A125127 -id:A125127 - cp's OEIS frontend

A125128 a(n) = 2^(n+1) - n - 2, or partial sums of main diagonal of array A125127 of k-step Lucas numbers.

1, 4, 11, 26, 57, 120, 247, 502, 1013, 2036, 4083, 8178, 16369, 32752, 65519, 131054, 262125, 524268, 1048555, 2097130, 4194281, 8388584, 16777191, 33554406, 67108837, 134217700, 268435427, 536870882, 1073741793, 2147483616, 4294967263, 8589934558

Offset: 1

Jonathan Vos Post, Nov 22 2006

Essentially a duplicate of A000295: a(n) = A000295(n+1).
Partial sums of main diagonal of array A125127 = L(k,n): k-step Lucas numbers, read by antidiagonals.
Equals row sums of triangle A130128. - Gary W. Adamson, May 11 2007
Row sums of triangle A130330 which is composed of (1,3,7,15,...) in every column, thus: row sums of (1; 3,1; 7,3,1; ...). - Gary W. Adamson, May 24 2007
Row sums of triangle A131768. - Gary W. Adamson, Jul 13 2007
Convolution A130321 * (1, 2, 3, ...). Binomial transform of (1, 3, 4, 4, 4, ...). - Gary W. Adamson, Jul 27 2007
Row sums of triangle A131816. - Gary W. Adamson, Jul 30 2007
A000975 convolved with [1, 2, 2, 2, ...]. - Gary W. Adamson, Jun 02 2009
The eigensequence of a triangle with the triangular series as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010

			a(1) = 1 because "1-step Lucas number"(1) = 1.
a(2) = 4 = a(1) + [2-step] Lucas number(2) = 1 + 3.
a(3) = 11 = a(2) + 3-step Lucas number(3) = 1 + 3 + 7.
a(4) = 26 = a(3) + 4-step Lucas number(4) = 1 + 3 + 7 + 15.
a(5) = 57 = a(4) + 5-step Lucas number(5) = 1 + 3 + 7 + 15 + 31.
a(6) = 120 = a(5) + 6-step Lucas number(6) = 1 + 3 + 7 + 15 + 31 + 63.
G.f. = x + 4*x^2 + 11*x^3 + 26*x^4 + 57*x^5 + 120*x^6 + 247*x^7 + 502*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See p. 18.
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A000295, A125127, A125129, A130103.

List([1..40], n-> 2^(n+1) -n-2); # G. C. Greubel, Jul 26 2019

I:=[1, 4, 11]; [n le 3 select I[n] else 4*Self(n-1)-5*Self(n-2)+2*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 28 2012

CoefficientList[Series[1/((1-x)^2*(1-2*x)),{x,0,40}],x] (* Vincenzo Librandi, Jun 28 2012 *)
LinearRecurrence[{4,-5,2},{1,4,11},40] (* Harvey P. Dale, Nov 16 2014 *)
a[ n_] := With[{m = n + 1}, If[ m < 0, 0, 2^m - (1 + m)]]; (* Michael Somos, Aug 17 2015 *)

PARI
```
A125128(n)=2<M. F. Hasler, Jul 30 2015
```

{a(n) = n++; if( n<0, 0, 2^n - (1+n))}; /* Michael Somos, Aug 17 2015 */

[2^(n+1) -n-2 for n in (1..40)] # G. C. Greubel, Jul 26 2019

a(n) = A000295(n+1) = 2^(n+1) - n - 2 = Sum_{i=1..n} A125127(i,i) = Sum_{i=1..n} ((2^i)-1). [Edited by M. F. Hasler, Jul 30 2015]
From Colin Barker, Jun 17 2012: (Start)
a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3).
G.f.: x/((1-x)^2*(1-2*x)). (End)
a(n) = A000225(n) + A000325(n) - 1. - Miquel Cerda, Aug 07 2016
a(n) = A095151(n) - A000225(n). - Miquel Cerda, Aug 12 2016
E.g.f.: 2*exp(2*x) - (2+x)*exp(x). - G. C. Greubel, Jul 26 2019

Edited by M. F. Hasler, Jul 30 2015

A125129 Partial sums of diagonals of array of k-step Lucas numbers as in A125127, read by antidiagonals.

Original entry on oeis.org

1, 1, 4, 1, 8, 11, 1, 12, 19, 26, 1, 19, 33, 45, 57, 1, 30, 58, 84, 102, 120, 1, 48, 101, 157, 197, 222, 247, 1, 77, 179, 292, 380, 436, 469, 502, 1, 124, 318, 546, 731, 855, 929, 971, 1013, 1, 200, 567, 1026, 1409, 1674, 1838, 1932, 1984, 2036

Offset: 1

Jonathan Vos Post, Nov 23 2006

Array of partial sums of diagonals of L(k,n) begins: 0.|.1...4..11...26...57..120..247..502.1013.2036.
|.1...8..19...45..102..222..469..971.1984.
|.1..12..33...84..197..436..929.1932.
|.1..19..58..157..380..855.1838.
|.1..30.101..292..731.1674.
|.1..48.179..546.1409.
|.1..77.318.1026.
|.1.124.567.
|.1.200.
|.1.

