A144112 -id:A144112 - cp's OEIS frontend

A007318 Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

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, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Offset: 0

N. J. A. Sloane and Mira Bernstein, Apr 28 1994

A. W. F. Edwards writes: "It [the triangle] was first written down long before 1654, the year in which Blaise Pascal wrote his Traité du triangle arithmétique, but it was this work that brought together all the different aspects of the numbers for the first time. In it Pascal developed the properties of the number as a piece of pure mathematics ... and then, in a series of appendices, showed how these properties were relevant to the study of the figurate numbers, to the theory of combinations, to the expansion of binomial expressions, and to the solution of an important problem in the theory of probability." (A. W. F. Edwards, Pascal's Arithmetical Triangle, Johns Hopkins University Press (2002), p. xiii)
Edwards reports that the naming of the triangle after Pascal was done first by Montmort in 1708 as the "Table de M. Pascal pour les combinaisons" and then by De Moivre in 1730 as the "Triangulum Arithmeticum PASCALANIUM". (Edwards, p. xiv)
In China, Yang Hui in 1261 listed the coefficients of (a+b)^n up to n=6, crediting the expansion to Chia Hsein's Shih-so suan-shu circa 1100. Another prominent early use was in Chu Shih-Chieh's Precious Mirror of the Four Elements in 1303. (Edwards, p. 51)
In Persia, Al-Karaji discovered the binomial triangle "some time soon after 1007", and Al-Samawal published it in the Al-bahir some time before 1180. (Edwards, p. 52)
In India, Halayuda's commentary (circa 900) on Pingala's treatise on syllabic combinations (circa 200 B.C.E.) contains a clear description of the additive computation of the triangle. (Amulya Kumar Bag, Binomial Theorem in Ancient India, p. 72)
Also in India, the multiplicative formula for C(n,k) was known to Mahavira in 850 and restated by Bhaskara in 1150. (Edwards, p. 27)
In Italy, Tartaglia published the triangle in his General trattato (1556), and Cardano published it in his Opus novum (1570). (Edwards, p. 39, 44) - Russ Cox, Mar 29 2022
Also sometimes called Omar Khayyam's triangle.
Also sometimes called Yang Hui's triangle.
C(n,k) = number of k-element subsets of an n-element set.
Row n gives coefficients in expansion of (1+x)^n.
Binomial(n+k-1,n-1) is the number of ways of placing k indistinguishable balls into n boxes (the "bars and stars" argument - see Feller).
Binomial(n-1,k-1) is the number of compositions (ordered partitions) of n with k summands.
Binomial(n+k-1,k-1) is the number of weak compositions (ordered weak partitions) of n into exactly k summands. - Juergen Will, Jan 23 2016
Binomial(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (1,1). - Joerg Arndt, Jul 01 2011
If thought of as an infinite lower triangular matrix, inverse begins:
  +1
  -1 +1
  +1 -2 +1
  -1 +3 -3 +1
  +1 -4 +6 -4 +1
All 2^n palindromic binomial coefficients starting after the A006516(n)-th entry are odd. - Lekraj Beedassy, May 20 2003
Binomial(n+k-1,n-1) is the number of standard tableaux of shape (n,1^k). - Emeric Deutsch, May 13 2004
Can be viewed as an array, read by antidiagonals, where the entries in the first row and column are all 1's and A(i,j) = A(i-1,j) + A(i,j-1) for all other entries. The determinant of each of its n X n subarrays starting at (0,0) is 1. - Gerald McGarvey, Aug 17 2004
Also the lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals j+1 (and all other entries are zero). - Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006
Binomial(n-3,k-1) counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 132 and k descents. Binomial(n-3,k-1) also counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 213 and k descents. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
Inverse of A130595 (as an infinite lower triangular matrix). - Philippe Deléham, Aug 21 2007
Consider integer lists LL of lists L of the form LL = [m#L] = [m#[k#2]] (where '#' means 'times') like LL(m=3,k=3) = [[2,2,2],[2,2,2],[2,2,2]]. The number of the integer list partitions of LL(m,k) is equal to binomial(m+k,k) if multiple partitions like [[1,1],[2],[2]] and [[2],[2],[1,1]] and [[2],[1,1],[2]] are counted only once. For the example, we find 4*5*6/3! = 20 = binomial(6,3). - Thomas Wieder, Oct 03 2007
The infinitesimal generator for Pascal's triangle and its inverse is A132440. - Tom Copeland, Nov 15 2007
Row n>=2 gives the number of k-digit (k>0) base n numbers with strictly decreasing digits; e.g., row 10 for A009995. Similarly, row n-1>=2 gives the number of k-digit (k>1) base n numbers with strictly increasing digits; see A009993 and compare A118629. - Rick L. Shepherd, Nov 25 2007
From Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008: (Start)
Binomial(n+k-1, k) is the number of ways a sequence of length k can be partitioned into n subsequences (see the Naish link).
Binomial(n+k-1, k) is also the number of n- (or fewer) digit numbers written in radix at least k whose digits sum to k. For example, in decimal, there are binomial(3+3-1,3)=10 3-digit numbers whose digits sum to 3 (see A052217) and also binomial(4+2-1,2)=10 4-digit numbers whose digits sum to 2 (see A052216). This relationship can be used to generate the numbers of sequences A052216 to A052224 (and further sequences using radix greater than 10). (End)
From Milan Janjic, May 07 2008: (Start)
Denote by sigma_k(x_1,x_2,...,x_n) the elementary symmetric polynomials. Then:
Binomial(2n+1,2k+1) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2(i*Pi/(2n+1)), (i=1,2,...,n).
Binomial(2n,2k+1) = 2n*sigma_{n-1-k}(x_1,x_2,...,x_{n-1}), where x_i = tan^2(i*Pi/(2n)), (i=1,2,...,n-1).
Binomial(2n,2k) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n)), (i=1,2,...,n).
Binomial(2n+1,2k) = (2n+1)sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n+2)), (i=1,2,...,n). (End)
Given matrices R and S with R(n,k) = binomial(n,k)*r(n-k) and S(n,k) = binomial(n,k)*s(n-k), then R*S = T where T(n,k) = binomial(n,k)*[r(.)+s(.)]^(n-k), umbrally. And, the e.g.f.s for the row polynomials of R, S and T are, respectively, exp(x*t)*exp[r(.)*x], exp(x*t)*exp[s(.)*x] and exp(x*t)*exp[r(.)*x]*exp[s(.)*x] = exp{[t+r(.)+s(.)]*x}. The row polynomials are essentially Appell polynomials. See A132382 for an example. - Tom Copeland, Aug 21 2008
As the rectangle R(m,n) = binomial(m+n-2,m-1), the weight array W (defined generally at A144112) of R is essentially R itself, in the sense that if row 1 and column 1 of W=A144225 are deleted, the remaining array is R. - Clark Kimberling, Sep 15 2008
If A007318 = M as an infinite lower triangular matrix, M^n gives A130595, A023531, A007318, A038207, A027465, A038231, A038243, A038255, A027466, A038279, A038291, A038303, A038315, A038327, A133371, A147716, A027467 for n=-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 respectively. - Philippe Deléham, Nov 11 2008
The coefficients of the polynomials with e.g.f. exp(x*t)*(cosh(t)+sinh(t)). - Peter Luschny, Jul 09 2009
The triangle or chess sums, see A180662 for their definitions, link Pascal's triangle with twenty different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 - Kn110 have been added. It is remarkable that all knight sums are related to the Fibonacci numbers, i.e., A000045, but none of the others. - Johannes W. Meijer, Sep 22 2010
Binomial(n,k) is also the number of ways to distribute n+1 balls into k+1 urns so that each urn gets at least one ball. See example in the example section below. - Dennis P. Walsh, Jan 29 2011
Binomial(n,k) is the number of increasing functions from {1,...,k} to {1,...,n} since there are binomial(n,k) ways to choose the k distinct, ordered elements of the range from the codomain {1,...,n}. See example in the example section below. - Dennis P. Walsh, Apr 07 2011
Central binomial coefficients: T(2*n,n) = A000984(n), T(n, floor(n/2)) = A001405(n). - Reinhard Zumkeller, Nov 09 2011
Binomial(n,k) is the number of subsets of {1,...,n+1} with k+1 as median element. To see this, note that Sum_{j=0..min(k,n-k)}binomial(k,j)*binomial(n-k,j) = binomial(n,k). See example in Example section below. - Dennis P. Walsh, Dec 15 2011
This is the coordinator triangle for the lattice Z^n, see Conway-Sloane, 1997. - N. J. A. Sloane, Jan 17 2012
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A046521 and A068555. For real r >= 0, C_r(n,k) := floor(r*n)!/(floor(r*k)!*floor(r*(n-k))!) is integral. See A211226 for the case r = 1/2. - Peter Bala, Apr 10 2012
Define a finite triangle T(m,k) with n rows such that T(m,0) = 1 is the left column, T(m,m) = binomial(n-1,m) is the right column, and the other entries are T(m,k) = T(m-1,k-1) + T(m-1,k) as in Pascal's triangle. The sum of all entries in T (there are A000217(n) elements) is 3^(n-1). - J. M. Bergot, Oct 01 2012
The lower triangular Pascal matrix serves as a representation of the operator exp(RLR) in a basis composed of a sequence of polynomials p_n(x) characterized by ladder operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x). See A132440, A218272, A218234, A097805, and A038207. The transposed and padded Pascal matrices can be associated to the special linear group SL2. - Tom Copeland, Oct 25 2012
See A193242. - Alexander R. Povolotsky, Feb 05 2013
A permutation p_1...p_n of the set {1,...,n} has a descent at position i if p_i > p_(i+1). Let S(n) denote the subset of permutations p_1...p_n of {1,...,n} such that p_(i+1) - p_i <= 1 for i = 1,...,n-1. Then binomial(n,k) gives the number of permutations in S(n+1) with k descents. Alternatively, binomial(n,k) gives the number of permutations in S(n+1) with k+1 increasing runs. - Peter Bala, Mar 24 2013
Sum_{n=>0} binomial(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where binomial(n,k) = 0. Also Sum_{n>=0} binomial(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
The square n X n submatrix (first n rows and n columns) of the Pascal matrix P(x) defined in the formulas below when multiplying on the left the Vandermonde matrix V(x_1,...,x_n) (with ones in the first row) translates the matrix to V(x_1+x,...,x_n+x) while leaving the determinant invariant. - Tom Copeland, May 19 2014
For k>=2, n>=k,  k/((k/(k-1) - Sum_{n=k..m} 1/binomial(n,k))) = m!/((m-k+1)!*(k-2)!). Note: k/(k-1) is the infinite sum. See A000217, A000292, A000332 for examples. - Richard R. Forberg, Aug 12 2014
Let G_(2n) be the subgroup of the symmetric group S_(2n) defined by G_(2n) = {p in S_(2n) | p(i) = i (mod n) for i = 1,2,...,2n}. G_(2n) has order 2^n. Binomial(n,k) gives the number of permutations in G_(2n) having n + k cycles. Cf. A130534 and A246117. - Peter Bala, Aug 15 2014
C(n,k) = the number of Dyck paths of semilength n+1, with k+1 "u"'s in odd numbered positions and k+1 returns to the x axis. Example: {U = u in odd position and  = return to x axis} binomial(3,0)=1 (Uudududd); binomial(3,1)=3 [(Uududd_Ud_), (Ud_Uududd_), (Uudd_Uudd_)]; binomial(3,2)=3 [(Ud_Ud_Uudd_), (Uudd_Ud_Ud_), (Ud_Uudd_Ud_)]; binomial(3,3)=1 (Ud_Ud_Ud_Ud_). - Roger Ford, Nov 05 2014
From Daniel Forgues, Mar 12 2015: (Start)
The binomial coefficients binomial(n,k) give the number of individuals of the k-th generation after n population doublings. For each doubling of population, each individual's clone has its generation index incremented by 1, and thus goes to the next row. Just tally up each row from 0 to 2^n - 1 to get the binomial coefficients.
       0   1       3               7                              15
    0: O | . | .   . | .   .   .   . | .   .   .   .   .   .   .   . |
    1:   | O | O   . | O   .   .   . | O   .   .   .   .   .   .   . |
    2:   |   |     O |     O   O   . |     O   O   .   O   .   .   . |
    3:   |   |       |             O |             O       O   O   . |
    4:   |   |       |               |                             O |
This is a fractal process: to get the pattern from 0 to 2^n - 1, append a shifted down (by one row) copy of the pattern from 0 to 2^(n-1) - 1 to the right of the pattern from 0 to 2^(n-1) - 1. (Inspired by the "binomial heap" data structure.)
Sequence of generation indices: 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n) (see A000120):
  {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, ...}
Binary expansion of 0 to 15:
  0   1   10   11  100  101  110  111 1000 1001 1010 1011 1100 1101 1111
(End)
A258993(n,k) = T(n+k,n-k), n > 0. - Reinhard Zumkeller, Jun 22 2015
T(n,k) is the number of set partitions w of [n+1] that avoid 1/2/3 with rb(w)=k. The same holds for ls(w)=k, where avoidance is in the sense of Klazar and ls,rb defined by Wachs and White.
Satisfies Benford's law [Diaconis, 1977] - N. J. A. Sloane, Feb 09 2017
Let {A(n)} be a set with exactly n identical elements, with {A(0)} being the empty set E. Let {A(n,k)} be the k-th iteration of {A(n)}, with {A(n,0)} = {A(n)}. {A(n,1)} = The set of all the subsets of A{(n)}, including {A(n)} and E. {A(n,k)} = The set of all subsets of {A(n,k-1)}, including all of the elements of {A(n,k-1)}. Let A(n,k) be the number of elements in {A(n,k)}. Then A(n,k) = C(n+k,k), with each successive iteration replicating the members of the k-th diagonal of Pascal's Triangle. See examples. - Gregory L. Simay, Aug 06 2018
Binomial(n-1,k) is also the number of permutations avoiding both 213 and 312 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n-1,k) is also the number of permutations avoiding both 132 and 213 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n,k) is the dimension of the k-th exterior power of a vector space of dimension n. - Stefano Spezia, Dec 22 2018
C(n,k-1) is the number of unoriented colorings of the facets (or vertices) of an n-dimensional simplex using exactly k colors. Each chiral pair is counted as one when enumerating unoriented arrangements. - Robert A. Russell, Oct 20 2020
From Dilcher and Stolarsky: "Two of the most ubiquitous objects in mathematics are the sequence of prime numbers and the binomial coefficients (and thus Pascal's triangle). A connection between the two is given by a well-known characterization of the prime numbers: Consider the entries in the k-th row of Pascal's triangle, without the initial and final entries. They are all divisible by k if and only if k is a prime." - Tom Copeland, May 17 2021
Named "Table de M. Pascal pour les combinaisons" by Pierre Remond de Montmort (1708) after the French mathematician, physicist and philosopher Blaise Pascal (1623-1662). - Amiram Eldar, Jun 11 2021
Consider the n-th diagonal of the triangle as a sequence b(n) with n starting at 0. From it form a new sequence by leaving the 0th term as is, and thereafter considering all compositions of n, taking the product of b(i) over the respective numbers i in each composition, adding terms corresponding to compositions with an even number of parts subtracting terms corresponding to compositions with an odd number of parts. Then the n-th row of the triangle is obtained, with every second term multiplied by -1, followed by infinitely many zeros. For sequences starting with 1, this operation is a special case of a self-inverse operation, and therefore the converse is true. - Thomas Anton, Jul 05 2021
C(n,k) is the number of vertices in an n-dimensional unit hypercube, at an L1 distance of k (or: with a shortest path of k 1d-edges) from a given vertex. - Eitan Y. Levine, May 01 2023
C(n+k-1,k-1) is the number of vertices at an L1 distance from a given vertex in an infinite-dimensional box, which has k sides of length 2^m for each m >= 0. Equivalently, given a set of tokens containing k distinguishable tokens with value 2^m for each m >= 0, C(n+k-1,k-1) is the number of subsets of tokens with a total value of n. - Eitan Y. Levine, Jun 11 2023
Numbers in the k-th column, i.e., numbers of the form C(n,k) for n >= k, are known as k-simplex numbers. - Pontus von Brömssen, Jun 26 2023
Let r(k) be the k-th row and c(k) the k-th column. Denote convolution by * and repeated convolution by ^. Then r(k)*r(m)=r(k+m) and c(k)*c(m)=c(k+m+1). This is because r(k) = r(1) ^ k and c(k) = c(0) ^ k+1. - Eitan Y. Levine, Jul 23 2023
Number of permutations of length n avoiding simultaneously the patterns 231 and 312(resp., 213 and 231; 213 and 312) with k descents (equivalently, with k ascents). An ascent (resp., descent) in a permutation a(1)a(2)...a(n) is position i such that a(i)a(i+1)). - Tian Han, Nov 25 2023
C(n,k) are generalized binomial coefficients of order m=0. Calculated by the formula C(n,k) = Sum_{i=0..n-k} binomial(n+1, n-k-i)*Stirling2(i+ m+ 1, i+1) *(-1)^i, where m = 0 for n>= 0, 0 <= k <= n. - Igor Victorovich Statsenko, Feb 26 2023
The Akiyama-Tanigawa algorithm applied to the diagonals, binomial(n+k,k), yields the powers of n. - Shel Kaphan, May 03 2024

