A214604 -id:A214604 - cp's OEIS frontend

A016754 Odd squares: a(n) = (2n+1)^2. Also centered octagonal numbers.

1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849, 2025, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 3969, 4225, 4489, 4761, 5041, 5329, 5625, 5929, 6241, 6561, 6889, 7225, 7569, 7921, 8281, 8649, 9025

Offset: 0

N. J. A. Sloane

The brown rat (rattus norwegicus) breeds very quickly. It can give birth to other rats 7 times a year, starting at the age of three months. The average number of pups is 8. The present sequence gives the total number of rats, when the intervals are 12/7 of a year and a young rat starts having offspring at 24/7 of a year. - Hans Isdahl, Jan 26 2008
Numbers n such that tau(n) is odd where tau(x) denotes the Ramanujan tau function (A000594). - Benoit Cloitre, May 01 2003
If Y is a fixed 2-subset of a (2n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
Binomial transform of [1, 8, 8, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 8, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
All terms of this sequence are of the form 8k+1. For numbers 8k+1 which aren't squares see A138393. Numbers 8k+1 are squares iff k is a triangular number from A000217. And squares have form 4n(n+1)+1. - Artur Jasinski, Mar 27 2008
Sequence arises from reading the line from 1, in the direction 1, 25, ... and the line from 9, in the direction 9, 49, ..., in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
Equals the triangular numbers convolved with [1, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
First differences: A008590(n) = a(n) - a(n-1) for n>0. - Reinhard Zumkeller, Nov 08 2009
Central terms of the triangle in A176271; cf. A000466, A053755. - Reinhard Zumkeller, Apr 13 2010
Odd numbers with odd abundance. Odd numbers with even abundance are in A088828. Even numbers with odd abundance are in A088827. Even numbers with even abundance are in A088829. - Jaroslav Krizek, May 07 2011
Appear as numerators in the non-simple continued fraction expansion of Pi-3: Pi-3 = K_{k>=1} (1-2*k)^2/6 = 1/(6+9/(6+25/(6+49/(6+...)))), see also the comment in A007509. - Alexander R. Povolotsky, Oct 12 2011
Ulam's spiral (SE spoke). - Robert G. Wilson v, Oct 31 2011
All terms end in 1, 5 or 9. Modulo 100, all terms are among { 1, 9, 21, 25, 29, 41, 49, 61, 69, 81, 89 }. - M. F. Hasler, Mar 19 2012
Right edge of both triangles A214604 and A214661: a(n) = A214604(n+1,n+1) = A214661(n+1,n+1). - Reinhard Zumkeller, Jul 25 2012
Also: Odd numbers which have an odd sum of divisors (= sigma = A000203). - M. F. Hasler, Feb 23 2013
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1) with hypotenuse c (A020882) and respective even leg b (A231100); sequence gives values c-b, sorted with duplicates removed. - K. G. Stier, Nov 04 2013
For n>1 a(n) is twice the area of the irregular quadrilateral created by the points ((n-2)*(n-1),(n-1)*n/2), ((n-1)*n/2,n*(n+1)/2), ((n+1)*(n+2)/2,n*(n+1)/2), and ((n+2)*(n+3)/2,(n+1)*(n+2)/2). - J. M. Bergot, May 27 2014
Number of pairs (x, y) of Z^2, such that max(abs(x), abs(y)) <= n. - Michel Marcus, Nov 28 2014
Except for a(1)=4, the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 737", based on the 5-celled von Neumann neighborhood. - Robert Price, May 23 2016
a(n) is the sum of 2n+1 consecutive numbers, the first of which is n+1. - Ivan N. Ianakiev, Dec 21 2016
a(n) is the number of 2 X 2 matrices with all elements in {0..n} with determinant = 2*permanent. - Indranil Ghosh, Dec 25 2016
Engel expansion of Pi*StruveL_0(1)/2 where StruveL_0(1) is A197037. - Benedict W. J. Irwin, Jun 21 2018
Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; the segments on the hypotenuse {p = a(n)/A001844(n), q = A060300(n)/A001844(n) = A001844(n) - p} and their ratio p/q = a(n)/A060300(n) are irreducible fractions in Q\Z. X values are A005408, Y values are A046092, Z values are A001844. - Ralf Steiner, Feb 25 2020
a(n) is the number of large or small squares that are used to tile primitive squares of type 2 (A344332). - Bernard Schott, Jun 03 2021
Also, positive odd integers with an odd number of odd divisors (for similar sequence with 'even', see A348005). - Bernard Schott, Nov 21 2021
a(n) is the least odd number k = x + y, with 0 < x < y, such that there are n distinct pairs (x,y) for which x*y/k is an integer; for example, a(2) = 25 and the two corresponding pairs are (5,20) and (10,15). The similar sequence with 'even' is A016742 (see Comment of Jan 26 2018). - Bernard Schott, Feb 24 2023
From Peter Bala, Jan 03 2024: (Start)
The sequence terms are the exponents of q in the series expansions of the following infinite products:
1) q*Product_{n >= 1} (1 - q^(16*n))*(1 + q^(8*n)) = q + q^9 + q^25 + q^49 + q^81 + q^121 + q^169 + ....
2) q*Product_{n >= 1} (1 + q^(16*n))*(1 - q^(8*n)) = q - q^9 - q^25 + q^49 + q^81 - q^121 - q^169 + + - - ....
3) q*Product_{n >= 1} (1 - q^(8*n))^3 = q - 3*q^9 + 5*q^25 - 7*q^49 + 9*q^81 - 11*q^121 + 13*q^169 - + ....
4) q*Product_{n >= 1} ( (1 + q^(8*n))*(1 - q^(16*n))/(1 + q^(16*n)) )^3 = q + 3*q^9 - 5*q^25 - 7*q^49 + 9*q^81 + 11*q^121 - 13*q^169 - 15*q^225 + + - - .... (End)

