A001106 -id:A001106 - cp's OEIS frontend

A152760 4 times 9-gonal numbers: a(n) = 2n(7*n-5).

0, 4, 36, 96, 184, 300, 444, 616, 816, 1044, 1300, 1584, 1896, 2236, 2604, 3000, 3424, 3876, 4356, 4864, 5400, 5964, 6556, 7176, 7824, 8500, 9204, 9936, 10696, 11484, 12300, 13144, 14016, 14916, 15844, 16800, 17784, 18796, 19836, 20904, 22000, 23124

Offset: 0

Omar E. Pol, Dec 14 2008

Sequence found by reading the line from 0, in the direction 0, 4, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. The square spiral is related to the primitive Pythagorean triple [3, 4, 5]. - Omar E. Pol, Oct 13 2011

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001106, A139268, A152759, A195019, A195020, A195021.

s=0;lst={s};Do[s+=n;AppendTo[lst,s],{n,4,8!,28}];lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
4*PolygonalNumber[9,Range[0,50]] (* Requires Mathematica version 10 or later *) (* or *) LinearRecurrence[{3,-3,1},{0,4,36},50] (* Harvey P. Dale, Aug 26 2019 *)

a(n)=2*n*(7*n-5) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 14*n^2 - 10*n = 4*A001106(n) = 2*A139268(n).
a(n) = a(n-1) + 28*n - 24 (with a(0)=0). - Vincenzo Librandi, Nov 26 2010
From Colin Barker, Apr 09 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 4*x*(1+6*x)/(1-x)^3. (End)
From Elmo R. Oliveira, Dec 27 2024: (Start)
E.g.f.: 2*exp(x)*x*(2 + 7*x).
a(n) = n + A195021(n). (End)

A218471 a(n) = n(7n-3)/2.

Original entry on oeis.org

0, 2, 11, 27, 50, 80, 117, 161, 212, 270, 335, 407, 486, 572, 665, 765, 872, 986, 1107, 1235, 1370, 1512, 1661, 1817, 1980, 2150, 2327, 2511, 2702, 2900, 3105, 3317, 3536, 3762, 3995, 4235, 4482, 4736, 4997, 5265, 5540, 5822, 6111, 6407, 6710, 7020, 7337

Offset: 0

Philippe Deléham, Mar 26 2013

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001106, A022264.

Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=7). - Bruno Berselli, Jun 10 2013

List([0..50], n-> n*(7*n-3)/2); # G. C. Greubel, Aug 31 2019

[n*(7*n-3)/2: n in [0..50]]; // G. C. Greubel, Aug 31 2019

seq(n*(7*n-3)/2, n=0..50); # G. C. Greubel, Aug 31 2019

Table[n*(7*n-3)/2, {n,0,50}] (* G. C. Greubel, Aug 23 2017 *)

a(n)=n*(7*n-3)/2 \\ Charles R Greathouse IV, Jun 17 2017

[n*(7*n-3)/2 for n in (0..50)] # G. C. Greubel, Aug 31 2019

G.f.: x*(2+5*x)/(1-x)^3.
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3) with a(0)=0, a(1)=2, a(2)=11.
a(n) = A001106(n) +  n.
a(n) = A022264(n) -  n.
a(n) = A022265(n) - 2*n.
a(n) = A186029(n) - 3*n.
a(n) = A179986(n) - 4*n.
a(n) = A024966(n) - 5*n.
a(n) = A174738(7*n+1).
E.g.f.: (x/2)*(7*x + 4)*exp(x). - G. C. Greubel, Aug 23 2017

A347263 Irregular triangle read by rows: T(n,k) is the sum of the subparts of the ziggurat diagram of n (described in A347186) that arise from the (2*k-1)-th double-staircase of the double-staircases diagram of n (described in A335616), n >= 1, k >= 1, and the first element of column k is in row A000384(k).

