A105488 -id:A105488 - cp's OEIS frontend

A000071 a(n) = Fibonacci(n) - 1.

0, 0, 1, 2, 4, 7, 12, 20, 33, 54, 88, 143, 232, 376, 609, 986, 1596, 2583, 4180, 6764, 10945, 17710, 28656, 46367, 75024, 121392, 196417, 317810, 514228, 832039, 1346268, 2178308, 3524577, 5702886, 9227464, 14930351, 24157816, 39088168, 63245985, 102334154

Offset: 1

N. J. A. Sloane

a(n) is the number of allowable transition rules for passing from one change to the next (on n-1 bells) in the English art of bell-ringing. This is also the number of involutions in the symmetric group S_{n-1} which can be represented as a product of transpositions of consecutive numbers from {1, 2, ..., n-1}. Thus for n = 6 we have a(6) from (12), (12)(34), (12)(45), (23), (23)(45), (34), (45), for instance. See my 1983 Math. Proc. Camb. Phil. Soc. paper. - Arthur T. White, letter to N. J. A. Sloane, Dec 18 1986
Number of permutations p of {1, 2, ..., n-1} such that max|p(i) - i| = 1. Example: a(4) = 2 since only the permutations 132 and 213 of {1, 2, 3} satisfy the given condition. - Emeric Deutsch, Jun 04 2003 [For a(5) = 4 we have 2143, 1324, 2134 and 1243. - Jon Perry, Sep 14 2013]
Number of 001-avoiding binary words of length n-3. a(n) is the number of partitions of {1, ..., n-1} into two blocks in which only 1- or 2-strings of consecutive integers can appear in a block and there is at least one 2-string. E.g., a(6) = 7 because the enumerated partitions of {1, 2, 3, 4, 5} are 124/35, 134/25, 14/235, 13/245, 1245/3, 145/23, 125/34. - Augustine O. Munagi, Apr 11 2005
Numbers for which only one Fibonacci bit-representation is possible and for which the maximal and minimal Fibonacci bit-representations (A104326 and A014417) are equal. For example, a(12) = 10101 because 8 + 3 + 1 = 12. - Casey Mongoven, Mar 19 2006
Beginning with a(2), the "Recamán transform" (see A005132) of the Fibonacci numbers (A000045). - Nick Hobson, Mar 01 2007
Starting with nonzero terms, a(n) gives the row sums of triangle A158950. - Gary W. Adamson, Mar 31 2009
a(n+2) is the minimum number of elements in an AVL tree of height n. - Lennert Buytenhek (buytenh(AT)wantstofly.org), May 31 2010
a(n) is the number of branch nodes in the Fibonacci tree of order n-1. A Fibonacci tree of order n (n >= 2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node (see the Knuth reference, p. 417). - Emeric Deutsch, Jun 14 2010
a(n+3) is the number of distinct three-strand positive braids of length n (cf. Burckel). - Maxime Bourrigan, Apr 04 2011
a(n+1) is the number of compositions of n with maximal part 2. - Joerg Arndt, May 21 2013
a(n+2) is the number of leafs of great-grandparent DAG (directed acyclic graph) of height n. A great-grandparent DAG of height n is a single node for n = 1; for n > 1 each leaf of ggpDAG(n-1) has two child nodes where pairs of adjacent new nodes are merged into single node if and only if they have disjoint grandparents and same great-grandparent. Consequence: a(n) = 2*a(n-1) - a(n-3). - Hermann Stamm-Wilbrandt, Jul 06 2014
2 and 7 are the only prime numbers in this sequence. - Emmanuel Vantieghem, Oct 01 2014
From Russell Jay Hendel, Mar 15 2015: (Start)
We can establish Gerald McGarvey's conjecture mentioned in the Formula section, however we require n > 4. We need the following 4 prerequisites.
