A128064 -id:A128064 - cp's OEIS frontend

A001844 Centered square numbers: a(n) = 2n(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

1, 5, 13, 25, 41, 61, 85, 113, 145, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2381, 2521, 2665, 2813, 2965, 3121, 3281, 3445, 3613, 3785, 3961, 4141, 4325, 4513

Offset: 0

N. J. A. Sloane

These are Hogben's central polygonal numbers denoted by
...2...
....P..
...4.n.
Numbers of the form (k^2+1)/2 for k odd.
(y(2x+1))^2 + (y(2x^2+2x))^2 = (y(2x^2+2x+1))^2. E.g., let y = 2, x = 1; (2(2+1))^2 + (2(2+2))^2 = (2(2+2+1))^2, (2(3))^2 + (2(4))^2 = (2(5))^2, 6^2 + 8^2 = 10^2, 36 + 64 = 100. - Glenn B. Cox (igloos_r_us(AT)canada.com), Apr 08 2002
a(n) is also the number of 3 X 3 magic squares with sum 3(n+1). - Sharon Sela (sharonsela(AT)hotmail.com), May 11 2002
For n > 0, a(n) is the smallest k such that zeta(2) - Sum_{i=1..k} 1/i^2 <= zeta(3) - Sum_{i=1..n} 1/i^3. - Benoit Cloitre, May 17 2002
Number of convex polyominoes with a 2 X (n+1) minimal bounding rectangle.
The prime terms are given by A027862. - Lekraj Beedassy, Jul 09 2004
First difference of a(n) is 4n = A008586(n). Any entry k of the sequence is followed by k + 2*(1 + sqrt(2k - 1)). - Lekraj Beedassy, Jun 04 2006
Integers of the form 1 + x + x^2/2 (generating polynomial is Schur's polynomial as in A127876). - Artur Jasinski, Feb 04 2007
If X is an n-set and Y and Z disjoint 2-subsets of X then a(n-4) is equal to the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Row sums of triangle A132778. - Gary W. Adamson, Sep 02 2007
Binomial transform of [1, 4, 4, 0, 0, 0, ...]; = inverse binomial transform of A001788: (1, 6, 24, 80, 240, ...). - Gary W. Adamson, Sep 02 2007
Narayana transform (A001263) of [1, 4, 0, 0, 0, ...]. Equals A128064 (unsigned) * [1, 2, 3, ...]. - Gary W. Adamson, Dec 29 2007
k such that the Diophantine equation x^3 - y^3 = x*y + k has a solution with y = x-1. If that solution is (x,y) = (m+1,m) then m^2 + (m+1)^2 = k. Note that this Diophantine equation is an elliptic curve and (m+1,m) is an integer point on it. - James R. Buddenhagen, Aug 12 2008
Numbers k such that (k, k, 2*k-2) are the sides of an isosceles triangle with integer area. Also, k such that 2*k-1 is a square. - James R. Buddenhagen, Oct 17 2008
a(n) is also the least weight of self-conjugate partitions having n+1 different odd parts. - Augustine O. Munagi, Dec 18 2008
Prefaced with a "1": (1, 1, 5, 13, 25, 41, ...) = A153869 * (1, 2, 3, ...). - Gary W. Adamson, Jan 03 2009
Prefaced with a "1": (1, 1, 5, 13, 25, 41, ...) where a(n) = 2n*(n-1)+1, all tuples of square numbers (X-Y, X, X+Y) are produced by ((m*(a(n)-2n))^2, (m*a(n))^2, (m*(a(n)+2n-2))^2) where m is a whole number. - Doug Bell, Feb 27 2009
Equals (1, 2, 3, ...) convolved with (1, 3, 4, 4, 4, ...). E.g., a(3) = 25 = (1, 2, 3, 4) dot (4, 4, 3, 1) = (4 + 8 + 9 + 4). - Gary W. Adamson, May 01 2009
The running sum of squares taken two at a time. - Al Hakanson (hawkuu(AT)gmail.com), May 18 2009
Equals the odd integers convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 25 2009
Equals the triangular numbers convolved with [1, 2, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
When the positive integers are written in a square array by diagonals as in A038722, a(n) gives the numbers appearing on the main diagonal. - Joshua Zucker, Jul 07 2009
The finite continued fraction [n,1,1,n] = (2n+1)/(2n^2 + 2n + 1) = (2n+1)/a(n); and the squares of the first two denominators of the convergents = a(n). E.g., the convergents and value of [4,1,1,4] = 1/4, 1/5, 2/9, 9/41 where 4^2 + 5^2 = 41. - Gary W. Adamson, Jul 15 2010
From Keith Tyler, Aug 10 2010: (Start)
Running sum of A008574.
Square open pyramidal number; that is, the number of elements in a square pyramid of height (n) with only surface and no bottom nodes. (End)
For k>0, x^4 + x^2 + k factors over the integers iff sqrt(k) is in this sequence. - James R. Buddenhagen, Aug 15 2010
Create the simple continued fraction from Pythagorean triples to get [2n + 1; 2n^2 + 2n, 2n^2 + 2n + 1]; its value equals the rational number 2n + 1 + a(n) / (4n^4 + 8n^3 + 6n^2 + 2n + 1). - J. M. Bergot, Sep 30 2011
a(n), n >= 1, has in its prime number factorization only primes of the form 4*k+1, i.e., congruent to 1 (mod 4) (see A002144). This follows from the fact that a(n) is a primitive sum of two squares and odd. See Theorem 3.20, p. 164, in the given Niven-Zuckerman-Montgomery reference. E.g., a(3) = 25 = 5^2, a(6) = 85 = 5*17. - Wolfdieter Lang, Mar 08 2012
From Ant King, Jun 15 2012: (Start)
a(n) is congruent to 1 (mod 4) for all n.
The digital roots of the a(n) form a purely periodic palindromic 9-cycle 1, 5, 4, 7, 5, 7, 4, 5, 1.
The units' digits of the a(n) form a purely periodic palindromic 5-cycle 1, 5, 3, 5, 1.
(End)
Number of integer solutions (x,y) of |x| + |y| <= n. Geometrically: number of lattice points inside a square with vertices (n,0), (0,-n), (-n,0), (0,n). - César Eliud Lozada, Sep 18 2012
(a(n)-1)/a(n) = 2*x / (1+x^2) where x = n/(n+1). Note that in this form, this is the velocity-addition formula according to the special theory of relativity (two objects traveling at 1/(n+1) slower than c relative to each other appear to travel at 1/a(n) less than c to a stationary observer). - Christian N. K. Anderson, May 20 2013 [Corrected by Rémi Guillaume, May 22 2025]
A geometric curiosity: the envelope of the circles x^2 + (y-a(n)/2)^2 = ((2n+1)/2)^2 is the parabola y = x^2, the n=0 circle being the osculating circle at the parabola vertex. - Jean-François Alcover, Dec 02 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; a(n-1) gives the number of internal regions into which the plane is divided (cf. A051890, A046092); a(n-1) = A051890(n) - 1 = A046092(n-1) + 1. - Jaroslav Krizek, Dec 27 2013
a(n) is also, of course, the scalar product of the 2-vector (n, n+1) (or (n+1, n)) with itself. The unique inverse of (n, n+1) as vector in the Clifford algebra over the Euclidean 2-space is (1/a(n))(0, n, n+1, 0) (similarly for the other vector). In general the unique inverse of such a nonzero vector v (odd element in Cl_2) is v^(-1) = (1/|v|^2) v. Note that the inverse with respect to the scalar product is not unique for any nonzero vector. See the P. Lounesto reference, sects. 1.7 - 1.12, pp. 7-14. See also the Oct 15 2014 comment in A147973. - Wolfdieter Lang, Nov 06 2014
Subsequence of A004431, for n >= 1. - Bob Selcoe, Mar 23 2016
Numbers k such that 2k - 1 is a perfect square. - Juri-Stepan Gerasimov, Apr 06 2016
The number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 574", based on the 5-celled von Neumann neighborhood. - Robert Price, May 13 2016
a(n) is the first integer in a sum of (2*n + 1)^2 consecutive integers that equals (2*n + 1)^4. - Patrick J. McNab, Dec 24 2016
Central elements of odd-length rows of the triangular array of positive integers. a(n) is the mean of the numbers in the (2*n + 1)-th row of this triangle. - David James Sycamore, Aug 01 2018
Intersection of A000982 and A080827. - David James Sycamore, Aug 07 2018
An off-diagonal of the array of Delannoy numbers, A008288, (or a row/column when the array is shown as a square). As such, this is one of the crystal ball sequences. - Jack W Grahl, Feb 15 2021 and Shel Kaphan, Jan 18 2023
a(n) appears as a solution to a "Riddler Express" puzzle on the FiveThirtyEight website. The Jan 21 2022 issue (problem) and the Jan 28 2022 issue (solution) present the following puzzle and include a proof. - Fold a square piece of paper in half, obtaining a rectangle. Fold again to obtain a square with 1/4 the size of the original square. Then make n cuts through the folded paper. a(n) is the greatest number of pieces of the unfolded paper after the cutting. - Manfred Boergens, Feb 22 2022
a(n) is (1/6) times the number of 2 X 2 triangles in the n-th order hexagram with 12*n^2 cells. - Donghwi Park, Feb 06 2024
If k is a centered square number, its index in this sequence is n = (sqrt(2k-1)-1)/2. - Rémi Guillaume, Mar 30 2025.
Row sums of the symmetric triangle of odd numbers [1]; [1, 3, 1]; [1, 3, 5, 3, 1]; [1, 3, 5, 7, 5, 3, 1]; .... - Marco Zárate, Jun 15 2025

