A195041 -id:A195041 - cp's OEIS frontend

A032528 Concentric hexagonal numbers: floor(3*n^2/2).

0, 1, 6, 13, 24, 37, 54, 73, 96, 121, 150, 181, 216, 253, 294, 337, 384, 433, 486, 541, 600, 661, 726, 793, 864, 937, 1014, 1093, 1176, 1261, 1350, 1441, 1536, 1633, 1734, 1837, 1944, 2053, 2166, 2281, 2400, 2521, 2646, 2773, 2904, 3037, 3174, 3313, 3456, 3601, 3750

Offset: 0

N. J. A. Sloane

From Omar E. Pol, Aug 20 2011: (Start)
Cellular automaton on the hexagonal net. The sequence gives the number of "ON" cells in the structure after n-th stage. A007310 gives the first differences. For a definition without words see the illustration of initial terms in the example section. Note that the cells become intermittent. A083577 gives the primes of this sequences.
A033581 and A003154 interleaved.
Row sums of an infinite square array T(n,k) in which column k lists 2*k-1 zeros followed by the numbers A008458 (see example).  (End)
Sequence found by reading the line from 0, in the direction 0, 1, ... and the same line from 0, in the direction 0, 6, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Main axis perpendicular to A045943 in the same spiral. - Omar E. Pol, Sep 08 2011

			From _Omar E. Pol_, Aug 20 2011: (Start)
Using the numbers A008458 we can write:
  0, 1, 6, 12, 18, 24, 30, 36, 42,  48,  54, ...
  0, 0, 0,  1,  6, 12, 18, 24, 30,  36,  42, ...
  0, 0, 0,  0,  0,  1,  6, 12, 18,  24,  30, ...
  0, 0, 0,  0,  0,  0,  0,  1,  6,  12,  18, ...
  0, 0, 0,  0,  0,  0,  0,  0,  0,   1,   6, ...
And so on.
===========================================
The sums of the columns give this sequence:
0, 1, 6, 13, 24, 37, 54, 73, 96, 121, 150, ...
...
Illustration of initial terms as concentric hexagons:
.
.                                         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      o   o o o   o
.                         o o o o        o         o
.                                         o o o o o
.
. 1    6        13           24               37
.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Index entries for sequences related to cellular automata.

Cf. A003154, A007310, A008458, A033581, A083577, A000326, A001318, A005449, A045943, A032527, A195041. Column 6 of A195040.
Cf. A004524, A004525, A033436, A212959, A238410, A227017, A282513.
Cf. A033581, A003154, A011934, A184533.

a032528 n = a032528_list !! n
a032528_list = scanl (+) 0 a007310_list
-- Reinhard Zumkeller, Jan 07 2012

[Floor(3*n^2/2): n in [0..50]]; // Vincenzo Librandi, Aug 21 2011

f[n_, m_] := Sum[Floor[n^2/k], {k, 1, m}]; t = Table[f[n, 2], {n, 1, 90}] (* Clark Kimberling, Apr 20 2012 *)

a(n)=3*n^2\2 \\ Charles R Greathouse IV, Sep 24 2015

From Joerg Arndt, Aug 22 2011: (Start)
G.f.: (x+4*x^2+x^3)/(1-2*x+2*x^3-x^4) = x*(1+4*x+x^2)/((1+x)*(1-x)^3).
a(n) = +2*a(n-1) -2*a(n-3) +1*a(n-4). (End)
a(n) = (6*n^2+(-1)^n-1)/4. - Bruno Berselli, Aug 22 2011
a(n) = A184533(n), n >= 2. - Clark Kimberling, Apr 20 2012
First differences of A011934: a(n) = A011934(n) - A011934(n-1) for n>0. - Franz Vrabec, Feb 17 2013
From Paul Curtz, Mar 31 2019: (Start)
a(-n) = a(n).
a(n) = a(n-2) + 6*(n-1) for n > 1.
a(2*n) = A033581(n).
a(2*n+1) = A003154(n+1). (End)
E.g.f.: (3*x*(x + 1)*cosh(x) + (3*x^2 + 3*x - 1)*sinh(x))/2. - Stefano Spezia, Aug 19 2022
Sum_{n>=1} 1/a(n) = Pi^2/36 + tan(Pi/(2*sqrt(3)))*Pi/(2*sqrt(3)). - Amiram Eldar, Jan 16 2023

New name and more terms a(41)-a(50) from Omar E. Pol, Aug 20 2011

A077221 a(0) = 0 and then alternately even and odd numbers in increasing order such that the sum of any two successive terms is a square.