L. Lorentzen and H. Waadeland, Continued Fractions with Applications, North-Holland 1992, p. 586.

Paolo Xausa, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe).
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Bruce C. Berndt and Ken Ono, Ramanujan's unpublished manuscript on the partition and tau functions with proofs and commentary, Séminaire Lotharingien de Combinatoire, B42c (1999), 63 pp.
John Elias, Illustration: 8-fold Square Progression of Ulam's Spiral.
Milan Janjic, Two Enumerative Functions.
Scientific American, Cover of the March 1964 issue.
Amelia Carolina Sparavigna, Groupoids of OEIS A002378 and A016754 Numbers (oblong and odd square numbers), Politecnico di Torino (Italy, 2019).
Leo Tavares, Illustration: Diamond Triangles.
Leo Tavares, Illustration: Diamond Stars.
Eric Weisstein's World of Mathematics, Moore Neighborhood.
R. Yin, J. Mu, and T. Komatsu, The p-Frobenius Number for the Triple of the Generalized Star Numbers, Preprints 2024, 2024072280. See p. 2.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A000384, A001263, A001539, A001844, A003881, A005408, A006752, A014105, A016742, A016802, A016814, A016826, A016838, A033996, A046092, A060300, A138393, A167661, A167700.
Cf. A000447 (partial sums).
Cf. A005917, A344330, A344332.
Cf. A348005, A379481 [= a(A048673(n)-1)].
Partial sums of A022144.
Positions of odd terms in A341528.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. A014634, A003154.

a016754 n = a016754_list !! n
a016754_list = scanl (+) 1 $ tail a008590_list
-- Reinhard Zumkeller, Apr 02 2012

[n^2: n in [1..100 by 2]]; // Vincenzo Librandi, Jan 03 2017

Range[1, 100, 2]^2 (* Paolo Xausa, Feb 13 2025 *)

A016754(n):=(n+n+1)^2$
makelist(A016754(n),n,0,20); /* Martin Ettl, Nov 12 2012 */

A016754(n)=(n<<1+1)^2 \\ Charles R Greathouse IV, Jun 16 2011, corrected and edited by M. F. Hasler, Apr 11 2023