			G.f.: 1 + 5*x + 13*x^2 + 25*x^3 + 41*x^4 + 61*x^5 + 85*x^6 + 113*x^7 + 145*x^8 + ...
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25), ...
The first four such partitions, corresponding to n = 0,1,2,3, i.e., to a(n) = 1,5,13,25, are 1, 3+1+1, 5+3+3+1+1, 7+5+5+3+3+1+1. - _Augustine O. Munagi_, Dec 18 2008

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
Pertti Lounesto, Clifford Algebras and Spinors, second edition, Cambridge University Press, 2001.
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 483.
Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.
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).
Travers et al., The Mysterious Lost Proof, Using Advanced Algebra, (1976), pp. 27.

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Ahmed, J. De Loera and R. Hemmecke, Polyhedral Cones of Magic Cubes and Squares, arXiv:math/0201108 [math.CO], 2002.
U. Alfred, n and n+1 consecutive integers with equal sums of squares, Math. Mag., 35 (1962), 155-164.
Bela Bajnok, Additive Combinatorics: A Menu of Research Problems, arXiv:1705.07444 [math.NT], May 2017. See Section 2.3.
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Matthias Beck, Moshe Cohen, Jessica Cuomo, and Paul Gribelyuk, The number of "magic" squares and hypercubes, arXiv:math/0201013 [math.CO], 2002-2005.
Arthur T. Benjamin and Doron Zeilberger, Pythagorean Primes and Palindromic Continued Fractions, Electronic Journal of Combinatorial Number Theory, 5(1) 2005, #A30.
J. A. De Loera, D. C. Haws and M. Koppe, Ehrhart Polynomials of Matroid Polytopes and Polymatroids, arXiv:0710.4346 [math.CO], 2007; Discrete Comput. Geom., 42 (2009), 670-702.
FiveThirtyEight, "Riddler Express" paper cutting problem and solution, Jan 28 2022.
D. C. Haws, Matroids [Broken link, Oct 30 2017]
D. C. Haws, Matroids [Copy on website of Matthias Koeppe]
D. C. Haws, Matroids [Cached copy, pdf file only]
L. Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, pp. 22 and 36.
Milan Janjic, Two Enumerative Functions. [Broken link; WayBackMachine archive.]
Milan Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7, 2004.
Ron Knott, Pythagorean Triples and Online Calculators.
G. Kreweras, Sur les hiérarchies de segments, Cahiers Bureau Universitaire Recherche Opérationnelle, Cahier 20, Inst. Statistiques, Univ. Paris, 1973.
G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973). (Annotated scanned copy)
A. O. Munagi, Pairing conjugate partitions by residue classes, Discrete Math., 308 (2008), 2492-2501.
Mitchell Paukner, Lucy Pepin, Manda Riehl, and Jarred Wieser, Pattern Avoidance in Task-Precedence Posets, arXiv:1511.00080 [math.CO], 2015-2016.
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
John A. Jr. Rochowicz, Harmonic Numbers: Insights, Approximations and Applications, Spreadsheets in Education (eJSiE), 2015, Vol. 8: Iss. 2, Article 4.
Amelia Carolina Sparavigna, Groupoid of OEIS A001844 Numbers (centered square numbers), Politecnico di Torino, Italy (2019).
R. G. Stanton and D. D. Cowan, Note on a "square" functional equation, SIAM Rev., 12 (1970), 277-279.
David James Sycamore, Triangular array
Leo Tavares, Illustration: Diamond Rows
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.
Eric Weisstein's World of Mathematics, Centered Polygonal Number, Centered Square Number, Diamond, Pythagorean Triple, and von Neumann Neighborhood.
Index entries for sequences related to centered polygonal numbers
Index entries for two-way infinite sequences.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Nicolay Avilov, Graphical representation of the sequence members

X values are A005408; Y values are A046092.
Cf. A008586 (first differences), A005900 (partial sums), A254373 (digital roots).
Cf. A069894, A000217, A000290, A001263, A001788, A001845, A002061, A002144, A003215, A005448, A005891, A005917, A008844 (square terms), A027862 (prime terms), A048395, A051890, A056106, A101096, A127876, A128064, A132778, A147973, A153869, A240876, A251599, A000982, A080827.
Subsequence of A004431.
Right edge of A055096; main diagonal of A069480, A078475, A129312.
Row n=2 (or column k=2) of A008288.
Cf. A016754.

a001844 n = 2 * n * (n + 1) + 1
a001844_list = zipWith (+) a000290_list $ tail a000290_list
-- Reinhard Zumkeller, Dec 04 2012

[2*n^2 + 2*n + 1: n in [0..50]]; // Vincenzo Librandi, Jan 19 2013

[n: n in [0..4400] | IsSquare(2*n-1)]; // Juri-Stepan Gerasimov, Apr 06 2016

A001844:=-(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation

Table[2n(n + 1) + 1, {n, 0, 50}]
FoldList[#1 + #2 &, 1, 4 Range@ 50] (* Robert G. Wilson v, Feb 02 2011 *)
maxn := 47; Flatten[Table[SeriesCoefficient[Series[(n + (n - 1)*x)/(1 - x)^2, {x, 0, maxn}], k], {n, maxn}, {k, n - 1, n - 1}]] (* L. Edson Jeffery, Aug 24 2014 *)
CoefficientList[ Series[-(x^2 + 2x + 1)/(x - 1)^3, {x, 0, 48}], x] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 5, 13}, 48] (* Robert G. Wilson v, Aug 01 2018 *)
Total/@Partition[Range[0,50]^2,2,1] (* Harvey P. Dale, Dec 05 2020 *)
Table[ j! Coefficient[Series[Exp[x]*(1 + 4*x + 2*x^2), {x, 0, 20}], x,
j], {j, 0, 20}] (* Nikolaos Pantelidis, Feb 07 2023 *)