(1) a(n) = F(n) - 1, with {F(n)}A000045.%20(2)%20(Binet%20form)%20F(n)%20=%20(d%5En%20-%20e%5En)/sqrt(5)%20with%20d%20=%20phi%20and%20e%20=%201%20-%20phi,%20de%20=%20-1%20and%20d%20+%20e%20=%201.%20It%20follows%20that%20a(n)%20=%20(d(n)%20-%20e(n))/sqrt(5)%20-%201.%20(3)%20To%20prove%20floor(x)%20=%20y%20is%20equivalent%20to%20proving%20that%20x%20-%20y%20lies%20in%20the%20half-open%20interval%20%5B0,%201).%20(4)%20The%20series%20%7Bs(n)%20=%20c1%20x%5En%20+%20c2%7D">{n >= 1} the Fibonacci numbers A000045. (2) (Binet form) F(n) = (d^n - e^n)/sqrt(5) with d = phi and e = 1 - phi, de = -1 and d + e = 1. It follows that a(n) = (d(n) - e(n))/sqrt(5) - 1. (3) To prove floor(x) = y is equivalent to proving that x - y lies in the half-open interval [0, 1). (4) The series {s(n) = c1 x^n + c2}{n >= 1}, with -1 < x < 0, and c1 and c2 positive constants, converges by oscillation with s(1) < s(3) < s(5) < ... < s(6) < s(4) < s(2). If follows that for any odd n, the open interval (s(n), s(n+1)) contains the subsequence {s(t)}_{t >= n + 2}. Using these prerequisites we can analyze the conjecture.
Using prerequisites (2) and (3) we see we must prove, for all n > 4, that d((d^(n-1) - e^(n-1))/sqrt(5) - 1) - (d^n - e^n)/sqrt(5) + 1 + c lies in the interval [0, 1). But de = -1, implying de^(n-1) = -e^(n-2). It follows that we must equivalently prove (for all n > 4) that E(n, c) = (e^(n-2) + e^n)/sqrt(5) + 1 - d + c = e^(n-2) (e^2 + 1)/sqrt(5) + e + c lies in [0, 1). Clearly, for any particular n, E(n, c) has extrema (maxima, minima) when c = 2*(1-d) and c = (1+d)*(1-d). Therefore, the proof is completed by using prerequisite (4). It suffices to verify E(5, 2*(1-d)) = 0, E(6, 2*(1-d)) = 0.236068, E(5, (1-d)*(1+d)) = 0.618034, E(6, (1-d)*(1+d)) = 0.854102, all lie in [0, 1).
(End)
a(n) can be shown to be the number of distinct nonempty matchings on a path with n vertices. (A matching is a collection of disjoint edges.) - Andrew Penland, Feb 14 2017
Also, for n > 3, the lexicographically earliest sequence of positive integers such that {phi*a(n)} is located strictly between {phi*a(n-1)} and {phi*a(n-2)}. - Ivan Neretin, Mar 23 2017
From Eric M. Schmidt, Jul 17 2017: (Start)
Number of sequences (e(1), ..., e(n-2)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) <= e(k). [Martinez and Savage, 2.5]
Number of sequences (e(1), ..., e(n-2)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) >= e(j) <= e(k) and e(i) != e(k). [Martinez and Savage, 2.5]
(End)
Numbers whose Zeckendorf (A014417) and dual Zeckendorf (A104326) representations are the same: alternating digits of 1 and 0. - Amiram Eldar, Nov 01 2019
a(n+2) is the length of the longest array whose local maximum element can be found in at most n reveals. See link to the puzzle by Alexander S. Kulikov. - Dmitry Kamenetsky, Aug 08 2020
a(n+2) is the number of nonempty subsets of {1,2,...,n} that contain no consecutive elements. For example, the a(6)=7 subsets of {1,2,3,4} are {1}, {2}, {3}, {4}, {1,3}, {1,4} and {2,4}. - Muge Olucoglu, Mar 21 2021
a(n+3) is the number of allowed patterns of length n in the even shift (that is, a(n+3) is the number of binary words of length n in which there are an even number of 0s between any two occurrences of 1). For example, a(7)=12 and the 12 allowed patterns of length 4 in the even shift are 0000, 0001, 0010, 0011, 0100, 0110, 0111, 1000, 1001, 1100, 1110, 1111. - Zoran Sunic, Apr 06 2022
Conjecture: for k a positive odd integer, the sequence {a(k^n): n >= 1} is a strong divisibility sequence; that is, for n, m >= 1, gcd(a(k^n), a(k^m)) = a(k^gcd(n,m)). - Peter Bala, Dec 05 2022
In general, the sum of a second-order linear recurrence having signature (c,d) will be a third-order recurrence having a signature (c+1,d-c,-d). - Gary Detlefs, Jan 05 2023
a(n) is the number of binary strings of length n-2 whose longest run of 1's is of length 1, for n >= 3. - Félix Balado, Apr 03 2025

