A030191 -id:A030191 - 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

A192232 Constant term of the reduction of n-th Fibonacci polynomial by x^2 -> x+1. (See Comments.)

Original entry on oeis.org

1, 0, 2, 1, 6, 7, 22, 36, 89, 168, 377, 756, 1630, 3353, 7110, 14783, 31130, 65016, 136513, 285648, 599041, 1254456, 2629418, 5508097, 11542854, 24183271, 50674318, 106173180, 222470009, 466131960, 976694489, 2046447180, 4287928678, 8984443769, 18825088134

Offset: 1

Clark Kimberling, Jun 26 2011

Polynomial reduction: an introduction
...
We begin with an example.  Suppose that p(x) is a polynomial, so that p(x)=(x^2)t(x)+r(x) for some polynomials t(x) and r(x), where r(x) has degree 0 or 1.  Replace x^2 by x+1 to get (x+1)t(x)+r(x), which is (x^2)u(x)+v(x) for some u(x) and v(x), where v(x) has degree 0 or 1.  Continuing in this manner results in a fixed polynomial w(x) of degree 0 or 1.  If p(x)=x^n, then w(x)=x*F(n)+F(n-1), where F=A000045, the sequence of Fibonacci numbers.
In order to generalize, write d(g) for the degree of an arbitrary polynomial g(x), and suppose that p, q, s are polynomials satisfying d(s)s in this manner until reaching w such that d(w)s.
The coefficients of (reduction of p by q->s) comprise a vector of length d(q)-1, so that a sequence p(n,x) of polynomials begets a sequence of vectors, such as (F(n), F(n-1)) in the above example.  We are interested in the component sequences (e.g., F(n-1) and F(n)) for various choices of p(n,x).
Following are examples of reduction by x^2->x+1:
n-th Fibonacci p(x) -> A192232+x*A112576
n-th cyclotomic p(x) -> A192233+x*A051258
n-th 1st-kind Chebyshev p(x) -> A192234+x*A071101
n-th 2nd-kind Chebyshev p(x) -> A192235+x*A192236
x(x+1)(x+2)...(x+n-1) -> A192238+x*A192239
(x+1)^n -> A001519+x*A001906
(x^2+x+1)^n -> A154626+x*A087635
(x+2)^n -> A020876+x*A030191
(x+3)^n -> A192240+x*A099453
...
Suppose that b=(b(0), b(1),...) is a sequence, and let p(n,x)=b(0)+b(1)x+b(2)x^2+...+b(n)x^n.  We define (reduction of sequence b by q->s) to be the vector given by (reduction of p(n,x) by q->s), with components in the order of powers, from 0 up to d(q)-1.  For k=0,1,...,d(q)-1, we then have the "k-sequence of (reduction of sequence b by q->s)".  Continuing the example, if b is the sequence given by b(k)=1 if k=n and b(k)=0 otherwise, then the 0-sequence of (reduction of b by x^2->x+1) is (F(n-1)), and the 1-sequence is (F(n)).
...
For selected sequences b, here are the 0-sequences and 1-sequences of (reduction of b by x^2->x+1):
b=A000045, Fibonacci sequence (1,1,2,3,5,8,...) yields
   0-sequence A166536 and 1-sequence A064831.
b=(1,A000045)=(1,1,1,2,3,5,8,...) yields
   0-sequence A166516 and 1-sequence A001654.
b=A000027, natural number sequence (1,2,3,4,...) yields
  0-sequence A190062 and 1-sequence A122491.
b=A000032, Lucas sequence (1,3,4,7,11,...) yields
  0-sequence A192243 and 1-sequence A192068.
b=A000217, triangular sequence (1,3,6,10,...) yields
  0-sequence A192244 and 1-sequence A192245.
b=A000290, squares sequence (1,4,9,16,...) yields
  0-sequence A192254 and 1-sequence A192255.
More examples:  A192245-A192257.
...
More comments:
(1) If s(n,x)=(reduction of x^n by q->s) and
    p(x)=p(0)x^n+p(1)x^(n-1)+...+p(n)x^0, then
    (reduction of p by q->s)=p(0)s(n,x)+p(1)s(n-1,x)
    +...+p(n-1)s(1,x)+p(n)s(0,x).  See A192744.
(2) For any polynomial p(x), let P(x)=(reduction of p(x)
    by q->s).  Then P(r)=p(r) for each zero r of
    q(x)-s(x).  In particular, if q(x)=x^2 and s(x)=x+1,
    then P(r)=p(r) if r=(1+sqrt(5))/2 (golden ratio) or
    r=(1-sqrt(5))/2.

			The first four Fibonacci polynomials and their reductions by x^2->x+1 are shown here:
F1(x)=1 -> 1 + 0x
F2(x)=x -> 0 + 1x
F3(x)=x^2+1 -> 2+1x
F4(x)=x^3+2x -> 1+4x
F5(x)=x^4+3x^2+1 -> (x+1)^2+3(x+1)+1 -> 6+6x.
From these, read A192232=(1,0,1,1,6,...) and A112576=(0,1,1,4,6,...).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-1,-1).

Cf. A168561, A192233 - A192240, A192744.

Cf. A206296, A265398, A265399, A265752, A265753.

q[x_] := x + 1;
reductionRules = {x^y_?EvenQ -> q[x]^(y/2),  x^y_?OddQ -> x q[x]^((y - 1)/2)};
t = Table[FixedPoint[Expand[#1 /. reductionRules] &, Fibonacci[n, x]], {n, 1, 40}];
Table[Coefficient[Part[t, n], x, 0], {n, 1, 40}]
  (* A192232 *)
Table[Coefficient[Part[t, n], x, 1], {n, 1, 40}]
(* A112576 *)
(* Peter J. C. Moses, Jun 25 2011 *)
LinearRecurrence[{1, 3, -1, -1}, {1, 0, 2, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

Vec((1-x-x^2)/(1-x-3*x^2+x^3+x^4)+O(x^99)) \\ Charles R Greathouse IV, Jan 08 2013