			Triangle T(n,k) begins:
   n\k 0   1   2   3   4   5   6   7   8   9  10  11 ...
   0   1
   1   1   1
   2   1   2   1
   3   1   3   3   1
   4   1   4   6   4   1
   5   1   5  10  10   5   1
   6   1   6  15  20  15   6   1
   7   1   7  21  35  35  21   7   1
   8   1   8  28  56  70  56  28   8   1
   9   1   9  36  84 126 126  84  36   9   1
  10   1  10  45 120 210 252 210 120  45  10   1
  11   1  11  55 165 330 462 462 330 165  55  11   1
  ...
There are C(4,2)=6 ways to distribute 5 balls BBBBB, among 3 different urns, < > ( ) [ ], so that each urn gets at least one ball, namely, <BBB>(B)[B], <B>(BBB)[B], <B>(B)[BBB], <BB>(BB)[B], <BB>(B)[BB], and <B>(BB)[BB].
There are C(4,2)=6 increasing functions from {1,2} to {1,2,3,4}, namely, {(1,1),(2,2)},{(1,1),(2,3)}, {(1,1),(2,4)}, {(1,2),(2,3)}, {(1,2),(2,4)}, and {(1,3),(2,4)}. - _Dennis P. Walsh_, Apr 07 2011
There are C(4,2)=6 subsets of {1,2,3,4,5} with median element 3, namely, {3}, {1,3,4}, {1,3,5}, {2,3,4}, {2,3,5}, and {1,2,3,4,5}. - _Dennis P. Walsh_, Dec 15 2011
The successive k-iterations of {A(0)} = E are E;E;E;...; the corresponding number of elements are 1,1,1,... The successive k-iterations of {A(1)} = {a} are (omitting brackets) a;a,E; a,E,E;...; the corresponding number of elements are 1,2,3,... The successive k-iterations of {A(2)} = {a,a} are aa; aa,a,E; aa, a, E and a,E and E;...; the corresponding number of elements are 1,3,6,... - _Gregory L. Simay_, Aug 06 2018
Boas-Buck type recurrence for column k = 4: T(8, 4) = (5/4)*(1 + 5 + 15 + 35) = 70. See the Boas-Buck comment above. - _Wolfdieter Lang_, Nov 12 2018