			Row 1 of the derived array is the partial sum of the diagonal above the main diagonal of array of k-step Lucas numbers as in A125127, hence the partial sums of: 1, 7, 11, 26, 57, 120, 247, 502, 103, ... are 1 = 1; 8 = 1 + 7; 19 = 1 + 7 + 11; 45 = 1 + 7 + 11 + 26; and so forth.

Cf. A000012, A000032, A000204, A001644, A001648, A048887, A048888, A074048, A074584, A092921, A104621, A105754, A105755, A125127, A000295.

Row 0 = SUM[i=1..n]L(i,i) = A127128 = partial sum of main diagonal of array of A125127. Row 1 = SUM[i=1..n]L(i,i+1) = partial sum of diagonal above main diagonal of array of A125127. Row 2 = SUM[i=1..n]L(i,i+2) = partial sum of diagonal 2 above main diagonal of array of A125127. .. Row m = SUM[i=1..n]L(i,i+m) = partial sum of diagonal 2 above main diagonal of array of A125127.

A000295 Eulerian numbers (Euler's triangle: column k=2 of A008292, column k=1 of A173018).

Original entry on oeis.org

0, 0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, 2036, 4083, 8178, 16369, 32752, 65519, 131054, 262125, 524268, 1048555, 2097130, 4194281, 8388584, 16777191, 33554406, 67108837, 134217700, 268435427, 536870882, 1073741793, 2147483616, 4294967263, 8589934558