Empirical G.f.: -x*(x^2+x-1)/(x^4+x^3-3*x^2-x+1). - Colin Barker, Sep 11 2012
The above formula is correct. - Charles R Greathouse IV, Jan 08 2013
a(n) = A265752(A206296(n)). - Antti Karttunen, Dec 15 2015
a(n) = A112576(n) -A112576(n-1) -A112576(n-2). - R. J. Mathar, Dec 16 2015

Example corrected by Clark Kimberling, Dec 18 2017

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

Original entry on oeis.org

1, 4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799

Offset: 0

N. J. A. Sloane

In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002
The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k), k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g., continued fraction for F(12)/F(9) is [4, 4,4]. - Benoit Cloitre, Apr 10 2003
See A135064 for a possible connection with Galois groups of quintics.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - Thomas Baruchel, Sep 15 2003
All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
Inverse binomial transform of A030191. - Philippe Deléham, Oct 04 2005
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) =  x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Let r = (2n+1), then a(n), n>0 = Product_{k=1..floor((r-1)/2)} (1 + sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). - Gary W. Adamson, Nov 26 2008
a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - Paul Barry, Apr 21 2009
a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - Alex Ratushnyak, May 06 2012
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... . - R. J. Mathar, Aug 10 2012
The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 05 2013
Solutions (x, y) = (a(n), a(n+1)) satisfying  x^2 + y^2 = 3xy + 5. - Michel Lagneau, Feb 01 2014
Conjecture: except for the number 3, a(n) are the numbers such that a(n)^2+2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
Comment on the preceding conjecture: It is clear that all a(n) satisfy a(n)^2 + 2 = L(2*(2*n+1)) due to the identity (17 c) of Vajda, p. 177: L(2*n) + 2*(-1)^n = L(n)^2 (take n -> 2*n+1). - Wolfdieter Lang, Oct 10 2014
Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - Derek Orr, Jun 18 2015
If d[k] denotes the sequence of k-th differences of this sequence, then d[0](0), d[1](1), d[2](2), d[3](3), ... = A048876, cf. message to SeqFan list by P. Curtz on March 2, 2016. - M. F. Hasler, Mar 03 2016
a(n-1) and a(n) are the least phi-antipalindromic numbers (A178482) with 2*n and 2*n+1 digits in base phi, respectively. - Amiram Eldar, Jul 07 2021
Triangulate (hyperbolic) 2-space such that around every vertex exactly 7 triangles touch. Call any 7 triangles having a common vertex the first layer and let the (n+1)-st layer be all triangles that do not appear in any of the first n layers and have a common vertex with the n-th layer. Then the n-th layer contains 7*a(n-1) triangles. E.g., the first layer (by definition) contains 7 triangles, the second layer (the "ring" of triangles around the first layer) consists of 28 triangles, the third layer (the next "ring") consists of 77 triangles, and so on. - Nicolas Nagel, Aug 13 2022

			G.f. = 1 + 4*x + 11*x^2 + 29*x^3 + 76*x^4 + 199*x^5 + 521*x^6 + ... - _Michael Somos_, Jan 13 2019

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
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).
Steven Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Marco Abrate, Stefano Barbero, Umberto Cerruti and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Kasper K. S. Andersen, Lisa Carbone and Diego Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq., Vol. 14 (2011), Article 11.4.3, page 9.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 25.
Nathan D. Cahill, John R. D'Errico and John P. Spence, Complex Factorizations of the Fibonacci and Lucas Numbers, Fibonacci Quarterly, 1(41):13-19, 2003.
L. Carlitz, Problem B-110, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; An Infinite Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 469-470.
Paul Curtz, A269500, SeqFan list, March 2, 2016.
Murray Elder and Arkadius Kalka, Logspace computations for rigid Garside groups, arXiv preprint arXiv:1310.0933 [math.GR], 2013.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, Vol. 5 (2014), pp. 2226-2234.
Alex Fink, Richard K. Guy and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol. 3, No. 2 (2008), pp. 76-114. See Section 13.
Dale Gerdemann, Collision of Digits "Also interesting are the two bisections of the Lucas numbers A005248 (digit minimizer) and A002878 (digit maximizer). I particularly like the multiples of A005248 because I have this image of the two digits piling up on top of each other and then spreading out like waves".
André Gougenheim, About the linear sequence of integers such that each term is the sum of the two preceding Part 1 Part 2, Fib. Quart., Vol. 9, No. 3 (1971), pp. 277-295, 298.
Tian-Xiao He and Louis W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic., Vol. 532 (2017) pp. 25-41, example p 34.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 13.
Seong Ju Kim, Ryan Stees and Laura Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.4.
Tanya Khovanova, Recursive Sequences.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
D. H. Lehmer, Recurrence formulas for certain divisor functions, Bull. Amer. Math. Soc., Vol. 49, No. 2 (1943), pp. 150-156.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum, Vol. 19 (2019), pp. 11-16.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics, Vol. 340, No. 2 (2017), pp. 160-174.
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
Serge Perrine, Some properties of the equation x^2=5y^2-4, The Fibonacci Quarterly, Vol. 54, No. 2 (2016) pp. 172-177.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000204. a(n) = A060923(n, 0), a(n)^2 = A081071(n).
Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence], A000045, A002315, A004146, A029907, A113224, A153387, A153416, A178482, A192425, A285992 (prime subsequence).
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

List([0..40], n-> Lucas(1,-1,2*n+1)[2] ); # G. C. Greubel, Jul 15 2019

a002878 n = a002878_list !! n
a002878_list = zipWith (+) (tail a001906_list) a001906_list
-- Reinhard Zumkeller, Jan 11 2012