Original entry on oeis.org

1, 4, 6, 16, 12, 36, 1, 20, 0, 64, 0, 30, 6, 90, 0, 42, 0, 144, 17, 56, 0, 156, 0, 72, 34, 1, 256, 0, 0, 90, 0, 0, 324, 10, 0, 110, 0, 0, 400, 0, 8, 132, 70, 0, 342, 0, 0, 156, 0, 0, 576, 121, 0, 182, 0, 25, 462, 0, 0, 210, 102, 0, 784, 0, 0, 1, 240, 0, 0, 0, 900, 24, 52, 0, 272, 0, 0, 0

Offset: 1

Omar E. Pol, Sep 05 2021

Conjecture 1: the number of nonzero terms in row n equals A082647(n).
Conjecture 2: column k lists positive integers interleaved with 2*k+2 zeros.
The subparts of the ziggurat diagram are the polygons formed by the cells that are under the staircases.
The connection of the subparts of the ziggurat diagram with the polygonal numbers is as follows:
The area under a double-staircase labeled with the number j is equal to the m-th (j+2)-gonal number plus the (m-1)-th (j+2)-gonal number, where m is the number of steps on one side of the ladder from the base to the top.
The area under a simple-staircase labeled with the number j is equal to the m-th (j+2)-gonal number, where m is the number of steps.
So the k-th column of the triangle is related to the (2*k+1)-gonal numbers, for example:
For the calculation of column 1 we use triangular numbers A000217.
For the calculation of column 2 we use pentagonal numbers A000326.
For the calculation of column 3 we use heptagonal numbers A000566.
For the calculation of column 4 we use enneagonal numbers A001106.
And so on.
More generally, for the calculation of column k we use the (2*k+1)-gonal numbers.
For further information about the ziggurat diagram see A347186.

			Triangle begins:
   n / k   1    2    3    4
------------------------------
   1 |     1;
   2 |     4;
   3 |     6;
   4 |    16;
   5 |    12;
   6 |    36,   1;
   7 |    20,   0;
   8 |    64,   0;
   9 |    30,   6;
  10 |    90,   0;
  11 |    42,   0;
  12 |   144,  17;
  13 |    56,   0;
  14 |   156,   0;
  15 |    72,  34,   1;
  16 |   256,   0,   0;
  17 |    90,   0,   0;
  18 |   324,  10,   0;
  19 |   110,   0    0;
  20 |   400,   0,   8;
  21 |   132,  70,   0;
  22 |   342,   0,   0;
  23 |   156,   0,   0;
  24 |   576, 121,   0;
  25 |   182,   0,  25;
  26 |   462,   0,   0;
  27 |   210, 102,   0;
  28 |   784,   0,   0,   1;
...
For n = 15 the calculation of the 15th row of the triangle (in accordance with the geometric algorithm described in A347186) is as follows:
Stage 1 (Construction):
We draw the diagram called "double-staircases" with 15 levels described in A335616.
Then we label the five double-staircases (j = 1..5) as shown below:
                               _
                             _| |_
                           _|  _  |_
                         _|   | |   |_
                       _|    _| |_    |_
                     _|     |  _  |     |_
                   _|      _| | | |_      |_
                 _|       |   | |   |       |_
               _|        _|  _| |_  |_        |_
             _|         |   |  _  |   |         |_
           _|          _|   | | | |   |_          |_
         _|           |    _| | | |_    |           |_
       _|            _|   |   | |   |   |_            |_
     _|             |     |  _| |_  |     |             |_
   _|              _|    _| |  _  | |_    |_              |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _|_|_|_|_ _|_ _ _|_ _ _ _ _ _ _ _|
  1               2     3   4 5
.
Stage 2 (Debugging):
We remove the fourth double-staircase as it does not have at least one step at level 1 of the diagram starting from the base, as shown below:
                               _
                             _| |_
                           _|  _  |_
                         _|   | |   |_
                       _|    _| |_    |_
                     _|     |  _  |     |_
                   _|      _| | | |_      |_
                 _|       |   | |   |       |_
               _|        _|  _| |_  |_        |_
             _|         |   |     |   |         |_
           _|          _|   |     |   |_          |_
         _|           |    _|     |_    |           |_
       _|            _|   |         |   |_            |_
     _|             |     |         |     |             |_
   _|              _|    _|    _    |_    |_              |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
  1               2     3     5
.
Stage 3 (Annihilation):
We delete the second double-staircase and the steps of the first double-staircase that are just above the second double-staircase.
The new diagram has two double-staircases and two simple-staircases as shown below:
                               _
                              | |
                 _            | |            _
               _| |          _| |_          | |_
             _|   |         |     |         |   |_
           _|     |         |     |         |     |_
         _|       |        _|     |_        |       |_
       _|         |       |         |       |         |_
     _|           |       |         |       |           |_
   _|             |      _|    _    |_      |             |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
  1                     3     5
.
The diagram is called "ziggurat of 15".
Now we calculate the area (or the number of cells) under the staircases with multiplicity using polygonal numbers as shown below:
The area under the staircase labeled 1 is equal to A000217(8) = 36. There is a pair of these staircases, so T(15,1) = 2*36 = 72.
The area under the double-staircase labeled 3 is equal to A000326(4) + A000326(3) = 22 + 12 = 34, so T(15,2) = 34.
The area under the double-staircase labeled 5 is equal to A000566(1) + A000566(0) = 1 + 0 = 1, so T(15,3) = 1.
Therefore the 15th row of the triangle is [72, 34, 1].