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Christian G. Bower, Table of n, a(n) for n = 1..500
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Michael Dairyko, Samantha Tyner, Lara Pudwell, and Casey Wynn, Non-contiguous pattern avoidance in binary trees. Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227. - From _N. J. A. Sloane_, Feb 01 2013
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Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000032, A000045, A000119, A001611, A001654, A001911, A005968, A005969.
Cf. A006327, A014417, A054761, A081006, A081007, A081008, A081009, A083368.
Cf. A098531, A098532, A098533, A101220, A104326, A105488, A105489, A119282.
Cf. A128697, A157725, A157726, A157727, A157728, A157729, A158950, A167616.
Cf. A001622, A228074.
Antidiagonal sums of array A004070.
Right-hand column 2 of triangle A011794.
Related to sum of Fibonacci(kn) over n. Cf. A099919, A058038, A138134, A053606.
Subsequence of A226538. Also a subsequence of A061489.

a000071 n = a000071_list !! n
a000071_list = map (subtract 1) $ tail a000045_list
-- Reinhard Zumkeller, May 23 2013

[Fibonacci(n)-1: n in [1..60]]; // Vincenzo Librandi, Apr 04 2011

A000071 := proc(n) combinat[fibonacci](n)-1 ; end proc; # R. J. Mathar, Apr 07 2011
a:= n-> (Matrix([[1, 1, 0], [1, 0, 0], [1, 0, 1]])^(n-1))[3, 2]; seq(a(n), n=1..50); # Alois P. Heinz, Jul 24 2008

Fibonacci[Range[40]] - 1 (* or *) LinearRecurrence[{2, 0, -1}, {0, 0, 1}, 40] (* Harvey P. Dale, Aug 23 2013 *)
Join[{0}, Accumulate[Fibonacci[Range[0, 39]]]] (* Alonso del Arte, Oct 22 2017, based on Giorgi Dalakishvili's formula *)

PARI
```
{a(n) = if( n<1, 0, fibonacci(n)-1)};
```

[fibonacci(n)-1 for n in range(1,60)] # G. C. Greubel, Oct 21 2024

a(n) = A000045(n) - 1.
a(0) = -1, a(1) = 0; thereafter a(n) = a(n-1) + a(n-2) + 1.
a(n) = A101220(1, 1, n-2), for n > 1.
G.f.: x^3/((1-x-x^2)*(1-x)). - Simon Plouffe in his 1992 dissertation, dropping initial 0's
a(n) = 2*a(n-1) - a(n-3). - R. H. Hardin, Apr 02 2011
Partial sums of Fibonacci numbers. - Wolfdieter Lang
a(n) = -1 + (A*B^n + C*D^n)/10, with A, C = 5 +- 3*sqrt(5), B, D = (1 +- sqrt(5))/2. - Ralf Stephan, Mar 02 2003
a(1) = 0, a(2) = 0, a(3) = 1, then a(n) = ceiling(phi*a(n-1)) where phi is the golden ratio (1 + sqrt(5))/2. - Benoit Cloitre, May 06 2003
Conjecture: for all c such that 2*(2 - Phi) <= c < (2 + Phi)*(2 - Phi) we have a(n) = floor(Phi*a(n-1) + c) for n > 4. - Gerald McGarvey, Jul 22 2004. This is true provided n > 3 is changed to n > 4, see proof in Comments section. - Russell Jay Hendel, Mar 15 2015
a(n) = Sum_{k = 0..floor((n-2)/2)} binomial(n-k-2, k+1). - Paul Barry, Sep 23 2004
a(n+3) = Sum_{k = 0..floor(n/3)} binomial(n-2*k, k)*(-1)^k*2^(n-3*k). - Paul Barry, Oct 20 2004
a(n+1) = Sum(binomial(n-r, r)), r = 1, 2, ... which is the case t = 2 and k = 2 in the general case of t-strings and k blocks: a(n+1, k, t) = Sum(binomial(n-r*(t-1), r)*S2(n-r*(t-1)-1, k-1)), r = 1, 2, ... - Augustine O. Munagi, Apr 11 2005
a(n) = Sum_{k = 0..n-2} k*Fibonacci(n - k - 3). - Ross La Haye, May 31 2006
a(n) = term (3, 2) in the 3 X 3 matrix [1, 1, 0; 1, 0, 0; 1, 0, 1]^(n-1). - Alois P. Heinz, Jul 24 2008
For n >= 4, a(n) = ceiling(phi*a(n-1)), where phi is the golden ratio. - Vladimir Shevelev, Jul 04 2010
Closed-form without two leading zeros g.f.: 1/(1 - 2*x - x^3); ((5 + 2*sqrt(5))*((1 + sqrt(5))/2)^n + (5 - 2*sqrt(5))*((1 - sqrt(5))/2)^n - 5)/5; closed-form with two leading 0's g.f.: x^2/(1 - 2*x - x^3); ((5 + sqrt(5))*((1 + sqrt(5))/2)^n + (5 - sqrt(5))*((1 - sqrt(5))/2)^n - 10)/10. - Tim Monahan, Jul 10 2011
A000119(a(n)) = 1. - Reinhard Zumkeller, Dec 28 2012
a(n) = A228074(n - 1, 2) for n > 2. - Reinhard Zumkeller, Aug 15 2013
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(4*k + 2 - x^2)/( x*(4*k + 4 - x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013
A083368(a(n+3)) = n. - Reinhard Zumkeller, Aug 10 2014
E.g.f.: 1 - exp(x) + 2*exp(x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Ilya Gutkovskiy, Jun 15 2016
a(n) = A000032(3+n) - 1 mod A000045(3+n). - Mario C. Enriquez, Apr 01 2017
a(n) = Sum_{i=0..n-2} Fibonacci(i). - Giorgi Dalakishvili (mcnamara_gio(AT)yahoo.com), Apr 02 2005 [corrected by Doug Bell, Jun 01 2017]
a(n+2) = Sum_{j = 0..floor(n/2)} Sum_{k = 0..j} binomial(n - 2*j, k+1)*binomial(j, k). - Tony Foster III, Sep 08 2017
From Peter Bala, Nov 12 2021: (Start)
a(4*n) = Fibonacci(2*n+1)*Lucas(2*n-1) = A081006(n);
a(4*n+1) = Fibonacci(2*n)*Lucas(2*n+1) = A081007(n);
a(4*n+2) = Fibonacci(2*n)*Lucas(2*n+2) = A081008(n);
a(4*n+3) = Fibonacci(2*n+2)*Lucas(2*n+1) = A081009(n). (End)
G.f.: x^3/((1 - x - x^2)*(1 - x)) = Sum_{n >= 0} (-1)^n * x^(n+3) *( Product_{k = 1..n} (k - x)/Product_{k = 1..n+2} (1 - k*x) )  (a telescoping series). - Peter Bala, May 08 2024
Product_{n>=4} (1 + (-1)^n/a(n)) = 3*phi/4, where phi is the golden ratio (A001622). - Amiram Eldar, Nov 28 2024