[Lucas(2*n+1): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011

A002878 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,4]);
    else
        3*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

a[n_]:= FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 40, 2}]
a[1]=1; a[2]=4; a[n_]:=a[n]= 3a[n-1] -a[n-2]; Array[a, 40]
LinearRecurrence[{3, -1}, {1, 4}, 41] (* Jean-François Alcover, Sep 23 2017 *)
Table[Sum[(-1)^Floor[k/2] Binomial[n -Floor[(k+1)/2], Floor[k/2]] 3^(n - k), {k, 0, n}], {n, 0, 40}] (* L. Edson Jeffery, Feb 26 2018 *)
a[ n_] := Fibonacci[2n] + Fibonacci[2n+2]; (* Michael Somos, Jul 31 2018 *)
a[ n_]:= LucasL[2n+1]; (* Michael Somos, Jan 13 2019 *)

a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ Charles R Greathouse IV, Jun 16 2011

for(n=1,40,q=((1+sqrt(5))/2)^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec((1+x)/(1-3*x+x^2) + O(x^40)) \\ Altug Alkan, Oct 26 2015

a002878 = [1, 4]
for n in range(30): a002878.append(3*a002878[-1] - a002878[-2])
print(a002878) # Gennady Eremin, Feb 05 2022

[lucas_number2(2*n+1,1,-1) for n in (0..40)] # G. C. Greubel, Jul 15 2019

a(n+1) = 3*a(n) - a(n-1).
G.f.: (1+x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(2*n, sqrt(5)) = S(n, 3) + S(n-1, 3); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 3) = A001906(n+1) (even-indexed Fibonacci numbers).
a(n) ~ phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -1) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = A005248(n+1) - A005248(n) = -1 + Sum_{k=0..n} A005248(k). - Lekraj Beedassy, Dec 31 2002
a(n) = 2^(-n)*A082762(n) = 4^(-n)*Sum_{k>=0} binomial(2*n+1, 2*k)*5^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = (-1)^n*Sum_{k=0..n} (-5)^k*binomial(n+k, n-k). - Benoit Cloitre, May 09 2004
From Paul Barry, May 27 2004: (Start)
Both bisection and binomial transform of A000204.
a(n) = Fibonacci(2n) + Fibonacci(2n+2). (End)
Sequence lists the numerators of sinh((2*n-1)*psi) where the denominators are 2; psi=log((1+sqrt(5))/2). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009
a(n) = A001906(n) + A001906(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = floor(phi^(2n+1)), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 10 2012
a(n) = A014217(2*n+1) = A014217(2*n+2) - A014217(2*n). - Paul Curtz, Jun 11 2013
Sum_{n >= 0} 1/(a(n) + 5/a(n)) = 1/2. Compare with A005248, A001906, A075796. - Peter Bala, Nov 29 2013
a(n) = lim_{m->infinity} Fibonacci(m)^(4n+1)*Fibonacci(m+2*n+1)/ Sum_{k=0..m} Fibonacci(k)^(4n+2). - Yalcin Aktar, Sep 02 2014
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 4, 0, 11, 0, 29, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -1, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
b(n) = (1/2)*((-1)^n - 1)*F(n) + (1 + (-1)^(n-1))*F(n+1), where F(n) is a Fibonacci number. The o.g.f. is x*(1 + x^2)/(1 - 3*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = sqrt(5*F(2*n+1)^2-4), where F(n) = A000045(n). - Derek Orr, Jun 18 2015
For n > 1, a(n) = 5*F(2*n-1) + L(2*n-3) with F(n) = A000045(n). - J. M. Bergot, Oct 25 2015
For n > 0, a(n) = L(n-1)*L(n+2) + 4*(-1)^n. - J. M. Bergot, Oct 25 2015
For n > 2, a(n) = a(n-2) + F(n+2)^2 + F(n-3)^2 = L(2*n-3) + F(n+2)^2 + F(n-3)^2. - J. M. Bergot, Feb 05 2016 and Feb 07 2016
E.g.f.: ((sqrt(5) - 5)*exp((3-sqrt(5))*x/2) + (5 + sqrt(5))*exp((3+sqrt(5))*x/2))/(2*sqrt(5)). - Ilya Gutkovskiy, Apr 24 2016
a(n) = Sum_{k=0..n} (-1)^floor(k/2)*binomial(n-floor((k+1)/2), floor(k/2))*3^(n-k). - L. Edson Jeffery, Feb 26 2018
a(n)*F(m+2n-1) = F(m+4n-2)-F(m), with Fibonacci number F(m), empirical observation. - Dan Weisz, Jul 30 2018
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jul 31 2018
Sum_{n>=0} 1/a(n) = A153416. - Amiram Eldar, Nov 11 2020
a(n) = Product_{k=1..n} (1 + 4*sin(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
Sum_{n>=0} (-1)^n/a(n) = (1/sqrt(5)) * A153387 (Carlitz, 1967). - Amiram Eldar, Feb 05 2022
The continued fraction [a(n);a(n),a(n),...] = phi^(2*n+1), with phi = A001622. - A.H.M. Smeets, Feb 25 2022
a(n) = 2*sinh((2*n + 1)*arccsch(2)). - Peter Luschny, May 25 2022
This gives the sequence with 2 1's prepended: b(1)=b(2)=1 and, for k >= 3, b(k) = Sum_{j=1..k-2} (2^(k-j-1) - 1)*b(j). - Neal Gersh Tolunsky, Oct 28 2022 (formula due to Jon E. Schoenfield)
For n > 0, a(n) = 1 + 1/(Sum_{k>=1} F(k)/phi^(2*n*k + k)). - Diego Rattaggi, Nov 08 2023
From Peter Bala, Apr 16 2025: (Start)
a(3*n+1) = a(n)^3 + 3*a(n).
a(5*n+2) = a(n)^5 + 5*a(n)^3 + 5*a(n).
a(7*n+3) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n).
For the coefficients see A034807.
The general result is: for k >= 0, a(k*n + (k-1)/2) = 2 * T(k, a(n)/2), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind and a(n) = ((1 + sqrt(5))/2)^(2*n+1) + ((1 - sqrt(5))/2)^(2*n+1).
Sum_{n >= 0} (-1)^n/a(n) = (1/4)* (theta_3(phi) - theta_3(phi^2)) = 0.815947983588122..., where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122) and phi = (sqrt(5) - 1)/2. See Borwein and Borwein, Exercise 3 a, p. 94 and Carlitz, 1967. (End)
From Peter Bala, May 15 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/5 (telescoping series: 5/(a(n) - 1/a(n)) = 1/A001906(n+1) + 1/A001906(n) ).
More generally, for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1) = Lucas(2*k) - 2.
For k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(n) + L(2*k)^2/a(n)) = (1/5) * A064170(k+2).
Sum_{n >= 1} 1/(a(n) + 9/a(n)) = 3/10 (follows from 1/(a(n) + 9/a(n)) = L(2*n)/A081076(n) - L(2*n+2)/A081076(n+1) ).
More generally, it appears that for k >= 1, Sum_{n >= 1} 1/(a(n) + L(2*k)^2/a(n)) is rational.
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) [telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5*(1 - 4/A240926(n+1)) ]. (End)

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A109466 Riordan array (1, x(1-x)).