Row sums give A347186.
Row n has length A351846(n).
Cf. A347529 (analog for the symmetric representation of sigma).
Cf. A000217, A000326, A000384, A000566, A001106, A082647, A139600, A139601, A237270, A237271, A237593, A279387, A279388, A279391, A335616, A346875.

A047346 Numbers that are congruent to {1, 4} mod 7.

Original entry on oeis.org

1, 4, 8, 11, 15, 18, 22, 25, 29, 32, 36, 39, 43, 46, 50, 53, 57, 60, 64, 67, 71, 74, 78, 81, 85, 88, 92, 95, 99, 102, 106, 109, 113, 116, 120, 123, 127, 130, 134, 137, 141, 144, 148, 151, 155, 158, 162, 165, 169, 172, 176, 179, 183, 186, 190, 193, 197, 200, 204, 207, 211, 214

Offset: 1

N. J. A. Sloane

Daniel Starodubtsev, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001106.

A047346:=n->4*n-3-floor(n/2); seq(A047346(k),k=1..100); # Wesley Ivan Hurt, Oct 16 2013

Table[4n-3-Floor[n/2],{n,100}] (* Wesley Ivan Hurt, Oct 16 2013 *)

a(n) = 4*n-n\2-3; \\ Altug Alkan, Sep 10 2016

a(n) = floor( (7n-5)/2 ). - Santi Spadaro, Jul 24 2001
G.f.: x*(1 + 3*x + 3*x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Dec 04 2011
a(n) = 4n - 3 - floor(n/2). - Wesley Ivan Hurt, Oct 16 2013
E.g.f.: 3 + ((14*x - 11)*exp(x) - exp(-x))/4. - David Lovler, Sep 01 2022

A125860 Rectangular table where column k equals row sums of matrix power A097712^k, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 5, 3, 1, 1, 17, 12, 4, 1, 1, 86, 69, 22, 5, 1, 1, 698, 612, 178, 35, 6, 1, 1, 9551, 8853, 2251, 365, 51, 7, 1, 1, 226592, 217041, 46663, 5990, 651, 70, 8, 1, 1, 9471845, 9245253, 1640572, 161525, 13131, 1057, 92, 9, 1, 1, 705154187

Offset: 0

Paul D. Hanna, Dec 13 2006

Triangle A097712 satisfies: A097712(n,k) = A097712(n-1,k) + [A097712^2](n-1,k-1) for n > 0, k > 0, with A097712(n,0)=A097712(n,n)=1 for n >= 0. Column 1 equals A016121, which counts the sequences (a_1, a_2, ..., a_n) of length n with a_1 = 1 satisfying a_i <= a_{i+1} <= 2*a_i.

T(2, n) = (n+1)*A005408(n) - Sum_{i=0..n} A001477(i) = (n+1)*(2*n+1) - A000217(n) = (n+1)*(3*n+2)/2; T(3, n) = (n+1)*A001106(n+1) - Sum_{i=0..n} A001477(i) = (n+1)*((n+1)*(7*n+2)/2) - A000217(n) = (n+1)*(7*n^2 + 8*n + 2)/2. - Bruno Berselli, Apr 25 2010