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Equals differences between consecutive terms of A102363. - David G. Williams (davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
The triangle of the antidiagonals is A011973.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
Cf. A008277, A132311, A132312, A052216, A052217, A052218, A052219, A052220, A052221, A052222, A052223, A144225, A202750, A211226, A047999, A026729, A052553, A051920, A193242.
Triangle sums (see the comments): A000079 (Row1); A000007 (Row2); A000045 (Kn11 & Kn21); A000071 (Kn12 & Kn22); A001924 (Kn13 & Kn23); A014162 (Kn14 & Kn24); A014166 (Kn15 & Kn25); A053739 (Kn16 & Kn26); A053295 (Kn17 & Kn27); A053296 (Kn18 & Kn28); A053308 (Kn19 & Kn29); A053309 (Kn110 & Kn210); A001519 (Kn3 & Kn4); A011782 (Fi1 & Fi2); A000930 (Ca1 & Ca2); A052544 (Ca3 & Ca4); A003269 (Gi1 & Gi2); A055988 (Gi3 & Gi4); A034943 (Ze1 & Ze2); A005251 (Ze3 & Ze4). - Johannes W. Meijer, Sep 22 2010
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074, A228196, A228576.
Cf. A137948, A245334.
Cf. A085478, A258993.
Cf. A115940 (pandigital binomial coefficients C(m,k) with k>1).
Cf. (simplex colorings) A325002 (oriented), [k==n+1] (chiral), A325003 (achiral), A325000 (k or fewer colors), A325009 (orthotope facets, orthoplex vertices), A325017 (orthoplex facets, orthotope vertices).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 2..12: A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

-- (start)
)set expose add constructor OutputForm
pascal(0,n) == 1
pascal(n,n) == 1
pascal(i,j | 0 < i and i < j) == pascal(i-1,j-1) + pascal(i,j-1)
pascalRow(n) == [pascal(i,n) for i in 0..n]
displayRow(n) == output center blankSeparate pascalRow(n)
for i in 0..20 repeat displayRow i -- (end)

Flat(List([0..12],n->List([0..n],k->Binomial(n,k)))); # Stefano Spezia, Dec 22 2018

a007318 n k = a007318_tabl !! n !! k
a007318_row n = a007318_tabl !! n
a007318_list = concat a007318_tabl
a007318_tabl = iterate (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1]
-- Cf. http://www.haskell.org/haskellwiki/Blow_your_mind#Mathematical_sequences
-- Reinhard Zumkeller, Nov 09 2011, Oct 22 2010

/* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 29 2015

Maple
```
A007318 := (n,k)->binomial(n,k);
```

Flatten[Table[Binomial[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 19 2004 *)
Flatten[CoefficientList[CoefficientList[Series[1/(1 - x - x*y), {x, 0, 12}], x], y]] (* Mats Granvik, Jul 08 2014 *)

create_list(binomial(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

C(n,k)=binomial(n,k) \\ Charles R Greathouse IV, Jun 08 2011

# See Hobson link. Further programs:
from math import prod,factorial
def C(n,k): return prod(range(n,n-k,-1))//factorial(k) # M. F. Hasler, Dec 13 2019, updated Apr 29 2022, Feb 17 2023

from math import comb, isqrt
def A007318(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Nov 11 2024

def C(n,k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014

a(n, k) = C(n,k) = binomial(n, k).
C(n, k) = C(n-1, k) + C(n-1, k-1).
The triangle is symmetric: C(n,k) = C(n,n-k).
a(n+1, m) = a(n, m) + a(n, m-1), a(n, -1) := 0, a(n, m) := 0, nC(n, k) = n!/(k!(n-k)!) if 0<=k<=n, otherwise 0.
C(n, k) = ((n-k+1)/k) * C(n, k-1) with C(n, 0) = 1. - Michael B. Porter, Mar 23 2025
G.f.: 1/(1-y-x*y) = Sum_(C(n, k)*x^k*y^n, n, k>=0)
G.f.: 1/(1-x-y) = Sum_(C(n+k, k)*x^k*y^n, n, k>=0).
G.f. for row n: (1+x)^n = Sum_{k=0..n} C(n, k)*x^k.
G.f. for column k: x^k/(1-x)^(k+1); [corrected by Werner Schulte, Jun 15 2022].
E.g.f.: A(x, y) = exp(x+x*y).
E.g.f. for column n: x^n*exp(x)/n!.
In general the m-th power of A007318 is given by: T(0, 0) = 1, T(n, k) = T(n-1, k-1) + m*T(n-1, k), where n is the row-index and k is the column; also T(n, k) = m^(n-k)*C(n, k).
Triangle T(n, k) read by rows; given by A000007 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
Let P(n+1) = the number of integer partitions of (n+1); let p(i) = the number of parts of the i-th partition of (n+1); let d(i) = the number of different parts of the i-th partition of (n+1); let m(i, j) = multiplicity of the j-th part of the i-th partition of (n+1). Define the operator Sum_{i=1..P(n+1), p(i)=k+1} as the sum running from i=1 to i=P(n+1) but taking only partitions with p(i)=(k+1) parts into account. Define the operator Product_{j=1..d(i)} = product running from j=1 to j=d(i). Then C(n, k) = Sum_{p(i)=(k+1), i=1..P(n+1)} p(i)! / [Product_{j=1..d(i)} m(i, j)!]. E.g., C(5, 3) = 10 because n=6 has the following partitions with m=3 parts: (114), (123), (222). For their multiplicities one has: (114): 3!/(2!*1!) = 3; (123): 3!/(1!*1!*1!) = 6; (222): 3!/3! = 1. The sum is 3 + 6 + 1 = 10 = C(5, 3). - Thomas Wieder, Jun 03 2005
C(n, k) = Sum_{j=0..k} (-1)^j*C(n+1+j, k-j)*A000108(j). - Philippe Deléham, Oct 10 2005
G.f.: 1 + x*(1 + x) + x^3*(1 + x)^2 + x^6*(1 + x)^3 + ... . - Michael Somos, Sep 16 2006
Sum_{k=0..floor(n/2)} x^(n-k)*T(n-k,k) = A000007(n), A000045(n+1), A002605(n), A030195(n+1), A057087(n), A057088(n), A057089(n), A057090(n), A057091(n), A057092(n), A057093(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. Sum_{k=0..floor(n/2)} (-1)^k*x^(n-k)*T(n-k,k) = A000007(n), A010892(n), A009545(n+1), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n), A084329(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, respectively. - Philippe Deléham, Sep 16 2006
C(n,k) <= A062758(n) for n > 1. - Reinhard Zumkeller, Mar 04 2008
C(t+p-1, t) = Sum_{i=0..t} C(i+p-2, i) = Sum_{i=1..p} C(i+t-2, t-1). A binomial number is the sum of its left parent and all its right ancestors, which equals the sum of its right parent and all its left ancestors. - Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008
From Paul D. Hanna, Mar 24 2011: (Start)
Let A(x) = Sum_{n>=0} x^(n*(n+1)/2)*(1+x)^n be the g.f. of the flattened triangle:
A(x) = 1 + (x + x^2) + (x^3 + 2*x^4 + x^5) + (x^6 + 3*x^7 + 3*x^8 + x^9) + ...
then A(x) equals the series Sum_{n>=0} (1+x)^n*x^n*Product_{k=1..n} (1-(1+x)*x^(2*k-1))/(1-(1+x)*x^(2*k));
also, A(x) equals the continued fraction 1/(1- x*(1+x)/(1+ x*(1-x)*(1+x)/(1- x^3*(1+x)/(1+ x^2*(1-x^2)*(1+x)/(1- x^5*(1+x)/(1+ x^3*(1-x^3)*(1+x)/(1- x^7*(1+x)/(1+ x^4*(1-x^4)*(1+x)/(1- ...))))))))).
These formulas are due to (1) a q-series identity and (2) a partial elliptic theta function expression. (End)
For n > 0: T(n,k) = A029600(n,k) - A029635(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Row n of the triangle is the result of applying the ConvOffs transform to the first n terms of the natural numbers (1, 2, 3, ..., n). See A001263 or A214281 for a definition of this transformation. - Gary W. Adamson, Jul 12 2012
From L. Edson Jeffery, Aug 02 2012: (Start)
Row n (n >= 0) of the triangle is given by the n-th antidiagonal of the infinite matrix P^n, where P = (p_{i,j}), i,j >= 0, is the production matrix
  0, 1,
  1, 0, 1,
  0, 1, 0, 1,
  0, 0, 1, 0, 1,
  0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 1, 0, 1,
  ... (End)
Row n of the triangle is also given by the n+1 coefficients of the polynomial P_n(x) defined by the recurrence P_0(x) = 1, P_1(x) = x + 1, P_n(x) = x*P_{n-1}(x) + P_{n-2}(x), n > 1. - L. Edson Jeffery, Aug 12 2013
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
(1+x)^n = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*Sum_{i=0..k} k^(n-i)*binomial(k,i)*x^(n-i)/(n-i)!. - Vladimir Kruchinin, Oct 21 2013
E.g.f.: A(x,y) = exp(x+x*y) = 1 + (x+y*x)/( E(0)-(x+y*x)), where E(k) = 1 + (x+y*x)/(1 + (k+1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 08 2013
E.g.f.:  E(0) -1, where E(k) = 2 + x*(1+y)/(2*k+1 - x*(1+y)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
G.f.: 1 + x*(1+x)*(1+x^2*(1+x)/(W(0)-x^2-x^3)), where W(k) = 1 + (1+x)*x^(k+2) - (1+x)*x^(k+3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
Sum_{n>=0} C(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where C(n,k) = 0. Also Sum_{n>=0} C(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
Sum_{n>=k} 1/C(n,k) = k/(k-1) for k>=1. - Richard R. Forberg, Feb 10 2014
From Tom Copeland, Apr 26 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result by A007318(x) = P(x). Then with :xD:^n = x^n*(d/dx)^n and B(n,x), the Bell polynomials (A008277),
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x)
   with dP = A132440, M = A238385-I, and I = identity matrix, and
B) P(:xD:) = exp(dP:xD:) = exp[(e^M-I):xD:] = exp[M*B(.,:xD:)] = exp[M*xD] = (I+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x] (cf. also A238363).
C) P(x)^y = P(y*x). P(2x) = A038207(x) = exp[M*B(.,2x)], the face vectors of the n-dim hypercubes.
D) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I+dP)^x*[St1]
E) = [St1]^(-1)*(I+dP)^x*[St1] = [St2]*(I+dP)^x*[St2]^(-1)
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 and exp(x*M) = (I+dP)^x = Sum_{k>=0} C(x,k) dP^k. (End)
T(n,k) = A245334(n,k) / A137948(n,k), 0 <= k <= n. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, Dec 21 2014: (Start)
Recurrence equation: T(n,k) = T(n-1,k)*(n + k)/(n - k) - T(n-1,k-1) for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1. Note, changing the minus sign in the recurrence to a plus sign gives a recurrence for the square of the binomial coefficients - see A008459.
There is a relation between the e.g.f.'s of the rows and the diagonals of the triangle, namely, exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 3*x + 3*x^2/2! + x^3/3!) = 1 + 4*x + 10*x^2/2! + 20*x^3/3! + 35*x^4/4! + .... This property holds more generally for the Riordan arrays of the form ( f(x), x/(1 - x) ), where f(x) is an o.g.f. of the form 1 + f_1*x + f_2*x^2 + .... See, for example, A055248 and A106516.
Let P denote the present triangle. For k = 0,1,2,... define P(k) to be the lower unit triangular block array
/I_k 0\
\ 0  P/ having the k X k identity matrix I_k as the upper left block; in particular, P(0) = P. The infinite product P(0)*P(1)*P(2)*..., which is clearly well-defined, is equal to the triangle of Stirling numbers of the second kind A008277. The infinite product in the reverse order, that is, ...*P(2)*P(1)*P(0), is equal to the triangle of Stirling cycle numbers A130534. (End)
C(a+b,c) = Sum_{k=0..a} C(a,k)*C(b,b-c+k). This is a generalization of equation 1 from section 4.2.5 of the Prudnikov et al. reference, for a=b=c=n: C(2*n,n) = Sum_{k=0..n} C(n,k)^2. See Links section for animation of new formula. - Hermann Stamm-Wilbrandt, Aug 26 2015
The row polynomials of the Pascal matrix P(n,x) = (1+x)^n are related to the Bernoulli polynomials Br(n,x) and their umbral compositional inverses Bv(n,x) by the umbral relation P(n,x) = (-Br(.,-Bv(.,x)))^n = (-1)^n Br(n,-Bv(.,x)), which translates into the matrix relation P = M * Br * M * Bv, where P is the Pascal matrix, M is the diagonal matrix diag(1,-1,1,-1,...), Br is the matrix for the coefficients of the Bernoulli polynomials, and Bv that for the umbral inverse polynomials defined umbrally by Br(n,Bv(.,x)) = x^n = Bv(n,Br(.,x)). Note M = M^(-1). - Tom Copeland, Sep 05 2015
1/(1-x)^k = (r(x) * r(x^2) * r(x^4) * ...) where r(x) = (1+x)^k. - Gary W. Adamson, Oct 17 2016
Boas-Buck type recurrence for column k for Riordan arrays (see the Aug 10 2017 remark in A046521, also for the reference) with the Boas-Buck sequence b(n) = {repeat(1)}. T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} T(j, k), for n >= 1, with T(n, n) = 1. This reduces, with T(n, k) = binomial(n, k), to a known binomial identity (e.g, Graham et al. p. 161). - Wolfdieter Lang, Nov 12 2018
C((p-1)/a, b) == (-1)^b * fact_a(a*b-a+1)/fact_a(a*b) (mod p), where fact_n denotes the n-th multifactorial, a divides p-1, and the denominator of the fraction on the right side of the equation represents the modular inverse. - Isaac Saffold, Jan 07 2019
C(n,k-1) = A325002(n,k) - [k==n+1] = (A325002(n,k) + A325003(n,k)) / 2 = [k==n+1] + A325003(n,k). - Robert A. Russell, Oct 20 2020
From Hermann Stamm-Wilbrandt, May 13 2021: (Start)
Binomial sums are Fibonacci numbers A000045:
Sum_{k=0..n} C(n + k, 2*k + 1) = F(2*n).
Sum_{k=0..n} C(n + k, 2*k) = F(2*n + 1). (End)
C(n,k) = Sum_{i=0..k} A000108(i) * C(n-2i-1, k-i), for 0 <= k <= floor(n/2)-1. - Tushar Bansal, May 17 2025