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 0, -2, 1, 0, 0, 1, -3, 1, 0, 0, 0, 3, -4, 1, 0, 0, 0, -1, 6, -5, 1, 0, 0, 0, 0, -4, 10, -6, 1, 0, 0, 0, 0, 1, -10, 15, -7, 1, 0, 0, 0, 0, 0, 5, -20, 21, -8, 1, 0, 0, 0, 0, 0, -1, 15, -35, 28, -9, 1, 0, 0, 0, 0, 0, 0, -6, 35, -56, 36, -10, 1, 0, 0, 0, 0, 0, 0, 1, -21, 70, -84, 45, -11, 1, 0, 0, 0, 0

Offset: 0

Philippe Deléham, Aug 28 2005

Inverse is Riordan array (1, xc(x)) (A106566).
Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, -1, 1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
Modulo 2, this sequence gives A106344. - Philippe Deléham, Dec 18 2008
Coefficient array of the polynomials Chebyshev_U(n, sqrt(x)/2)*(sqrt(x))^n. - Paul Barry, Sep 28 2009

			Rows begin:
  1;
  0,  1;
  0, -1,  1;
  0,  0, -2,  1;
  0,  0,  1, -3,  1;
  0,  0,  0,  3, -4,   1;
  0,  0,  0, -1,  6,  -5,   1;
  0,  0,  0,  0, -4,  10,  -6,   1;
  0,  0,  0,  0,  1, -10,  15,  -7,  1;
  0,  0,  0,  0,  0,   5, -20,  21, -8,  1;
  0,  0,  0,  0,  0,  -1,  15, -35, 28, -9, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0,    1,
  0,   -1,    1,
  0,   -1,   -1,   1,
  0,   -2,   -1,  -1,   1,
  0,   -5,   -2,  -1,  -1,  1,
  0,  -14,   -5,  -2,  -1, -1,  1,
  0,  -42,  -14,  -5,  -2, -1, -1,  1,
  0, -132,  -42, -14,  -5, -2, -1, -1,  1,
  0, -429, -132, -42, -14, -5, -2, -1, -1, 1 (End)

Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Tom Copeland, Addendum to Elliptic Lie Triad

Cf. A026729 (unsigned version), A000108, A030528, A124644.

/* As triangle */ [[(-1)^(n-k)*Binomial(k, n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jan 14 2016

(* The function RiordanArray is defined in A256893. *)
RiordanArray[1&, #(1-#)&, 13] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

Number triangle T(n, k) = (-1)^(n-k)*binomial(k, n-k).
T(n, k)*2^(n-k) = A110509(n, k); T(n, k)*3^(n-k) = A110517(n, k).
Sum_{k=0..n} T(n,k)*A000108(k)=1. - Philippe Deléham, Jun 11 2007
From Philippe Deléham, Oct 30 2008: (Start)
Sum_{k=0..n} T(n,k)*A144706(k) = A082505(n+1).
Sum_{k=0..n} T(n,k)*A002450(k) = A100335(n).
Sum_{k=0..n} T(n,k)*A001906(k) = A100334(n).
Sum_{k=0..n} T(n,k)*A015565(k) = A099322(n).
Sum_{k=0..n} T(n,k)*A003462(k) = A106233(n). (End)
Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = -12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12 respectively. - Philippe Deléham, Oct 27 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A010892(n), A099087(n), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n) for x = 0,1,2,3,4,5,6,7,8,9,10 respectively. - Philippe Deléham, Oct 28 2008
G.f.: 1/(1-y*x+y*x^2). - Philippe Deléham, Dec 15 2011
T(n,k) = T(n-1,k-1) - T(n-2,k-1), T(n,0) = 0^n. - Philippe Deléham, Feb 15 2012
Sum_{k=0..n} T(n,k)*x^(n-k) = F(n+1,-x) where F(n,x)is the n-th Fibonacci polynomial in x defined in A011973. - Philippe Deléham, Feb 22 2013
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 26 2013
Sum_{k=0..n} T(n,k)*T(n+1,k) = -A110320(n). - Philippe Deléham, Feb 26 2013
For T(0,0) = 0, the signed triangle below has the o.g.f. G(x,t) = [t*x(1-x)]/[1-t*x(1-x)] = L[t*Cinv(x)] where L(x) = x/(1-x) and Cinv(x)=x(1-x) with the inverses Linv(x) = x/(1+x) and C(x)= [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108, so the inverse o.g.f. is Ginv(x,t) = C[Linv(x)/t] = [1-sqrt[1-4*x/(t(1+x))]]/2 (cf. A124644 and A030528). - Tom Copeland, Jan 19 2016

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..190
Index entries for linear recurrences with constant coefficients, signature (2,-10).