Edited by N. J. A. Sloane, Apr 04 2011

A285362 Sum T(n,k) of the entries in the k-th blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 4, 2, 15, 12, 3, 60, 58, 28, 4, 262, 273, 185, 55, 5, 1243, 1329, 1094, 495, 96, 6, 6358, 6839, 6293, 3757, 1148, 154, 7, 34835, 37423, 36619, 26421, 11122, 2380, 232, 8, 203307, 217606, 219931, 180482, 96454, 28975, 4518, 333, 9, 1257913, 1340597, 1376929, 1230737, 787959, 308127, 67898, 7995, 460, 10

Offset: 1

Alois P. Heinz, Apr 17 2017

			T(3,2) = 12 because the sum of the entries in the second blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 0+3+2+5+2 = 12.
Triangle T(n,k) begins:
      1;
      4,     2;
     15,    12,     3;
     60,    58,    28,     4;
    262,   273,   185,    55,     5;
   1243,  1329,  1094,   495,    96,    6;
   6358,  6839,  6293,  3757,  1148,  154,   7;
  34835, 37423, 36619, 26421, 11122, 2380, 232, 8;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
Wikipedia, Partition of a set

Columns k=1-10 give: A285363, A285364, A285365, A285366, A285367, A285368, A285369, A285370, A285371, A285372.
Row sums give A000110(n) * A000217(n) = A105488(n+3).
Main diagonal and first lower diagonal give: A000027, A006000 (for n>0).
T(2n+1,n+1) gives A285410.
Cf. A270236, A285439, A285595.

T:= proc(h) option remember; local b; b:=
      proc(n, m) option remember; `if`(n=0, [1, 0], add((p-> p
        +[0, (h-n+1)*p[1]*x^j])(b(n-1, max(m, j))), j=1..m+1))
      end: (p-> seq(coeff(p, x, i), i=1..n))(b(h, 0)[2])
    end:
seq(T(n), n=1..12);

T[h_] := T[h] = Module[{b}, b[n_, m_] := b[n, m] = If[n == 0, {1, 0}, Sum[# + {0, (h - n + 1)*#[[1]]*x^j}&[b[n - 1, Max[m, j]]], {j, 1, m + 1}]]; Table[Coefficient[#, x, i], {i, 1, n}]&[b[h, 0][[2]]]];
Table[T[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, Apr 30 2018, after Alois P. Heinz *)

A105489 Number of partitions of {1...n} containing 3 detached pairs of consecutive integers, i.e., partitions in which only 1- or 2-strings of consecutive integers can appear in a block and there are exactly three 2-strings.

Original entry on oeis.org

2, 20, 150, 1040, 7105, 49112, 347760, 2537640, 19135875, 149285400, 1205088742, 10062575068, 86859191510, 774456785200, 7126496659960, 67617733760064, 660932425168071, 6649326113764980, 68793130453044760, 731299516881396540

Offset: 6

Augustine O. Munagi, Apr 10 2005

Number of partitions enumerated by A105480 in which the maximal length of consecutive integers in a block is 2.