def A016754(n): return ((n<<1)|1)**2 # Chai Wah Wu, Jul 06 2023

a(n) = 1 + Sum_{i=1..n} 8*i = 1 + 8*A000217(n). - Xavier Acloque, Jan 21 2003; Zak Seidov, May 07 2006; Robert G. Wilson v, Dec 29 2010
O.g.f.: (1+6*x+x^2)/(1-x)^3. - R. J. Mathar, Jan 11 2008
a(n) = 4*n*(n + 1) + 1 = 4*n^2 + 4*n + 1. - Artur Jasinski, Mar 27 2008
a(n) = A061038(2+4n). - Paul Curtz, Oct 26 2008
Sum_{n>=0} 1/a(n) = Pi^2/8 = A111003. - Jaume Oliver Lafont, Mar 07 2009
a(n) = A000290(A005408(n)). - Reinhard Zumkeller, Nov 08 2009
a(n) = a(n-1) + 8*n with n>0, a(0)=1. - Vincenzo Librandi, Aug 01 2010
a(n) = A033951(n) + n. - Reinhard Zumkeller, May 17 2009
a(n) = A033996(n) + 1. - Omar E. Pol, Oct 03 2011
a(n) = (A005408(n))^2. - Zak Seidov, Nov 29 2011
From George F. Johnson, Sep 05 2012: (Start)
a(n+1) = a(n) + 4 + 4*sqrt(a(n)).
a(n-1) = a(n) + 4 - 4*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 8.
a(n+1) = 3*a(n) - 3*a(n-1) + a(n-2).
(a(n+1) - a(n-1))/8 = sqrt(a(n)).
a(n+1)*a(n-1) = (a(n)-4)^2.
a(n) = 2*A046092(n) + 1 = 2*A001844(n) - 1 = A046092(n) + A001844(n).
Limit_{n -> oo} a(n)/a(n-1) = 1. (End)
a(n) = binomial(2*n+2,2) + binomial(2*n+1,2). - John Molokach, Jul 12 2013
E.g.f.: (1 + 8*x + 4*x^2)*exp(x). - Ilya Gutkovskiy, May 23 2016
a(n) = A101321(8,n). - R. J. Mathar, Jul 28 2016
Product_{n>=1} A033996(n)/a(n) = Pi/4. - Daniel Suteu, Dec 25 2016
a(n) = A014105(n) + A000384(n+1). - Bruce J. Nicholson, Nov 11 2017
a(n) = A003215(n) + A002378(n). - Klaus Purath, Jun 09 2020
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=0} a(n)/n! = 13*e.
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 3/e. (End)
Sum_{n>=0} (-1)^n/a(n) = A006752. - Amiram Eldar, Oct 10 2020
From Amiram Eldar, Jan 28 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = Pi/4 (A003881). (End)
From Leo Tavares, Nov 24 2021: (Start)
a(n) = A014634(n) - A002943(n). See Diamond Triangles illustration.
a(n) = A003154(n+1) - A046092(n). See Diamond Stars illustration. (End)
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(9 - 3/(17 - 60/(33 - 315/(57 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*2^2 ))))).
3/2 - 2*log(2) = Sum_{k >= 1} 1/(k*a(k)*a(k-1)) = 1/(9 - 3/(17 - 60/(33 - 315/(57 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*2^2 - ... ))))).
Row 2 of A142992. (End)
From Peter Bala, Mar 26 2024: (Start)
8*a(n) = (2*n + 1)*(a(n+1) - a(n-1)).
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 1/2 - Pi/8 = 1/(9 + (1*3)/(8 + (3*5)/(8 + ... + (4*n^2 - 1)/(8 + ... )))). For the continued fraction use Lorentzen and Waadeland, p. 586, equation 4.7.9 with n = 1. Cf. A057813. (End)

Additional description from Terrel Trotter, Jr., Apr 06 2002

A028387 a(n) = n + (n+1)^2.

Original entry on oeis.org

1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549, 2651

Offset: 0

Patrick De Geest

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005
Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006
A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007
Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007
Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008
a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009
sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009
When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010
a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010
The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011
Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012
Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012
Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013
a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014
A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015
An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015
Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015
Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017
The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017
From Klaus Purath, Mar 18 2019: (Start)
Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with
x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.
But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)
a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019
a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021
Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021
(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022
The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