Original entry on oeis.org

0, 1, 8, 17, 32, 49, 72, 97, 128, 161, 200, 241, 288, 337, 392, 449, 512, 577, 648, 721, 800, 881, 968, 1057, 1152, 1249, 1352, 1457, 1568, 1681, 1800, 1921, 2048, 2177, 2312, 2449, 2592, 2737, 2888, 3041, 3200, 3361, 3528, 3697, 3872, 4049, 4232

Offset: 0

Amarnath Murthy, Nov 03 2002

This sequence arises from reading the line from 0, in the direction 0, 1, ... and the same line from 0, in the direction 0, 8, ..., in the square spiral whose vertices are the triangular numbers A000217. Cf. A139591, etc. - Omar E. Pol, May 03 2008
The general formula for alternating sums of powers of odd integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,0)-(-1)^k*P(n,2*k))/2. Here n=2, thus a(k) = |(P(2,0)-(-1)^k*P(2,2*k))/2|. - Peter Luschny, Jul 12 2009
Axis perpendicular to A046092 in the square spiral whose vertices are the triangular numbers A000217. See the comment above. - Omar E. Pol, Sep 14 2011
Column 8 of A195040. - Omar E. Pol, Sep 28 2011
Concentric octagonal numbers. A139098 and A069129 interleaved. - Omar E. Pol, Sep 17 2011
Subsequence of A194274. - Bruno Berselli, Sep 22 2011
Partial sums of A047522. - Reinhard Zumkeller, Jan 07 2012
Alternating sum of the first n odd squares in decreasing order, n >= 1. Also number of "ON" cells at n-th stage in simple 2-dimensional cellular automaton. The rules are: on the infinite square grid, start with all cells OFF, so a(0) = 0. Turn a single cell to the ON state, so a(1) = 1. At each subsequent step, the neighbor cells of each cell of the old generation are turned ON, and the cells of the old generation are turned OFF. Here "neighbor" refers to the eight adjacent cells of each ON cell. See example. - Omar E. Pol, Feb 16 2014

			From _Omar E. Pol_, Feb 16 2014: (Start)
Illustration of initial terms as a cellular automaton:
.
.                                   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
.
.     1       8           17              32
.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Bruno Berselli, An origin of A077221 (illustration) (see Pol's comment).
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Index entries for sequences related to cellular automata.

Cf. A032528, A069129, A077222, A131925, A139098, A194274, A195040, A195041, A195042, A195142.

a077221 n = a077221_list !! n
a077221_list = scanl (+) 0 a047522_list
-- Reinhard Zumkeller, Jan 07 2012

[2*n^2 - (n mod 2): n in [0..50]]; // Vincenzo Librandi, Sep 22 2011

a := n -> 2*n^2 - (n mod 2); # Peter Luschny, Jul 12 2009

a=1;lst={a};Do[b=n^2-a;AppendTo[lst,b];a=b,{n,3,6!,2}];lst (* Vladimir Joseph Stephan Orlovsky, May 18 2009 *)

a(2n) = 8*n^2, a(2n+1) = 8*n(n+1) + 1.
From Ralf Stephan, Mar 31 2003: (Start)
a(n) = 2*n^2 + 4*n + 1 [+1 if n is odd] with a(0)=1.
G.f.: x*(x^2+6*x+1)/(1-x)^3/(1+x). (End)
Row sums of triangle A131925; binomial transform of (1, 7, 2, 4, -8, 16, -32, ...). - Gary W. Adamson, Jul 29 2007
a(n) = a(-n); a(n+1) = A195605(n) - (-1)^n. - Bruno Berselli, Sep 22 2011
a(n) = 2*n^2 + ((-1)^n-1)/2. - Omar E. Pol, Sep 28 2011
Sum_{n>=1} 1/a(n) = Pi^2/48 + tan(Pi/(2*sqrt(2)))*Pi /(4*sqrt(2)). - Amiram Eldar, Jan 16 2023

Extended by Ralf Stephan, Mar 31 2003

A047336 Numbers that are congruent to {1, 6} mod 7.