Offset: 0

N. J. A. Sloane

There are 2 versions of Euler's triangle:
* A008292 Classic version of Euler's triangle used by Comtet (1974).
* A173018 Version of Euler's triangle used by Graham, Knuth and Patashnik in Concrete Math. (1990).
Euler's triangle rows and columns indexing conventions:
* A008292 The rows and columns of the Eulerian triangle are both indexed starting from 1. (Classic version: used in the classic books by Riordan and Comtet.)
* A173018 The rows and columns of the Eulerian triangle are both indexed starting from 0. (Graham et al.)
Number of Dyck paths of semilength n having exactly one long ascent (i.e., ascent of length at least two). Example: a(4)=11 because among the 14 Dyck paths of semilength 4, the paths that do not have exactly one long ascent are UDUDUDUD (no long ascent), UUDDUUDD and UUDUUDDD (two long ascents). Here U=(1,1) and D=(1,-1). Also number of ordered trees with n edges having exactly one branch node (i.e., vertex of outdegree at least two). - Emeric Deutsch, Feb 22 2004
Number of permutations of {1,2,...,n} with exactly one descent (i.e., permutations (p(1),p(2),...,p(n)) such that #{i: p(i)>p(i+1)}=1). E.g., a(3)=4 because the permutations of {1,2,3} with one descent are 132, 213, 231 and 312.
a(n+1) is the convolution of nonnegative integers (A001477) and powers of two (A000079). - Graeme McRae, Jun 07 2006
Partial sum of main diagonal of A125127. - Jonathan Vos Post, Nov 22 2006
Number of partitions of an n-set having exactly one block of size > 1. Example: a(4)=11 because, if the partitioned set is {1,2,3,4}, then we have 1234, 123|4, 124|3, 134|2, 1|234, 12|3|4, 13|2|4, 14|2|3, 1|23|4, 1|24|3 and 1|2|34. - Emeric Deutsch, Oct 28 2006
k divides a(k+1) for k in A014741. - Alexander Adamchuk, Nov 03 2006
(Number of permutations avoiding patterns 321, 2413, 3412, 21534) minus one. - Jean-Luc Baril, Nov 01 2007, Mar 21 2008
The chromatic invariant of the prism graph P_n for n >= 3. - Jonathan Vos Post, Aug 29 2008
Decimal integer corresponding to the result of XORing the binary representation of 2^n - 1 and the binary representation of n with leading zeros. This sequence and a few others are syntactically similar. For n > 0, let D(n) denote the decimal integer corresponding to the binary number having n consecutive 1's. Then D(n).OP.n represents the n-th term of a sequence when .OP. stands for a binary operator such as '+', '-', '*', 'quotentof', 'mod', 'choose'. We then get the various sequences A136556, A082495, A082482, A066524, A000295, A052944. Another syntactically similar sequence results when we take the n-th term as f(D(n)).OP.f(n). For example if f='factorial' and .OP.='/', we get (A136556)(A000295) ; if f='squaring' and .OP.='-', we get (A000295)(A052944). - K.V.Iyer, Mar 30 2009
Chromatic invariant of the prism graph Y_n.
Number of labelings of a full binary tree of height n-1, such that each path from root to any leaf contains each label from {1,2,...,n-1} exactly once. - Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009
Also number of nontrivial equivalence classes generated by the weak associative law X((YZ)T)=(X(YZ))T on words with n open and n closed parentheses. Also the number of join (resp. meet)-irreducible elements in the pruning-grafting lattice of binary trees with n leaves. - Jean Pallo, Jan 08 2010
Nonzero terms of this sequence can be found from the row sums of the third sub-triangle extracted from Pascal's triangle as indicated below by braces:
                     1;
                  1,    1;
              {1},   2,    1;
           {1,    3},   3,    1;
        {1,    4,    6},   4,    1;
     {1,    5,   10,   10},   5,    1;
  {1,    6,   15,   20,   15},   6,    1;
  ... - L. Edson Jeffery, Dec 28 2011
For integers a, b, denote by a<+>b the least c >= a, such that the Hamming distance D(a,c) = b (note that, generally speaking, a<+>b differs from b<+>a). Then for n >= 3, a(n) = n<+>n. This has a simple explanation: for n >= 3 in binary we have a(n) = (2^n-1)-n = "anti n". - Vladimir Shevelev, Feb 14 2012
a(n) is the number of binary sequences of length n having at least one pair 01. - Branko Curgus, May 23 2012
Nonzero terms are those integers k for which there exists a perfect (Hamming) error-correcting code. - L. Edson Jeffery, Nov 28 2012