Sequences of the form a(n) = c*a(n-1) - d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A128834,A107920,A106852,A106853,A106854,A145934,A145976,A145978,A146078,A146080
.A000027,A009545,A088137,A088138,A045873,A088139,A089181,A090591,A127357,A190958
.A001906,A000225,A057083,A049072,A190959,A190960,A190961,A190962,A190963,A190964
.A001353,A007070,A003462,A001787,A099456,A190965,A168175,A190966,A190967,A190968
.A004254,A107839,A116415,A002450,A030191,A001047,A099450,A190969,A190970,A190971
.A001109,A154244,A138395,A084326,A003463,A030192,A081179,A006516,A027471,A106392
.A004187,A186446,A190972,A190973,A190974,A003464,A030240,A152268,A099459,A016127
.A001090,A190975,A190976,A099156,A190977,A190978,A023000,A057084,A154245,A154235
.A018913,A190979,A190980,A190981,A190982,A190983,A190984,A023001,A057085,A178869
A004189,A190869,A190985,A190986,A190987,A190988,A190989,A190990,A002452,A057086

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011

Mathematica
```
LinearRecurrence[{2,-10}, {0,1}, 50]
```

a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A094440 Triangular array read by rows: T(n,k) = Fibonacci(n+1-k)*C(n,k-1), k = 1..n; n >= 1.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 8, 6, 4, 5, 15, 20, 10, 5, 8, 30, 45, 40, 15, 6, 13, 56, 105, 105, 70, 21, 7, 21, 104, 224, 280, 210, 112, 28, 8, 34, 189, 468, 672, 630, 378, 168, 36, 9, 55, 340, 945, 1560, 1680, 1260, 630, 240, 45, 10, 89, 605, 1870, 3465, 4290, 3696, 2310, 990, 330, 55, 11

Offset: 1

Clark Kimberling, May 03 2004

Row sums yield the even-subscripted Fibonacci numbers (A001906).

Row n shows the coefficients of the numerator of the n-th derivative of c(n)/(x^2+x-1), where c(n) = ((-1)^(n + 1))/n!; see the Mathematica program. - Clark Kimberling, Oct 22 2019

			Triangle starts:
   1;
   1,  2;
   2,  3,   3;
   3,  8,   6,   4;
   5, 15,  20,  10,  5;
   8, 30,  45,  40, 15,  6;
  13, 56, 105, 105, 70, 21, 7;
  ...
T(4,3) = F(2)*C(4,2) = 1*6 = 6.

G. C. Greubel, Rows n = 1..100 of triangle, flattened
M. Norfleet, Characterization of second-order strong divisibility sequences of polynomials, The Fibonacci Quarterly, 43(2) (2005), 166-169.

Cf. A000032, A000045, A001906, A030191, A081574, A094435, A132148.
Cf. A094435, A094436, A094437, A094438, A094439, A094441, A094442, A094443, A094444.
Cf. A367208, A367209, A367210, A367211.

Flat(List([1..12], n-> List([1..n], k-> Binomial(n,k-1)* Fibonacci(n-k+1) ))); # G. C. Greubel, Oct 30 2019

/* As triangle */ [[Fibonacci(n+1-k)*Binomial(n,k-1): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Aug 15 2017

with(combinat): T:=(n,k)->binomial(n,k-1)*fibonacci(n+1-k): for n from 1 to 11 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form # Emeric Deutsch

Table[Fibonacci[n+1-k]Binomial[n,k-1],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Sep 14 2016 *)
(* Next program outputs polynomials having coefficients T(n,k) *)
g[x_, n_] := Numerator[(-1)^(n + 1) Factor[D[1/(1 - x - x^2), {x, n}]]]
Column[Expand[Table[g[x, n]/n!, {n, 0, 12}]]] (* Clark Kimberling, Oct 22 2019 *)

T(n,k) = binomial(n,k-1)*fibonacci(n-k+1);
for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 30 2019

[[binomial(n,k-1)*fibonacci(n-k+1) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Oct 30 2019

From Peter Bala, Aug 17 2007: (Start)
With an offset of 0, the row polynomials F(n,x) = Sum_{k = 0..n} C(n,k)* Fibonacci(n-k)*x^k satisfy F(n,x)*L(n,x) = F(2*n,x), where L(n,x) = Sum_{k = 0..n} C(n,k)*Lucas(n-k)*x^k.
Other identities and formulas include:
F(n+1,x)^2 - F(n,x)*F(n+2,x) = (x^2 + x - 1)^n;
Sum_{k = 0..n} C(n,k)*F(n-k,x)*L(k,x) = (2^n)*F(n,x);
F(n,2*x) = Sum_{k = 0..n} C(n,k)*F(n-k,x)*x^k;
F(n,3*x) = Sum_{k = 0..n} C(n,k)*F(n-k,2*x)*x^k, etc.
The sequence {F(n,r)}n>=1 gives the r-th binomial transform of the Fibonacci numbers: r = 1 gives A001906, r = 2 gives A030191, r = 3 gives A099453, r = 4 gives A081574, r = 5 gives A081575.
F(n,1/phi) = (-1)^(n-1)*F(n,-phi) = sqrt(5)^(n-1) for n >= 1, where phi = (1 + sqrt(5))/2.
The polynomials F(n,-x) satisfy a Riemann hypothesis: the zeros of F(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
G.f.: t/(1 - (2*x + 1)*t + (x^2 + x - 1)*t^2) = t + (1 + 2*x)*t^2 + (2 + 3*x + 3*x^2)*t^3 + (3 + 8*x + 6*x^2 + 4*x^3)*t^4 + ... . (End)
From Peter Bala, Jun 29 2016: (Start)
Working with an offset of 0, the n-th row polynomial F(n,x) = 1/sqrt(5)*( (x + phi)^n - (x - 1/phi)^n ), where phi = (1 + sqrt(5))/2.
d/dx(F(n,x)) = n*F(n-1,x).
F(-n,x) = -F(n,x)/(x^2 + x - 1)^n.
F(n,x - 1) = (-1)^(n-1)*F(n,-x).
F(n,x) is a divisibility sequence of polynomials, that is, if n divides m then F(n,x) divides F(m,x) in the polynomial ring Z[x]. (End)
From G. C. Greubel, Oct 30 2019: (Start)
Sum_{k = 1..n} T(n,k) = Fibonacci(2*n).
Sum_{k = 1..n} (-1)^k * T(n,k) = (-1)^n * Fibonacci(n). (End)
From Clark Kimberling, Oct 30 2019: (Start)
F(n,x) is a strong divisibility sequence of polynomials in Z[x]; that is,
gcd(F(x,h),F(x,k)) = F(x,gcd(h,k)) for h,k >= 1.  Thus, if x is an integer, then F(n,x) is a strong divisibility sequence of integers; e.g., for x=3, we have A099453. (End)
Let p(n) denote the polynomial F(x,n). Then p(n) = k(b^n - c^n), where k = -1/sqrt(5), b = (1/2)(2x + 1 - sqrt(5)), c = (1/2)(2x + 1 + sqrt(5)), and for n >=3, p(n) = u*p(n - 1) + v*p(n - 2), where u = 1 + 2 x, v = 1 - x - x^2. - Clark  Kimberling, Nov 11 2023

Error in expansion of generating function corrected by Peter Bala, Sep 24 2008

A081567 Second binomial transform of F(n+1).