PARI
```
{a(n) = 2*n*(n+1) + 1};
```

x='x+O('x^200); Vec((1+x)^2/(1-x)^3) \\ Altug Alkan, Mar 23 2016

print([2*n*(n+1)+1 for n in range(48)]) # Michael S. Branicky, Jan 05 2021

[i**2 + (i + 1)**2 for i in range(46)] # Zerinvary Lajos, Jun 27 2008

a(n) = 2*n^2 + 2*n + 1 = n^2 + (n+1)^2.
a(n) = 1 + 3 + 5 + ... + 2*n-1 + 2*n+1 + 2*n-1 + ... + 3 + 1. - Amarnath Murthy, May 28 2001
a(n) = 1/real(z(n+1)) where z(1)=i, (i^2=-1), z(k+1) = 1/(z(k)+2i). - Benoit Cloitre, Aug 06 2002
Nearest integer to 1/Sum_{k>n} 1/k^3. - Benoit Cloitre, Jun 12 2003
G.f.: (1+x)^2/(1-x)^3.
E.g.f.: exp(x)*(1+4x+2x^2).
a(n) = a(n-1) + 4n.
a(-n) = a(n-1).
a(n) = A064094(n+3, n) (fourth diagonal).
a(n) = 1 + Sum_{j=0..n} 4*j. - Xavier Acloque, Oct 08 2003
a(n) = A046092(n)+1 = (A016754(n)+1)/2. - Lekraj Beedassy, May 25 2004
a(n) = Sum_{k=0..n+1} (-1)^k*binomial(n, k)*Sum_{j=0..n-k+1} binomial(n-k+1, j)*j^2. - Paul Barry, Dec 22 2004
a(n) = ceiling((2n+1)^2/2). - Paul Barry, Jul 16 2006
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=1, a(1)=5, a(2)=13. - Jaume Oliver Lafont, Dec 02 2008
a(n)*a(n-1) = 4*n^4 + 1 for n > 0. - Reinhard Zumkeller, Feb 12 2009
Prefaced with a "1" (1, 1, 5, 13, 25, 41, ...): a(n) = 2*n*(n-1)+1. - Doug Bell, Feb 27 2009
a(n) = sqrt((A056220(n)^2 + A056220(n+1)^2) / 2). - Doug Bell, Mar 08 2009
a(n) = floor(2*(n+1)^3/(n+2)). - Gary Detlefs, May 20 2010
a(n) = A000330(n) - A000330(n-2). - Keith Tyler, Aug 10 2010
a(n) = A069894(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = 2*a(n-1) - a(n-2) + 4. - Ant King, Jun 12 2012
Sum_{n>=0} 1/a(n) = (Pi/2)*tanh(Pi/2) = 1.4406595199775... = A228048. - Ant King, Jun 15 2012
a(n) = A209297(2*n+1,n+1). - Reinhard Zumkeller, Jan 19 2013
a(n)^3 = A048395(n)^2 + A048395(-n-1)^2. - Vincenzo Librandi, Jan 19 2013
a(n) = A000217(2n+1) - n. - Ivan N. Ianakiev, Nov 08 2013
a(n) = A251599(3*n+1). - Reinhard Zumkeller, Dec 13 2014
a(n) = A101321(4,n). - R. J. Mathar, Jul 28 2016
From Ilya Gutkovskiy, Jul 30 2016: (Start)
a(n) = Sum_{k=0..n} A008574(k).
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = exp(-1) = A068985. (End)
a(n) = 4 * A000217(n) + 1. - Bruce J. Nicholson, Jul 10 2017
a(n) = A002522(n) + A005563(n) = A002522(n+1) + A005563(n-1). - Bruce J. Nicholson, Aug 05 2017
Sum_{n>=0} a(n)/n! = 7*e. Sum_{n>=0} 1/a(n) = A228048. - Amiram Eldar, Jun 20 2020
a(n) = A000326(n+1) + A000217(n-1). - Charlie Marion, Nov 16 2020
a(n) = Integral_{x=0..2n+2} |1-x| dx. - Pedro Caceres, Dec 29 2020
From Amiram Eldar, Feb 17 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)*sech(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = Pi*csch(Pi)*sinh(Pi/2). (End)
a(n) = A001651(n+1) + 1 - A028242(n). - Charlie Marion, Apr 05 2022
a(n) = A016754(n) - A046092(n). - Leo Tavares, Sep 16 2022
For n>0, a(n) = A101096(n+2) / 30. - Andy Nicol, Feb 06 2025
From Rémi Guillaume, Apr 21 2025: (Start)
a(n) = (2*A003215(n)+1)/3.
a(n) = (4*A005448(n+1)-1)/3.
a(n) + a(n-1) = A001845(n) - A001845(n-1), for n >= 1.
a(n) = (A005917(n+1))/(2n+1). (End)

Partially edited by Joerg Arndt, Mar 11 2010

A001792 a(n) = (n+2)*2^(n-1).

Original entry on oeis.org

1, 3, 8, 20, 48, 112, 256, 576, 1280, 2816, 6144, 13312, 28672, 61440, 131072, 278528, 589824, 1245184, 2621440, 5505024, 11534336, 24117248, 50331648, 104857600, 218103808, 452984832, 939524096, 1946157056, 4026531840, 8321499136, 17179869184, 35433480192

Offset: 0

N. J. A. Sloane

Number of parts in all compositions (ordered partitions) of n + 1. For example, a(2) = 8 because in 3 = 2 + 1 = 1 + 2 = 1 + 1 + 1 we have 8 parts. Also number of compositions (ordered partitions) of 2n + 1 with exactly 1 odd part. For example, a(2) = 8 because the only compositions of 5 with exactly 1 odd part are 5 = 1 + 4 = 2 + 3 = 3 + 2 = 4 + 1 = 1 + 2 + 2 = 2 + 1 + 2 = 2 + 2 + 1. - Emeric Deutsch, May 10 2001
Binomial transform of natural numbers [1, 2, 3, 4, ...].
For n >= 1 a(n) is also the determinant of the n X n matrix with 3's on the diagonal and 1's elsewhere. - Ahmed Fares (ahmedfares(AT)my-deja.com), May 06 2001
The arithmetic mean of first n terms of the sequence is 2^(n-1). - Amarnath Murthy, Dec 25 2001, corrected by M. F. Hasler, Dec 17 2016
Also the number of "winning paths" of length n across an n X n Hex board. Satisfies the recursion a(n) = 2a(n-1) + 2^(n-2). - David Molnar (molnar(AT)stolaf.edu), Apr 10 2002
Diagonal in A053218. - Benoit Cloitre, May 08 2002
Let M_n be the n X n matrix m_(i, j) = 1 + abs(i-j) then det(M_n) = (-1)^(n-1)*a(n-1). - Benoit Cloitre, May 28 2002
Absolute value of determinant of n X n matrix of form: [1 2 3 4 5 / 2 1 2 3 4 / 3 2 1 2 3 / 4 3 2 1 2 / 5 4 3 2 1]. - Benoit Cloitre, Jun 20 2002
Number of ones in all (n+1)-bit integers (cf. A000120). - Ralf Stephan, Aug 02 2003
This sequence also emerges as a floretion force transform of powers of 2 (see program code). Define a(-1) = 0 (as the sequence is returned by FAMP). Then a(n-1) + A098156(n+1) = 2*a(n) (conjecture). - Creighton Dement, Mar 14 2005
This sequence gives the absolute value of the determinant of the Toeplitz matrix with first row containing the first n integers. - Paul Max Payton, May 23 2006
Equals sums of rows right of left edge of A102363 divided by three, + 2^K. - David G. Williams (davidwilliams(AT)paxway.com), Oct 08 2007
If X_1, X_2, ..., X_n are 2-blocks of a (2n+1)-set X then, for n >= 1, a(n) is the number of (n+1)-subsets of X intersecting each X_i, (i = 1, 2, ..., n). - Milan Janjic, Nov 18 2007
Also, a(n-1) is the determinant of the n X n matrix with A[i, j] = n - |i-j|. - M. F. Hasler, Dec 17 2008
1/2 the number of permutations of 1 .. (n+2) arranged in a circle with exactly one local maximum. - R. H. Hardin, Apr 19 2009
The first corrector line for transforming 2^n offset 0 with a leading 1 into the Fibonacci sequence. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
a(n) is the number of runs of consecutive 1's in all binary sequences of length (n+1). - Geoffrey Critzer, Jul 02 2009
Let X_j (0 < j <= 2^n) all the subsets of N_n; m(i, j) := if {i} in X_j then 1 else 0. Let A = transpose(M).M; Then a(i, j) = (number of elements)|X_i intersect X_j|. Determinant(X*I-A) = (X-(n+1)*2^(n-2))*(X-2^(n-2))^(n-1)*X^(2^n-n).
  Eigenvector for (n+1)*2^(n-2) is V_i=|X_i|.
  Sum_{k=1..2^n} |X_i intersect X_k|*|X_k| = (n+1)*2^(n-2)*|X_i|.
  Eigenvectors for 2^(n-2) are {line(M)[i] - line(M)[j], 1 <= i, j <= n}. - CLARISSE Philippe (clarissephilippe(AT)yahoo.fr), Mar 24 2010
The sequence b(n) = 2*A001792(n), for n >= 1 with b(0) = 1, is an elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 187, 190, 250 and 442, lead to the b(n) sequence. For the corner squares these vectors lead to the companion sequence A134401. - Johannes W. Meijer, Aug 15 2010
Equals partial sums of A045623: (1, 2, 5, 12, 28, ...); where A045623 = the convolution square of (1, 1, 2, 4, 8, 16, 32, ...). - Gary W. Adamson, Oct 26 2010
Equals (1, 2, 4, 8, 16, ...) convolved with (1, 1, 2, 4, 8, 16, ...); e.g., a(3) = 20 = (1, 1, 2, 4) dot (8, 4, 2, 1) = (8 + 4 + 4 + 4). - Gary W. Adamson, Oct 26 2010
This sequence seems to give the first x+1 nonzero terms in the sequence derived by subtracting the m-th term in the x_binacci sequence (where the first term is one and the y-th term is the sum of x terms immediately preceding it) from 2^(m-2). - Dylan Hamilton, Nov 03 2010
Recursive formulas for a(n) are in many cases derivable from its property wherein delta^k(a(n)) - a(n) = k*2^n where delta^k(a(n)) represents the k-th forward difference of a(n). Provable with a difference table and a little induction. - Ethan Beihl, May 02 2011
Let f(n,k) be the sum of numbers in the subsets of size k of {1, 2, ..., n}. Then a(n-1) is the average of the numbers f(n, 0), ... f(n, n). Example: (f(3, 1), f(3, 2), f(3, 3)) = (6, 12, 6), with average (6+12+6)/3. - Clark Kimberling, Feb 24 2012
a(n) is the number of length-2n binary sequences that contain a subsequence of ones with length n or more. To derive this result, note that there are 2^n sequences where the initial one of the subsequence occurs at entry one. If the initial one of the subsequence occurs at entry 2, 3, ..., or n + 1, there are 2^(n-1) sequences since a zero must precede the initial one. Hence a(n) = 2^n + n*2^(n-1)=(n+2)2^(n-1). An example is given in the example section below. - Dennis P. Walsh, Oct 25 2012
As the total number of parts in all compositions of n+1 (see the first line in Comments) the equivalent sequence for partitions is A006128. On the other hand, as the first differences of A001787 (see the first line in Crossrefs) the equivalent sequence for partitions is A138879. - Omar E. Pol, Aug 28 2013
a(n) is the number of spanning trees of the complete tripartite graph K_{n,1,1}. - James Mahoney, Oct 24 2013
a(n-1) = denominator of the mean (2n/(n+1), after reduction), of the compositions of n; numerator is given by A022998(n). - Clark Kimberling, Mar 11 2014
From Tom Copeland, Nov 09 2014: (Start)
The shifted array belongs to an interpolated family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the interpolating o.g.f. (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse x(1-x)/(1+(t-1)x(1-x)). See A091867 for more info on this family. Here the interpolation is t=-3 (mod signs in the results).
Let C(x) = (1 - sqrt(1-4x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t).
Shifted o.g.f: G(x) = x*(1-x)/(1 - 4x*(1-x)) = P[Cinv(x),-4].
Inverse o.g.f: Ginv(x) = [1 - sqrt(1 - 4*x/(1+4x))]/2 = C[P(x, 4)] (signed shifted A001700). Cf. A030528. (End)
For n > 0, element a(n) of the sequence is equal to the gradients of the (n-1)-th row of Pascal triangle multiplied with the square of the integers from n+1,...,1. I.e., row 3 of Pascal's triangle 1,3,3,1 has gradients 1,2,0,-2,-1, so a(4) = 1*(5^2) + 2*(4^2) + 0*(3^2) - 2*(2^2) - 1*(1^2) = 48. - Jens Martin Carlsson, May 18 2017
Number of self-avoiding paths connecting all the vertices of a convex (n+2)-gon. - Ivaylo Kortezov, Jan 19 2020
a(n-1) is the total number of elements of subsets of {1,2,..,n} that contain n.  For example, for n = 3, a(2) = 8, and the subsets of {1,2,3} that contain 3 are {3}, {1,3}, {2,3}, {1,2,3}, with a total of 8 elements. - Enrique Navarrete, Aug 01 2020