			From _Ilya Gutkovskiy_, Apr 13 2016: (Start)
Illustration of initial terms:
                                        o               o
                        o           o   o o           o o
            o       o   o o       o o   o o o       o o o
    o   o   o o   o o   o o o   o o o   o o o o   o o o o
o   o o o   o o o o o   o o o o o o o   o o o o o o o o o
n=0  n=1       n=2           n=3               n=4
(End)
From _Klaus Purath_, Mar 18 2019: (Start)
Examples:
a(0) = 1: 1^1-0*1 = 1, 0+1 = 1 (Fibonacci A000045).
a(1) = 5: 3^2-1*4 = 5, 1+3 = 4 (Lucas A000032).
a(2) = 11: 4^2-1*5 = 11, 1+4 = 5 (A000285); 5^2-2*7 = 11, 2+5 = 7 (A001060).
a(3) = 19: 5^2-1*6 = 19, 1+5 = 6 (A022095); 7^2-3*10 = 19, 3+7 = 10 (A022120).
a(4) = 29: 6^2-1*7 = 29, 1+6 = 7 (A022096); 9^2-4*13 = 29, 4+9 = 13 (A022130).
a(11)/5 = 31: 7^2-2*9 = 31, 2+7 = 9 (A022113); 8^2-3*11 = 31, 3+8 = 11 (A022121).
a(24)/11 = 59: 9^2-2*11 = 59, 2+9 = 11 (A022114); 12^2-5*17 = 59, 5+12 = 17 (A022137).
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Patrick De Geest, World!Of Numbers, Palindromes of the form n+(n+1)^2.
Adalbert Kerber, A matrix of combinatorial numbers related to the symmetric groups, Discrete Math., 21 (1978), 319-321. [Annotated scanned copy]
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
Nandini Nilakantan and Anurag Singh, Homotopy type of neighborhood complexes of Kneser graphs, KG_{2,k}, Proceeding-Mathematical Sciences, 128, Article number: 53(2018).
Yanni Pei and Jiang Zeng, Counting signed derangements with right-to-left minima and excedances, arXiv:2206.11236 [math.CO], 2022.
Popular Computing (Calabasas, CA), The CSR Function, Vol. 4 (No. 34, Jan 1976), pages PC34-10 to PC34-11. Annotated and scanned copy.
Zdzislaw Skupień and Andrzej Żak, Pair-sums packing and rainbow cliques, in Topics In Graph Theory, A tribute to A. A. and T. E. Zykovs on the occasion of A. A. Zykov's 90th birthday, ed. R. Tyshkevich, Univ. Illinois, 2013, pages 131-144, (in English and Russian).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Complement of A028392. Third column of array A094954.
Cf. A000217, A002522, A062392, A062786, A127701, A135223, A143596, A052905, A162997, A062938 (squares of this sequence).
A110331 and A165900 are signed versions.
Cf. A002327 (primes), A094210.
Frobenius number for k successive numbers: this sequence (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).
Cf. also A135223, A176271, A214604, A105728, A005408, A002378, A084990, A253607, A260910, A089270.

a028387 n = n + (n + 1) ^ 2  -- Reinhard Zumkeller, Jul 17 2014

[n + (n+1)^2: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

FoldList[## + 2 &, 1, 2 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
Table[n + (n + 1)^2, {n, 0, 100}] (* Vincenzo Librandi, Oct 17 2012 *)
Table[ FrobeniusNumber[{n, n + 1}], {n, 2, 30}] (* Zak Seidov, Jan 14 2015 *)

a(n)=n^2+3*n+1 \\ Charles R Greathouse IV, Jun 10 2011

def a(n): return (n**2+3*n+1) # Torlach Rush, May 07 2024

[n+(n+1)^2 for n in range(0,48)] # Zerinvary Lajos, Jul 03 2008

a(n) = sqrt(A062938(n)). - Floor van Lamoen, Oct 08 2001
a(0) = 1, a(1) = 5, a(n) = (n+1)*a(n-1) - (n+2)*a(n-2) for n > 1. - Gerald McGarvey, Sep 24 2004
a(n) = A105728(n+2, n+1). - Reinhard Zumkeller, Apr 18 2005
a(n) = A109128(n+2, 2). - Reinhard Zumkeller, Jun 20 2005
a(n) = 2*T(n+1) - 1, where T(n) = A000217(n). - Gary W. Adamson, Aug 15 2007
a(n) = A005408(n) + A002378(n); A084990(n+1) = Sum_{k=0..n} a(k). - Reinhard Zumkeller, Aug 20 2007
Binomial transform of [1, 4, 2, 0, 0, 0, ...] = (1, 5, 11, 19, ...). - Gary W. Adamson, Sep 20 2007
G.f.: (1+2*x-x^2)/(1-x)^3. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - R. J. Mathar, Jul 11 2009
a(n) = (n + 2 + 1/phi) * (n + 2 - phi); where phi = 1.618033989... Example: a(3) = 19 = (5 + .6180339...) * (3.381966...). Cf. next to leftmost column in A162997 array. - Gary W. Adamson, Jul 23 2009
a(n) = a(n-1) + 2*(n+1), with n > 0, a(0) = 1. - Vincenzo Librandi, Nov 18 2010
For k < n, a(n) = (k+1)*a(n-k) - k*a(n-k-1) + k*(k+1); e.g., a(5) = 41 = 4*11 - 3*5 + 3*4. - Charlie Marion, Jan 13 2011
a(n) = lower right term in M^2, M = the 2 X 2 matrix [1, n; 1, (n+1)]. - Gary W. Adamson, Jun 29 2011
G.f.: (x^2-2*x-1)/(x-1)^3 = G(0) where G(k) = 1 + x*(k+1)*(k+4)/(1 - 1/(1 + (k+1)*(k+4)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 16 2012
Sum_{n>0} 1/a(n) = 1 + Pi*tan(sqrt(5)*Pi/2)/sqrt(5). - Enrique Pérez Herrero, Oct 11 2013
E.g.f.: exp(x) (1+4*x+x^2). - Tom Copeland, Dec 02 2013
a(n) = A005408(A000217(n)). - Tony Foster III, May 31 2016
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2)/6. (End)
a(5*n+1)/5 = A062786(n+1). - Torlach Rush, Jun 05 2024