Original entry on oeis.org

1, 3, 10, 35, 125, 450, 1625, 5875, 21250, 76875, 278125, 1006250, 3640625, 13171875, 47656250, 172421875, 623828125, 2257031250, 8166015625, 29544921875, 106894531250, 386748046875, 1399267578125, 5062597656250, 18316650390625, 66270263671875, 239768066406250

Offset: 0

Paul Barry, Mar 22 2003

Binomial transform of F(2*n-1), index shifted by 1, where F is A000045. - corrected by Richard R. Forberg, Aug 12 2013
Case k=2 of family of recurrences a(n) = (2k+1)*a(n-1) - A028387(k-1)*a(n-2), a(0)=1, a(1)=k+1.
Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2*n+1, s(0) = 3, s(2*n+1) = 4.
a(n+1) gives the number of periodic multiplex juggling sequences of length n with base state <2>. - Steve Butler, Jan 21 2008
a(n) is also the number of idempotent order-preserving partial transformations (of an n-element chain) of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Sep 14 2008
Counts all paths of length (2*n+1), n>=0, starting at the initial node on the path graph P_9, see the Maple program. - Johannes W. Meijer, May 29 2010
Given the 3 X 3 matrix M = [1,1,1; 1,1,0; 1,1,3], a(n) = term (1,1) in M^(n+1). - Gary W. Adamson, Aug 06 2010
Number of nonisomorphic graded posets with 0 and 1 of rank n+2, with exactly 2 elements of each rank level between 0 and 1. Also the number of nonisomorphic graded posets with 0 of rank n+1, with exactly 2 elements of each rank level above 0. (This is by Stanley's definition of graded, that all maximal chains have the same length.) - David Nacin, Feb 26 2012
a(n) = 3^n a(n;1/3) = Sum_{k=0..n} C(n,k) * F(k-1) * (-1)^k * 3^(n-k), which also implies the Deleham formula given below and where a(n;d), n=0,1,...,d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also the papers of Witula et al.). - Roman Witula, Jul 12 2012
The limiting ratio a(n)/a(n-1) is 1 + phi^2. - Bob Selcoe, Mar 17 2014
a(n) counts closed walks on K_2 containing 3 loops on the index vertex and 2 loops on the other. Equivalently the (1,1) entry of A^n where the adjacency matrix of digraph is A=(3,1; 1,2). - David Neil McGrath, Nov 18 2014

			a(4)=125: 35*(3 + (35 mod 10 - 10 mod 3)/(10-3)) = 35*(3 + 4/7) = 125. - _Bob Selcoe_, Mar 17 2014

R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pages 96-100.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Santiago Alzate, Oscar Correa, and Rigoberto Flórez, Fibonacci identities from Jordan Identities, arXiv:2009.02639 [math.NT], 2020.
Carolina Benedetti, Christopher R. H. Hanusa, Pamela E. Harris, Alejandro H. Morales, and Anthony Simpson, Kostant's partition function and magic multiplex juggling sequences, arXiv:2001.03219 [math.CO], 2020. See Table 1 p. 12.
S. Butler and R. Graham, Enumerating (multiplex) juggling sequences, arXiv:0801.2597 [math.CO], 2008.
P. E. Harris, E. Insko, and L. K. Williams, The adjoint representation of a Lie algebra and the support of Kostant's weight multiplicity formula, arXiv preprint arXiv:1401.0055 [math.RT], 2013.
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
A. Laradji and A. Umar, A. Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), #04.3.8.
Mircea Merca, A Note on Cosine Power Sums J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
D. Nacin, The Minimal Non-Koszul A(Gamma), arXiv preprint arXiv:1204.1534 [math.QA], 2012. - From _N. J. A. Sloane_, Oct 05 2012
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009) 310-329, MR2555042.
Index entries for linear recurrences with constant coefficients, signature (5,-5).

a(n) = 5*A052936(n-1), n > 1.
Row sums of A114164.
Cf. A000045, A007051 (INVERTi transform), A007598, A028387, A030191, A039717, A049310, A081568 (binomial transform), A086351 (INVERT transform), A090041, A093129, A094441, A111776, A147748, A178381, A189315.

I:=[1, 3]; [n le 2 select I[n] else 5*Self(n-1)-5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 27 2012

with(GraphTheory):G:=PathGraph(9): A:= AdjacencyMatrix(G): nmax:=23; n2:=nmax*2+2: for n from 0 to n2 do B(n):=A^n; a(n):=add(B(n)[1,k],k=1..9); od: seq(a(2*n+1),n=0..nmax); # Johannes W. Meijer, May 29 2010

Table[MatrixPower[{{2,1},{1,3}},n][[2]][[2]],{n,0,44}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{5,-5},{1,3},30] (* Vincenzo Librandi, Feb 27 2012 *)