			a(0) = 1, a(1) = 2*1 + 1 = 3, a(2) = 2*3 + 2 = 8, a(3) = 2*8 + 4 = 20, a(4) = 2*20 + 8 = 48, a(5) = 2*48 + 16 = 112, a(6) = 2*112 + 32 = 256, ... - _Philippe Deléham_, Apr 19 2009
a(2) = 8 since there are 8 length-4 binary sequences with a subsequence of ones of length 2 or more, namely, 1111, 1110, 1101, 1011, 0111, 1100, 0110, and 0011. - _Dennis P. Walsh_, Oct 25 2012
G.f. = 1 + 3*x + 8*x^2 + 20*x^3 + 48*x^4 + 112*x^5 + 256*x^6 + 576*x^7 + ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
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).
A. M. Stepin and A. T. Tagi-Zade, Words with restrictions, pp. 67-74 of Kvant Selecta: Combinatorics I, Amer. Math. Soc., 2001 (G_n on p. 70).

T. D. Noe, Table of n, a(n) for n = 0..500
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Marco Abrate, Stefano Barbero, Umberto Cerruti and Nadir Murru, Colored compositions, Invert operator and elegant compositions with the "black tie", Discrete Math., Vol. 335 (2014), pp. 1-7. MR3248794.
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Colored compositions, Invert operator and elegant compositions with the "black tie", arXiv:1409.6454 [math.NT], 2014.
Milica Andelic, C. M. da Fonseca and A. Pereira, The mu-permanent, a new graph labeling, and a known integer sequence, arXiv preprint arXiv:1609.04208 [math.CO], 2016.
Félix Balado and Guénolé C. M. Silvestre, Runs of Ones in Binary Strings, arXiv:2302.11532 [math.CO], 2023. See pp. 6-7.
Jean-Luc Baril and Nathanaël Hassler, Intervals in a family of Fibonacci lattices, Univ. de Bourgogne (France, 2024). See p. 7.
Neil J. Calkin, A curious binomial identity, Discr. Math., Vol. 131, No. 1-3 (1994), pp. 335-337.
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs., Vol. 3 (2000), #00.1.5.
Filippo Disanto and Simone Rinaldi, Symmetric convex permutominoes and involutions, PU. M. A., Vol. 22, No. 1 (2011), pp. 39-60.
Frank Ellermann, Illustration of binomial transforms.
David Eppstein, Non-crossing Hamiltonian Paths and Cycles in Output-Polynomial Time, arXiv:2303.00147 [cs.CG], 2023, pp. 2, 19.
Guillermo Esteban, Clemens Huemer and Rodrigo I. Silveira, New production matrices for geometric graphs, arXiv:2003.00524 [math.CO], 2020.
M. Hirschhorn, Calkin's binomial identity, Discr. Math., Vol. 159, No. 1-3 (1996), pp. 273-278.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 146.
Milan Janjić, Two Enumerative Functions
Milan Janjić and Boris Petković, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
Milan Janjić and Boris Petković, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
C. W. Jones, J. C. P. Miller, J. F. C. Conn and R. C. Pankhurst, Tables of Chebyshev polynomials, Proc. Roy. Soc. Edinburgh. Sect. A., Vol. 62, No. 2 (1946), pp. 187-203.
Sergey Kitaev and Jeffrey B. Remmel, A note on p-Ascent Sequences, Preprint, 2016.
Sergey Kitaev and Jeffrey Remmel, p-Ascent Sequences, arXiv preprint arXiv:1503.00914 [math.CO], 2015.
Ivaylo Kortezov, problem 8.4 ("Задача 8.4" in Bulgarian) in National Math Contest "Atanas Radev" 2020.
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Maohua Le, Two Classes of Smarandache Determinants, Scientia Magna, Vol. 2, No. 1 (2006), pp. 20-25.
Donatella Merlini and Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
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.
Silvana Ramaj, New Results on Cyclic Compositions and Multicompositions, Master's Thesis, Georgia Southern Univ., 2021.
John Riordan and N. J. A. Sloane, Correspondence, 1974.
N. J. A. Sloane, Transforms.
I. Tasoulas, K. Manes, and A. Sapounakis, Hamiltonian intervals in the lattice of binary paths, Elect. J. Comb. (2024) Vol. 31, Issue 1, P1.39.
Jun Wang and Zhizheng Zhang, On extensions of Calkin's binomial identities, Discrete Math., Vol. 274 (2004), 331-342.
Index entries for linear recurrences with constant coefficients, signature (4,-4).
Index entries for sequences related to Chebyshev polynomials.

First differences of A001787.
a(n) = A049600(n, 1), a(n) = A030523(n + 1, 1).
Cf. A053113.
Row sums of triangles A008949 and A055248.
a(n) = -A039991(n+2, 2).
Cf. A130584, A125092, A128064, A000079, A164910, A045623, A000120, A053120.
Cf. A000108, A005043, A091867, A001700, A030528.
If the exponent E in a(n) = Sum_{m=0..n} (Sum_{k=0..m} C(n,k))^E is 1, 2, 3, 4, 5 we get A001792, A003583, A007403, A294435, A294436 respectively.