With offset 3t, number of partitions of {1...N} containing 3 detached strings of t consecutive integers, where N = n + 3j, t = 2 + j, j = 0, 1, 2, ..., i.e., partitions of {1,..,N} in which only v-strings of consecutive integers can appear in a block, where v=1 or v=t and there are exactly three t-strings.

			a(6) = 2 because the partitions of {1,2,3,4,5,6} with 3 detached pairs of consecutive integers are 12/34/56, 1256/34.

A. O. Munagi, Set Partitions with Successions and Separations, Int. J. Math and Math. Sc. 2005, no. 3 (2005), 451-463.

Cf. A105480, A105485, A105488, A105490.

seq(binomial(n-3,3)*combinat[bell](n-4),n=6..25);
a:=n->sum(numbcomb (n,2)*bell(n)/3, j=0..n): seq(a(n), n=2..21); # Zerinvary Lajos, Apr 25 2007

a(n) = binomial(n-3, 3)*Bell(n-4), which is the case r=3 in the general case of r pairs, d(n,r) = binomial(n-r, r)*Bell(n-r-1), which is the case t=2 of the general formula d(n,r,t) = binomial(n-r*(t-1), r)*Bell(n-r*(t-1)-1).

A285595 Sum T(n,k) of the k-th entries in all blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 4, 2, 17, 10, 3, 76, 52, 18, 4, 362, 274, 111, 28, 5, 1842, 1500, 675, 200, 40, 6, 9991, 8614, 4185, 1380, 325, 54, 7, 57568, 51992, 26832, 9568, 2510, 492, 70, 8, 351125, 329650, 178755, 67820, 19255, 4206, 707, 88, 9, 2259302, 2192434, 1239351, 494828, 149605, 35382, 6629, 976, 108, 10

Offset: 1

Alois P. Heinz, Apr 22 2017

T(n,k) is also k times the number of blocks of size >k in all set partitions of [n+1]. T(3,2) = 10 = 2 * 5 because there are 5 blocks of size >2 in all set partitions of [4], namely in 1234, 123|4, 124|3, 134|2, 1|234.

			T(3,2) = 10 because the sum of the second entries in all blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 2+2+3+3+0  = 10.
Triangle T(n,k) begins:
      1;
      4,     2;
     17,    10,     3;
     76,    52,    18,    4;
    362,   274,   111,   28,    5;
   1842,  1500,   675,  200,   40,   6;
   9991,  8614,  4185, 1380,  325,  54,  7;
  57568, 51992, 26832, 9568, 2510, 492, 70, 8;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
Wikipedia, Partition of a set

Column k=1 gives A124325(n+1).
Row sums give A000110(n) * A000217(n) = A105488(n+3).
Main diagonal and first lower diagonal give: A000027, A028552.
Cf. A007318, A175757, A278677, A283424, A285362, A285793, A286897.

T:= proc(h) option remember; local b; b:=
      proc(n, l) option remember; `if`(n=0, [1, 0],
        (p-> p+[0, (h-n+1)*p[1]*x^1])(b(n-1, [l[], 1]))+
         add((p-> p+[0, (h-n+1)*p[1]*x^(l[j]+1)])(b(n-1,
         sort(subsop(j=l[j]+1, l), `>`))), j=1..nops(l)))
      end: (p-> seq(coeff(p, x, i), i=1..n))(b(h, [])[2])
    end:
seq(T(n), n=1..12);
# second Maple program:
b:= proc(n) option remember; `if`(n=0, [1, 0],
      add((p-> p+[0, p[1]*add(x^k, k=1..j-1)])(
         b(n-j)*binomial(n-1, j-1)), j=1..n))
    end:
T:= n-> (p-> seq(coeff(p, x, i)*i, i=1..n))(b(n+1)[2]):
seq(T(n), n=1..12);

b[n_] := b[n] = If[n == 0, {1, 0}, Sum[# + {0, #[[1]]*Sum[x^k, {k, 1, j-1} ]}&[b[n - j]*Binomial[n - 1, j - 1]], {j, 1, n}]];
T[n_] := Table[Coefficient[#, x, i]*i, {i, 1, n}] &[b[n + 1][[2]]];
Table[T[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, May 23 2018, translated from 2nd Maple program *)

T(n,k) = k * Sum_{j=k+1..n+1} binomial(n+1,j)*A000110(n+1-j).
T(n,k) = k * Sum_{j=k+1..n+1} A175757(n+1,j).
Sum_{k=1..n} T(n,k)/k = A278677(n-1).