			Recurrence is illustrated by:
  T(4,1) = T(3,1) + T(3,2) = 17 + 69 = 86;
  T(4,2) = T(3,2) + T(3,3) + T(3,4) = 69 + 178 + 365 = 612;
  T(4,3) = T(3,3) + T(3,4) + T(3,5) + T(3,6) = 178 + 365 + 651 + 1057 = 2251.
Rows of this table begin:
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...;
  1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,...;
  1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, 247, 287, 330, ...;
  1, 17, 69, 178, 365, 651, 1057, 1604, 2313, 3205, 4301, 5622, 7189,..;
  1, 86, 612, 2251, 5990, 13131, 25291, 44402, 72711, 112780, 167486,..;
  1, 698, 8853, 46663, 161525, 435801, 996583, 2025458, 3768273, ...;
  1, 9551, 217041, 1640572, 7387640, 24530016, 66593821, 156664796, ...;
  1, 226592, 9245253, 100152049, 586285040, 2394413286, 7713533212, ...;
  1, 9471845, 695682342, 10794383587, 82090572095, 412135908606, ...;
  1, 705154187, 93580638024, 2079805452133, 20540291522675, ...;
  1, 94285792211, 22713677612832, 723492192295786, 9278896006526795,...;
  1, 22807963405043, 10025101876435413, 458149292979837523, ...;
  ...
where column k equals the row sums of matrix power A097712^k for k >= 0.
Triangle A097712 begins:
  1;
  1,      1;
  1,      3,       1;
  1,      8,       7,       1;
  1,     25,      44,      15,       1;
  1,    111,     346,     208,      31,      1;
  1,    809,    4045,    3720,     912,     63,     1;
  1,  10360,   77351,   99776,   35136,   3840,   127,   1;
  1, 236952, 2535715, 4341249, 2032888, 308976, 15808, 255; ...
where A097712(n,k) = A097712(n-1,k) + [A097712^2](n-1,k-1);
e.g., A097712(5,2) = A097712(4,2) + [A097712^2](4,1) = 44 + 302 = 346.
Matrix square A097712^2 begins:
     1;
     2,     1;
     5,     6,     1;
    17,    37,    14,     1;
    86,   302,   193,    30,    1;
   698,  3699,  3512,   881,   62,   1;
  9551, 73306, 96056, 34224, 3777, 126, 1; ...
Matrix cube A097712^3 begins:
       1;
       3,      1;
      12,      9,      1;
      69,     87,     21,      1;
     612,   1146,    447,     45,    1;
    8853,  22944,  12753,   2019,   93,   1;
  217041, 744486, 549453, 120807, 8595, 189, 1; ...

Olivier Danvy, Summa Summarum: Moessner's Theorem without Dynamic Programming, arXiv:2412.03127 [cs.DM], 2024. See p. 16.

Cf. A097712; columns: A016121, A125862, A125863, A125864, A125865; A125861 (diagonal), A125859 (antidiagonal sums). Variants: A125790, A125800.

Cf. for recursive method [Ar(m) is the m-th term of a sequence in the OEIS] a(n) = n*Ar(n) - A000217(n-1) or a(n) = (n+1)*Ar(n+1) - A000217(n) and similar: A081436, A005920, A005945, A006003. - Bruno Berselli, Apr 25 2010

T[n_, k_] := T[n, k] = If[Or[n == 0, k == 0], 1, Sum[T[n - 1, j + k], {j, 0, k}]];
Table[T[#, k] &[n - k + 1], {n, 0, 9}, {k, 0, n + 1}] (* Michael De Vlieger, Dec 10 2024, after PARI *)

T(n,k)=if(n==0 || k==0,1,sum(j=0,k,T(n-1,j+k)))

T(n,k) = Sum_{j=0..k} T(n-1, j+k) for n > 0, with T(0,n)=T(n,0)=1 for n >= 0.

A226449 a(n) = n(5n^2-8*n+5)/2.

Original entry on oeis.org

0, 1, 9, 39, 106, 225, 411, 679, 1044, 1521, 2125, 2871, 3774, 4849, 6111, 7575, 9256, 11169, 13329, 15751, 18450, 21441, 24739, 28359, 32316, 36625, 41301, 46359, 51814, 57681, 63975, 70711, 77904, 85569, 93721, 102375, 111546, 121249, 131499, 142311, 153700

Offset: 0

Bruno Berselli, Jun 07 2013

Sequences of the type b(m)+m*b(m-1), where b is a polygonal number:
A006003(n) = A000217(n) + n*A000217(n-1)      (b = triangular numbers);
A069778(n) = A000290(n+1) + (n+1)*A000290(n)  (b = square numbers);
A143690(n) = A000326(n+1) + (n+1)*A000326(n)  (b = pentagonal numbers);
A212133(n) = A000384(n) + n*A000384(n-1)      (b = hexagonal numbers);
a(n)       = A000566(n) + n*A000566(n-1)      (b = heptagonal numbers);
A226450(n) = A000567(n) + n*A000567(n-1)      (b = octagonal numbers);
A226451(n) = A001106(n) + n*A001106(n-1)      (b = nonagonal numbers);
A204674(n) = A001107(n+1) + (n+1)*A001107(n)  (b = decagonal numbers).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. (see the comment) A000566, A006003, A069778, A143690, A204674, A212133, A226450, A226451.