List([0..35],n->(n+2)*2^(n-1)); # Muniru A Asiru, Sep 25 2018

a001792 n = a001792_list !! n
a001792_list = scanl1 (+) a045623_list
-- Reinhard Zumkeller, Jul 21 2013

[(n+2)*2^(n-1): n in [0..40]]; // Vincenzo Librandi, Nov 10 2014

A001792 := n-> (n+2)*2^(n-1);
spec := [S, {B=Set(Z, 0 <= card), S=Prod(Z, B, B)}, labeled]: seq(combstruct[count](spec, size=n)/4, n=2..30); # Zerinvary Lajos, Oct 09 2006
A001792:=-(-3+4*z)/(2*z-1)^2; # Simon Plouffe in his 1992 dissertation, which gives the sequence without the initial 1
G(x):=1/exp(2*x)*(1-x): f[0]:=G(x): for n from 1 to 54 do f[n]:=diff(f[n-1],x) od: x:=0: seq(abs(f[n]),n=0..28 ); # Zerinvary Lajos, Apr 17 2009
a := n -> hypergeom([-n, 2], [1], -1);
seq(round(evalf(a(n),32)), n=0..31); # Peter Luschny, Aug 02 2014

matrix[n_Integer /; n >= 1] := Table[Abs[p - q] + 1, {q, n}, {p, n}]; a[n_Integer /; n >= 1] := Abs[Det[matrix[n]]] (* Josh Locker (joshlocker(AT)macfora.com), Apr 29 2004 *)
g[n_,m_,r_] := Binomial[n - 1, r - 1] Binomial[m + 1, r] r; Table[1 + Sum[g[n, k - n, r], {r, 1, k}, {n, 1, k - 1}], {k, 1, 29}] (* Geoffrey Critzer, Jul 02 2009 *)
a[n_] := (n + 2)*2^(n - 1); a[Range[0, 40]] (* Vladimir Joseph Stephan Orlovsky, Feb 09 2011 *)
LinearRecurrence[{4, -4}, {1, 3}, 40] (* Harvey P. Dale, Aug 29 2011 *)
CoefficientList[Series[(1 - x) / (1 - 2 x)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *)
b[i_]:=i; a[n_]:=Abs[Det[ToeplitzMatrix[Array[b, n], Array[b, n]]]]; Array[a, 40] (* Stefano Spezia, Sep 25 2018 *)
a[n_]:=Hypergeometric2F1[2,-n+1,1,-1];Array[a,32] (* Giorgos Kalogeropoulos, Jan 04 2022 *)

A001792(n)=(n+2)<<(n-1) \\ M. F. Hasler, Dec 17 2008

for n in range(0,40): print(int((n+2)*2**(n-1)), end=' ') # Stefano Spezia, Oct 16 2018

a(n) = (n+2)*2^(n-1).
G.f.: (1 - x)/(1 - 2*x)^2 = 2F1(1,3;2;2x).
a(n) = 4*a(n-1) - 4*a(n-2).
G.f. (-1 + (1-2*x)^(-2))/(x*2^2). - Wolfdieter Lang
a(n) = A018804(2^n). - Matthew Vandermast, Mar 01 2003
a(n) = Sum_{k=0..n+2} binomial(n+2, 2k)*k. - Paul Barry, Mar 06 2003
a(n) = (1/4)*A001787(n+2). - Emeric Deutsch, May 24 2003
With a leading 0, this is ((n+1)2^n - 0^n)/4 = Sum_{m=0..n} binomial(n - 1, m - 1)*m, the binomial transform of A004526(n+1). - Paul Barry, Jun 05 2003
a(n) = Sum_{k=0..n} binomial(n, k)*(k + 1). - Lekraj Beedassy, Jun 24 2004
a(n) = A000244(n) - A066810(n). - Ross La Haye, Apr 29 2006
Row sums of triangle A130585. - Gary W. Adamson, Jun 06 2007
Equals A125092 * [1/1, 1/2, 1/3, ...]. - Gary W. Adamson, Nov 16 2007
a(n) = (n+1)*2^n - n*2^(n-1). Equals A128064 * A000079. - Gary W. Adamson, Dec 28 2007
G.f.: F(3, 1; 2; 2x). - Paul Barry, Sep 03 2008
a(n) = 1 + Sum_{k=1..n} (n - k + 4)2^(n - k - 1). This follows from the result that the number of parts equal to k in all compositions of n is (n - k + 3)2^(n - k - 2) for 0 < k < n. - Geoffrey Critzer, Sep 21 2008
a(n) = 2^(n-1) + 2 a(n-1) ; a(n-1) = det(n - |i - j|){i, j = 1..n}. - _M. F. Hasler, Dec 17 2008
a(n) = 2*a(n-1) + 2^(n-1). - Philippe Deléham, Apr 19 2009
a(n) = A164910(2^n). - Gary W. Adamson, Aug 30 2009
a(n) = Sum_{i=1..2^n} gcd(i, 2^n) = A018804(2^n). So we have: 2^0 * phi(2^n) + ... + 2^n * phi(2^0) = (n + 2)*2^(n-1), where phi is the Euler totient function. - Jeffrey R. Goodwin, Nov 11 2011
a(n) = Sum_{j=0..n} Sum_{i=0..n} binomial(n, i + j). - Yalcin Aktar, Jan 17 2012
Eigensequence of an infinite lower triangular matrix with 2^n as the left border and the rest 1's. - Gary W. Adamson, Jan 30 2012
G.f.: 1 + 2*x*U(0) where U(k) =  1 + (k + 1)/(2 - 8*x/(4*x + (k + 1)/U(k + 1))); (continued fraction, 3 - step). - Sergei N. Gladkovskii, Oct 19 2012
a(n) = Sum_{k=0..n} Sum_{j=0..k} binomial(n,j). - Peter Luschny, Dec 03 2013
a(n) = Hyper2F1([-n, 2], [1], -1). - Peter Luschny, Aug 02 2014
G.f.: 1 / (1 - 3*x / (1 + x / (3 - 4*x))). - Michael Somos, Aug 26 2015
a(n) = -A053120(2+n, n), n >= 0, the negative of the third (sub)diagonal of the triangle of Chebyshev's T polynomials. - Wolfdieter Lang, Nov 26 2019
From Amiram Eldar, Jan 12 2021: (Start)
Sum_{n>=0} 1/a(n) = 8*log(2) - 4.
Sum_{n>=0} (-1)^n/a(n) = 4 - 8*log(3/2). (End)
E.g.f.: exp(2*x)*(1 + x). - Stefano Spezia, Jun 11 2021

A003506 Triangle of denominators in Leibniz's Harmonic Triangle a(n,k), n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 2, 2, 3, 6, 3, 4, 12, 12, 4, 5, 20, 30, 20, 5, 6, 30, 60, 60, 30, 6, 7, 42, 105, 140, 105, 42, 7, 8, 56, 168, 280, 280, 168, 56, 8, 9, 72, 252, 504, 630, 504, 252, 72, 9, 10, 90, 360, 840, 1260, 1260, 840, 360, 90, 10, 11, 110, 495, 1320, 2310, 2772, 2310, 1320, 495, 110, 11

Offset: 1

N. J. A. Sloane

Array 1/Beta(n,m) read by antidiagonals. - Michael Somos, Feb 05 2004
a(n,3) = A027480(n-2); a(n,4) = A033488(n-3). - Ross La Haye, Feb 13 2004
a(n,k) = total size of all of the elements of the family of k-size subsets of an n-element set. For example, a 2-element set, say, {1,2}, has 3 families of k-size subsets: one with 1 0-size element, one with 2 1-size elements and one with 1 2-size element; respectively, {{}}, {{1},{2}}, {{1,2}}. - Ross La Haye, Dec 31 2006
Second slice along the 1-2-plane in the cube a(m,n,o) = a(m-1,n,o) + a(m,n-1,o) + a(m,n,o-1) with a(1,0,0)=1 and a(m<>1=0,n>=0,0>=o)=0, for which the first slice is Pascal's triangle (slice read by antidiagonals). - Thomas Wieder, Aug 06 2006
Triangle, read by rows, given by [2,-1/2,1/2,0,0,0,0,0,0,...] DELTA [2,-1/2,1/2,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 07 2007
This sequence * [1/1, 1/2, 1/3, ...] = (1, 3, 7, 15, 31, ...). - Gary W. Adamson, Nov 14 2007
n-th row = coefficients of first derivative of corresponding Pascal's triangle row. Example: x^4 + 4x^3 + 6x^2 + 4x + 1 becomes (4, 12, 12, 4). - Gary W. Adamson, Dec 27 2007
From Paul Curtz, Jun 03 2011: (Start)
Consider
   1     1/2   1/3    1/4    1/5
  -1/2  -1/6  -1/12  -1/20  -1/30
   1/3   1/12  1/30   1/60   1/105
  -1/4  -1/20 -1/60  -1/140 -1/280
   1/5   1/30  1/105  1/280  1/630
This is an autosequence (the inverse binomial transform is the sequence signed) of the second kind: the main diagonal is 2 times the first upper diagonal.
Note that 2, 12, 60, ... = A005430(n+1), Apery numbers = 2*A002457(n). (End)
From Louis Conover (for the 9th grade G1c mathematics class at the Chengdu Confucius International School), Mar 02 2015: (Start)
The i-th order differences of n^-1 appear in the (i+1)th row.
  1,    1/2,   1/3,   1/4,    1/5,    1/6,    1/7,     1/8, ...
  1/2,  1/6,  1/12,  1/20,   1/30,   1/42,   1/56,    1/72, ...
  1/3, 1/12,  1/30,  1/60,  1/105,  1/168,  1/252,   1/360, ...
  1/4, 1/20,  1/60, 1/140,  1/280,  1/504,  1/840,  1/1320, ...
  1/5, 1/30, 1/105, 1/280,  1/630, 1/1260, 1/2310,  1/3960, ...
  1/6, 1/42, 1/168, 1/504, 1/1260, 1/2772, 1/5544, 1/12012, ...
(End)
T(n,k) is the number of edges of distance k from a fixed vertex in the n-dimensional hypercube. - Simon Burton, Nov 04 2022