A105484 Number of partitions of {1...n} containing 2 strings of 3 consecutive integers, where each string is counted within a block and a string of more than 3 consecutive integers are counted three at a time.

Original entry on oeis.org

1, 2, 9, 38, 177, 882, 4711, 26795, 161583, 1028992, 6896067, 48487476, 356703531, 2738868784, 21901044795, 182022288438, 1569519971934, 14017732109520, 129480496353104, 1235228480628932, 12154988981496309, 123229919746398894, 1285758785855488107

Offset: 4

Augustine O. Munagi, Apr 10 2005

nonn

			a(6)=9 because the partitions of {1,...,6} with 2 strings of 3 consecutive integers are 12346/5, 13456/2, 16/2345, 1234/56, 123/456, 12/3456, 1234/5/6, 1/2345/6, 1/2/3456.

Augustine O. Munagi, Set Partitions with Successions and Separations, Int. J. Math and Math. Sc., 2005:3 (2005), 451-463.

Cf. A105483, A105485, A105488, A105492.

c := proc(n,k,r) option remember ; local j ; if r =0 then add(binomial(n-j,j)*combinat[stirling2](n-j-1,k-1),j=0..floor(n/2)) ; else if r <0 or r > n-k-1 then RETURN(0) fi ; if n <1 then RETURN(0) fi ; if k <1 then RETURN(0) fi ; RETURN( c(n-1,k-1,r)+(k-1)*c(n-1,k,r)+c(n-2,k-1,r)+(k-1)*c(n-2,k,r) +c(n-1,k,r-1)-c(n-2,k-1,r-1)-(k-1)*c(n-2,k,r-1) ) ; fi ; end: A105484 := proc(n) local k ; add(c(n,k,2),k=1..n) ; end: for n from 4 to 27 do printf("%d, ",A105484(n)) ; od ; # R. J. Mathar, Feb 20 2007

S2[_, -1] = 0;
S2[n_, k_] = StirlingS2[n, k];
c[n_, k_, r_] := c[n, k, r] = Which[r == 0, Sum[Binomial[n - j, j]*S2[n - j - 1, k - 1], {j, 0, Floor[n/2]}], r < 0 || r > n - k - 1, 0, n < 1, 0, k < 1, 0, True, c[n - 1, k - 1, r] + (k - 1)*c[n - 1, k, r] + c[n - 2, k - 1, r] + (k - 1)*c[n - 2, k, r] + c[n - 1, k, r - 1] - c[n - 2, k - 1, r - 1] - (k - 1)*c[n - 2, k, r - 1]];
A105484[n_] := Sum[c[n, k, 2], {k, 1, n}];
Table[A105484[n], {n, 4, 27}] (* Jean-François Alcover, May 10 2023, after R. J. Mathar *)

a(n) = Sum_{k=1..n} c(n, k, 2), where c(n, k, 2) is the case r =2 of c(n, k, r) given by c(n, k, r)=c(n-1, k-1, r)+(k-1)c(n-1, k, r)+c(n-2, k-1, r)+(k-1)c(n-2, k, r)+c(n-1, k, r-1)-c(n-2, k-1, r-1)-(k-1)c(n-2, k, r-1), r=0, 1, .., n-k-1, k=1, 2, .., n-2r, c(n, k, 0) = Sum_{j= 0..floor(n/2)} binomial(n-j, j)*S2(n-j-1, k-1).

More terms from R. J. Mathar, Feb 20 2007

A286232 Sum T(n,k) of the entries in the k-th last blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 5, 1, 19, 10, 1, 75, 57, 17, 1, 323, 285, 145, 26, 1, 1512, 1421, 975, 317, 37, 1, 7630, 7395, 5999, 2865, 616, 50, 1, 41245, 40726, 36183, 22411, 7315, 1094, 65, 1, 237573, 237759, 221689, 163488, 72581, 16630, 1812, 82, 1, 1451359, 1468162, 1405001, 1160764, 649723, 206249, 34425, 2840, 101, 1

Offset: 1

Alois P. Heinz, May 04 2017

			T(3,2) = 10 because the sum of the entries in the second last blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 0+3+4+1+2 = 10.
Triangle T(n,k) begins:
      1;
      5,     1;
     19,    10,     1;
     75,    57,    17,     1;
    323,   285,   145,    26,    1;
   1512,  1421,   975,   317,   37,    1;
   7630,  7395,  5999,  2865,  616,   50,  1;
  41245, 40726, 36183, 22411, 7315, 1094, 65, 1;
  ...