Checked all links, deleted 8 that seemed lost forever and were probably not of great importance. - N. J. A. Sloane, May 08 2018

A002260 Triangle read by rows: T(n,k) = k for n >= 1, k = 1..n.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

Offset: 1

Angele Hamel (amh(AT)maths.soton.ac.uk)

Old name: integers 1 to k followed by integers 1 to k+1 etc. (a fractal sequence).
Start counting again and again.
This is a "doubly fractal sequence" - see the Franklin T. Adams-Watters link.
The PARI functions t1, t2 can be used to read a square array T(n,k) (n >= 1, k >= 1) by antidiagonals downwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002
Reading this sequence as the antidiagonals of a rectangular array, row n is (n,n,n,...); this is the weight array (Cf. A144112) of the array A127779 (rectangular). - Clark Kimberling, Sep 16 2008
The upper trim of an arbitrary fractal sequence s is s, but the lower trim of s, although a fractal sequence, need not be s itself. However, the lower trim of A002260 is A002260. (The upper trim of s is what remains after the first occurrence of each term is deleted; the lower trim of s is what remains after all 0's are deleted from the sequence s-1.) - Clark Kimberling, Nov 02 2009
Eigensequence of the triangle = A001710 starting (1, 3, 12, 60, 360, ...). - Gary W. Adamson, Aug 02 2010
The triangle sums, see A180662 for their definitions, link this triangle of natural numbers with twenty-three different sequences, see the crossrefs. The mirror image of this triangle is A004736. - Johannes W. Meijer, Sep 22 2010
A002260 is the self-fission of the polynomial sequence (q(n,x)), where q(n,x) = x^n + x^(n-1) + ... + x + 1. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements of the sequence A. Sequence A002260 is reluctant sequence of sequence 1,2,3,... (A000027). - Boris Putievskiy, Dec 12 2012
This is the maximal sequence of positive integers, such that once an integer k has occurred, the number of k's always exceeds the number of (k+1)'s for the remainder of the sequence, with the first occurrence of the integers being in order. - Franklin T. Adams-Watters, Oct 23 2013
A002260 are the k antidiagonal numerators of rationals in Cantor's proof of 1-to-1 correspondence between rationals and naturals; the denominators are k-numerator+1. - Adriano Caroli, Mar 24 2015
T(n,k) gives the distance to the largest triangular number < n. - Ctibor O. Zizka, Apr 09 2020

			First six rows:
  1
  1   2
  1   2   3
  1   2   3   4
  1   2   3   4   5
  1   2   3   4   5   6

Clark Kimberling, "Fractal sequences and interspersions," Ars Combinatoria 45 (1997) 157-168. (Introduces upper trimming, lower trimming, and signature sequences.)
M. Myers, Smarandache Crescendo Subsequences, R. H. Wilde, An Anthology in Memoriam, Bristol Banner Books, Bristol, 1998, p. 19.
F. Smarandache, Sequences of Numbers Involved in Unsolved Problems, Hexis, Phoenix, 2006.

N. J. A. Sloane, Table of n, a(n) for n = 1..11325
Franklin T. Adams-Watters, Doubly Fractal Sequences
Matin Amini and Majid Jahangiri, A Novel Proof for Kimberling's Conjecture on Doubly Fractal Sequences, arXiv:1612.09481 [math.NT], 2017.
Bruno Berselli, Illustration of the initial terms
Jerry Brown et al., Problem 4619, School Science and Mathematics (USA), Vol. 97(4), 1997, pp. 221-222.
Glen Joyce C. Dulatre, Jamilah V. Alarcon, Vhenedict M. Florida, and Daisy Ann A. Disu, On Fractal Sequences, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 109-113.
Clark Kimberling, Fractal sequences
Clark Kimberling, Numeration systems and fractal sequences, Acta Arithmetica 73 (1995) 103-117.
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
F. Smarandache, Sequences of Numbers Involved in Unsolved Problems.
Aaron Snook, Augmented Integer Linear Recurrences, 2012. - _N. J. A. Sloane_, Dec 19 2012
Michael Somos, Sequences used for indexing triangular or square arrays
Eric Weisstein's World of Mathematics, Smarandache Sequences
Eric Weisstein's World of Mathematics, Unit Fraction

Cf. A000217, A001710, A002262, A003056, A004736 (ordinal transform), A025581, A056534, A094727, A127779.
Cf. A140756 (alternating signs).
Triangle sums (see the comments): A000217 (Row1, Kn11); A004526 (Row2); A000096 (Kn12); A055998 (Kn13); A055999 (Kn14); A056000 (Kn15); A056115 (Kn16); A056119 (Kn17); A056121 (Kn18); A056126 (Kn19); A051942 (Kn110); A101859 (Kn111); A132754 (Kn112); A132755 (Kn113); A132756 (Kn114); A132757 (Kn115); A132758 (Kn116); A002620 (Kn21); A000290 (Kn3); A001840 (Ca2); A000326 (Ca3); A001972 (Gi2); A000384 (Gi3).
Cf. A108872.