Original entry on oeis.org

1, 6, 8, 13, 15, 20, 22, 27, 29, 34, 36, 41, 43, 48, 50, 55, 57, 62, 64, 69, 71, 76, 78, 83, 85, 90, 92, 97, 99, 104, 106, 111, 113, 118, 120, 125, 127, 132, 134, 139, 141, 146, 148, 153, 155, 160, 162, 167, 169, 174, 176, 181, 183, 188, 190, 195, 197, 202, 204, 209

Offset: 1

N. J. A. Sloane

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2-1 == 0 (mod 7). - Bruno Berselli, Nov 17 2010

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

Cf. A007310, A019674, A047522, A045472 (primes), A195041 (partial sums), A005408, A047209, A056020, A090771, A091998, A113801, A160389, A175885, A175886, A175887, A178818, A371858.

a047336 n = a047336_list !! (n-1)
a047336_list = 1 : 6 : map (+ 7) a047336_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..210]| n mod 7 in {1,6}]; // Bruno Berselli, Feb 22 2011

Rest[Flatten[Table[{7i-1,7i+1},{i,0,40}]]] (* Harvey P. Dale, Nov 20 2010 *)

a(n)=n\2*7-(-1)^n \\ Charles R Greathouse IV, May 02 2016

a(1) = 1; a(n) = 7(n-1) - a(n-1). - Rolf Pleisch, Jan 31 2008 (corrected by Jon E. Schoenfield, Dec 22 2008)
a(n) = (7/2)*(n-(1-(-1)^n)/2) - (-1)^n. - Rolf Pleisch, Nov 02 2010
From Bruno Berselli, Nov 17 2010: (Start)
G.f.: x*(1+5*x+x^2)/((1+x)*(1-x)^2).
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = a(n-2)+7.
a(n) = 7*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
a(n) = 7*floor(n/2)+(-1)^(n+1). - Gary Detlefs, Dec 29 2011
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/7)*cot(Pi/7) = A019674 * A178818. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((14*x - 7)*exp(x) + 3*exp(-x))/4. - David Lovler, Sep 01 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/7) (A160389).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/7) * cosec(Pi/7) (A371858). (End)

More terms from Jon E. Schoenfield, Jan 18 2009

A032527 Concentric pentagonal numbers: floor( 5*n^2 / 4 ).

Original entry on oeis.org

0, 1, 5, 11, 20, 31, 45, 61, 80, 101, 125, 151, 180, 211, 245, 281, 320, 361, 405, 451, 500, 551, 605, 661, 720, 781, 845, 911, 980, 1051, 1125, 1201, 1280, 1361, 1445, 1531, 1620, 1711, 1805, 1901, 2000, 2101, 2205, 2311, 2420, 2531, 2645, 2761, 2880, 3001

Offset: 0

N. J. A. Sloane

Also A033429 and A062786 interleaved. - Omar E. Pol, Sep 28 2011
Partial sums of A047209. - Reinhard Zumkeller, Jan 07 2012
From Wolfdieter Lang, Aug 06 2013: (Start)
a(n) = -N(-floor(n/2),n) with the N(a,b) = ((2*a+b)^2 - b^2*5)/4, the norm for integers a + b*omega(5), a, b rational integers, in the quadratic number field Q(sqrt(5)), where omega(5) = (1 + sqrt(5))/2 (golden section).
a(n) = max({|N(a,n)|,a = -n..+n}) = |N(-floor(n/2),n)| = n^2 + n*floor(n/2) - floor(n/2)^2 = floor(5*n^2/4) (the last eq. checks for even and odd n). (End)

