A254963 -id:A254963 - cp's OEIS frontend

A000566 Heptagonal numbers (or 7-gonal numbers): n(5n-3)/2.

0, 1, 7, 18, 34, 55, 81, 112, 148, 189, 235, 286, 342, 403, 469, 540, 616, 697, 783, 874, 970, 1071, 1177, 1288, 1404, 1525, 1651, 1782, 1918, 2059, 2205, 2356, 2512, 2673, 2839, 3010, 3186, 3367, 3553, 3744, 3940, 4141, 4347, 4558, 4774, 4995, 5221, 5452, 5688

Offset: 0

N. J. A. Sloane

Binomial transform of (0, 1, 5, 0, 0, 0, ...). Binomial transform is A084899. - Paul Barry, Jun 10 2003
Row sums of triangle A131413. - Gary W. Adamson, Jul 08 2007
Sequence starting (1, 7, 18, 34, ...) = binomial transform of (1, 6, 5, 0, 0, 0, ...). Also row sums of triangle A131896. - Gary W. Adamson, Jul 24 2007
Also the partial sums of A016861, a zero added in front; therefore a(n) = n (mod 5). - R. J. Mathar, Mar 19 2008
Also sequence found by reading the line from 0, in the direction 0, 7, ..., and the line from 1, in the direction 1, 18, ..., in the square spiral whose edges have length A195013 and whose vertices are the numbers A195014. These parallel lines are the semi-axes perpendicular to the main axis A195015 in the same spiral. - Omar E. Pol, Oct 14 2011
Also sequence found by reading the line from 0, in the direction 0, 7, ... and the parallel line from 1, in the direction 1, 18, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Jul 18 2012
Partial sums give A002413. - Omar E. Pol, Jan 12 2013
The n-th heptagonal number equals the sum of the n consecutive integers starting at 2*n-1; for example, 1, 3+4, 5+6+7, 7+8+9+10, etc. In general, the n-th (2k+1)-gonal number is the sum of the n consecutive integers starting at (k-1)*n - (k-2). When k = 1 and 2, this result generates the triangular numbers, A000217, and the pentagonal numbers, A000326, respectively. - Charlie Marion, Mar 02 2022

			G.f. = x + 7*x^2 + 18*x^3 + 34*x^4 + 55*x^5 + 81*x^6 + 112*x^7 + ... - _Michael Somos_, Jan 25 2019

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 38.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
Leonard E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 123.

Daniel Mondot, Table of n, a(n) for n = 0..10000 (first 1000 terms by T. D. Noe)
S. Barbero, U. Cerruti, and N. Murru, Transforming Recurrent Sequences by Using the Binomial and Invert Operators, J. Int. Seq. 13 (2010) # 10.7.7., section 4.4.
C. K. Cook and M. R. Bacon, Some polygonal number summation formulas, Fib. Q., 52 (2014), 336-343.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 341.
Bir Kafle, Florian Luca, and Alain Togbé, Pentagonal and heptagonal repdigits, Annales Mathematicae et Informaticae, pp. 137-145.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Omar E. Pol, Illustration of initial terms of A000217, A000290, A000326, A000384, A000566, A000567.
B. Srinivasa Rao, Heptagonal Numbers in the Pell Sequence and Diophantine Equations 2x^2 = y^2(5y - 3)^2 ± 2, Fib. Quarterly, 43 (2005), 194-201.
B. Srinivasa Rao, Heptagonal numbers in the associated Pell sequence and Diophantine equations x^2(5x - 3)^2 = 8y^2 ± 4, Fib. Quarterly, 43 (2005), 302-306.
Leo Tavares, Illustration.
Eric Weisstein's World of Mathematics, Heptagonal Number.
Index to sequences related to polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A014637, A014640, A014773, A014792, A069099, A131413, A131896, A134483, A000384.
a(n)= A093562(n+1, 2), (5, 1)-Pascal column.
Cf. A006564, A147875, A244639.
Cf. sequences listed in A254963.

a000566 n = n * (5 * (n - 1) + 2) `div` 2
a000566_list = scanl (+) 0 a016861_list  -- Reinhard Zumkeller, Jun 16 2013

a000566:=func< n | n*(5*n-3) div 2 >; [ a000566(n): n in [0..50] ];

A000566 := proc(n)
        n*(5*n-3)/2 ;
end proc:
seq(A000566(n),n=0..30); # R. J. Mathar, Oct 02 2020