Minor edits by N. J. A. Sloane, Jul 04 2010, following suggestions from the Sequence Fans Mailing List

A025581 Triangle read by rows: T(n, k) = n-k, for 0 <= k <= n.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 2, 1, 0, 4, 3, 2, 1, 0, 5, 4, 3, 2, 1, 0, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5, 4, 3, 2, 1, 0, 8, 7, 6, 5, 4, 3, 2, 1, 0, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3

Offset: 0

David W. Wilson

Decreasing integers m to 0 followed by decreasing integers m+1 to 0, etc.
The point with coordinates (x = A025581(n), y = A002262(n)) sweeps out the first quadrant by upwards antidiagonals. N. J. A. Sloane, Jul 17 2018
The PARI functions t1, t2 can be used to read a square array T(n,k) (n >= 0, k >= 0) by antidiagonals upwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002
Riordan array (x/(1-x)^2, x). - Philippe Deléham, Feb 18 2012
a(n,k) = (A214604(n,k) - A214661(n,k)) / 2. - Reinhard Zumkeller, Jul 25 2012
Sequence B is called a reverse reluctant sequence of sequence A if B is a triangular array read by rows such that row number k lists the first k terms of the sequence A in reverse order. This sequence is the reverse reluctant sequence of sequence 0,1,2,3,..., the nonnegative integers A001477. - Boris Putievskiy, Dec 13 2012
A problem posed by François Viète (Vieta) in his book Zeteticorum liber quinque (1593), liber 2, problem 19 (quoted in the Alten et al. reference, on p. 292) is to find for a rectangle (a >= b >= 1) with given a^3 - b^3, name it C, and a*b, name it F, the difference a-b, name it x. This is a simple exercise which Viète found remarkable. It reduces to a standard cubic equation for x, namely  x^3 + 3*F*x = C. Proof: Use the square of the diagonal d^2 = a^2 + b^2. Then (i) C = a^3 - b^3 = (a - b)*(a^2 + b^2 + a*b) = x*(d^2 + F). (ii) use the trivial relation d^2 = (a-b)^2 + 2*a*b = x^2 + 2*F, to eliminate d^2 in (i). End of the Proof. Here for positive integers a = n and b = k: (T(n, k)^2 + 3*A079904(n, k))*T(n, k) = A257238(n, k) (also true for n = k = 0). - Wolfdieter Lang, May 12 2015
See a comment on A051162 on the cubic equation for S = a+b in terms of Cplus = a^3 + b^3 and D = a - b. This equation leads to a - b = sqrt((4*Cplus -S^3)/(3*S)). - Wolfdieter Lang, May 15 2015
The entries correspond to the first of the 2 coordinates of the Cantor Pairs, specifically x=w-(CPKey-(w^2+w)/2), where w=floor((sqrt(8*CPKey+1)-1)/2) and CPKey=Cantor Pair key (A001477). The second of the coordinate pairs is A002262. - Bill McEachen, Sep 12 2015

			The triangle T(n, k) begins (note that one could use l <= k <= n, for any integer l, especially 1):
  n\k  0 1 2 3 4 5 6 7 8 9 10 ...
   0:  0
   1:  1 0
   2:  2 1 0
   3:  3 2 1 0
   4:  4 3 2 1 0
   5:  5 4 3 2 1 0
   6:  6 5 4 3 2 1 0
   7:  7 6 5 4 3 2 1 0
   8:  8 7 6 5 4 3 2 1 0
   9:  9 8 7 6 5 4 3 2 1 0
  10: 10 9 8 7 6 5 4 3 2 1 0
  ... [formatted by _Wolfdieter Lang_, May 12 2015]

H.-W. Alten et al., 4000 Jahre Algebra, 2. Auflage, Springer, 2014, p. 203.

Reinhard Zumkeller, Rows n = 0..100 of triangle, flattened
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Michael Somos, Sequences used for indexing triangular or square arrays
Eric Weisstein's World of Mathematics, Pairing Function.