a(n) is the number of length n binary words constructed in the following manner:  Select two positions in which to place the first two 0's of the word.  Fill in all (possibly none) of the positions before the second 0 with 1's and then complete the word with an arbitrary string of 0's or 1's. So a(n) = Sum_{k=2..n} (k-1)*2^(n-k). - Geoffrey Critzer, Dec 12 2013
Without first 0: a(n)/2^n equals Sum_{k=0..n} k/2^k. For example: a(5)=57, 57/32 = 0/1 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32. - Bob Selcoe, Feb 25 2014
The first barycentric coordinate of the centroid of the first n rows of Pascal's triangle, assuming the numbers are weights, is A000295(n+1)/A000337(n). See attached figure. - César Eliud Lozada, Nov 14 2014
Starting (0, 1, 4, 11, ...), this is the binomial transform of (0, 1, 2, 2, 2, ...). - Gary W. Adamson, Jul 27 2015
Also the number of (non-null) connected induced subgraphs in the n-triangular honeycomb rook graph. - Eric W. Weisstein, Aug 27 2017
a(n) is the number of swaps needed in the worst case to transform a binary tree with n full levels into a heap, using (bottom-up) heapify. - Rudy van Vliet, Sep 19 2017
The utility of large networks, particularly social networks, with n participants is given by the terms a(n) of this sequence. This assertion is known as Reed's Law, see the Wikipedia link. - Johannes W. Meijer, Jun 03 2019
a(n-1) is the number of subsets of {1..n} in which the largest element of the set exceeds by at least 2 the next largest element. For example, for n = 5, a(4) = 11 and the 11 sets are {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}, {1,2,4}, {1,2,5}, {1,3,5}, {2,3,5}, {1,2,3,5}. - Enrique Navarrete, Apr 08 2020
a(n-1) is also the number of subsets of {1..n} in which the second smallest element of the set exceeds by at least 2 the smallest element. For example, for n = 5, a(4) = 11 and the 11 sets are {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,4,5}, {1,3,4,5}. - Enrique Navarrete, Apr 09 2020
a(n+1) is the sum of the smallest elements of all subsets of {1..n}. For example, for n=3, a(4)=11; the subsets of {1,2,3} are {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}, and the sum of smallest elements is 11. - Enrique Navarrete, Aug 20 2020
Number of subsets of an n-set that have more than one element. - Eric M. Schmidt, Mar 13 2021
Number of individual bets in a "full cover" bet on n-1 horses, dogs, etc. in different races. Each horse, etc. can be bet on or not, giving 2^n bets.  But, by convention, singles (a bet on only one race) are not included, reducing the total number bets by n.  It is also impossible to bet on no horses at all, reducing the number of bets by another 1.  A full cover on 4 horses, dogs, etc. is therefore 6 doubles, 4 trebles and 1 four-horse etc. accumulator.  In British betting, such a bet on 4 horses etc. is a Yankee; on 5, a super-Yankee. - Paul Duckett, Nov 17 2021
From Enrique Navarrete, May 25 2022: (Start)
Number of binary sequences of length n with at least two 1's.
a(n-1) is the number of ways to choose an odd number of elements greater than or equal to 3 out of n elements.
a(n+1) is the number of ways to split [n] = {1,2,...,n} into two (possibly empty) complementary intervals {1,2,...,i} and {i+1,i+2,...,n} and then select a subset from the first interval (2^i choices, 0 <= i <= n), and one block/cell (i.e., subinterval) from the second interval (n-i choices, 0 <= i <= n).
(End)
Number of possible conjunctions in a system of n planets; for example, there can be 0 conjunctions with one planet, one with two planets, four with three planets (three pairs of planets plus one with all three) and so on. - Wendy Appleby, Jan 02 2023
Largest exponent m such that 2^m divides (2^n-1)!. - Franz Vrabec, Aug 18 2023
It seems that a(n-1) is the number of odd r with 0 < r < 2^n for which there exist u,v,w in the x-independent beginning of the Collatz trajectory of 2^n x + r with u+v = w+1, as detailed in the link "Collatz iteration and Euler numbers?". A better understanding of this might also give a formula for A374527. - Markus Sigg, Aug 02 2024
This sequence has a connection to consecutively halved positional voting (CHPV); see Mendenhall and Switkay. - Hal M. Switkay, Feb 25 2025
a(n) is the number of subsets of size 2 and more of an n-element set. Equivalently, a(n) is the number of (hyper)edges of size 2 and more in a complete hypergraph of n vertices. - Yigit Oktar, Apr 05 2025

