A022114 -id:A022114 - cp's OEIS frontend

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

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

A356988 a(n) = n - a^[2](n - a^[3](n-1)) with a(1) = 1, where a^[2](n) = a(a(n)) and a^[3](n) = a(a(a(n))).

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 34, 34, 34, 34, 34, 34, 34, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 55, 55, 55, 55, 55

Offset: 1

Peter Bala, Sep 08 2022

This is the second sequence in a family of nested-recurrent sequences with apparently similar structure defined as follows. Given a sequence s = {s(n) : n >= 1} we define the k-th iterated sequence s^[k] by putting s^[1](n) = s(n) and setting s^[k](n) = s^[k-1](s(n)) for k >= 2. For k >= 1, we define a nested-recurrent sequence, dependent on k, by putting u(1) = 1 and setting u(n) = n - u^[k](n - u^[k+1](n-1)) for n >= 2. This is the case k = 2. For other cases see A006165 (k = 1), A356989 (k = 3) and A356990 (k = 4).
The sequence is slow, that is, for n >= 1, a(n+1) - a(n) is either 0 or 1. The sequence is unbounded.
The line graph of the sequence {a(n)} thus consists of a series of plateaus (where the value of the ordinate a(n) remains constant as n increases) joined by lines of slope 1.
The sequence of plateau heights begins 3, 5, 8, 13, 21, 34, 55, ..., the Fibonacci numbers A000045.
The plateaus start at abscissa values n = 4, 7, 11, 18, 29, 47, 76, ..., the Lucas numbers A000032, and finish at abscissa values n = 5, 8, 13, 21, 34, 55, 89, ..., the Fibonacci numbers. The sequence of plateau lengths 1, 1, 2, 3, 5, 8, 13, ... is thus the Fibonacci sequence.
The iterated sequences{a^[k](n) : n >= 1}, k = 2, 3,..., share similar properties to the present sequence. See the Example section below.

			Related sequences:
1) The square of the sequence: {a^[2](n) : n >= 1} = {a(a(n)) : n >= 1}. The first few terms are
  1, 1, 1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 7, 8, 8, 8, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 35, 36, 37, 38, 39, ...
The sequence is slow. The line graph of the sequence has plateaus of height Fibonacci(k), k >= 2, starting at abscissa value 2*Fibonacci(k) and ending at abscissa Fibonacci(k+2).
2) The cube of the sequence: {a^[3](n) : n >= 1} = {a(a(a(n))) : n >= 1}. The first few terms are
  1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, ...
The line graph of the sequence has plateaus of height Fibonacci(k), k >= 2, starting at abscissa value 3*Fibonacci(k) and ending at abscissa Fibonacci(k+3).

Peter Bala, Notes on A356988

Cf. A000032, A000045, A001611, A006165, A022114, A022120, A022367, A104449, A356989, A356990, A356991 - A356999.

a := proc(n) option remember; if n = 1 then 1 else n - a(a(n - a(a(a(n-1))))) end if; end proc:
seq(a(n), n = 1..100);

a(n+1) - a(n) = 0 or 1.
The terms of the sequence are completely determined by the following two results:
a) for n >= 2, a(L(n-1) + j) = F(n) for 0 <= j <= F(n-3), where F(n) = A000045(n), the n-th Fibonacci number with F(-1) = 1 and L(n) = A000032(n), the n-th Lucas number;
b) for n >= 2, a(F(n+1) + j) = F(n) + j for 0 <= j <= F(n-1).
Hence a(F(n+2)) = a(F(n+1)) + a(F(n)) for n >= 2 and a(L(n+2)) = a(L(n+1)) + a(L(n)) for n >= 0.
a(2*F(n)) = Lucas(n-1) for n >= 2;
a(3*F(n)) = 2*F(n) for n >= 1;
a(4*F(n)) = F(n+2) for n >= 2;
a(5*F(n)) = 4*F(n) - F(n-1) = A022120(n-2) for n >= 2.
a(2*L(n)) = F(n) + 3*F(n-1) = A104449(n) for n >= 0;
a(3*L(n)) = F(n+3) for n >= 3;
a(4*L(n)) = F(n+4) - L(n-3) = A022114(n-1) for n >= 3;
a(5*L(n)) = 11*F(n-1) + F(n-4) = A022367(n-1) for n >= 4.
For n >= 1, m >= 2, a(F(m*n)) = F(m*n-1) and a(L(m*n)) = F(m*n+1). Hence
a(L(m*n)) + a(F(m*n)) = L(m*n) and a(L(m*n)) - a(F(m*n)) = F(m*n).
Conjectures:
1) a(n) + a^[2](n - a^[2](n - a^[2](n))) = n for n >= 2.
2) If k >= 2 and  m = 2*k - 1 then a(m*n - a(k*n)) = a(m*n - a(m*n - a(m*n - a(k*n)))).

