = T(n,3) = (n-1)! for n>=2.
Sum_{k=1..n*(n+1)/2} k * T(n,k) = n! * n = A001563(n).
Generating polynomial of row n is t(t^2+1)(t^3+2)...(t^n+n-1).
Sum_{k=1..n*(n+1)/2} (n*(n+1)/2-k) * T(n,k) = A001804(n). - Alois P. Heinz, Dec 19 2023

A212598 a(n) = n - m!, where m is the largest number such that m! <= n.

Original entry on oeis.org

0, 0, 1, 2, 3, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66

Offset: 1

N. J. A. Sloane, May 22 2012

nonn

The m in definition is given by A084558(n).

Sequence consists of numbers 0..A001563(n)-1 followed by numbers 0..A001563(n+1)-1, and so on. - Antti Karttunen, Dec 17 2012

A. Karttunen, Table of n, a(n) for n = 1..5040

Cf. A136437, A212266, A220656.

f:=proc(n) local i; for i from 0 to n do if i! > n then break; fi; od; n-(i-1)!; end;
[seq(f(n),n=1..70)];

a(n)=my(m); while(m++!<=n,); n-(m-1)! \\ Charles R Greathouse IV, Sep 02 2015

(define (A212598 n) (- n (A000142 (A084558 n))))

a(n) = n-A000142(A084558(n)). - Antti Karttunen, Dec 17 2012

A227550 A triangle formed like Pascal's triangle, but with factorial(n) on the borders instead of 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 6, 4, 4, 6, 24, 10, 8, 10, 24, 120, 34, 18, 18, 34, 120, 720, 154, 52, 36, 52, 154, 720, 5040, 874, 206, 88, 88, 206, 874, 5040, 40320, 5914, 1080, 294, 176, 294, 1080, 5914, 40320, 362880, 46234, 6994, 1374, 470, 470, 1374, 6994, 46234, 362880, 3628800

Offset: 0

Vincenzo Librandi, Aug 04 2013

A003422 gives the second column (after 0).

			Triangle begins:
       1;
       1,     1;
       2,     2,    2;
       6,     4,    4,    6;
      24,    10,    8,   10,  24;
     120,    34,   18,   18,  34, 120;
     720,   154,   52,   36,  52, 154,  720;
    5040,   874,  206,   88,  88, 206,  874, 5040;
   40320,  5914, 1080,  294, 176, 294, 1080, 5914, 40320;
  362880, 46234, 6994, 1374, 470, 470, 1374, 6994, 46234, 362880;

Vincenzo Librandi, Rows n = 0..70, flattened

Cf. similar triangles with t on the borders: A007318 (t = 1), A028326 (t = 2), A051599 (t = prime(n)), A051601 (t = n), A051666 (t = n^2), A108617 (t = fibonacci(n)), A134636 (t = 2n+1), A137688 (t = 2^n), A227075 (t = 3^n).
Cf. A003422.
Cf. A227791 (central terms), A001563, A074911.

a227550 n k = a227550_tabl !! n !! k
a227550_row n = a227550_tabl !! n
a227550_tabl = map fst $ iterate
   (\(vs, w:ws) -> (zipWith (+) ([w] ++ vs) (vs ++ [w]), ws))
   ([1], a001563_list)
-- Reinhard Zumkeller, Aug 05 2013

function T(n,k)
  if k eq 0 or k eq n then return Factorial(n);
  else return T(n-1,k-1) + T(n-1,k);
  end if; return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 02 2021

t = {}; Do[r = {}; Do[If[k == 0||k == n, m = n!, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

def T(n,k): return factorial(n) if (k==0 or k==n) else T(n-1, k-1) + T(n-1, k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 02 2021

From G. C. Greubel, May 02 2021: (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = T(n, n) = n!.
Sum_{k=0..n} T(n, k) = 2^n * (1 +Sum_{j=1..n-1} j*j!/2^j) = A140710(n). (End)

cp's OEIS Frontend

A240796 Total number of occurrences of the pattern 1<2 in all preferential arrangements (or ordered partitions) of n elements.

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

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A239069 Decimal expansion of gamma - Ei(-1).

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A255566 a(0) = 0; after which, a(2n) = A255411(a(n)), a(2n+1) = A256450(a(n)).

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A285793 Sum T(n,k) of the k-th entries in all cycles of all permutations of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

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A318215 Expansion of e.g.f. exp(x/(1 + x)^2).

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Maple

Mathematica

Formula

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

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A143946 Triangle read by rows: T(n,k) is the number of permutations of [n] for which the sum of the positions of the left-to-right maxima is k (1 <= k <= n(n+1)/2).

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