Cf. A025669, A025676, A025683, A002262, A004736, A001477.
Cf. A141418 (partial sums per row).
Cf. A079904, A257238, A051162, A105125.

a025581 n k = n - k
a025581_row n = [n, n-1 .. 0]
a025581_tabl = iterate (\xs@(x:_) -> (x + 1) : xs) [0]
-- Reinhard Zumkeller, Aug 04 2014, Jul 22 2012, Mar 07 2011

/* As triangle */ [[(n-k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Sep 13 2015

A025581 := n -> binomial(1+floor((1/2)+sqrt(2*(1+n))),2) - (n+1): seq(A025581(n), n=0..100);

Flatten[NestList[Prepend[#, #[[1]]+1]&, {0}, 13]] (* Jean-François Alcover, May 17 2011 *)
With[{nn=20},Flatten[Table[Join[{0},Reverse[Range[i]]],{i,nn}]]] (* Harvey P. Dale, Dec 31 2014 *)
Table[Range[n,0,-1],{n,0,15}]//Flatten (* Harvey P. Dale, Aug 01 2020 *)

a(n)=binomial(1+floor(1/2+sqrt(2+2*n)),2)-(n+1) /* produces a(n) */

t1(n)=binomial(floor(3/2+sqrt(2+2*n)),2)-(n+1) /* A025581 */

t2(n)=n-binomial(floor(1/2+sqrt(2+2*n)),2) /* A002262 */

apply( {A025581(n)=binomial(sqrtint(8*n+1)\/2+1,2)-n-1}, [0..90]) \\ M. F. Hasler, Dec 06 2019

from math import isqrt, comb
def A025581(n): return comb((m:=isqrt(k:=n+1<<1))+(k>m*(m+1))+1,2)-n-1 # Chai Wah Wu, Nov 08 2024

T(n, k) = n-k, for 0 <= k <= n.
As a sequence: a(n) = (((trinv(n)-1)*(((1/2)*trinv(n))+1))-n), with trinv(n) = floor((1+sqrt(1+8*n))/2). Cf. A002262.
a(n) = A004736(n+1) - 1.
G.f. for T(n,k): y / ((1-x)^2 * (1-x*y)). - Ralf Stephan, Jan 25 2005
For the cubic equation satisfied by T(n, k) see the comment on a problem by Viète above. - Wolfdieter Lang, May 12 2015
G.f. for a(n): -(1-x)^(-2) + (1-x)^(-1) * Sum_{n>=0} (n+1)*x^(n*(n+1)/2). The sum is related to Jacobi theta functions. - Robert Israel, May 12 2015
T(n, k) = sqrt((4*A105125(n, k) - A051162(n, k)^3)/(3*A051162(n, k))). See a comment above. - Wolfdieter Lang, May 15 2015
a(n) = (1/2)*(t^2 + t - 2*n - 2), where t = floor(sqrt(2*n+1) + 1/2) = round(sqrt(2*n+1)). - Ridouane Oudra, Dec 01 2019
a(n) = ((1/2) * ceiling((-1 + sqrt(9 + 8 * n))/2) * ceiling((1 + sqrt(9 + 8 * n))/2)) - n - 1. - Ryan Jean, Apr 22 2022

Typo in definition corrected by Arkadiusz Wesolowski, Nov 24 2011

Edited (part of name moved to first comment; definition of trinv added in formula) by Wolfdieter Lang, May 12 2015

A176271 The odd numbers as a triangle read by rows.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131

Offset: 1

Reinhard Zumkeller, Apr 13 2010

A108309(n) = number of primes in n-th row.

			From _Philippe Deléham_, Oct 03 2011: (Start)
Triangle begins:
   1;
   3,  5;
   7,  9, 11;
  13, 15, 17, 19;
  21, 23, 25, 27, 29;
  31, 33, 35, 37, 39, 41;
  43, 45, 47, 49, 51, 53, 55;
  57, 59, 61, 63, 65, 67, 69, 71;
  73, 75, 77, 79, 81, 83, 85, 87, 89; (End)

G. C. Greubel, Rows n = 1..100 of the triangle, flattened
Eric Weisstein's World of Mathematics, Nicomachus's Theorem
Wikipedia, Nikomachos von Gerasa

Cf. A000466, A000537, A000578, A001105, A002061, A005408, A014209.
Cf. A016754, A027688, A027690, A027692, A027694, A028387, A048058.
Cf. A053755, A065599, A082111, A108195, A108309, A214604, A214661.

a176271 n k = a176271_tabl !! (n-1) !! (k-1)
a176271_row n = a176271_tabl !! (n-1)
a176271_tabl = f 1 a005408_list where
   f x ws = us : f (x + 1) vs where (us, vs) = splitAt x ws
-- Reinhard Zumkeller, May 24 2012