			The triangle begins:
  1;
  1/2, 1/2;
  1/3, 1/6, 1/3;
  1/4, 1/12, 1/12, 1/4;
  1/5, 1/20, 1/30, 1/20, 1/5;
  ...
The triangle of denominators begins:
   1
   2   2
   3   6   3
   4  12  12    4
   5  20  30   20    5
   6  30  60   60   30    6
   7  42 105  140  105   42    7
   8  56 168  280  280  168   56    8
   9  72 252  504  630  504  252   72   9
  10  90 360  840 1260 1260  840  360  90  10
  11 110 495 1320 2310 2772 2310 1320 495 110 11

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, see 130.
B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8. English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 38.
G. Boole, A Treatise On The Calculus of Finite Differences, Dover, 1960, p. 26.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25.
M. Elkadi and B. Mourrain, Symbolic-numeric methods for solving polynomial equations and applications, Chap 3. of A. Dickenstein and I. Z. Emiris, eds., Solving Polynomial Equations, Springer, 2005, pp. 126-168. See p. 152.
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 35.

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened
F. S. Al-Kharousi, A. Umar, and M. M. Zubairu, On injective partial Catalan monoids, arXiv:2501.00285 [math.GR], 2024. See pp. 8-9.
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993.
H. J. Brothers, Pascal's Prism: Supplementary Material.
D. Dumont, Matrices d'Euler-Seidel, Sem. Loth. Comb. B05c (1981) 59-78.
Yilmaz Simsek, Construction of a generalization of the Leibnitz numbers and their properties, arXiv:2011.13701 [math.NT], 2020.
Eric Weisstein's World of Mathematics, Leibniz Harmonic Triangle

Cf. A007622, A128064, A164555/A027642, A356546.
Cf. A094305, A121547, A121306, A119800, A002378, A007318.
Row sums are in A001787. Central column is A002457. Half-diagonal is in A090816. A116071, A215652.
Denominators of i-th order differences of n^-1 are given in: (1st) A002378, (2nd) A027480, (3rd) A033488, (4th) A174002, (5th) A253946. - Louis Conover, Mar 02 2015
Columns k >= 1 (offset 1): A000027, A002378, A027480, A033488, A174002, A253946(n+4), ..., with sum of reciprocals: infinity, 1, 1/2, 1/3, 1/4, 1/5, ..., respectively. - Wolfdieter Lang, Jul 20 2022

a003506 n k = a003506_tabl !! (n-1) !! (n-1)
a003506_row n = a003506_tabl !! (n-1)
a003506_tabl = scanl1 (\xs ys ->
   zipWith (+) (zipWith (+) ([0] ++ xs) (xs ++ [0])) ys) a007318_tabl
a003506_list = concat a003506_tabl
-- Reinhard Zumkeller, Nov 14 2013, Nov 17 2011

with(combstruct):for n from 0 to 11 do seq(m*count(Combination(n), size=m), m = 1 .. n) od; # Zerinvary Lajos, Apr 09 2008
A003506 := (n,k) -> k*binomial(n,k):
seq(print(seq(A003506(n,k),k=1..n)),n=1..7); # Peter Luschny, May 27 2011

L[n_, 1] := 1/n; L[n_, m_] := L[n, m] = L[n - 1, m - 1] - L[n, m - 1]; Take[ Flatten[ Table[ 1 / L[n, m], {n, 1, 12}, {m, 1, n}]], 66]
t[n_, m_] = Gamma[n]/(Gamma[n - m]*Gamma[m]); Table[Table[t[n, m], {m, 1, n - 1}], {n, 2, 12}]; Flatten[%] (* Roger L. Bagula and Gary W. Adamson, Sep 14 2008 *)
Table[k*Binomial[n,k],{n,1,7},{k,1,n}] (* Peter Luschny, May 27 2011 *)
t[n_, k_] := Denominator[n!*k!/(n+k+1)!]; Table[t[n-k, k] , {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 28 2013 *)

A(i,j)=if(i<1||j<1,0,1/subst(intformal(x^(i-1)*(1-x)^(j-1)),x,1))

A(i,j)=if(i<1||j<1,0,1/sum(k=0,i-1,(-1)^k*binomial(i-1,k)/(j+k)))

{T(n, k) = (n + 1 - k) * binomial( n, k - 1)} /* Michael Somos, Feb 06 2011 */

T_row = lambda n: (n*(x+1)^(n-1)).list()
for n in (1..10): print(T_row(n)) # Peter Luschny, Feb 04 2017
# Assuming offset 0:
def A003506(n, k):
    return falling_factorial(n+1,n)//(factorial(k)*factorial(n-k))
for n in range(9): print([A003506(n, k) for k in range(n+1)]) # Peter Luschny, Aug 13 2022

a(n, 1) = 1/n; a(n, k) = a(n-1, k-1) - a(n, k-1) for k > 1.
Considering the integer values (rather than unit fractions): a(n, k) = k*C(n, k) = n*C(n-1, k-1) = a(n, k-1)*a(n-1, k-1)/(a(n, k-1) - a(n-1, k-1)) = a(n-1, k) + a(n-1, k-1)*k/(k-1) = (a(n-1, k) + a(n-1, k-1))*n/(n-1) = k*A007318(n, k) = n*A007318(n-1, k-1). Row sums of integers are n*2^(n-1) = A001787(n); row sums of the unit fractions are A003149(n-1)/A000142(n). - Henry Bottomley, Jul 22 2002
From Vladeta Jovovic, Nov 01 2003: (Start)
G.f.: x*y/(1-x-y*x)^2.
E.g.f.: x*y*exp(x+x*y). (End)
T(n,k) = n*binomial(n-1,k-1) = n*A007318(n-1,k-1). - Philippe Deléham, Aug 04 2006
Binomial transform of A128064(unsigned). - Gary W. Adamson, Aug 29 2007
From Roger L. Bagula and Gary W. Adamson, Sep 14 2008: (Start)
t(n,m) = Gamma(n)/(Gamma(n - m)*Gamma(m)).
f(s,n) = Integral_{x=0..oo} exp(-s*x)*x^n dx = Gamma(n)/s^n; t(n,m) = f(s,n)/(f(s,n-m)*f(s,m)) = Gamma(n)/(Gamma(n - m)*Gamma(m)); the powers of s cancel out. (End)
From Reinhard Zumkeller, Mar 05 2010: (Start)
T(n,5) = T(n,n-4) = A174002(n-4) for n > 4.
T(2*n,n) = T(2*n,n+1) = A005430(n). (End)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-2,k-2), T(1,1) = 1 and, for n > 1, T(n,k) = 0 if k <= 1 or if k > n. - Philippe Deléham, Mar 17 2012
T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n+1-k,k+1-i). - Mircea Merca, Apr 11 2012
If we include a main diagonal of zeros so that the array is in the form
  0
  1   0
  2   2   0
  3   6   3   0
  4  12  12   4   0
...
then we obtain the exponential Riordan array [x*exp(x),x], which factors as [x,x]*[exp(x),x] = A132440*A007318. This array is the infinitesimal generator for A116071. A signed version of the array is the infinitesimal generator for A215652. - Peter Bala, Sep 14 2012
a(n,k) = (n-1)!/((n-k)!(k-1)!) if k > n/2 and a(n,k) = (n-1)!/((n-k-1)!k!) otherwise. [Forms 'core' for Pascal's recurrence; gives common term of RHS of T(n,k) = T(n-1,k-1) + T(n-1,k)]. - Jon Perry, Oct 08 2013
Assuming offset 0: T(n, k) = FallingFactorial(n + 1, n) / (k! * (n - k)!). The counterpart using the rising factorial is A356546. - Peter Luschny, Aug 13 2022

Edited by N. J. A. Sloane, Oct 07 2007

A051924 a(n) = binomial(2n,n) - binomial(2n-2,n-1); or (3n-2)*C(n-1), where C = Catalan numbers (A000108).