Magma
```
[n*(5*n^2-8*n+5)/2: n in [0..40]];
```

I:=[0,1,9,39]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Aug 18 2013

Table[n (5 n^2 - 8 n + 5)/2, {n, 0, 40}]
CoefficientList[Series[x (1 + 5 x + 9 x^2)/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{4,-6,4,-1},{0,1,9,39},50] (* Harvey P. Dale, May 19 2017 *)

a(n)=n*(5*n^2-8*n+5)/2 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: x*(1+5*x+9*x^2)/(1-x)^4.

a(n) - a(-n) = A008531(n) for n>0.

A051740 Partial sums of A007584.

Original entry on oeis.org

1, 11, 45, 125, 280, 546, 966, 1590, 2475, 3685, 5291, 7371, 10010, 13300, 17340, 22236, 28101, 35055, 43225, 52745, 63756, 76406, 90850, 107250, 125775, 146601, 169911, 195895, 224750, 256680, 291896, 330616, 373065, 419475, 470085, 525141

Offset: 0

Barry E. Williams, Dec 07 1999

Convolution of A000027 with A001106 (excluding 0). - Bruno Berselli, Dec 07 2012

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
Murray R.Spiegel, Calculus of Finite Differences and Difference Equations, "Schaum's Outline Series", McGraw-Hill, 1971, pp. 10-20, 79-94.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A001106, A007584.
Cf. A093564 ((7, 1) Pascal, column m=4).
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.

List([0..40], n-> (7*n+4)*Binomial(n+3,3)/4); # G. C. Greubel, Aug 29 2019

/* A000027 convolved with A001106 (excluding 0): */ A001106:=func; [&+[(n-i+1)*A001106(i): i in [1..n]]: n in [1..36]]; // Bruno Berselli, Dec 07 2012

seq((7*n+4)*binomial(n+3,3)/4, n=0..40); # G. C. Greubel, Aug 29 2019

Table[(7*n+4)*Binomial[n+3,3]/4, {n,0,40}] (* G. C. Greubel, Aug 29 2019 *)
LinearRecurrence[{5,-10,10,-5,1},{1,11,45,125,280},40] (* Harvey P. Dale, May 18 2023 *)

vector(40, n, (7*n-3)*binomial(n+2,3)/4) \\ G. C. Greubel, Aug 29 2019

[(7*n+4)*binomial(n+3,3)/4 for n in (0..40)] # G. C. Greubel, Aug 29 2019

a(n) = binomial(n+3, 3)*(7*n+4)/4.
a(n) = (7*n+4)*binomial(n+3, 3)/4.
G.f.: (1+6*x)/(1-x)^5.
a(n) = A080852(7,n). - R. J. Mathar, Jul 28 2016
E.g.f.: (4! + 240*x + 288*x^2 + 88*x^3 + 7*x^4)*exp(x)/4!. - G. C. Greubel, Aug 29 2019

A036411 9-gonal square numbers.

Original entry on oeis.org

1, 9, 1089, 8281, 978121, 7436529, 878351769, 6677994961, 788758910641, 5996832038649, 708304623404049, 5385148492712041, 636056763057925561, 4835857349623374369, 571178264921393749929, 4342594514813297471521

Offset: 1

Jean-Francois Chariot (jeanfrancois.chariot(AT)afoc.alcatel.fr)

From Ant King, Nov 17 2011: (Start)
lim_{n -> infinity} a(2n+1)/a(2n) = 1/625 * (36913 + 9864 * sqrt(14));
lim_{n -> infinity} a(2n)/a(2n-1) = 1/625 * (2417 + 624 * sqrt(14)).
(End)

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Nonagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (1,898,-898,-1,1)

Cf. A001106, A048911, A048919.