[n^2-n+2*k-1: k in [1..n], n in [1..15]]; // G. C. Greubel, Mar 10 2024

A176271 := proc(n,k)
    n^2-n+2*k-1 ;
end proc: # R. J. Mathar, Jun 28 2013

Table[n^2-n+2*k-1, {n,15}, {k,n}]//Flatten (* G. C. Greubel, Mar 10 2024 *)

flatten([[n^2-n+2*k-1 for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Mar 10 2024

T(n, k) = n^2 - n + 2*k - 1 for 1 <= k <= n.
T(n, k) = A005408(n*(n-1)/2 + k - 1).
T(2*n-1, n) = A016754(n-1) (main diagonal).
T(2*n, n) = A000466(n).
T(2*n, n+1) = A053755(n).
T(n, k) + T(n, n-k+1) = A001105(n), 1 <= k <= n.
T(n, 1)   = A002061(n), central polygonal numbers.
T(n, 2)   = A027688(n-1) for n > 1.
T(n, 3)   = A027690(n-1) for n > 2.
T(n, 4)   = A027692(n-1) for n > 3.
T(n, 5)   = A027694(n-1) for n > 4.
T(n, 6)   = A048058(n-1) for n > 5.
T(n, n-3) = A108195(n-2) for n > 3.
T(n, n-2) = A082111(n-2) for n > 2.
T(n, n-1) = A014209(n-1) for n > 1.
T(n, n) = A028387(n-1).
Sum_{k=1..n} T(n, k) = A000578(n) (Nicomachus's theorem).
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = (-1)^(n-1)*A065599(n) (alternating sign row sums).
Sum_{j=1..n} (Sum_{k=1..n} T(j, k)) = A000537(n) (sum of first n rows).

A094727 Triangle read by rows: T(n,k) = n + k, 0 <= k < n, n >= 1.

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 4, 5, 6, 7, 5, 6, 7, 8, 9, 6, 7, 8, 9, 10, 11, 7, 8, 9, 10, 11, 12, 13, 8, 9, 10, 11, 12, 13, 14, 15, 9, 10, 11, 12, 13, 14, 15, 16, 17, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23

Offset: 1

Reinhard Zumkeller, May 24 2004

All numbers m occur ceiling(m/2) times, see A004526.

The LCM of the n-th row is A076100. - Michel Marcus, Mar 18 2018

			Triangle begins:
  1;
  2,  3;
  3,  4,  5;
  4,  5,  6,  7;
  5,  6,  7,  8,  9;
  6,  7,  8,  9, 10, 11;
  7,  8,  9, 10, 11, 12, 13;
  8,  9, 10, 11, 12, 13, 14, 15;
  9, 10, 11, 12, 13, 14, 15, 16, 17;
  ... - _Philippe Deléham_, Mar 30 2013

Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened
Bruno Berselli, Illustration of the initial terms.
László Németh, On the Binomial Interpolated Triangles, Journal of Integer Sequences, Vol. 20 (2017), Article 17.7.8.
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A002024, A002260, A003056, A004526, A004736, A005408, A005843.
Cf. A016777, A016813, A016861, A037213, A076100, A078112, A093005.
Cf. A094728, A214604, A214661, A319556.
Cf. A128076 (rows reversed).

a094727 n k = n + k
a094727_row n = a094727_tabl !! (n-1)
a094727_tabl = iterate (\row@(h:_) -> (h + 1) : map (+ 2) row) [1]
-- Reinhard Zumkeller, Jul 22 2012

z:=12; &cat[ [m+n-1: m in [1..n] ]: n in [1..z] ];

Table[n + Range[0, n-1], {n, 12}]//Flatten (* Michael De Vlieger, Dec 16 2016 *)

from math import isqrt
def A094727(n): return ((a:=(m:=isqrt(k:=n<<1))+(k>m*(m+1)))*(3-a)>>1)+n-1 # Chai Wah Wu, Jun 19 2025

flatten([[n+k for k in range(n)] for n in range(1,16)]) # G. C. Greubel, Mar 10 2024