			G.f. = x^2 + 4*x^3 + 11*x^4 + 26*x^5 + 57*x^6 + 120*x^7 + 247*x^8 + 502*x^9 + ...

O. Bottema, Problem #562, Nieuw Archief voor Wiskunde, 28 (1980) 115.
L. Comtet, "Permutations by Number of Rises; Eulerian Numbers." Section 6.5 in Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. enl. ed. Dordrecht, Netherlands: Reidel, pp. 51 and 240-246, 1974.
F. N. David and D. E. Barton, Combinatorial Chance. Hafner, NY, 1962, p. 151.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, p. 34.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
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).

T. D. Noe, Table of n, a(n) for n = 0..500
Joerg Arndt and N. J. A. Sloane, Counting Words that are in "Standard Order"
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See pp. 9, 17, 18.
E. Banaian, S. Butler, C. Cox, J. Davis, J. Landgraf and S. Ponce A generalization of Eulerian numbers via rook placements, arXiv:1508.03673 [math.CO], 2015.
Jean-Luc Baril and J. M. Pallo, The pruning-grafting lattice of binary trees, Theoretical Computer Science, 409, 2008, 382-393.
Jean-Luc Baril and José L. Ramírez, Some distributions in increasing and flattened permutations, arXiv:2410.15434 [math.CO], 2024. See p. 7.
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Peter J. Cameron, Maximilien Gadouleau, James D. Mitchell, and Yann Peresse, Chains of subsemigroups, arXiv preprint arXiv:1501.06394 [math.GR], 2015. See Table 4.
Matteo Cervetti and Luca Ferrari, Pattern avoidance in the matching pattern poset, arXiv:2009.01024 [math.CO], 2020.
Shelby Cox, Pratik Misra, and Pardis Semnani, Homaloidal Polynomials and Gaussian Models of Maximum Likelihood Degree One, arXiv:2402.06090 [math.AG], 2024.
Benjamin Hellouin de Menibus and Yvan Le Borgne, Asymptotic behaviour of the one-dimensional "rock-paper-scissors" cyclic cellular automaton, arXiv:1903.12622 [math.PR], 2019.
Karl Dilcher and Maciej Ulas, Arithmetic properties of polynomial solutions of the Diophantine equation P(x)x^(n+1)+Q(x)(x+1)^(n+1) = 1, arXiv:1909.11222 [math.NT], 2019.
Filippo Disanto, Some Statistics on the Hypercubes of Catalan Permutations, Journal of Integer Sequences, Vol. 18 (2015), Article 15.2.2.
J. M. Dusel, Balanced parabolic quotients and branching rules for Demazure crystals, J Algebr Comb (2016) 44: 363. DOI: 10.1007/s10801-016-0673-y.
Pascal Floquet, Serge Domenech and Luc Pibouleau, Combinatorics of Sharp Separation System synthesis : Generating functions and Search Efficiency Criterion, Industrial Engineering and Chemistry Research, 33, pp. 440-443, 1994.
Pascal Floquet, Serge Domenech, Luc Pibouleau and Said Aly, Some Complements in Combinatorics of Sharp Separation System Synthesis, American Institute of Chemical Engineering Journal, 39(6), pp. 975-978, 1993.
E. T. Frankel, A calculus of figurate numbers and finite differences, American Mathematical Monthly, 57 (1950), 14-25. [Annotated scanned copy]
Joël Gay, Representation of Monoids and Lattice Structures in the Combinatorics of Weyl Groups, Doctoral Thesis, Discrete Mathematics [cs.DM], Université Paris-Saclay, 2018.
R. K. Guy, Letter to N. J. A. Sloane.
Ryota Inagaki, Tanya Khovanova, and Austin Luo, On Chip-Firing on Undirected Binary Trees, Ann. Comb. (2025). See pp. 24-25.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 388.
Wayne A. Johnson, An Euler operator approach to Ehrhart series, arXiv:2303.16991 [math.CO], 2023.
Lucas Kang, Investigation of Rule 73 as Case Study of Class 4 Long-Distance Cellular Automata, arXiv preprint arXiv:1310.3311 [nlin.CG], 2013.
Oliver Kullmann and Xishun Zhao, Bounds for variables with few occurrences in conjunctive normal forms, arXiv preprint arXiv:1408.0629 [math.CO], 2014.
César Eliud Lozada, Centroids of Pascal triangles
Candice A. Marshall, Construction of Pseudo-Involutions in the Riordan Group, Dissertation, Morgan State University, 2017.
Peter Charles Mendenhall and Hal M. Switkay, Consecutively Halved Positional Voting: A Special Case of Geometric Voting, Social Sciences vol. 12 no. 2 (2023), 47-59.
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Ronald Orozco López, Solution of the Differential Equation y^(k)= e^(a*y), Special Values of Bell Polynomials and (k,a)-Autonomous Coefficients, Universidad de los Andes (Colombia 2021).
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Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A008292 (classic version of Euler's triangle used by Comtet (1974)).
Cf. A173018 (version of Euler's triangle used by Graham, Knuth and Patashnik in Concrete Math. (1990)).
Cf. A002662 (partial sums).
Partial sums of A000225.
Row sums of A014473 and of A143291.
Second column of triangles A112493 and A112500.
Sequences A125128 and A130103 are essentially the same.
Column k=1 of A124324.
Cf. A000079, A000108, A000217, A000325, A000975, A001511, A002663.
Cf. A002664, A007814, A008292, A008949, A014741, A016031, A028366.
Cf. A035039, A035040, A035041, A035042, A079583, A107907, A130321.
Cf. A130128, A130330, A131768, A131816.