a002260 n k = k
a002260_row n = [1..n]
a002260_tabl = iterate (\row -> map (+ 1) (0 : row)) [1]
-- Reinhard Zumkeller, Aug 04 2014, Jul 03 2012

at:=0; for n from 1 to 150 do for i from 1 to n do at:=at+1; lprint(at,i); od: od: # N. J. A. Sloane, Nov 01 2006
seq(seq(i,i=1..k),k=1..13); # Peter Luschny, Jul 06 2009

FoldList[{#1, #2} &, 1, Range[2, 13]] // Flatten (* Robert G. Wilson v, May 10 2011 *)
Flatten[Table[Range[n],{n,20}]] (* Harvey P. Dale, Jun 20 2013 *)

T(n,k):=sum((i+k)*binomial(i+k-1,i)*binomial(k,n-i-k+1)*(-1)^(n-i-k+1),i,max(0,n+1-2*k),n-k+1); /* Vladimir Kruchinin, Oct 18 2013 */

t1(n)=n-binomial(floor(1/2+sqrt(2*n)),2) /* this sequence */

A002260(n)=n-binomial((sqrtint(8*n)+1)\2,2) \\ M. F. Hasler, Mar 10 2014

from math import isqrt, comb
def A002260(n): return n-comb((m:=isqrt(k:=n<<1))+(k>m*(m+1)),2) # Chai Wah Wu, Nov 08 2024

a(n) = 1 + A002262(n).
n-th term is n - m*(m+1)/2 + 1, where m = floor((sqrt(8*n+1) - 1) / 2).
The above formula is for offset 0; for offset 1, use a(n) = n-m*(m+1)/2 where m = floor((-1+sqrt(8*n-7))/2). - Clark Kimberling, Jun 14 2011
a(k * (k + 1) / 2 + i) = i for k >= 0 and 0 < i <= k + 1. - Reinhard Zumkeller, Aug 14 2001
a(n) = (2*n + round(sqrt(2*n)) - round(sqrt(2*n))^2)/2. - Brian Tenneson, Oct 11 2003
a(n) = n - binomial(floor((1+sqrt(8*n))/2), 2). - Paul Barry, May 25 2004
T(n,k) = A001511(A118413(n,k)); T(n,k) = A003602(A118416(n,k)). - Reinhard Zumkeller, Apr 27 2006
a(A000217(n)) = A000217(n) - A000217(n-1), a(A000217(n-1) + 1) = 1, a(A000217(n) - 1) = A000217(n) - A000217(n-1) - 1. - Alexander R. Povolotsky, May 28 2008
a(A169581(n)) = A038566(n). - Reinhard Zumkeller, Dec 02 2009
T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n-k,n-i) (regarded as triangle, see the example). - Mircea Merca, Apr 11 2012
T(n,k) = Sum_{i=max(0,n+1-2*k)..n-k+1} (i+k)*binomial(i+k-1,i)*binomial(k,n-i-k+1)*(-1)^(n-i-k+1). - Vladimir Kruchinin, Oct 18 2013
G.f.: x*y / ((1 - x) * (1 - x*y)^2) = Sum_{n,k>0} T(n,k) * x^n * y^k. - Michael Somos, Sep 17 2014
a(n) = n - S(n) where S(n) = sum of distinct terms in {a(1), a(2), ..., a(n-1)}. - David James Sycamore, Mar 10 2025

More terms from Reinhard Zumkeller, Apr 27 2006
Incorrect program removed by Franklin T. Adams-Watters, Mar 19 2010
New name from Omar E. Pol, Jul 15 2012

A002024 k appears k times; a(n) = floor(sqrt(2n) + 1/2).

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13

Offset: 1

N. J. A. Sloane

Integer inverse function of the triangular numbers A000217. The function trinv(n) = floor((1+sqrt(1+8n))/2), n >= 0, gives the values 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, ..., that is, the same sequence with offset 0. - N. J. A. Sloane, Jun 21 2009
Array T(k,n) = n+k-1 read by antidiagonals.
Eigensequence of the triangle = A001563. - Gary W. Adamson, Dec 29 2008
Can apparently also be defined via a(n+1)=b(n) for n >= 2 where b(0)=b(1)=1 and b(n) = b(n-b(n-2))+1. Tested to be correct for all n <= 150000. - José María Grau Ribas, Jun 10 2011
For any n >= 0, a(n+1) is the least integer m such that A000217(m)=m(m+1)/2 is larger than n. This is useful when enumerating representations of n as difference of triangular numbers; see also A234813. - M. F. Hasler, Apr 19 2014
Number of binary digits of A023758, i.e., a(n) = ceiling(log_2(A023758(n+2))). - Andres Cicuttin, Apr 29 2016
a(n) and A002260(n) give respectively the x(n) and y(n) coordinates of the sorted sequence of points in the integer lattice such that x(n) > 0, 0 < y(n) <= x(n), and min(x(n), y(n)) < max(x(n+1), y(n+1)) for n > 0. - Andres Cicuttin, Dec 25 2016
Partial sums (A060432) are given by S(n) = (-a(n)^3 + a(n)*(1+6n))/6. - Daniel Cieslinski, Oct 23 2017
As an array, T(k,n) is the number of digits columns used in carryless multiplication between a k-digit number and an n-digit number. - Stefano Spezia, Sep 24 2022
a(n) is the maximum number of possible solutions to an n-statement Knights and Knaves Puzzle, where each statement is of the form "x of us are knights" for some 1 <= x <= n, knights can only tell the truth and knaves can only lie. - Taisha Charles and Brittany Ohlinger, Jul 29 2023

			From _Clark Kimberling_, Sep 16 2008: (Start)
As a rectangular array, a northwest corner:
  1 2 3 4 5 6
  2 3 4 5 6 7
  3 4 5 6 7 8
  4 5 6 7 8 9
This is the weight array (cf. A144112) of A107985 (formatted as a rectangular array). (End)
G.f. = x + 2*x^2 + 2*x^3 + 3*x^4 + 3*x^5 + 3*x^6 + 4*x^7 + 4*x^9 + 4*x^9 + 4*x^10 + ...

Edward S. Barbeau, Murray S. Klamkin, and William O. J. Moser, Five Hundred Mathematical Challenges, Prob. 441, pp. 41, 194. MAA 1995.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 97.
K. Hardy and K. S. Williams, The Green Book of Mathematical Problems, p. 59, Solution to Prob. 14, Dover NY, 1985
R. Honsberger, Mathematical Morsels, pp. 133-134, MAA 1978.
J. F. Hurley, Litton's Problematical Recreations, pp. 152; 313-4 Prob. 22, VNR Co., NY, 1971.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 43.
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).
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 129.

Franklin T. Adams-Watters, Table of n, a(n) for n = 1..5050
Jaegug Bae and Sungjin Choi, A generalization of a subset-sum-distinct sequence, J. Korean Math. Soc. 40 (2003), no. 5, 757--768. MR1996839 (2004d:05198). See b(n).
Jonathan H. B. Deane and Guido Gentile, A diluted version of the problem of the existence of the Hofstadter sequence, arXiv:2311.13854 [math.NT], 2023. See p. 10.
Nathan Fox, Connecting Slow Solutions to Nested Recurrences with Linear Recurrent Sequences, arXiv:2203.09340 [math.CO], 2022.
H. T. Freitag and H. W. Gould, Solution to Problem 571, Math. Mag., 38 (1965), 185-187.
H. T. Freitag (Proposer) and H. W. Gould (Solver), Problem 571: An Ordering of the Rationals, Math. Mag., 38 (1965), 185-187 [Annotated scanned copy]
Mikel Garcia-de-Andoin, Álvaro Saiz, Pedro Pérez-Fernández, Lucas Lamata, Izaskun Oregi, and Mikel Sanz, Digital-Analog Quantum Computation with Arbitrary Two-Body Hamiltonians, arXiv:2307.00966 [quant-ph], 2023.
S. W. Golomb, Discrete chaos: sequences satisfying "strange" recursions, unpublished manuscript, circa 1990 [cached copy, with permission (annotated)]
Henry W. Gould, Letters to N. J. A. Sloane, Oct 1973 and Jan 1974.
Abraham Isgur, Vitaly Kuznetsov, and Stephen Tanny, A combinatorial approach for solving certain nested recursions with non-slow solutions, arXiv:1202.0276 [math.CO], 2012.
Stanley Wu-Wei Liu, Closed form expressions for A002024(n).
M. A. Nyblom, Some curious sequences involving floor and ceiling functions, Am. Math. Monthly 109 (#6, 200), 559-564.
Boris Putievskiy, Transformations [Of] Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.
Raphael Schumacher, Extension of Summation Formulas involving Stirling series, arXiv:1605.09204 [math.NT], 2016.
Raphael Schumacher, The self-counting identity, Fib. Quart., 55 (No. 2 2017), 157-167.
Raphael Schumacher, The Self-Counting Flow, Fibonacci Quart. 60 (2022), no. 5, 324-343.
N. J. A. Sloane, Handwritten notes on Self-Generating Sequences, 1970. (note that A1148 has now become A005282)
Michael Somos, Sequences used for indexing triangular or square arrays.
L. J. Upton, Letter to N. J. A. Sloane, May 22 1991.
Eric Weisstein's World of Mathematics, Self-Counting Sequence.
Index entries for Hofstadter-type sequences

a(n+1) = 1+A003056(n), A022846(n)=a(n^2), a(n+1)=A002260(n)+A025581(n).
Cf. A001462, A002262, A003881, A004736, A127899, A107985, A001563, A014132, A000194, A005145, A131507, A093995, A060432 (partial sums).
A123578 is an essentially identical sequence.