T(n+1, k) = T(n, k) + 1 = T(n, k+1); T(n+1, k+1) = T(n, k) + 2.
T(n, n - A005843(k)) = A005843(n-k) for 0 <= k <= n/2.
T(n, n - A005408(k)) = A005408(n-k) for 0 <= k < n/2.
T(A005408(n), n) = A016777(n), n >= 0.
Sum_{k=1..n} T(n, k) = A000326(n) (row sums).
T(n, k) = A002024(n,k) + A002260(n,k) - 1. - Reinhard Zumkeller, Apr 27 2006
As a sequence rather than as a table: If m = floor((sqrt(8n-7)+1)/2), a(n) = n - m*(m-3)/2 - 1. - Carl R. White, Jul 30 2009
T(n, k) = n+k-1, n >= k >= 1. - Vincenzo Librandi, Nov 23 2009 [corrected by Klaus Brockhaus, Nov 23 2009]
T(n,k) = A037213((A214604(n,k) + A214661(n,k)) / 2). - Reinhard Zumkeller, Jul 25 2012
From Boris Putievskiy, Jan 16 2013: (Start)
a(n) = A002260(n) + A003056(n).
a(n) = i+t, where i=n-t*(t+1)/2, t=floor((-1+sqrt(8*n-7))/2). (End)
From G. C. Greubel, Mar 10 2024: (Start)
T(3*n-3, n) = A016813(n-1).
T(4*n-4, n) = A016861(n-1).
Sum_{k=0..n-1} (-1)^k*T(n, k) = A319556(n).
Sum_{k=0..floor((n-1)/2)} T(n-k, k) = A093005(n).
Sum_{k=0..floor((n-1)/2)} (-1)^k*T(n-k, k) = A078112(n-1).
Sum_{j=1..n} (Sum_{k=0..n-1} T(j, k)) = A002411(n) (sum of n rows). (End)

A214661 Odd numbers obtained by transposing the left half of A176271 into rows of a triangle: T(n,k) = A176271(n - 1 + k, k), 1 <= k <= n.

Original entry on oeis.org

1, 3, 9, 7, 15, 25, 13, 23, 35, 49, 21, 33, 47, 63, 81, 31, 45, 61, 79, 99, 121, 43, 59, 77, 97, 119, 143, 169, 57, 75, 95, 117, 141, 167, 195, 225, 73, 93, 115, 139, 165, 193, 223, 255, 289, 91, 113, 137, 163, 191, 221, 253, 287, 323, 361, 111, 135, 161, 189, 219, 251, 285, 321, 359, 399, 441

Offset: 1

Reinhard Zumkeller, Jul 25 2012

			.     Take the first n elements of the n-th diagonal (northwest to
.     southeast) of the triangle on the left side
.     and write this as n-th row on the triangle of the right side.
. 1:                1                    1
. 2:              3   _                  3  9
. 3:            7   9  __                7 15 25
. 4:         13  15  __  __             13 23 35 49
. 5:       21  23  25  __  __           21 33 47 63 ..
. 6:     31  33  35  __  __  __         31 45 61 .. .. ..
. 7:   43  45  47  49  __  __  __       43 59 .. .. .. .. ..
. 8: 57  59  61  63  __  __  __  __     57 .. .. .. .. .. .. .. .

Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened

Cf. A000290, A002061, A016754, A025581, A094727, A176271, A214604.

Cf. A051673 (row sums), A214675 (main diagonal).

import Data.List (transpose)
a214661 n k = a214661_tabl !! (n-1) !! (k-1)
a214661_row n = a214661_tabl !! (n-1)
a214661_tabl = zipWith take [1..] $ transpose $ map reverse a176271_tabl

[(n+k)^2-3*n-k+1: k in [1..n], n in [1..15]]; // G. C. Greubel, Mar 10 2024

Table[(n+k)^2-3*n-k+1, {n,15}, {k,n}]//Flatten (* G. C. Greubel, Mar 10 2024 *)

flatten([[(n+k)^2-3*n-k+1 for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Mar 10 2024

T(n, k) = (n+k)^2 - 3*n - k + 1.
T(n,k) = A176271(n+k-1, k).
T(n, k) = A214604(n,k) - 2*A025581(n,k).
T(n, k) = 2*A000290(A094727(n,k)) - A214604(n,k).
T(2*n-1, n) = A214675() (main diagonal).
T(n,1) = A002061(n).
T(n,n) = A016754(n-1).
Sum_{k=1..n} T(n, k) = A051673(n) (row sums).

cp's OEIS Frontend

A016754 Odd squares: a(n) = (2n+1)^2. Also centered octagonal numbers.

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A028387 a(n) = n + (n+1)^2.

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A025581 Triangle read by rows: T(n, k) = n-k, for 0 <= k <= n.

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A176271 The odd numbers as a triangle read by rows.

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A094727 Triangle read by rows: T(n,k) = n + k, 0 <= k < n, n >= 1.

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A214659 a(n) = n(7n^2 - 3*n - 1)/3.

A214660 a(n) = 9n^2 - 11n + 3.