A127830 a(n) = Sum_{k=0..n} (binomial(floor(k/2),n-k) mod 2).

Original entry on oeis.org

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

Offset: 0

Paul Barry, Feb 01 2007

Row sums of number triangle A127829.
From Johannes W. Meijer, Jun 05 2011: (Start)
The Ze3 and Ze4 triangle sums, see A180662 for their definitions, of Sierpinski's triangle A047999 equal this sequence.
The sequences A127830(2^n-p), p>=0, are apparently all Fibonacci like sequences, i.e., the next term is the sum of the two nonzero terms that precede it; see the crossrefs. (End)

Cf.: A000045 (p=0), A000204 (p=7), A001060 (p=13), A000285 (p=14), A022095 (p=16), A022120 (p=24), A022121 (p=25), A022113 (p=28), A022096 (p=30), A022097 (p=31), A022098 (p=32), A022130 (p=44), A022137 (p=48), A022138 (p=49), A022122 (p=52), A022114 (p=53), A022123 (p=56), A022115 (p=60), A022100 (p=62), A022101 (p=63), A022103 (p=64), A022136 (p=79), A022388 (p=80), A022389 (p=88). - Johannes W. Meijer, Jun 05 2011

A127830 := proc(n) local k: option remember: add(binomial(floor(k/2), n-k) mod 2, k=0..n) end: seq(A127830(n), n=0..80); # Johannes W. Meijer, Jun 05 2011

Table[Sum[Mod[Binomial[Floor[k/2],n-k],2],{k,0,n}],{n,0,80}] (* James C. McMahon, Jan 04 2025 *)

def A127830(n): return sum(not ~(k>>1)&n-k for k in range(n+1)) # Chai Wah Wu, Jul 29 2025

a(2^n) = F(n); a(2^(n+1)+1) = L(n).

a(n) mod 2 = A000931(n+5) mod 2 = A011656(n+4).

A294116 Fibonacci sequence beginning 2, 21.

Original entry on oeis.org

2, 21, 23, 44, 67, 111, 178, 289, 467, 756, 1223, 1979, 3202, 5181, 8383, 13564, 21947, 35511, 57458, 92969, 150427, 243396, 393823, 637219, 1031042, 1668261, 2699303, 4367564, 7066867, 11434431, 18501298, 29935729, 48437027, 78372756, 126809783, 205182539, 331992322, 537174861

Offset: 0

Bruno Berselli, Oct 23 2017

Steven Vajda, Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications, Dover Publications (2008), page 24 (formula 8).

Colin Barker, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000032, A000045.
Subsequence of A047201, A047592, A113763.
Sequences of the type g(2,k;n): A118658 (k=0), A000032 (k=1), 2*A000045 (k=2,4), A020695 (k=3), A001060 (k=5), A022112 (k=6), A022113 (k=7), A294157 (k=8), A022114 (k=9), A022367 (k=10), A022115 (k=11), A022368 (k=12), A022116 (k=13), A022369 (k=14), A022117 (k=15), A022370 (k=16), A022118 (k=17), A022371 (k=18), A022119 (k=19), A022372 (k=20), this sequence (k=21), A022373 (k=22); A022374 (k=24); A022375 (k=26); A022376 (k=28), A190994 (k=29), A022377 (k=30); A022378 (k=32).