a002024 n k = a002024_tabl !! (n-1) !! (k-1)
a002024_row n = a002024_tabl !! (n-1)
a002024_tabl = iterate (\xs@(x:_) -> map (+ 1) (x : xs)) [1]
a002024_list = concat a002024_tabl
a002024' = round . sqrt . (* 2) . fromIntegral
-- Reinhard Zumkeller, Jul 05 2015, Feb 12 2012, Mar 18 2011

a002024_list = [1..] >>= \n -> replicate n n

a002024 = (!!) $ [1..] >>= \n -> replicate n n
-- Sascha Mücke, May 10 2016

[Floor(Sqrt(2*n) + 1/2): n in [1..80]]; // Vincenzo Librandi, Nov 19 2014

A002024 := n-> ceil((sqrt(1+8*n)-1)/2); seq(A002024(n), n=1..100);

a[1] = 1; a[n_] := a[n] = a[n - a[n - 1]] + 1 (* Branko Curgus, May 12 2009 *)
Table[n, {n, 13}, {n}] // Flatten (* Robert G. Wilson v, May 11 2010 *)
Table[PadRight[{},n,n],{n,15}]//Flatten (* Harvey P. Dale, Jan 13 2019 *)

t1(n)=floor(1/2+sqrt(2*n)) /* A002024 = this sequence */

t2(n)=n-binomial(floor(1/2+sqrt(2*n)),2) /* A002260(n-1) */

t3(n)=binomial(floor(3/2+sqrt(2*n)),2)-n+1 /* A004736 */

t4(n)=n-1-binomial(floor(1/2+sqrt(2*n)),2) /* A002260(n-1)-1 */

A002024(n)=(sqrtint(n*8)+1)\2 \\ M. F. Hasler, Apr 19 2014

PARI
```
a(n)=(sqrtint(8*n-7)+1)\2
```

a(n)=my(k=1);while(binomial(k+1,2)+1<=n,k++);k \\ R. J. Cano, Mar 17 2014

from math import isqrt
def A002024(n): return (isqrt(8*n)+1)//2 # Chai Wah Wu, Feb 02 2022

[floor(sqrt(2*n) +1/2) for n in (1..80)] # G. C. Greubel, Dec 10 2018

a(n) = floor(1/2 + sqrt(2n)). Also a(n) = ceiling((sqrt(1+8n)-1)/2). [See the Liu link for a large collection of explicit formulas. - N. J. A. Sloane, Oct 30 2019]
a((k-1)*k/2 + i) = k for k > 0 and 0 < i <= k. - Reinhard Zumkeller, Aug 30 2001
a(n) = a(n - a(n-1)) + 1, with a(1)=1. - Ian M. Levitt (ilevitt(AT)duke.poly.edu), Aug 18 2002
a(n) = round(sqrt(2n)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002
T(n,k) = A003602(A118413(n,k)); = T(n,k) = A001511(A118416(n,k)). - Reinhard Zumkeller, Apr 27 2006
G.f.: (x/(1-x))*Product_{k>0} (1-x^(2*k))/(1-x^(2*k-1)). - Vladeta Jovovic, Oct 06 2003
Equals A127899 * A004736. - Gary W. Adamson, Feb 09 2007
Sum_{i=1..n} Sum_{j=i..n+i-1} T(j,i) = A000578(n); Sum_{i=1..n} T(n,i) = A000290(n). - Reinhard Zumkeller, Jun 24 2007
a(n) + n = A014132(n). - Vincenzo Librandi, Jul 08 2010
a(n) = ceiling(-1/2 + sqrt(2n)). - Branko Curgus, May 12 2009
a(A169581(n)) = A038567(n). - Reinhard Zumkeller, Dec 02 2009
a(n) = round(sqrt(2*n)) = round(sqrt(2*n-1)); there exist a and b greater than zero such that 2*n = 2+(a+b)^2 -(a+3*b) and a(n)=(a+b-1). - Fabio Civolani (civox(AT)tiscali.it), Feb 23 2010
A005318(n+1) = 2*A005318(n) - A205744(n), A205744(n) = A005318(A083920(n)), A083920(n) = n - a(n). - N. J. A. Sloane, Feb 11 2012
Expansion of psi(x) * x / (1 - x) in powers of x where psi() is a Ramanujan theta function. - Michael Somos, Mar 19 2014
G.f.: (x/(1-x)) * Product_{n>=1} (1 + x^n) * (1 - x^(2*n)). - Paul D. Hanna, Feb 27 2016
a(n) = 1 + Sum_{i=1..n/2} ceiling(floor(2(n-1)/(i^2+i))/(2n)). - José de Jesús Camacho Medina, Jan 07 2017
a(n) = floor((sqrt(8*n-7)+1)/2). - Néstor Jofré, Apr 24 2017
a(n) = floor((A000196(8*n)+1)/2). - Pontus von Brömssen, Dec 10 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/4 (A003881). - Amiram Eldar, Oct 01 2022
G.f. as array: (x^2*(1 - y)^2 + y^2 + x*y*(1 - 2*y))/((1 - x)^2*(1 - y)^2). - Stefano Spezia, Apr 22 2024

A051340 A simple 2-dimensional array, read by antidiagonals: T[i,j] = 1 for j>0, T[i,0] = i+1; i,j = 0,1,2,3,...

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13

Offset: 0

N. J. A. Sloane

Warning: contributions from Kimberling refer to an alternate version indexed by 1 instead of 0. Other contributors (Adamson in A125026/A130301/A130295) refer to this considering the upper right triangle set to zero, T[i,j]=0 for j>i. - M. F. Hasler, Aug 15 2015
From Clark Kimberling, Feb 05 2011: (Start)
A member of the accumulation chain:
... < A051340 < A141419 < A185874 < A185875 < A185876 < ...
(See A144112 for the definition of accumulation array.)
In the m-th accumulation array of A051340,
row_1 = C(m,1) and column_1 = C(1,m+1), for m>=0. (End)

			Northwest corner:
  1...1...1...1...1...1...1
  2...1...1...1...1...1...1
  3...1...1...1...1...1...1
  4...1...1...1...1...1...1
  5...1...1...1...1...1...1
  6...1...1...1...1...1...1
The Mathematica code shows that the weight array of this array (i.e., the array of which this array is the accumulation array), has northwest corner
  1....0...0...0...0...0...0
  1...-1...0...0...0...0...0
  1...-1...0...0...0...0...0
  1...-1...0...0...0...0...0
  1...-1...0...0...0...0...0. - _Clark Kimberling_, Feb 05 2011

G. C. Greubel, Antidiagonals n = 0..100, flattened
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 24.
A. V. Mikhalev and A. A. Nechaev, Linear recurring sequences over modules, Acta Applic. Math., 42 (1996), 161-202.

Cf. A144112, A141419, A185874, A185875, A185876.

[k eq n select n+1 else 1: k in [0..n], n in [0..15]]; // G. C. Greubel, Mar 18 2023

A051340 := proc(n, k) if k=0 then n+1; else 1; end if; end proc: # R. J. Mathar, Jul 16 2015

(* This program generates A051340, then its accumulation array A141419, then its weight array described under Example. *)
f[n_,0]:=0; f[0,k_]:=0;  (* needed for the weight array *)
f[n_,1]:=n; f[n_,k_]:=1/;k>1;
TableForm[Table[f[n,k],{n,1,10},{k,1,15}]] (* A051340 *)
Table[f[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten
s[n_,k_]:=Sum[f[i,j],{i,1,n},{j,1,k}]; (* accumulation array of {f(n,k)} *)
TableForm[Table[s[n,k],{n,1,10},{k,1,15}]] (* A141419 *)
Table[s[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten
w[m_,n_]:=f[m,n]+f[m-1,n-1]-f[m,n-1]-f[m-1,n]/;Or[m>0,n>0];
TableForm[Table[w[n,k],{n,1,10},{k,1,15}]] (* weight array *)
Table[w[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten (* Clark Kimberling, Feb 05 2011 *)
f[n_] := Join[ Table[1, {n - 1}], {n}]; Array[ f, 14] // Flatten (* Robert G. Wilson v, Mar 04 2012 *)
Table[PadLeft[{n},n,1],{n,15}]//Flatten (* Harvey P. Dale, Jun 17 2025 *)

from math import comb, isqrt
def A051340(n):
    a = (m:=isqrt(k:=n+2<<1))+(k>m*(m+1))
    return 1 if n-comb(a,2)+1 else a-1 # Chai Wah Wu, Jun 21 2025

def A051340(n,k): return n+1 if (k==n) else 1
flatten([[A051340(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Mar 18 2023

For n>0, a(n(n+3)/2)=n+1, and if k is not of the form n*(n+3)/2, then a(k)=1. - Benoit Cloitre, Oct 31 2002, corrected by M. F. Hasler, Aug 15 2015

T(n,0) = n+1 and T(n,k) = 1 if k >= 0, for n >= 0. - Clark Kimberling, Feb 05 2011

Edited by M. F. Hasler, Aug 15 2015

A086270 Rectangular array T(k,n) of polygonal numbers, by antidiagonals.

Original entry on oeis.org

1, 3, 1, 6, 4, 1, 10, 9, 5, 1, 15, 16, 12, 6, 1, 21, 25, 22, 15, 7, 1, 28, 36, 35, 28, 18, 8, 1, 36, 49, 51, 45, 34, 21, 9, 1, 45, 64, 70, 66, 55, 40, 24, 10, 1, 55, 81, 92, 91, 81, 65, 46, 27, 11, 1, 66, 100, 117, 120, 112, 96, 75, 52, 30, 12, 1, 78, 121, 145, 153, 148, 133, 111

Offset: 1

Clark Kimberling, Jul 14 2003

The antidiagonal sums 1, 4, 11, 25, 50, ... are the numbers A006522(n) for n >= 3.
This is the accumulation array (cf. A144112) of A144257 (which is the weight array of this sequence). - Clark Kimberling, Sep 16 2008
By rows, the sequence beginning (1, N, ...) is the binomial transform of (1, (N-1), (N-2), 0, 0, 0, ...); and is the second partial sum of (1, (N-2), (N-2), (N-2), ...). Example: The sequence (1, 4, 9, 16, 25, ...) is the binomial transform of (1, 3, 2, 0, 0, 0, ...) and the second partial sum of (1, 2, 2, 2, ...). - Gary W. Adamson, Aug 23 2015

			First 6 rows:
=========================================
n\k|  1   2    3    4    5    6     7
---|-------------------------------------
1  |  1   3    6   10   15   21    28 ... (A000217, triangular numbers)
2  |  1   4    9   16   25   36    49 ... (A000290, squares)
3  |  1   5   12   22   35   51    70 ... (A000326, pentagonal numbers)
4  |  1   6   15   28   45   66    91 ... (A000384, hexagonal numbers)
5  |  1   7   18   34   55   81   112 ... (A000566, heptagonal numbers)
6  |  1   8   21   40   65   96   133 ... (A000567, octagonal numbers)
...
The array formed by the complements: A183225.