			From _Omar E. Pol_, Sep 28 2011 (Start):
Illustration of initial terms (In a precise representation the pentagons should appear strictly concentric):
.
.                                             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 o o o
.
. 1    5        11          20                31
.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000290, A032528, A077043, A195041. Column 5 of A195040. [Omar E. Pol, Sep 28 2011]

a032527 n = a032527_list !! n
a032527_list = scanl (+) 0 a047209_list
-- Reinhard Zumkeller, Jan 07 2012

[5*n^2/4+((-1)^n-1)/8: n in [0..50]]; // Vincenzo Librandi, Sep 29 2011

A032527:=n->5*n^2/4+((-1)^n-1)/8: seq(A032527(n), n=0..100); # Wesley Ivan Hurt, Mar 12 2015

Table[Round[5n^2/4], {n, 0, 39}] (* Alonso del Arte, Sep 28 2011 *)

a(n)=5*n^2>>2 \\ Charles R Greathouse IV, Sep 28 2011

def A032527(n): return 5*n**2>>2  # Chai Wah Wu, Jul 30 2022

a(n) = 5*n^2/4+((-1)^n-1)/8. - Omar E. Pol, Sep 28 2011
G.f.: x*(1+3*x+x^2)/(1-2*x+2*x^3-x^4). - Colin Barker, Jan 06 2012
a(n) = a(-n); a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>0, a(-1) = 1, a(0) = 0, a(1) = 1, a(2) = 5, n >= 3. (See the Bruno Berselli recurrence and a general comment for primes 1 (mod 4) under A227541). - Wolfdieter Lang, Aug 08 2013
a(n) = Sum_{j=1..n} Sum{i=1..n} ceiling((i+j-n+1)/2). - Wesley Ivan Hurt, Mar 12 2015
Sum_{n>=1} 1/a(n) = Pi^2/30 + tan(Pi/(2*sqrt(5)))*Pi/sqrt(5). - Amiram Eldar, Jan 16 2023

New name from Omar E. Pol, Sep 28 2011

A195040 Square array read by antidiagonals with T(n,k) = kn^2/4+(k-4)((-1)^n-1)/8, n>=0, k>=0.

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 3, 2, 1, 0, 1, 4, 5, 3, 1, 0, 0, 7, 8, 7, 4, 1, 0, 1, 9, 13, 12, 9, 5, 1, 0, 0, 13, 18, 19, 16, 11, 6, 1, 0, 1, 16, 25, 27, 25, 20, 13, 7, 1, 0, 0, 21, 32, 37, 36, 31, 24, 15, 8, 1, 0, 1, 25, 41, 48, 49, 45, 37, 28, 17, 9, 1, 0

Offset: 0

Omar E. Pol, Sep 27 2011

Also, if k >= 2 and m = 2*k, then column k lists the numbers of the form k*n^2 and the centered m-gonal numbers interleaved.
For k >= 3, this is also a table of concentric polygonal numbers. Column k lists the concentric k-gonal numbers.
It appears that the first differences of column k are the numbers that are congruent to {1, k-1} mod k, if k >= 3.

			Array begins:
  0,   0,   0,   0,   0,   0,   0,   0,   0,   0, ...
  1,   1,   1,   1,   1,   1,   1,   1,   1,   1, ...
  0,   1,   2,   3,   4,   5,   6,   7,   8,   9, ...
  1,   3,   5,   7,   9,  11,  13,  15,  17,  19, ...
  0,   4,   8,  12,  16,  20,  24,  28,  32,  36, ...
  1,   7,  13,  19,  25,  31,  37,  43,  49,  55, ...
  0,   9,  18,  27,  36,  45,  54,  63,  72,  81, ...
  1,  13,  25,  37,  49,  61,  73,  85,  97, 109, ...
  0,  16,  32,  48,  64,  80,  96, 112, 128, 144, ...
  1,  21,  41,  61,  81, 101, 121, 141, 161, 181, ...
  0,  25,  50,  75, 100, 125, 150, 175, 200, 225, ...
  ...

Muniru A Asiru, Rows n=0..100, flattened

Rows n: A000004 (n=0), A000012 (n=1), A001477 (n=2), A005408 (n=3), A008586 (n=4), A016921 (n=5), A008591 (n=6), A017533 (n=7), A008598 (n=8), A215145 (n=9), A008607 (n=10).