I:=[1, 9, 1089, 8281]; [n le 4 select I[n] else 898*Self(n-2)-Self(n-4)+200: n in [1..20]]; // Vincenzo Librandi, Nov 18 2011

a(0):=1:a(1):=9:a(2):=1089:a(3):=8281: a(4):=978121:for n from 0 to 20 do a(n+5):=a(n+4)+898*a(n+3)-898*a(n+2)-a(n+1)+a(n):od:seq(a(n),n=0..20); # Richard Choulet, May 08 2009

LinearRecurrence[ {1, 898, - 898, - 1, 1 }, { 1, 9, 1089, 8281, 978121 }, 16] (* Ant King, Nov 17 2011 *)

O.g.f. f(z) = 1 + 9*z + ... = ((1 + 8*z + 182*z^2 + 8*z^3 + z^4)/((1-z)*(1 - 898*z^2 + z^4))). With the first values, for n > 0: a(n+5) = a(n+4) + 898*a(n+3) - 898*a(n+2) - a(n+1) + a(n). On every bisection modulo 2: a(n+2) = 30*a(n+1) - a(n) + 200. On every bisection modulo 2: a(n+1) = 449*a(n) + 100 + 60*sqrt(56*a(n)^2 + 25*a(n)). a(n) = -25/112 + (11/28 + (11/112)*sqrt(14))*(15 + 4*sqrt(14))^n + (11/28 - (11/112)*sqrt(14))*(15 - 4*sqrt(14))^n + (7/32 - (1/16)*sqrt(14))*(-15 + 4*sqrt(14))^n + (7/32 + (1/16)*sqrt(14))*(-15 - 4*sqrt(14))^n. - Richard Choulet, May 08 2009

a(n) = 898 * a(n-2) - a(n-4) + 200. - Ant King, Nov 17 2011

More terms from Eric W. Weisstein

More terms from Richard Choulet, May 08 2009

A082723 Palindromic nonagonal (or 9-gonal or enneagonal) numbers.

Original entry on oeis.org

1, 9, 111, 474, 969, 6666, 18981, 67276, 4411144, 6964696, 15444451, 57966975, 448707844, 460595064, 579696975, 931929139, 994040499, 1227667221, 9698998969, 61556965516, 664248842466, 699030030996, 99451743334715499

Offset: 1

Patrick De Geest, Apr 13 2003

P. De Geest, Palindromic nonagonals

Cf. A001106, A055560.

palQ[n_]:=Module[{idn=IntegerDigits[n]},idn==Reverse[idn]]; Select[Table[ (n(7n-5))/2,{n,170000000}],palQ] (* Harvey P. Dale, Nov 15 2011 *)

A214675 a(n) = 9n^2 - 13n + 5.

Original entry on oeis.org

1, 15, 47, 97, 165, 251, 355, 477, 617, 775, 951, 1145, 1357, 1587, 1835, 2101, 2385, 2687, 3007, 3345, 3701, 4075, 4467, 4877, 5305, 5751, 6215, 6697, 7197, 7715, 8251, 8805, 9377, 9967, 10575, 11201, 11845, 12507, 13187, 13885, 14601, 15335, 16087

Offset: 1

Reinhard Zumkeller, Jul 25 2012

Central terms of triangle A214661.

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

Cf. A000567, A001106, A214661.

Cf. A220083 for a list of numbers of the form n*P(s,n)-(n-1)*P(s,n-1), where P(s,n) is the n-th polygonal number with s sides.

Haskell
```
a214675 n = (9 * n - 13) * n + 5
```

[(3*n-2)^2-(n-1): n in [1..50]]; // G. C. Greubel, Mar 08 2024

Table[9n^2-13n+5,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,15,47}, 50] (* Harvey P. Dale, Nov 09 2019 *)

a(n)=9*n^2-13*n+5 \\ Charles R Greathouse IV, Oct 07 2015

[(3*n-2)^2-(n-1) for n in range(1,51)] # G. C. Greubel, Mar 08 2024

G.f.: x*(1+12*x+5*x^2)/(1-x)^3. - Bruno Berselli, Dec 10 2012
a(n) = n*A000567(n)-(n-1)*A000567(n-1). - Bruno Berselli, Dec 10 2012
E.g.f.: -5 + (5 - 4*x + 9*x^2)*exp(x). - G. C. Greubel, Mar 08 2024

cp's OEIS Frontend

A152760 4 times 9-gonal numbers: a(n) = 2*n*(7*n-5).

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A218471 a(n) = n*(7*n-3)/2.

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A047346 Numbers that are congruent to {1, 4} mod 7.

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A125860 Rectangular table where column k equals row sums of matrix power A097712^k, read by antidiagonals.

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A226449 a(n) = n*(5*n^2-8*n+5)/2.

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A051740 Partial sums of A007584.

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A152760 4 times 9-gonal numbers: a(n) = 2n(7*n-5).

A218471 a(n) = n(7n-3)/2.

A226449 a(n) = n(5n^2-8*n+5)/2.

A214675 a(n) = 9n^2 - 13n + 5.