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 76 at p. 189.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 123.

Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7, 2004.
Wikipedia, Polygonal number: Table of values.

Cf. A006522, A057145, A086271, A086272, A086273, A139601, A183225.
Cf. A000217, A000290, A000326, A000384, A000566, A000567.
Cf. A114112, A144257 .

T:=func; [T(k,n-k+1): k in [1..n], n in [1..12]]; // Bruno Berselli, Dec 19 2014

t[n_, k_] := n*Binomial[k, 2] + k; Table[ t[k, n - k + 1], {n, 12}, {k, n}] // Flatten

T(n, k) = n*binomial(k, 2) + k = A057145(n+2,k).
2*T(n, k) = T(n+r, k) + T(n-r, k), where r = 0, 1, 2, 3, ..., n-1 (see table in Example field). - Bruno Berselli, Dec 19 2014
From Stefano Spezia, Sep 02 2022: (Start)
G.f.: x*y*(1 - x + x*y)/((1 - x)^2*(1 - y)^3).
G.f. of k-th column: k*(1 + k - 2*x)*x/(2*(1 - x)^2). (End)

Extended by Clark Kimberling, Jan 01 2011

A185787 Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

Original entry on oeis.org

1, 7, 25, 62, 125, 221, 357, 540, 777, 1075, 1441, 1882, 2405, 3017, 3725, 4536, 5457, 6495, 7657, 8950, 10381, 11957, 13685, 15572, 17625, 19851, 22257, 24850, 27637, 30625, 33821, 37232, 40865, 44727, 48825, 53166, 57757, 62605, 67717, 73100, 78761, 84707, 90945, 97482, 104325, 111481, 118957, 126760, 134897, 143375

Offset: 1

Clark Kimberling, Feb 03 2011

This is one of many interesting sequences and arrays that stem from the natural number array A000027, of which a northwest corner is as follows:
  1....2.....4.....7...11...16...22...29...
  3....5.....8....12...17...23...30...38...
  6....9....13....18...24...31...39...48...
  10...14...19....25...32...40...49...59...
  15...20...26....33...41...50...60...71...
  21...27...34....42...51...61...72...84...
  28...35...43....52...62...73...85...98...
Blocking out all terms below the main diagonal leaves columns whose sums comprise A185787.  Deleting the main diagonal and then summing give A185787.  Analogous treatments to the left of the main diagonal give A100182 and A101165.  Further sequences obtained directly from this array are easily obtained using the following formula for the array: T(n,k)=n+(n+k-2)(n+k-1)/2.
Examples:
row 1:  A000124
row 2:  A022856
row 3:  A016028
row 4:  A145018
row 5:  A077169
col 1:  A000217
col 2:  A000096
col 3:  A034856
col 4:  A055998
col 5:  A046691
col 6:  A052905
col 7:  A055999
diag. (1,5,...) ...... A001844
diag. (2,8,...) ...... A001105
diag. (4,12,...)...... A046092
diag. (7,17,...)...... A056220
diag. (11,23,...) .... A132209
diag. (16,30,...) .... A054000
diag. (22,38,...) .... A090288
diag. (3,9,...) ...... A058331
diag. (6,14,...) ..... A051890
diag. (10,20,...) .... A005893
diag. (15,27,...) .... A097080
diag. (21,35,...) .... A093328
antidiagonal sums:  (1,5,15,34,...)=A006003=partial sums of A002817.
Let S(n,k) denote the n-th partial sum of column k.  Then
S(n,k)=n*(n^2+3k*n+3*k^2-6*k+5)/6.
S(n,1)=n(n+1)(n+2)/6
S(n,2)=n(n+1)(n+5)/6
S(n,3)=n(n+2)(n+7)/6
S(n,4)=n(n^2+12n+29)/6
S(n,5)=n(n+5)(n+10)/6
S(n,6)=n(n+7)(n+11)/6
S(n,7)=n(n+10)(n+11)/6
Weight array of T: A144112
Accumulation array of T: A185506
Second rectangular sum array of T: A185507
Third rectangular sum array of T: A185508
Fourth rectangular sum array of T: A185509

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000027, A185788, A100182, A101165, A079824.

[n*(7*n^2-6*n+5)/6: n in [1..50]]; // Vincenzo Librandi, Jul 04 2012

f[n_,k_]:=n+(n+k-2)(n+k-1)/2;
s[k_]:=Sum[f[n,k],{n,1,k}];
Factor[s[k]]
Table[s[k],{k,1,70}]  (* A185787 *)
CoefficientList[Series[(3*x^2+3*x+1)/(1-x)^4,{x,0,50}],x] (* Vincenzo Librandi, Jul 04 2012 *)

a(n)=n*(7*n^2-6*n+5)/6.

G.f.: x*(3*x^2+3*x+1)/(1-x)^4. - Vincenzo Librandi, Jul 04 2012

Edited by Clark Kimberling, Feb 25 2023

A380661 Rectangular array neg(i,j,1,2) read by descending antidiagonals: pos() and neg() denote the positive part and negative part of a determinant; see Comments.

Original entry on oeis.org

6, 20, 30, 56, 72, 90, 132, 156, 182, 210, 272, 306, 342, 380, 420, 506, 552, 600, 650, 702, 756, 870, 930, 992, 1056, 1122, 1190, 1260, 1406, 1482, 1560, 1640, 1722, 1806, 1892, 1980, 2162, 2256, 2352, 2450, 2550, 2652, 2756, 2862, 2970, 3192, 3306, 3422

Offset: 1

Clark Kimberling, Feb 04 2025

Suppose that (m(i,j)) is a rectangular array of infinitely many rows and infinitely many columns. For integers s>=1 and n>=1, let M(i,j,s,n) be the nXn matrix (m(i+h*s,j+k*s)), where h=0..n-1, k=0..n-1.
Let D(i,j,s,n) and P(i,j,s,n) denote the determinant and permanent of M(i,j,s,n), respectively. Define arrays pos(i,j,s,n) and neg(i,j,s,n) by pos(i,j,s,n) = (P(i,j,s,n)+D(i,j,s,n))/2 and neg(i,j,s,n) = (P(i,j,s,n)-D(i,j,s,n))/2, so that P(i,j,s,n) = pos(i,j,s,n)+neg(i,j,s,n) and D(i,j,s,n) = pos(i,j,s,n)-neg(i,j,s,n).
A definition of determinant of an nXn matrix (a(i,j)) is the sum of the products (-1)^p(u) a(1,j(1))*a(2,j(2))*...*a(n,j(n)) over the n! permutations u = (j(1),j(2),...,j(n)) of (1,2,...,n), where p(u) is the parity of u; i.e., p(u) = 0 or 1 according as u is an even or odd permutation; see Lang, pp. 452-3, especially Proposition 4.8.
We have:
  pos(i,j,s,n) is the sum of the n!/2 products for which p(u) = 0, and
  neg(i,j,s,n) is the sum of the n!/2 products for which p(u) = 1.
Here, the foundational array (m(i,j)) is the natural number array (see A000027, A185787, A144112). The row sequences of pos(i,j,s,n) and neg(i,j,s,n) are linearly recurrent with signature (5, -10, 10, -5, 1).

			Corner of neg(i,j,1,2):
     6     20    56     132    272    506    870   1406   2162   3192
    30     72    156    306    552    930   1482   2256   3306   4692
    90    182    342    600    992   1560   2352   3422   4830   6642
   210    380    650   1056   1640   2450   3540   4970   6806   9120
   420    702   1122   1722   2550   3660   5112   6972   9312  12210
   756   1190   1806   2652   3782   5256   7140   9506  12432  16002
  1260   1892   2756   3906   5402   7310   9702  12656  16256  20592
  1980   2862   4032   5550   7482   9900  12882  16512  20880  26082
  2970   4160   5700   7656  10100  13110  16770  21170  26406  32580
  4290   5852   7832  10302  13340  17030  21462  26732  32942  40200
  6006   8010  10506  13572  17292  21756  27060  33306  40602  49062
  8190  10712  13806  17556  22052  27390  33672  41006  49506  59292
M(1,1,1,2) is the matrix with (row 1) = (1,2), (row 2) =(3,5), so that
pos(1,1,1,2) = 1*5 = 5; neg(1,1,1,2) = 2*3 = 6; D(1,1,1,2) = -1; P(1,1,1,2) = 11.