Columns k: A000035 (k=0), A004652 (k=1), A000982 (k=2), A077043 (k=3), A000290 (k=4), A032527 (k=5), A032528 (k=6), A195041 (k=7), A077221 (k=8), A195042 (k=9), A195142 (k=10), A195043 (k=11), A195143 (k=12), A195045 (k=13), A195145 (k=14), A195046 (k=15), A195146 (k=16), A195047 (k=17), A195147 (k=18), A195048 (k=19), A195148 (k=20), A195049 (k=21), A195149 (k=22), A195058 (k=23), A195158 (k=24).

nmax:=13;; T:=List([0..nmax],n->List([0..nmax],k->k*n^2/4+(k-4)*((-1)^n-1)/8));; b:=List([2..nmax],n->OrderedPartitions(n,2));;
a:=Flat(List([1..Length(b)],i->List([1..Length(b[i])],j->T[b[i][j][2]][b[i][j][1]]))); # Muniru A Asiru, Jul 19 2018

A195040 := proc(n,k)
        k*n^2/4+((-1)^n-1)*(k-4)/8 ;
end proc:
for d from 0 to 12 do
        for k from 0 to d do
                printf("%d,",A195040(d-k,k)) ;
        end do:
end do; # R. J. Mathar, Sep 28 2011

t[n_, k_] := k*n^2/4+(k-4)*((-1)^n-1)/8; Flatten[ Table[ t[n-k, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, Dec 14 2011 *)

A033582 a(n) = 7*n^2.

Original entry on oeis.org

0, 7, 28, 63, 112, 175, 252, 343, 448, 567, 700, 847, 1008, 1183, 1372, 1575, 1792, 2023, 2268, 2527, 2800, 3087, 3388, 3703, 4032, 4375, 4732, 5103, 5488, 5887, 6300, 6727, 7168, 7623, 8092, 8575, 9072, 9583, 10108, 10647, 11200, 11767, 12348, 12943, 13552, 14175

Offset: 0

N. J. A. Sloane

From Roberto E. Martinez II, Jan 07 2002: (Start)
Number of edges of the complete bipartite graph of order 8n, K_n,7n.
Number of edges of the complete tripartite graph of order 5n, K_n,n,3n. (End)

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

Cf. A000290, A008589, A132111, A195041.

7Range[0, 49]^2 (* Alonso del Arte, Jun 30 2013 *)

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

Central terms of the triangle in A132111: a(n) = A132111(2*n,n). - Reinhard Zumkeller, Aug 10 2007
a(n) = 7*A000290(n). - Omar E. Pol, Dec 11 2008
a(n) = 14*n + a(n-1) - 7 (with a(0) = 0). - Vincenzo Librandi, Aug 05 2010
G.f.: -7*x*(1+x)/(x-1)^3. - R. J. Mathar, Feb 06 2017
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/42.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/84.
Product_{n>=1} (1 + 1/a(n)) = sqrt(7)*sinh(Pi/sqrt(7))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(7)*sin(Pi/sqrt(7))/Pi. (End)
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 7*exp(x)*x*(1 + x).
a(n) = n*A008589(n) = A195041(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A195042 Concentric 9-gonal numbers.

Original entry on oeis.org

0, 1, 9, 19, 36, 55, 81, 109, 144, 181, 225, 271, 324, 379, 441, 505, 576, 649, 729, 811, 900, 991, 1089, 1189, 1296, 1405, 1521, 1639, 1764, 1891, 2025, 2161, 2304, 2449, 2601, 2755, 2916, 3079, 3249, 3421, 3600, 3781, 3969, 4159, 4356, 4555, 4761, 4969, 5184, 5401, 5625

Offset: 0

Omar E. Pol, Sep 27 2011

Also concentric enneagonal numbers or concentric nonagonal numbers.
A016766 and A069131 interleaved.
Partial sums of A056020. - Reinhard Zumkeller, Jan 07 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Column 9 of A195040.