Original entry on oeis.org

1, 4, 14, 50, 182, 672, 2508, 9438, 35750, 136136, 520676, 1998724, 7696444, 29716000, 115000920, 445962870, 1732525830, 6741529080, 26270128500, 102501265020, 400411345620, 1565841089280, 6129331763880, 24014172955500, 94163002754652, 369507926510352

Offset: 1

Barry E. Williams, Dec 19 1999

Number of partitions with Ferrers plots that fit inside an n X n box, but not in an n-1 X n-1 box. - Wouter Meeussen, Dec 10 2001
From Benoit Cloitre, Jan 29 2002: (Start)
Let m(1,j)=j, m(i,1)=i and m(i,j) = m(i-1,j) + m(i,j-1); then a(n) = m(n,n):
  1  2  3  4 ...
  2  4  7 11 ...
  3  7 14 25 ...
  4 11 25 50 ... (End)
This sequence also gives the number of clusters and non-crossing partitions of type D_n. - F. Chapoton, Jan 31 2005
If Y is a 2-subset of a 2n-set X then a(n) is the number of (n+1)-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007
Prefaced with a 1: (1, 1, 4, 14, 50, ...) and convolved with the Catalan sequence = A097613: (1, 2, 7, 25, 91, ...). - Gary W. Adamson, May 15 2009
Total number of up steps before the second return in all Dyck n-paths. - David Scambler, Aug 21 2012
Conjecture: a(n) mod n^2 = n+2 iff n is an odd prime. - Gary Detlefs, Feb 19 2013
First differences of A000984 and A030662. - J. M. Bergot, Jun 22 2013
From R. J. Mathar, Jun 30 2013: (Start)
Equivalent to the Meeussen comment and the Bergot comment: The array view of A007318 is
  1,   1,   1,   1,   1,   1,
  1,   2,   3,   4,   5,   6,
  1,   3,   6,  10,  15,  21,
  1,   4,  10,  20,  35,  56,
  1,   5,  15,  35,  70, 126,
  1,   6,  21,  56, 126, 252,
and a(n) are the hook sums Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). (End)
From Gus Wiseman, Apr 12 2019: (Start)
Equivalent to Wouter Meeussen's comment, a(n) is the number of integer partitions (of any positive integer) such that the maximum of the length and the largest part is n. For example, the a(1) = 1 through a(3) = 14 partitions are:
  (1)  (2)   (3)
       (11)  (31)
       (21)  (32)
       (22)  (33)
             (111)
             (211)
             (221)
             (222)
             (311)
             (321)
             (322)
             (331)
             (332)
             (333)
(End)
Coxeter-Catalan numbers for Coxeter groups of type D_n [Armstrong]. - N. J. A. Sloane, Mar 09 2022
a(n+1) is the number of ways that a best of n pairs contest with early termination can go. For example, the first stage of an association football (soccer) penalty-kick shoot out has n=5 pairs of shots and there are a(6)=672 distinct ways it can go. For n=2 pairs, writing G for goal and M for miss, and listing the up-to-four shots in chronological order with teams alternating shots, the n(3)=14 possibilities are MMMM, MMMG, MMGM, MMGG, MGM, MGGM, MGGG, GMMM, GMMG, GMG, GGMM, GGMG, GGGM, and GGGG. Not all four shots are taken in two cases because it becomes impossible for one team to overcome the lead of the other team. - Lee A. Newberg, Jul 20 2024

			Sums of {1}, {2, 1, 1}, {2, 2, 3, 3, 2, 1, 1}, {2, 2, 4, 5, 7, 6, 7, 5, 5, 3, 2, 1, 1}, ...

Drew Armstrong, Generalized Noncrossing Partitions and Combinatorics of Coxeter Groups, Mem. Amer. Math. Soc. 202 (2009), no. 949, x+159. MR 2561274 16; See Table 2.8.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Drew Armstrong, Generalized Noncrossing Partitions and Combinatorics of Coxeter Groups, arXiv:math/0611106 [math.CO], 2006-2007.
Jean-Luc Baril, Pamela E. Harris, Kimberly J. Harry, Matt McClinton, and José L. Ramírez, Enumerating runs, valleys, and peaks in Catalan words, arXiv:2404.05672 [math.CO], 2024. See p. 10.
F. Chapoton, Clusters.
FindStat, St000784: The maximum of the length and the largest part of the integer partition.
Sergey Fomin and Andrei Zelevinsky, Y-systems and generalized associahedra, Ann. of Math. (2) 158 (2003), no. 3, 977-1018.
Joël Gay and Vincent Pilaud, The weak order on Weyl posets, arXiv:1804.06572 [math.CO], 2018.
Milan Janjic, Two Enumerative Functions.
Joshua Marsh and Nathan Williams, Nesting Nonpartitions, J. Int. Seq., Vol. 25 (2022), Article 22.8.8.
Sanjay Moudgalya, Abhinav Prem, Rahul Nandkishore, Nicolas Regnault, and B. Andrei Bernevig, Thermalization and its absence within Krylov subspaces of a constrained Hamiltonian, arXiv:1910.14048 [cond-mat.str-el], 2019.
Hugh Thomas, Tamari Lattices and Non-Crossing Partitions in Types B and D, arXiv:math/0311334 [math.CO], 2003-2005.
Lin Yang and Shengliang Yang, Protected Branches in Ordered Trees, J. Math. Study, Vol. 56, No. 1 (2023), 1-17.

Left-central elements of the (1, 2)-Pascal triangle A029635.
Column sums of A096771.
Cf. A000108, A024482 (diagonal from 2), A076540 (diagonal from 3), A000124 (row from 2), A004006 (row from 3), A006522 (row from 4).
Cf. A128064; first differences of A000984.
Cf. A097613.
Cf. A115720, A252464, A257990, A263297, A325189, A325192, A325193.

a051924 n = a051924_list !! (n-1)
a051924_list = zipWith (-) (tail a000984_list) a000984_list
-- Reinhard Zumkeller, May 25 2013

[Binomial(2*n, n)-Binomial(2*n-2, n-1): n in [1..28]]; // Vincenzo Librandi, Dec 21 2016

C:= n-> binomial(2*n, n)/(n+1): seq((n+1)*C(n)-n*C(n-1), n=1..25); # Emeric Deutsch, Jan 08 2008
Z:=(1-z-sqrt(1-4*z))/sqrt(1-4*z): Zser:=series(Z, z=0, 32): seq(coeff(Zser, z, n), n=1..24); # Zerinvary Lajos, Jan 01 2007
a := n -> 2^(-2+2*n)*GAMMA(-1/2+n)*(3*n-2)/(sqrt(Pi)*GAMMA(1+n)):
seq(simplify(a(n)), n=1..24); # Peter Luschny, Dec 14 2015

Table[Binomial[2n,n]-Binomial[2n-2,n-1],{n,30}] (* Harvey P. Dale, Jan 15 2012 *)

a(n)=binomial(2*n,n)-binomial(2*n-2,n-1) \\ Charles R Greathouse IV, Jun 25 2013

{a(n)=polcoeff((1-x) / sqrt(1-4*x +x*O(x^n)) - 1,n)}
for(n=1,30,print1(a(n),", ")) \\ Paul D. Hanna, Nov 08 2014

{a(n)=polcoeff( sum(m=1, n, x^m * sum(k=0, m, binomial(m, k)^2 * x^k) / (1-x +x*O(x^n))^(2*m)), n)}
for(n=1, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Nov 08 2014

a = lambda n: 2^(-2+2*n)*gamma(n-1/2)*(3*n-2)/(sqrt(pi)*gamma(1+n))
[a(n) for n in (1..120)] # Peter Luschny, Dec 14 2015