S. Lang, Algebra, 2nd ed., Addison-Wesley, 1984, 452-453.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000027, A380649, A380660.

s = 1; n = 2; z = 12;
r[n_, k_] := n + (n + k - 2)*(n + k - 1)/2 (* Array A000027 *)
Grid[Table[r[n, k], {n, 1, z}, {k, 1, z}]]
t[i_, j_] := Table[r[i, j + k*s], {k, 0, n - 1}];
d[i_, j_] := Det[Table[t[i + k*s, j], {k, 0, n - 1}]];  (* D(i,j,s,n) *)
p[i_, j_] := Permanent[Table[t[i + k*s, j], {k, 0, n - 1}]];  (* P(i,j,s,n) *)
pos[i_, j_] := (p[i, j] + d[i, j])/2;
neg[i_, j_] := (p[i, j] - d[i, j])/2;
Grid[Table[pos[i, j], {i, 1, z}, {j, 1, z}]]  (* A380660 array *)
Grid[Table[neg[i, j], {i, 1, z}, {j, 1, z}]]  (* A380661 array *)
FindLinearRecurrence[Table[pos[1, k], {k, 1, 20}]] (* row recurrence, all rows *)
FindLinearRecurrence[Table[neg[7, k], {k, 1, 20}]] (* row recurrence, all rows *)
Table[pos[k, m - k], {m, 2, z}, {k, 1, m - 1}] // Flatten (* A380660 sequence *)
Table[neg[k, m - k], {m, 2, z}, {k, 1, m - 1}] // Flatten (* A380661 sequence *)

A141419 Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 4, 7, 9, 10, 5, 9, 12, 14, 15, 6, 11, 15, 18, 20, 21, 7, 13, 18, 22, 25, 27, 28, 8, 15, 21, 26, 30, 33, 35, 36, 9, 17, 24, 30, 35, 39, 42, 44, 45, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55

Offset: 1

Roger L. Bagula, Aug 05 2008

As a rectangle, the accumulation array of A051340.
From Clark Kimberling, Feb 05 2011: (Start)
Here all the weights are divided by two where they aren't in Cahn.
As a rectangle, A141419 is in the accumulation chain
... < A051340 < A141419 < A185874 < A185875 < A185876 < ...
(See A144112 for the definition of accumulation array.)
row 1: A000027
col 1: A000217
diag (1,5,...): A000326 (pentagonal numbers)
diag (2,7,...): A005449 (second pentagonal numbers)
diag (3,9,...): A045943 (triangular matchstick numbers)
diag (4,11,...): A115067
diag (5,13,...): A140090
diag (6,15,...): A140091
diag (7,17,...): A059845
diag (8,19,...): A140672
(End)
Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n](1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery, Nov 20 2011
Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x)  (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N](i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N](k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery, Jan 20 2012
Row sums: n*(n+1)*(2*n+1)/6. - L. Edson Jeffery, Jan 25 2013
n-th row = partial sums of n-th row of A004736. - Reinhard Zumkeller, Aug 04 2014
T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - Derek Orr, Nov 26 2014
A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - Hartmut F. W. Hoft, Apr 14 2016
Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - L. Edson Jeffery, Jan 30 2018
From Peter Munn, Aug 21 2019 in respect of the sequence read as a triangle: (Start)
A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).
The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.
Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A270877; this is the only one that contains its own row numbers only once.
(End)

			As a triangle:
   1,
   2,  3,
   3,  5,  6,
   4,  7,  9, 10,
   5,  9, 12, 14, 15,
   6, 11, 15, 18, 20, 21,
   7, 13, 18, 22, 25, 27, 28,
   8, 15, 21, 26, 30, 33, 35, 36,
   9, 17, 24, 30, 35, 39, 42, 44, 45,
  10, 19, 27, 34, 40, 45, 49, 52, 54, 55;
As a rectangle:
   1   2   3   4   5   6   7   8   9  10
   3   5   7   9  11  13  15  17  19  21
   6   9  12  15  18  21  24  27  30  33
  10  14  18  22  26  30  34  38  42  46
  15  20  25  30  35  40  45  50  55  60
  21  27  33  39  45  51  57  63  69  75
  28  35  42  49  56  63  70  77  84  91
  36  44  52  60  68  76  84  92 100 108
  45  54  63  72  81  90  99 108 117 126
  55  65  75  85  95 105 115 125 135 145
Since the odd divisors of 15 are 1, 3, 5 and 15, number 15 appears four times in the triangle at t(3+(5-1)/2, 5) in column 5 since 5+1 <= 2*3, t(5+(3-1)/2, 3), t(1+(15-1)/2, 2*1) in column 2 since 15+1 > 2*1, and t(15+(1-1)/2, 1). - _Hartmut F. W. Hoft_, Apr 14 2016

R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 24.
Carlton Gamer, David W. Roeder, and John J. Watkins, Trapezoidal Numbers, Mathematics Magazine 58:2 (1985), pp. 108-110.
L. E. Jeffery, Unit-primitive matrices
M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263.

Cf. A000330 (row sums), A004736, A057059, A070543.
A144112, A051340, A141419, A185874, A185875, A185876 are accumulation chain related.
A141418 is a variant.
Cf. A001227, A209260. - Hartmut F. W. Hoft, Apr 14 2016
A109814, A212652, A270877, A286013 relate to where each natural number appears in this sequence.
A000027, A000217, A000326, A005449, A045943, A059845, A115067, A140090, A140091, A140672 are rows, columns or diagonals - refer to comments.

a141419 n k =  k * (2 * n - k + 1) `div` 2
a141419_row n = a141419_tabl !! (n-1)
a141419_tabl = map (scanl1 (+)) a004736_tabl
-- Reinhard Zumkeller, Aug 04 2014

a:=(n,k)->k*n-binomial(k,2): seq(seq(a(n,k),k=1..n),n=1..12); # Muniru A Asiru, Oct 14 2018

T[n_, m_] = m*(2*n - m + 1)/2; a = Table[Table[T[n, m], {m, 1, n}], {n, 1, 10}]; Flatten[a]

t(n,m) = m*(2*n - m + 1)/2.
t(n,m) = A000217(n) - A000217(n-m). - L. Edson Jeffery, Jan 16 2013
Let v = d*h with h odd be an integer factorization, then v = t(d+(h-1)/2, h) if h+1 <= 2*d, and v = t(d+(h-1)/2, 2*d) if h+1 > 2*d; see A209260. - Hartmut F. W. Hoft, Apr 14 2016
G.f.: y*(-x + y)/((-1 + x)^2*(-1 + y)^3). - Stefano Spezia, Oct 14 2018
T(n, 2) = A060747(n) for n > 1. T(n, 3) = A008585(n - 1) for n > 2. T(n, 4) = A016825(n - 2) for n > 3. T(n, 5) = A008587(n - 2) for n > 4. T(n, 6) = A016945(n - 3) for n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7.r n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7. T(n, 9) = A008591(n - 4) for n > 8. T(n, 10) = A017329(n - 5) for n > 9. T(n, 11) = A008593(n - 5) for n > 10.  T(n, 12) = A017593(n - 6) for n > 11. T(n, 13) = A008595(n - 6) for n > 12. T(n, 14) = A147587(n - 7) for n > 13. T(n, 15) = A008597(n - 7) for n > 14. T(n, 16) = A051062(n - 8) for n > 15. T(n, 17) = A008599(n - 8) for n > 16. - Stefano Spezia, Oct 14 2018
T(2*n-k, k) = A070543(n, k). - Peter Munn, Aug 21 2019

Simpler name by Stefano Spezia, Oct 14 2018

A107985 Triangle read by rows: T(n,k) = (k+1)(n+2)(n-k+1)/2 for 0 <= k <= n.

Original entry on oeis.org

1, 3, 3, 6, 8, 6, 10, 15, 15, 10, 15, 24, 27, 24, 15, 21, 35, 42, 42, 35, 21, 28, 48, 60, 64, 60, 48, 28, 36, 63, 81, 90, 90, 81, 63, 36, 45, 80, 105, 120, 125, 120, 105, 80, 45, 55, 99, 132, 154, 165, 165, 154, 132, 99, 55, 66, 120, 162, 192, 210, 216, 210, 192, 162, 120, 66

Offset: 0

Emeric Deutsch, Jun 12 2005

Kekulé numbers for certain benzenoids.
T(n,k) is the number of Dyck (n+3)-paths with 3 peaks (UDs) and last descent of length k+1. For example, T(1,1)=3 counts UUDUDUDD, UDUUDUDD, UDUDUUDD. The number of Dyck n-paths containing k peaks and with last descent of length j is (j/n)*binomial(n,k-1)*binomial(n-j-1,k-2) (where as usual binomial(a,b)=0 for b < 0 except that binomial(-1,-1):=1). - David Callan, Jun 26 2006
As a rectangular array, this is the accumulation array (cf. A144112) of the rectangular array W given by w(i,j)=i+j-1; i.e., W=A002024 as a rectangular array. - Clark Kimberling, Sep 16 2008
T(n,k) gives the dimension of an irreducible representation of SU(3) whose Young diagram (n,k) has two rows of length n and k, respectively. - Dimitris Cardaris, May 10 2025

			Triangle begins:
   1;
   3,  3;
   6,  8,  6;
  10, 15, 15, 10;
  15, 24, 27, 24, 15;
  ...

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 237, K{B(n,2,-l)}).

Volodymyr Mazorchuk and Xiaoyu Zhu, Combinatorics of infinite rank module categories over finite dimensional sl3-modules in Lie-algebraic context, arXiv:2501.00291 [math.RT], 2024. See page 9.

Cf. A000217 (column 0 and main diagonal), A002024, A002415 (row sums), A098737, A144112.

T:=proc(n,k) if k<=n then (k+1)*(n+2)*(n-k+1)/2 else 0 fi end: for n from 0 to 11 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[n_,k_]:= (k+1)(n+2)(n-k+1)/2; Table[T[n,k],{n,0,10},{k,0,n}]//Flatten (* Stefano Spezia, Jan 06 2025 *)

from math import isqrt, comb
def A107985(n):
    a = (m:=isqrt(k:=n+1<<1))+(k>m*(m+1))
    b = n-comb(a,2)
    return (b+1)*(a+1)*(a-b)>>1 # Chai Wah Wu, Jun 14 2025

T(n,n-k) = T(n,k); T(2n,n) = (n+1)^3.

G.f.: (1 - x^2*y)/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Oct 01 2023

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A000012 The simplest sequence of positive numbers: the all 1's sequence.

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A007318 Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

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A002260 Triangle read by rows: T(n,k) = k for n >= 1, k = 1..n.

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A002024 k appears k times; a(n) = floor(sqrt(2n) + 1/2).

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