a0:=2; a1:=21; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]];

Mathematica
```
LinearRecurrence[{1, 1}, {2, 21}, 40]
```

Vec((2 + 19*x)/(1 - x - x^2) + O(x^40)) \\ Colin Barker, Oct 25 2017

a = BinaryRecurrenceSequence(1, 1, 2, 21)
print([a(n) for n in range(38)]) # Peter Luschny, Oct 25 2017

G.f.: (2 + 19*x)/(1 - x - x^2).
a(n) = a(n-1) + a(n-2).
Let g(r,s;n) be the n-th generalized Fibonacci number with initial values r, s. We have:
a(n) =     Lucas(n) + g(0,20;n), see A022354;
a(n) = Fibonacci(n) + g(2,20;n), see A022372;
a(n) =  2*g(1,21;n) - g(0,21;n);
a(n) =     g(1,k;n) + g(1,21-k;n) for all k in Z.
a(h+k) = a(h)*Fibonacci(k-1) + a(h+1)*Fibonacci(k) for all h, k in Z (see S. Vajda in References section). For h=0 and k=n:
a(n) = 2*Fibonacci(n-1) + 21*Fibonacci(n).
Sum_{j=0..n} a(j) = a(n+2) - 21.
a(n) = (2^(-n)*((1-sqrt(5))^n*(-20+sqrt(5)) + (1+sqrt(5))^n*(20+sqrt(5)))) / sqrt(5). - Colin Barker, Oct 25 2017

A172185 (9,11) Pascal triangle.

Original entry on oeis.org

1, 9, 11, 9, 20, 11, 9, 29, 31, 11, 9, 38, 60, 42, 11, 9, 47, 98, 102, 53, 11, 9, 56, 145, 200, 155, 64, 11, 9, 65, 201, 345, 355, 219, 75, 11, 9, 74, 266, 546, 700, 574, 294, 86, 11, 9, 83, 340, 812, 1246, 1274, 868, 380, 97, 11, 9, 92, 423, 1152, 2058, 2520, 2142, 1248, 477, 108, 11

Offset: 0

Mark Dols, Jan 28 2010

Sums of NW-SE diagonals give A022114 (apart from first two terms).
Triangle T(n,k), read by rows, given by (9,-8,0,0,0,0,0,0,0,...) DELTA (11,-10,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 09 2011
Row n: Expansion of (9+11x)*(1+x)^(n-1), n > 0. - Philippe Deléham, Oct 09 2011

			Triangle begins:
  1;
  9, 11;
  9, 20,  11;
  9, 29,  31,   11;
  9, 38,  60,   42,   11;
  9, 47,  98,  102,   53,   11;
  9, 56, 145,  200,  155,   64,   11;
  9, 65, 201,  345,  355,  219,   75,   11;
  9, 74, 266,  546,  700,  574,  294,   86,  11;
  9, 83, 340,  812, 1246, 1274,  868,  380,  97,  11;
  9, 92, 423, 1152, 2058, 2520, 2142, 1248, 477, 108, 11;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A022114, A093644, A172179.

T[n_, k_]:= If[n==0, 1, (9 + 2*k/n)*Binomial[n, k]]
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 28 2022 *)

def A172185(n,k): return 9*binomial(n,k) +2*binomial(n-1,k-1) -8*bool(n==0)
flatten([[A172185(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 28 2022

T(n,k) = T(n-1,k-1) + T(n-1,k) with T(0,0)=1, T(1,0)=9, T(1,1)=11. - Philippe Deléham, Oct 09 2011
G.f.: (1+8*x+10*y*x)/(1-x-y*x). - Philippe Deléham, Apr 13 2012
From G. C. Greubel, Apr 28 2022: (Start)
T(n, k) = 9*binomial(n, k) + 2*binomial(n-1, k-1) with T(0, 0) = 1.
Sum_{k=0..n} T(n, k) = 10*2^n - 9*[n=0]. (End)

Corrected and extended by Philippe Deléham, Oct 09 2011

cp's OEIS Frontend

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

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