G.f.: (1-x) / sqrt(1-4*x) - 1. - Paul D. Hanna, Nov 08 2014
G.f.: Sum_{n>=1} x^n/(1-x)^(2*n) * Sum_{k=0..n} C(n,k)^2 * x^k. - Paul D. Hanna, Nov 08 2014
a(n+1) = binomial(2*n, n) + 2*Sum_{i=0..n-1} binomial(n+i, i) (V's in Pascal's Triangle). - Jon Perry Apr 13 2004
a(n) = n*C(n-1) - (n-1)*C(n-2), where C(n) = A000108(n) = Catalan(n). For example, a(5) = 50 = 5*C(4) - 4*C(3) - 5*14 - 3*5 = 70 - 20. Triangle A128064 as an infinite lower triangular matrix * A000108 = A051924 prefaced with a 1: (1, 1, 4, 14, 50, 182, ...). - Gary W. Adamson, May 15 2009
Sum of 3 central terms of Pascal's triangle: 2*C(2+2*n, n)+C(2+2*n, 1+n). - Zerinvary Lajos, Dec 20 2005
a(n+1) = A051597(2n,n). - Philippe Deléham, Nov 26 2006
The sequence 1,1,4,... has a(n) = C(2*n,n)-C(2*(n-1),n-1) = 0^n+Sum_{k=0..n} C(n-1,k-1)*A002426(k), and g.f. given by (1-x)/(1-2*x-2*x^2/(1-2*x-x^2/(1-2*x-x^2/(1-2*x-x^2/(1-.... (continued fraction). - Paul Barry, Oct 17 2009
a(n) = (3*n-2)*(2*n-2)!/(n*(n-1)!^2) = A001700(n) + A001791(n-1). - David Scambler, Aug 21 2012
D-finite with recurrence: a(n) = 2*(3*n-2)*(2*n-3)*a(n-1)/(n*(3*n-5)). - Alois P. Heinz, Apr 25 2014
a(n) = 2^(-2+2*n)*Gamma(-1/2+n)*(3*n-2)/(sqrt(Pi)*Gamma(1+n)). - Peter Luschny, Dec 14 2015
a(n) ~ (3/4)*4^n*(1-(7/24)/n-(7/128)/n^2-(85/3072)/n^3-(581/32768)/n^4-(2611/262144)/n^5)/sqrt(n*Pi). - Peter Luschny, Dec 16 2015
E.g.f.: ((1 - x)*BesselI(0,2*x) + x*BesselI(1,2*x))*exp(2*x) - 1. - Ilya Gutkovskiy, Dec 20 2016
a(n) = 2 * A097613(n) for n > 1. - Bruce J. Nicholson, Jan 06 2019
Sum_{n>=1} a(n)/8^n = 7/(4*sqrt(2)) - 1. - Amiram Eldar, May 06 2023

Edited by N. J. A. Sloane, May 03 2008, at the suggestion of R. J. Mathar

A002418 4-dimensional figurate numbers: a(n) = (5n-1)binomial(n+2,3)/4.

Original entry on oeis.org

0, 1, 9, 35, 95, 210, 406, 714, 1170, 1815, 2695, 3861, 5369, 7280, 9660, 12580, 16116, 20349, 25365, 31255, 38115, 46046, 55154, 65550, 77350, 90675, 105651, 122409, 141085, 161820, 184760, 210056, 237864, 268345, 301665, 337995

Offset: 0

N. J. A. Sloane

Partial sums of A002413.
Principal diagonal of the convolution array A213550, for n>0. - Clark Kimberling, Jun 17 2012
Convolution of A000027 with A000566. - Bruno Berselli, Dec 06 2012
Coefficients in the hypergeometric series identity 1 - 9*(x - 1)/(4*x + 1) + 35*(x - 1)*(x - 2)/((4*x + 1)*(4*x + 2)) - 95*(x - 1)*(x - 2)*(x - 3)/((4*x + 1)*(4*x + 2)*(4*x + 3)) + ... = 0, valid for Re(x) > 1. Cf. A000326 and A002412. Column 4 of A103450. - Peter Bala, Mar 14 2019

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 195.
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).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
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.
Luis Verde-Star, A Matrix Approach to Generalized Delannoy and Schröder Arrays, J. Int. Seq., Vol. 24 (2021), Article 21.4.1.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5, 1).
Index to sequences related to pyramidal numbers.

Cf. A093562 ((5, 1) Pascal, column m=4).
Cf. A000332, A000566, A001622, A080852, A128064.
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.

List([0..40],n->(5*n-1)*Binomial(n+2,3)/4); # Muniru A Asiru, Mar 18 2019

[(5*n - 1)*Binomial(n + 2, 3)/4: n in [0..40]]; // Vincenzo Librandi, Oct 17 2012

/* A000027 convolved with A000566: */ A000566:=func; [&+[(n-i+1)*A000566(i): i in [0..n]]: n in [0..35]]; // Bruno Berselli, Dec 06 2012

Table[(5n-1) Binomial[n+2,3]/4,{n,0,40}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{0,1,9,35,95},40] (* Harvey P. Dale, Oct 16 2012 *)
CoefficientList[Series[x*(1 + 4*x)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Oct 17 2012 *)

a(n)=(5*n-1)*binomial(n+2,3)/4 \\ Charles R Greathouse IV, Sep 24 2015

[(5*n-1)*binomial(n+2,3)/4 for n in (0..40)] # G. C. Greubel, Jul 03 2019

G.f.: x*(1+4*x)/(1-x)^5. - Simon Plouffe in his 1992 dissertation.
Starting (1, 9, 35, 95, ...), = A128064 * A000332, (A000332 starting 1, 5, 15, 35, 70, ...), such that a(n) = n*C(n+3,4) - (n-1)*C(n+2,4). E.g., a(5) = 210 = 5*C(8,4) - 4*C(7,4) = 5*70 - 4*35. - Gary W. Adamson, Dec 28 2007
Unit digit, A010879(a(n)), is one of {0,1,9,5,6,4} [Eric Desbiaux] because a(n) mod 5 = 0,1,4,0,0, periodic with period 5. [Proof: A002413(n) mod 5 = 1,3,1,0,0 with period 5 and a(n) are the partial sums of A002413.] - R. J. Mathar, Mar 19 2008
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Oct 16 2012
a(n) = A080852(5,n-1). - R. J. Mathar, Jul 28 2016
a(n) = Sum_{i=0..n} (n-i) * Sum_{j=i..n} j. - J. M. Bergot, May 30 2017
E.g.f.: x*(24 + 84*x + 44*x^2 + 5*x^3)*exp(x)/4!. - G. C. Greubel, Jul 03 2019
Sum_{n>=1} 1/a(n) = (50*sqrt(5)*log(phi) + 125*log(5) - 50*sqrt(1+2/sqrt(5))*Pi - 26)/11, where phi is the golden ratio (A001622). - Amiram Eldar, Feb 11 2022

From Boris Putievskiy, Jan 24 2013: (Start)
Table T(n,k) = n+2*k-2 n, k > 0, read by antidiagonals.
General case A209304. Let m be natural number. The first column of the table T(n,1) is the sequence of the natural numbers A000027. Every next column is formed from previous shifted by m elements.
For m=0 the result is A002260,
for m=1 the result is A002024,
for m=2 the result is A128076,
for m=3 the result is A131914,
for m=4 the result is A209304. (End)

Row sums of triangle A135852. - Gary W. Adamson, Dec 01 2007

Binomial transform of [1, 4, 6, 8, 10, 12, 14, 16, ...]. Equals A128064 * A000225, (A000225 starting 1, 3, 7, 15, ...). - Gary W. Adamson, Dec 28 2007

This triangle * [1/1, 1/2, 1/3, ...] = (1, 2, 4, 8, 16, 32, ...). - Gary W. Adamson, Nov 15 2007

Triangle read by rows: T(n,k) = (k+1)*binomial(n,k), 0 <= k <= n. - Philippe Deléham, Apr 20 2009

For n>2, mod 2 = (0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, ...), i.e., two evens followed by four odds (repeating).

Inverse binomial transform of A117202: (1, 4, 15, 52, ...). - Gary W. Adamson, Sep 03 2008

cp's OEIS Frontend

A128584 A054521 * A128064 (unsigned).

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A001844 Centered square numbers: a(n) = 2*n*(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

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A001792 a(n) = (n+2)*2^(n-1).

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A003506 Triangle of denominators in Leibniz's Harmonic Triangle a(n,k), n >= 1, 1 <= k <= n.

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A051924 a(n) = binomial(2*n,n) - binomial(2*n-2,n-1); or (3n-2)*C(n-1), where C = Catalan numbers (A000108).

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A001844 Centered square numbers: a(n) = 2n(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

A051924 a(n) = binomial(2n,n) - binomial(2n-2,n-1); or (3n-2)*C(n-1), where C = Catalan numbers (A000108).

A002418 4-dimensional figurate numbers: a(n) = (5n-1)binomial(n+2,3)/4.

A136376 a(n) = nF(n) + (n-1)F(n-1).