Alois P. Heinz, Rows n = 1..100, flattened
Wikipedia, Partition of a set

Column k=1 gives A285424.
Main diagonal and first lower diagonal give: A000012, A002522.
Row sums give A000110(n) * A000217(n) = A105488(n+3).
Cf. A285362, A286231, A286416.

app[P_, n_] := Module[{P0}, Table[P0 = Append[P, {}]; AppendTo[P0[[i]], n]; If[Last[P0] == {}, Most[P0], P0], {i, 1, Length[P]+1}]];
setPartitions[n_] := setPartitions[n] = If[n == 1, {{{1}}}, Flatten[app[#, n]& /@ setPartitions[n-1], 1]];
T[n_, k_] := Select[setPartitions[n], Length[#] >= k&][[All, -k]] // Flatten // Total;
Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Aug 21 2021 *)

A346772 Total sum of block indices of the elements over all partitions of [n].

Original entry on oeis.org

0, 1, 5, 22, 100, 482, 2475, 13527, 78476, 481687, 3117962, 21218851, 151387882, 1129430737, 8790433999, 71222812912, 599577147056, 5235054113412, 47331036294905, 442462325254995, 4270909302907430, 42514043248222709, 435920900603529954, 4599155199953703373

Offset: 0

Alois P. Heinz, Aug 02 2021

nonn

			a(3) = 22 = 3 + 4 + 4 + 5 + 6, summing block indices 111, 112, 121, 122, 123 of the 5 partitions of [3]: 123, 12|3, 13|2, 1|23, 1|2|3.

Alois P. Heinz, Table of n, a(n) for n = 0..574
Wikipedia, Partition of a set

Row sums of A120057.

Cf. A000110, A070071, A105488, A126347, A270236.

b:= proc(n, m) option remember; `if`(n=0, [1, 0], add(
     (p-> p+[0, p[1]*j])(b(n-1, max(m, j))), j=1..m+1))
    end:
a:= n-> b(n, 0)[2]:
seq(a(n), n=0..25);

b[n_, m_] := b[n, m] = If[n == 0, {1, 0}, Sum[
     Function[p, p+{0, p[[1]]*j}][b[n-1, Max[m, j]]], {j, 1, m+1}]];
a[n_] := b[n, 0][[2]];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Apr 27 2022, after Alois P. Heinz *)

a(n) = Sum_{k=1..n} A120057(n,k).
a(n) = Sum_{k=0..n*(n-1)/2} (n+k) * A126347(n,k).
a(n) = Sum_{k=1..n} k * A270236(n,k).

A271023 Number T(n,k) of set partitions of [n] having exactly k pairs (i,j) with i=0, 0<=k<=n*(n-1)/2 read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 0, 1, 1, 6, 3, 4, 0, 0, 1, 1, 10, 15, 10, 10, 0, 5, 0, 0, 0, 1, 1, 15, 45, 35, 60, 0, 25, 15, 0, 0, 6, 0, 0, 0, 0, 1, 1, 21, 105, 140, 210, 105, 105, 105, 0, 35, 21, 21, 0, 0, 0, 7, 0, 0, 0, 0, 0, 1, 1, 28, 210, 476, 665, 840, 350, 700, 210

Offset: 0

Alois P. Heinz, Mar 28 2016

			T(3,0) = 1: 1|2|3.
T(3,1) = 3: 12|3, 13|2, 1|23.
T(3,3) = 1: 123.
Triangle T(n,k) begins:
  1;
  1;
  1,  1;
  1,  3,  0,  1;
  1,  6,  3,  4,  0, 0,  1;
  1, 10, 15, 10, 10, 0,  5,  0, 0, 0, 1;
  1, 15, 45, 35, 60, 0, 25, 15, 0, 0, 6, 0, 0, 0, 0, 1;

Alois P. Heinz, Rows n = 0..40, flattened

Columns k=0-2 give: A000012, A161680, A050534(n-1) for n>0.
Row sums give A000110.
Cf. A105488, A161680, A271024.

b:= proc(n, l) option remember; `if`(n=0, x^
      add(j*(j-1)/2, j=l), b(n-1, [l[], 1])+
      add(b(n-1, subsop(j=l[j]+1, l)), j=1..nops(l)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, [])):
seq(T(n), n=0..10);

b[n_, l_] := b[n, l] = If[n == 0, x^Sum[j*(j-1)/2, {j, l}], b[n-1, Append[l, 1]] + Sum[b[n-1, ReplacePart[l, j -> l[[j]]+1]], {j, 1, Length[l]}]];
T[n_] := CoefficientList[b[n, {}], x];
Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Apr 29 2022, after Alois P. Heinz *)

T(n,k) = A271024(n,n*(n-1)/2-k).