a000295 n = 2^n - n - 1  -- Reinhard Zumkeller, Nov 25 2013

[2^n-n-1: n in [0..40]]; // Vincenzo Librandi, Jul 29 2015

[EulerianNumber(n, 1): n in [0..40]]; // G. C. Greubel, Oct 02 2024

[ seq(2^n-n-1, n=1..50) ];
A000295 := -z/(2*z-1)/(z-1)**2; # Simon Plouffe in his 1992 dissertation
# Grammar specification:
spec := [S, { B = Set(Z, 1 <= card), C = Sequence(B, 2 <= card), S = Prod(B, C) }, unlabeled]:
struct := n -> combstruct[count](spec, size = n+1);
seq(struct(n), n = 0..33); # Peter Luschny, Jul 22 2014

a[n_] = If[n==0, 0, n*(HypergeometricPFQ[{1, 1-n}, {2}, -1] - 1)];
Table[a[n], {n,0,40}] (* Olivier Gérard, Mar 29 2011 *)
LinearRecurrence[{4, -5, 2}, {0, 0, 1}, 40] (* Vincenzo Librandi, Jul 29 2015 *)
Table[2^n -n-1, {n,0,40}] (* Eric W. Weisstein, Nov 16 2017 *)

a(n)=2^n-n-1 \\ Charles R Greathouse IV, Jun 10 2011

[2^n -(n+1) for n in range(41)] # G. C. Greubel, Oct 02 2024

a(n) = 2^n - n - 1.
G.f.: x^2/((1-2*x)*(1-x)^2).
A107907(a(n+2)) = A000079(n+2). - Reinhard Zumkeller, May 28 2005
E.g.f.: exp(x)*(exp(x)-1-x). - Emeric Deutsch, Oct 28 2006
a(0)=0, a(1)=0, a(n) = 3*a(n-1) - 2*a(n-2) + 1. - Miklos Kristof, Mar 09 2005
a(0)=0, a(n) = 2*a(n-1) + n - 1 for all n in Z.
a(n) = Sum_{k=2..n} binomial(n, k). - Paul Barry, Jun 05 2003
a(n+1) = Sum_{i=1..n} Sum_{j=1..i} C(i, j). - Benoit Cloitre, Sep 07 2003
a(n+1) = 2^n*Sum_{k=0..n} k/2^k. - Benoit Cloitre, Oct 26 2003
a(0)=0, a(1)=0, a(n) = Sum_{i=0..n-1} i+a(i) for i > 1. - Gerald McGarvey, Jun 12 2004
a(n+1) = Sum_{k=0..n} (n-k)*2^k. - Paul Barry, Jul 29 2004
a(n) = Sum_{k=0..n} binomial(n, k+2); a(n+2) = Sum_{k=0..n} binomial(n+2, k+2). - Paul Barry, Aug 23 2004
a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-k-1, k+1)*2^(n-k-2)*(-1/2)^k. - Paul Barry, Oct 25 2004
a(0) = 0; a(n) = Stirling2(n,2) + a(n-1) = A000225(n-1) + a(n-1). - Thomas Wieder, Feb 18 2007
a(n) = A000325(n) - 1. - Jonathan Vos Post, Aug 29 2008
a(0) = 0, a(n) = Sum_{k=0..n-1} 2^k - 1. - Doug Bell, Jan 19 2009
a(n) = A000217(n-1) + A002662(n) for n>0. - Geoffrey Critzer, Feb 11 2009
a(n) = A000225(n) - n. - Zerinvary Lajos, May 29 2009
a(n) = n*(2F1([1,1-n],[2],-1) - 1). - Olivier Gérard, Mar 29 2011
Column k=1 of A173018 starts a'(n) = 0, 1, 4, 11, ... and has the hypergeometric representation n*hypergeom([1, -n+1], [-n], 2). This can be seen as a formal argument to prefer Euler's A173018 over A008292. - Peter Luschny, Sep 19 2014
E.g.f.: exp(x)*(exp(x)-1-x); this is U(0) where U(k) = 1 - x/(2^k - 2^k/(x + 1 - x^2*2^(k+1)/(x*2^(k+1) - (k+1)/U(k+1)))); (continued fraction, 3rd kind, 4-step). - Sergei N. Gladkovskii, Dec 01 2012
a(n) = A079583(n) - A000225(n+1). - Miquel Cerda, Dec 25 2016
a(0) = 0; a(1) = 0; for n > 1: a(n) = Sum_{i=1..2^(n-1)-1} A001511(i). - David Siegers, Feb 26 2019
a(n) = A007814(A028366(n)). - Franz Vrabec, Aug 18 2023
a(n) = Sum_{k=1..floor((n+1)/2)} binomial(n+1, 2*k+1). - Taras Goy, Jan 02 2025

A322057 Array read by upwards antidiagonals: T(i,n) is the number of binary necklaces of length n that avoid 00...0 (i 0's).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 3, 1, 1, 2, 3, 4, 3, 1, 1, 2, 3, 5, 5, 5, 1, 1, 2, 3, 5, 6, 9, 5, 1, 1, 2, 3, 5, 7, 11, 11, 8, 1, 1, 2, 3, 5, 7, 12, 15, 19, 10, 1, 1, 2, 3, 5, 7, 13, 17, 27, 29, 15, 1, 1, 2, 3, 5, 7, 13, 18, 31, 43, 48, 19, 1

Offset: 1

N. J. A. Sloane, Dec 25 2018

T(i,n) is the number of necklace compositions with sum n and parts at most i. For example, the T(3,5) = 5 compositions up to cyclic equivalence are 11111, 1112, 113, 122, 23. - Andrew Howroyd, Jan 08 2024

			The first few antidiagonals are:
  1;
  1,  1;
  1,  2,  1;
  1,  2,  2,  1;
  1,  2,  3,  3,  1;
  1,  2,  3,  4,  3,  1;
  1,  2,  3,  5,  5,  5,  1;
  1,  2,  3,  5,  6,  9,  5,  1;
  1,  2,  3,  5,  7, 11, 11,  8,  1;
  1,  2,  3,  5,  7, 12, 15, 19, 10,  1;
  ...
From _Petros Hadjicostas_, Jan 16 2019: (Start)
In the above triangle (first few antidiagonals, read upwards), the j-th row corresponds to T(j,1), T(j-1,2), T(j-2,3), ..., T(1,j).
This, however, is not the j-th row of the square array (see the scanned page above).
For example, the sixth row of the square array is as follows:
T(6,1) = 1, T(6,2) = 2, T(6,3) = 3, T(6,4) = 5, T(6, 5) = 7, T(6, 6) = 13, ...
To generate these numbers, we use T(6, n) = (1/n)*Sum_{d|n} phi(n/d)*L(6,d), where
L(6,1) = 1, L(6,2) = 3, L(6,3) = 7, L(6,4) = 15, L(6,5) = 31, L(6,6) = 63, ...
See the sixth row of A125127. See also the Sage program below by Freddy Barrera.
(End)

Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 520.

Freddy Barrera, Rows n = 1..100 of triangle, flattened.
Miklos Bona, Handbook of Enumerative Combinatorics, CRC Press, 2015, annotated scan of page 520.
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 89-96.

Rows 2, 3, 4 and 5 are A000358, A093305, A280218, A280303.
The rows converge to A008965.
Cf. A119458.