Cf. A016766, A069131, A077221, A195041, A195142, A195043.

a195042 n = a195042_list !! n
a195042_list = scanl (+) 0 a056020_list
-- Reinhard Zumkeller, Jan 07 2012

[(9*n^2+5/2*((-1)^n-1))/4: n in [0..50]]; // Vincenzo Librandi, Sep 29 2011

LinearRecurrence[{2,0,-2,1},{0,1,9,19},60] (* Harvey P. Dale, Nov 24 2019 *)

a(n)=(9*n^2+5/2*((-1)^n-1))/4 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = (9*n^2 + 5/2*((-1)^n - 1))/4.
From R. J. Mathar, Sep 28 2011: (Start)
G.f.: -x*(1+7*x+x^2) / ( (1+x)*(x-1)^3 ).
a(n) + a(n+1) = A060544(n+1). (End)
Sum_{n>=1} 1/a(n) = Pi^2/54 + tan(sqrt(5)*Pi/6)*Pi/(3*sqrt(5)). - Amiram Eldar, Jan 16 2023

A270693 Alternating sum of centered 25-gonal numbers.

Original entry on oeis.org

1, -25, 51, -100, 151, -225, 301, -400, 501, -625, 751, -900, 1051, -1225, 1401, -1600, 1801, -2025, 2251, -2500, 2751, -3025, 3301, -3600, 3901, -4225, 4551, -4900, 5251, -5625, 6001, -6400, 6801, -7225, 7651, -8100, 8551, -9025, 9501, -10000, 10501

Offset: 0

Ilya Gutkovskiy, Mar 21 2016

The absolute value alternating sum of centered k-gonal numbers gives concentric k-gonal numbers.

More generally, the ordinary generating function for the alternating sum of centered k-gonal numbers is (1 - (k - 2)*x + x^2)/((1 - x)*(1 + x)^3).

OEIS Wiki, Centered polygonal numbers
Eric Weisstein's World of Mathematics, Centered Polygonal Number
Index entries for linear recurrences with constant coefficients, signature (-2,0,2,1).

Cf. A262221 (centered 25-gonal numbers).

Cf. A032527, A032528, A077043, A077221, A195041, A195042, A195045, A195046, A195047, A195048, A195049, A195058, A195142, A195043, A195143, A195145, A195146, A195147, A195148, A195149, A195158.

[((-1)^n*(50*n^2 + 100*n + 29) - 21)/8 : n in [0..40]]; // Wesley Ivan Hurt, Mar 21 2016

A270693:=n->((-1)^n*(50*n^2 + 100*n + 29) - 21)/8: seq(A270693(n), n=0..100); # Wesley Ivan Hurt, Sep 18 2017

LinearRecurrence[{-2, 0, 2, 1}, {1, -25, 51, -100}, 41]
Table[((-1)^n (50 n^2 + 100 n + 29) - 21)/8, {n, 0, 40}]

x='x+O('x^100); Vec((1-23*x+x^2)/((1-x)*(1+x)^3)) \\ Altug Alkan, Mar 21 2016

G.f.: (1 - 23*x + x^2)/((1 - x)*(1 + x)^3).
E.g.f.: (1/8)*(-21*exp(x) + (29 - 150*x + 50*x^2)*exp(-x)).
a(n) = -2*a(n-1) + 2*a(n-3) + a(n-4).
a(n) = ((-1)^n*(50*n^2 + 100*n + 29) - 21)/8.

cp's OEIS Frontend

A032528 Concentric hexagonal numbers: floor(3*n^2/2).

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A077221 a(0) = 0 and then alternately even and odd numbers in increasing order such that the sum of any two successive terms is a square.

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A047336 Numbers that are congruent to {1, 6} mod 7.

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A032527 Concentric pentagonal numbers: floor( 5*n^2 / 4 ).

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A195040 Square array read by antidiagonals with T(n,k) = k*n^2/4+(k-4)*((-1)^n-1)/8, n>=0, k>=0.

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A033582 a(n) = 7*n^2.

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A195040 Square array read by antidiagonals with T(n,k) = kn^2/4+(k-4)((-1)^n-1)/8, n>=0, k>=0.