Vec((1-2*x)/(1-5*x+5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Mar 18 2014

def a(n, adict={0:1, 1:3}):
    if n in adict:
        return adict[n]
    adict[n]=5*a(n-1) - 5*a(n-2)
    return adict[n] # David Nacin, Mar 04 2012

a(n) = 5*a(n-1) - 5*a(n-2) for n >= 2, with a(0) = 1 and a(1) = 3.
a(n) = (1/2 - sqrt(5)/10) * (5/2 - sqrt(5)/2)^n + (sqrt(5)/10 + 1/2) * (sqrt(5)/2 + 5/2)^n.
G.f.: (1 - 2*x)/(1 - 5*x + 5*x^2).
a(n-1) = Sum_{k=1..n} binomial(n, k)*F(k)^2. - Benoit Cloitre, Oct 26 2003
a(n) = A090041(n)/2^n. - Paul Barry, Mar 23 2004
The sequence 0, 1, 3, 10, ... with a(n) = (5/2 - sqrt(5)/2)^n/5 + (5/2 + sqrt(5)/2)^n/5 - 2(0)^n/5 is the binomial transform of F(n)^2 (A007598). - Paul Barry, Apr 27 2004
From Paul Barry, Nov 15 2005: (Start)
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n, j)*binomial(j+k, 2k);
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n, k+j)*binomial(k, k-j)2^(n-k-j);
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} binomial(n+k-j, n-k-j)*binomial(k, j)(-1)^j*2^(n-k-j). (End)
a(n) = A111776(n, n). - Abdullahi Umar, Sep 14 2008
a(n) = Sum_{k=0..n} A094441(n,k)*2^k. - Philippe Deléham, Dec 14 2009
a(n+1) = Sum_{k=-floor(n/5)..floor(n/5)} ((-1)^k*binomial(2*n, n+5*k)/2). -Mircea Merca, Jan 28 2012
a(n) = A030191(n) - 2*A030191(n-1). - R. J. Mathar, Jul 19 2012
G.f.: Q(0,u)/x - 1/x, where u=x/(1-2*x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k+1 + u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
For n>=3: a(n) = a(n-1)*(3+(a(n-1) mod a(n-2) - a(n-2) mod a(n-3))/(a(n-2) - a(n-3))). - Bob Selcoe, Mar 17 2014
a(n) = sqrt(5)^(n-1)*(3*S(n-1, sqrt(5)) - sqrt(5)*S(n-2, sqrt(5))) with Chebyshev's S-polynomials (see A049310), where S(-1, x) = 0 and S(-2, x) = -1. This is the (1,1) entry of A^n with the matrix A=(3,1;1,2). See the comment by David Neil McGrath, Nov 18 2014. - Wolfdieter Lang, Dec 04 2014
Conjecture: a(n) = 2*a(n-1) + A039717(n). - Benito van der Zander, Nov 20 2015
a(n) = A189315(n+1) / 10. - Tom Copeland, Dec 08 2015
a(n) = A093129(n) + A030191(n-1). - Gary W. Adamson, Apr 24 2023
E.g.f.: exp(5*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024

A063967 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-2,k) + T(n-1,k-1) + T(n-2,k-1) and T(0,0) = 1.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 3, 7, 5, 1, 5, 15, 16, 7, 1, 8, 30, 43, 29, 9, 1, 13, 58, 104, 95, 46, 11, 1, 21, 109, 235, 271, 179, 67, 13, 1, 34, 201, 506, 705, 591, 303, 92, 15, 1, 55, 365, 1051, 1717, 1746, 1140, 475, 121, 17, 1, 89, 655, 2123, 3979, 4759, 3780, 2010, 703, 154, 19, 1

Offset: 0

Henry Bottomley, Sep 05 2001

			T(3,1) = T(2,1) + T(1,1) + T(2,0) + T(1,0) = 3 + 1 + 2 + 1 = 7.
Triangle begins:
   1,
   1,   1,
   2,   3,   1,
   3,   7,   5,   1,
   5,  15,  16,   7,   1,
   8,  30,  43,  29,   9,   1,
  13,  58, 104,  95,  46,  11,  1,
  21, 109, 235, 271, 179,  67, 13,  1,
  34, 201, 506, 705, 591, 303, 92, 15, 1

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
E. Deutsch, L. Ferrari and S. Rinaldi, Production Matrices, Advances in Mathematics, 34 (2005) pp. 101-122.
Emanuele Munarini, A generalization of André-Jeannin's symmetric identity, Pure Mathematics and Applications (2018) Vol. 27, No. 1, 98-118.

Row sums are A002605.
Columns include: A000045(n+1), A023610(n-1).
Main diagonal: A000012, a(n, n-1) = A005408(n-1).
Matrix inverse: A091698, matrix square: A091700.
Cf. A321620.
Sum_{k=0..n} x^k*T(n,k) is (-1)^n*A057086(n) (x=-11), (-1)^n*A057085(n+1) (x=-10), (-1)^n*A057084(n) (x=-9), (-1)^n*A030240(n) (x=-8), (-1)^n*A030192(n) (x=-7), (-1)^n*A030191(n) (x=-6), (-1)^n*A001787(n+1) (x=-5), A000748(n) (x=-4), A108520(n) (x=-3), A049347(n) (x=-2), A000007(n) (x=-1), A000045(n) (x=0), A002605(n) (x=1), A030195(n+1) (x=2), A057087(n) (x=3), A057088(n) (x=4), A057089(n) (x=5), A057090(n) (x=6), A057091(n) (x=7), A057092(n) (x=8), A057093(n) (x=9). - Philippe Deléham, Nov 03 2006

a063967_tabl = [1] : [1,1] : f [1] [1,1] where
   f us vs = ws : f vs ws where
     ws = zipWith (+) ([0] ++ us ++ [0]) $
          zipWith (+) (us ++ [0,0]) $ zipWith (+) ([0] ++ vs) (vs ++ [0])
-- Reinhard Zumkeller, Apr 17 2013

T[n_, k_] := Sum[Binomial[j, n - j]*Binomial[j, k], {j, 0, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 11 2017, after Paul Barry *)
(* Function RiordanSquare defined in A321620. *)
RiordanSquare[1/(1 - x - x^2), 11] // Flatten (* Peter Luschny, Nov 27 2018 *)

G.f.: 1/(1-x*(1+x)*(1+y)). - Vladeta Jovovic, Oct 11 2003
Riordan array (1/(1-x-x^2), x(1+x)/(1-x-x^2)). The inverse of the signed version (1/(1+x-x^2),x(1-x)/(1+x-x^2)) is abs(A091698). - Paul Barry, Jun 10 2005
T(n, k) = Sum_{j=0..n} C(j, n-j)C(j, k). - Paul Barry, Nov 09 2005
Diagonal sums are A002478. - Paul Barry, Nov 09 2005
A026729*A007318 as infinite lower triangular matrices. - Philippe Deléham, Dec 11 2008
Central coefficients T(2*n,n) are A137644. - Paul Barry, Apr 15 2010
Product of Riordan arrays (1, x(1+x))*(1/(1-x), x/(1-x)), that is, A026729*A007318. - Paul Barry, Mar 14 2011
Triangle T(n,k), read by rows, given by (1,1,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 12 2011

A020876 a(n) = ((5+sqrt(5))/2)^n + ((5-sqrt(5))/2)^n.