T(i,n) = {my(p=1/(1 - x*(1 - x^i)/(1 - x))); polcoef(sum(d=1, n, eulerphi(d)*log(subst(p + O(x*x^(n\d)), x, x^d))/d), n)} \\ Andrew Howroyd, Jan 08 2024

# uses the L method from A125127
def T(i, n):
    return sum(euler_phi(n//d)*L(i, d) for d in n.divisors()) // n
[T(i, n) for d in (1..12) for i, n in zip((d..1, step=-1), (1..d))] # Freddy Barrera, Jan 15 2019

T(i,n) = (1/n) * Sum_{d divides n} totient(n/d)*L(i,d), where L(i,d) = A125127(i,d). See Zhang and Hadjicostas link. - Freddy Barrera, Jan 15 2019

G.f. for row i: Sum_{k>=1} (phi(k)/k) * log(1/(1-B(i, x^k))) where B(i, x) = x*(1 + x + x^2 + ... + x^(i-1)). (This is a generalization of Joerg Arndt's formulae for the g.f.'s of rows 2 and 3.) - Petros Hadjicostas, Jan 24 2019

A212634 Triangle read by rows: T(n,k) is the number of dominating subsets with cardinality k of the cycle C_n (n >= 1, 1 <= k <= n).

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 0, 6, 4, 1, 0, 5, 10, 5, 1, 0, 3, 14, 15, 6, 1, 0, 0, 14, 28, 21, 7, 1, 0, 0, 8, 38, 48, 28, 8, 1, 0, 0, 3, 36, 81, 75, 36, 9, 1, 0, 0, 0, 25, 102, 150, 110, 45, 10, 1, 0, 0, 0, 11, 99, 231, 253, 154, 55, 11, 1

Offset: 1

Emeric Deutsch, Jun 14 2012

The entries in row n are the coefficients of the domination polynomial of the cycle C_n (see the Alikhani and Peng reference in Opuscula Math).
Sum of entries in row n = A001644(n) (number of dominating subsets; tribonacci numbers with initial values 1,3,7).
From Petros Hadjicostas, Jan 26 2019: (Start)
Let L(n, r, K) be the number of r-combinations of the n consecutive integers 1, 2, ..., n displaced on a circle, with no K integers consecutive (assuming that n and 1 are consecutive integers).
If on the circle where the n integers 1, 2, ..., n are displaced we put 1 for any integer that is included in the r-combination and 0 otherwise, we get a marked cyclic sequence of r ones and n-r zeros with no K consecutive ones.
Let L_{n,K}(x) = Sum_{r=0..floor(n-(n/K))} L(n, r, K)*x^(n-r) for n, K >= 1. Charalambides (1991) proved that L_{n,K}(x) = x*(n + Sum_{j=1..n-1} L_{n-j, K}(x)) for K >= 1 and 1 <= n <= K, and L_{n,K}(x) = x*Sum_{j=1..K} L_{n-j, K}(x) for K >= 1 and n >= K + 1. See Theorem 2.1 (p. 291) in his paper.
He also proved that L_{n,K}(x) = -1 + Sum_{j=0..floor(n/(K + 1))} (-1)^j*(n/(n-j*K))*binomial(n - j*K, j)*x^j*(1+x)^(n-j*(K+1)). See Theorem 2.3 (p. 293) in his paper.
He also produced other formulas for L(n, r, K) and L_{n,K}(x) including the explicit formulas of Moser and Abramson (1969) (regarding L(n, r, K)).
For the current sequence, we have T(n, k) = L(n, r = n-k, K = 3) for 1 <= n <= k. In other words, T(n, n-k) is the number of k-combinations of the n consecutive integers 1, 2, ..., n displaced on a circle, with no K = 3 consecutive integers (assuming n and 1 are consecutive).
(End)

			Row 4 is [0,6,4,1] because the cycle A-B-C-D-A has dominating subsets AB, AC, AD, BC, BD, CD, ABC, ABD, ACD, BCD, and ABCD.
Triangle starts:
1;
2, 1;
3, 3,  1;
0, 6,  4,  1;
0, 5, 10,  5, 1;
0, 3, 14, 15, 6, 1;
...
From _Petros Hadjicostas_, Jan 27 2019: (Start)
Let n=6 and 1 <= k <= 6. Then T(n, k) is the number of (n-k)-combinations of the integers 1, 2, 3, 4, 5, 6 displaced on a circle with no K=3 consecutive integers (assuming 6 and 1 are consecutive). Equivalently, T(n, k) is the number of marked cyclic sequences consisting of n-k ones and k zeros with no K=3 consecutive ones.
For each k below we give the corresponding (n-k)-combinations and the equivalent marked sequences of 0's and 1's.
k=1, n-k = 5: none; T(n=6, k=1) = 0.
k=2, n-k = 4: 1245 <-> 110110, 2356 <-> 011011, 1346 <-> 101101; T(n=6, k=2) = 3.
k=3, n-k = 3: 124 <-> 110100, 125 <-> 110010, 134 <-> 101100, 135 <-> 101010, 136 <-> 101001, 145 <-> 100110, 146 <-> 100101, 235 <-> 011010, 236 <-> 011001, 245 <-> 010110, 246 <-> 010101, 256 <-> 010011, 346 <-> 001101, 356 <-> 001011; T(n=6, k=3) = 14.
k=4, n-k=2: all 2-combinations of the integers 1,2,3,4,5,6 and all marked cyclic sequences with exactly 2 ones and 4 zeros; hence, T(n=6, k=4) = binomial(6, 2) = 15.
k=5, n-k=1: all 1-combinations of the integers 1,2,3,4,5,6 and all marked cyclic sequences with exactly 1 one and 5 zeros; hence, T(n=6, k=5) = binomial(6, 1) = 6.
k=6, n-k=0: empty combination <-> 000000; T(n=6, k=6) = 1.
(End)