Sum_{k=0..n*(n-1)/2} k * T(n,k) = A105488(n+2) for n > 1.

A286897 Sum T(n,k) of the k-th last entries in all blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 5, 1, 23, 6, 1, 109, 33, 7, 1, 544, 182, 45, 8, 1, 2876, 1034, 284, 59, 9, 1, 16113, 6122, 1815, 420, 75, 10, 1, 95495, 37927, 11931, 2987, 595, 93, 11, 1, 597155, 246030, 81205, 21620, 4665, 814, 113, 12, 1, 3929243, 1669941, 573724, 160607, 36900, 6979, 1082, 135, 13, 1

Offset: 1

Alois P. Heinz, May 15 2017

			T(3,2) = 6 because the sum of the second last entries in all blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 2+1+1+2 = 6.
Triangle T(n,k) begins:
      1;
      5,     1;
     23,     6,     1;
    109,    33,     7,    1;
    544,   182,    45,    8,   1;
   2876,  1034,   284,   59,   9,  1;
  16113,  6122,  1815,  420,  75, 10,  1;
  95495, 37927, 11931, 2987, 595, 93, 11, 1;
  ...

Alois P. Heinz, Row n = 1..50, flattened
Wikipedia, Partition of a set

Column k=1 gives A278677(n-1).
Row sums give A000110(n) * A000217(n) = A105488(n+3).
Cf. A285595.

b:= proc(n, l) option remember; `if`(n=0, [1, 0],
      (p-> p+[0, n*p[1]*x^1])(b(n-1, [l[], 1]))+
       add((p-> p+[0, n*p[1]*x^(l[j]+1)])(b(n-1,
       sort(subsop(j=l[j]+1, l), `>`))), j=1..nops(l)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n, [])[2]):
seq(T(n), n=1..14);

b[0, ] = {1, 0}; b[n, l_] := b[n, l] = Function[p, p + {0, n*p[[1]]*x^1} ][b[n - 1, Append[l, 1]]] + Sum[Function[p, p + {0, n*p[[1]]*x^(l[[j]] + 1)}][b[n - 1, Reverse @ Sort[ReplacePart[l, j -> l[[j]] + 1]]]], {j, 1, Length[l]}];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, n}]][b[n, {}][[2]]];
Table[T[n], {n, 1, 14}] // Flatten (* Jean-François Alcover, May 26 2018, from Maple *)

A105492 Number of partitions of {1,...,n} containing 2 strings of 3 consecutive integers such that only v-strings of consecutive integers can appear in a block, where v = 1,2,3.

Original entry on oeis.org

1, 6, 36, 210, 1260, 7833, 50701, 342126, 2406645, 17633820, 134427468, 1064801442, 8751834839, 74540800014

Offset: 6

Augustine O. Munagi, Apr 11 2005

Partitions enumerated by A105484 in which the maximal length of consecutive integers in a block is 3.

			a(7)=6; the enumerated partitions are 123567/4, 1237/456, 1567/234, 123/456/7, 123/4/567, 1/234/567.

A. O. Munagi, Set Partitions with Successions and Separations, Int. J. Math and Math. Sc. 2005, no. 3 (2005), 451-463

A. O. Munagi, Set Partitions with Successions and Separations,IJMMS 2005:3 (2005), 451-463.

Cf. A105484, A105488, A105493.

a(n)=Sum(w(n, k, 2), k=1...n), where w(n, k, 2) is the case r=2 of w(n, k, r) given by w(m, k, r)=w(m-1, k-1, r)+(k-1)w(m-1, k, r)+w(m-2, k-1, r)+(k-1)w(m-2, k, r) +w(m-3, k-1, r-1)+(k-1)w(m-3, k, r-1) r=0, 1, ..., floor(n/3), k=1, 2, ..., n-2r, w(n, k, 0)=sum(binomial(n-j, j)*S2(n-j-1, k-1), j=0..floor(n/2)).

cp's OEIS Frontend

A000071 a(n) = Fibonacci(n) - 1.

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A285362 Sum T(n,k) of the entries in the k-th blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

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A105489 Number of partitions of {1...n} containing 3 detached pairs of consecutive integers, i.e., partitions in which only 1- or 2-strings of consecutive integers can appear in a block and there are exactly three 2-strings.

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A285595 Sum T(n,k) of the k-th entries in all blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

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A105484 Number of partitions of {1...n} containing 2 strings of 3 consecutive integers, where each string is counted within a block and a string of more than 3 consecutive integers are counted three at a time.

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A286232 Sum T(n,k) of the entries in the k-th last blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

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A346772 Total sum of block indices of the elements over all partitions of [n].

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