Original entry on oeis.org

2, 5, 15, 50, 175, 625, 2250, 8125, 29375, 106250, 384375, 1390625, 5031250, 18203125, 65859375, 238281250, 862109375, 3119140625, 11285156250, 40830078125, 147724609375, 534472656250, 1933740234375, 6996337890625, 25312988281250, 91583251953125

Offset: 0

N. J. A. Sloane

Number of no-leaf edge-subgraphs in Moebius ladder M_n.

			G.f. = 2 + 5*x + 15*x^2 + 50*x^3 + 175*x^4 + 625*x^5 + 2250*x^6 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Santiago Alzate, Oscar Correa, and Rigoberto Flórez, Fibonacci identities from Jordan Identities, arXiv:2009.02639 [math.NT], 2020.
Noah Giansiracusa, Fibonacci, Golden Ratio, and Vector Bundles, Mathematics (2021) Vol. 9, Issue 4, 426.
J. P. McSorley, Counting structures in the Moebius ladder, Discrete Math., 184 (1998), 137-164.
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Index entries for linear recurrences with constant coefficients, signature (5,-5).

Cf. A020876, A093131, A005248.

Appears in A109106. - Johannes W. Meijer, Jul 01 2010

[Floor(((5+Sqrt(5))/2)^n+((5-Sqrt(5))/2)^n): n in [0..30]]; // Vincenzo Librandi, Aug 08 2014

G:=(x,n)-> cos(x)^n+cos(3*x)^n:
seq(simplify(4^n*G(Pi/10,2*n)), n=0..22); # Gary Detlefs, Dec 05 2010

Table[Sum[LucasL[2*i] Binomial[n, i], {i, 0, n}], {n, 0, 50}] (* T. D. Noe, Sep 10 2011 *)
CoefficientList[Series[(2 - 5 x)/(1 - 5 x + 5 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 08 2014 *)
LinearRecurrence[{5,-5},{2,5},30] (* Harvey P. Dale, Mar 13 2016 *)

[lucas_number2(n,5,5) for n in range(0,24)] # Zerinvary Lajos, Jul 08 2008

Also, a(n) = (sqrt(5)*phi)^n + (sqrt(5)/phi)^n, where phi = golden ratio. - N. J. A. Sloane, Aug 08 2014
Let S(n, m)=sum(k=0, n, binomial(n, k)*fibonacci(m*k)), then for n>0 a(n)= S(2*n, 2)/S(n, 2). - Benoit Cloitre, Oct 22 2003
From R. J. Mathar, Feb 06 2010: (Start)
a(n)= 5*a(n-1) - 5*a(n-2).
G.f.: (2-5*x)/(1-5*x+5*x^2). (End)
From Johannes W. Meijer, Jul 01 2010: (Start)
Lim_{k->infinity} a(n+k)/a(k) = (A020876(n) + A093131(n)*sqrt(5))/2.
Lim_{k->infinity} A020876(n)/A093131(n) = sqrt(5). (End)
Binomial transform of A005248. - Carl Najafi, Sep 10 2011
a(n) = 2*A030191(n) - 5*A030191(n-1). - R. J. Mathar, Mar 02 2012
From Kai Wang, Dec 22 2019: (Start)
a((2*m+1)*k)/a(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*a(2*(m-i)*k) + 5^(m*k).
A093131(m+r)*A093131(n+s) + A093131(m+s)*A093131(n+r) = (2*a(m+n+r+s) - 5^(n+s)*a(m-n)*a(r-s))/5.
a(m+r)*a(n+s) - a(m+s)*a(n+r) = 5^(n+s+1)*A093131(m-n)*A093131(r-s).
a(m+r)*a(n+s) + a(m+s)*a(n+r) = 2*a(m+n+r+s) + 5^(n+s)*a(m-n)*a(r-s).
a(m+r)*a(n+s) - 5*A093131(m+s)*A093131(n+r) = 5^(n+s)*a(m-n)*a(r-s).
a(m+r)*a(n+s) + 5*A093131(m+s)*A093131(n+r) = 2*a(m+n+r+s)+ 5^(n+s+1)*A093131(m-n)*A093131(r-s).
A093131(m-n) = (A093131(m)*a(n) - a(m)*A093131(n))/(2*5^n).
A093131(m+n) = (A093131(m)*a(n) + a(m)*A093131(n))/2.
a(n)^2 - a(n+1)*a(n-1) = -5^n.
a(n)^2 - a(n+r)*a(n-r) = -5^(n-r+1)*A093131(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = -5^(n+1)*A093131(m-n).
a(m+n) - 5^(n)*a(m-n) = 5*A093131(m)*A093131(n).
a(m+n) + 5^(n)*a(m-n) =  a(m)*a(n).
a(m-n) = (a(m)*a(n) - 5*A093131(m)*A093131(n))/(2*5^n).
a(m+n) = (a(m)*a(n) + 5*A093131(m)*A093131(n))/2.  (End)
E.g.f.: 2*exp(5*x/2)*cosh(sqrt(5)*x/2). - Stefano Spezia, Dec 27 2019

Definition simplified by N. J. A. Sloane, Aug 08 2014

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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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A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

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A109466 Riordan array (1, x(1-x)).

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A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.