S. Alikhani and Y. H. Peng, Introduction to domination polynomial of a graph, arXiv:0905.2251 [math.CO], 2009.
S. Alikhani and Y. H. Peng, Dominating sets and domination polynomials of paths, International J. Math. and Math. Sci., Vol. 2009, Article ID542040.
S. Alikhani and Y. H. Peng, Dominating sets and domination polynomials of certain graphs, II, Opuscula Math., 30, No. 1, 2010, 37-51.
J. L. Arocha, B. Llano, The number of dominating k-sets of paths, cycles and wheels, arXiv preprint arXiv:1601.01268 [math.CO], 2016.
C. A. Charalambides, Lucas numbers and polynomials of order k and the length of the longest circular success run, The Fibonacci Quarterly, 29 (1991), 290-297.
W. O. J. Moser and M. Abramson, Enumeration of combinations with restricted differences and cospan, J. Combin. Theory, 7 (1969), 162-170.
Eric Weisstein's World of Mathematics, Cycle Graph
Eric Weisstein's World of Mathematics, Domination Polynomial

Cf. A001644, A125127, A284966, A322057.

Cf. A212635 (corresponding sequence for wheel graphs). - Eric W. Weisstein, Apr 06 2017

p := proc (n) if n = 1 then x elif n = 2 then x^2+2*x elif n = 3 then x^3+3*x^2+3*x else sort(expand(x*(p(n-1)+p(n-2)+p(n-3)))) end if end proc: for n to 15 do seq(coeff(p(n), x, k), k = 1 .. n) end do; # yields sequence in triangular form

CoefficientList[LinearRecurrence[{x, x, x}, {1, 2 + x, 3 + 3 x + x^2}, 10], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)

If p(n)=p(n,x) denotes the generating polynomial of row n (called the domination polynomial of the cycle C_n), then p(1)=x, p(2) = 2x + x^2 , p(3) = 3x + 3x^2 + x^3 and p(n) = x*[p(n-1) + p(n-2) + p(n-3)] for n >= 4 (see Theorem 4.5 in the Alikhani & Peng reference in Opuscula Math.).
From Petros Hadjicostas, Jan 26 2019: (Start)
T(n, k) = binomial(n, k) for 1 <= k <= n and n = 1, 2.
T(n, k) = Sum_{j=0..floor(n/3)-1} (-1)^j*binomial(k, j)*(n/(n - 3*j))*binomial(n - 3*j, k) for n - floor(2*n/3) <= k <= n and n >= 3. (This is a special case of a corrected version of formula (2.1) in Charalambides (1991) and equation (14) in Moser and Abramson (1969).)
T(n, k) = 0 for 1 <= k < n - floor(2*n/3) and n >= 4. (Thus, the number of initial zeros in row n is n - floor(2*n/3) - 1.)
G.f. for row n: p(n, x) = -1 + Sum_{j=0..floor(n/4)} (-1)^j*(n/(n-3*j))*binomial(n - 3*j, j)*x^j*(1+x)^(n-4*j). It satisfies the recurrence given above (found in Alikhani and Peng (2010) and Charalambides (1991)).
(End)

cp's OEIS Frontend

A125128 a(n) = 2^(n+1) - n - 2, or partial sums of main diagonal of array A125127 of k-step Lucas numbers.

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A125129 Partial sums of diagonals of array of k-step Lucas numbers as in A125127, read by antidiagonals.

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A000295 Eulerian numbers (Euler's triangle: column k=2 of A008292, column k=1 of A173018).

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A322057 Array read by upwards antidiagonals: T(i,n) is the number of binary necklaces of length n that avoid 00...0 (i 0's).

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A212634 Triangle read by rows: T(n,k) is the number of dominating subsets with cardinality k of the cycle C_n (n >= 1, 1 <= k <= n).

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A247505 Generalized Lucas numbers: square array A(n,k) read by antidiagonals, A(n,k)=(-1)^(k+1)*k*[x^k](-log((1+sum_{j=1..n}(-1)^(j+1)*x^j)^(-1))), (n>=0, k>=0).

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A247505 Generalized Lucas numbers: square array A(n,k) read by antidiagonals, A(n,k)=(-1)^(k+1)k[x^k](-log((1+sum_{j=1..n}(-1)^(j+1)*x^j)^(-1))), (n>=0, k>=0).