Table[n (5n - 3)/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 1, 7}, 50] (* Harvey P. Dale, Oct 13 2011 *)
Join[{0},Accumulate[Range[1,315,5]]] (* Harvey P. Dale, Mar 26 2016 *)
(* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[7], n], {n, 0, 48}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
PolygonalNumber[7,Range[0,50]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 23 2021 *)

makelist(n*(5*n-3)/2, n, 0, 20); /* Martin Ettl, Dec 11 2012 */

PARI
```
a(n) = n * (5*n - 3) / 2
```

# Intended to compute the initial segment of the sequence, not isolated terms.
def aList():
     x, y = 1, 1
     yield 0
     while True:
         yield x
         x, y = x + y + 5, y + 5
A000566 = aList()
print([next(A000566) for i in range(49)]) # Peter Luschny, Aug 04 2019

[n*(5*n-3)//2 for n in range(50)] # Gennady Eremin, Mar 24 2022

G.f.: x*(1 + 4*x)/(1 - x)^3. Simon Plouffe in his 1992 dissertation.
a(n) = C(n, 1) + 5*C(n, 2). - Paul Barry, Jun 10 2003
a(n) = Sum_{k = 1..n} (4*n - 3*k). - Paul Barry, Sep 06 2005
a(n) = n + 5*A000217(n-1) - Floor van Lamoen, Oct 14 2005
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for a(0) = 0, a(1) = 1, a(2) = 7. -  Jaume Oliver Lafont, Dec 02 2008
a(n+1) = A153126(n) + n mod 2; a(2*n+1) = A033571(n); a(2*(n+1)) = A153127(n) + 1. - Reinhard Zumkeller, Dec 20 2008
40*a(n)+ 9 = A017354(n-1). - Ken Rosenbaum, Dec 02 2009.
a(n) = 2*a(n-1) - a(n-2) + 5, with a(0) = 0 and a(1) = 1. - Mohamed Bouhamida, May 05 2010
a(n) = A000217(n) + 4*A000217(n-1). - Vincenzo Librandi, Nov 20 2010
a(n) = a(n-1) + 5*n - 4, with a(0) = 0. - Vincenzo Librandi, Nov 20 2010
a(n) = A130520(5*n). - Philippe Deléham, Mar 26 2013
a(5*a(n) + 11*n + 1) = a(5*a(n) + 11*n) + a(5*n + 1). - Vladimir Shevelev, Jan 24 2014
Sum_{n>=1} 1/a(n) = sqrt(1 - 2/sqrt(5))*Pi/3 + 5*log(5)/6 - sqrt(5)*log((1 + sqrt(5))/2)/3 = 1.32277925312238885674944226131... . See A244639. - Vaclav Kotesovec, Apr 27 2016
E.g.f.: x*(2 + 5*x)*exp(x)/2. - Ilya Gutkovskiy, Aug 27 2016
From Charlie Marion, May 02 2017: (Start)
a(n+m) = a(n) + 5*n*m + a(m);
a(n-m) = a(n) - 5*n*m + a(m) + 3*m;
a(n) - a(m) = (5*(n + m) - 3)*(n - m)/2.
In general, let P(k,n) be the n-th k-gonal number. Then
P(k,n+m) = P(k,n) + (k - 2)*n*m + P(k,m);
P(k,n-m) = P(k,n) - (k - 2)*n*m + P(k,m) + (k - 4)*m;
P(k,n) - P(k,m) = ((k-2)*(n + m) + 4 - k)*(n - m)/2.
(End)
a(n) = A147875(-n) for all n in Z. - Michael Somos, Jan 25 2019
a(n) = A000217(n-1) + A000217(2*n-1). - Charlie Marion, Dec 19 2019
Product_{n>=2} (1 - 1/a(n)) = 5/7. - Amiram Eldar, Jan 21 2021
a(n) + a(n+1) = (2*n+1)^2 + n^2 - 2*n. In general, if we let P(k,n) = the n-th k-gonal number, then P(k^2-k+1,n)+ P(k^2-k+1,n+1) = ((k-1)*n+1)^2 + (k-2)*(n^2-2*n) = ((k-1)*n+1)^2 + (k-2)*A005563(n-2). When k = 2, this formula reduces to the well-known triangular number formula: T(n) + T(n+1) = (n+1)^2. - Charlie Marion, Jul 01 2021

Partially edited by Joerg Arndt, Mar 11 2010

A016742 Even squares: a(n) = (2*n)^2.

Original entry on oeis.org

0, 4, 16, 36, 64, 100, 144, 196, 256, 324, 400, 484, 576, 676, 784, 900, 1024, 1156, 1296, 1444, 1600, 1764, 1936, 2116, 2304, 2500, 2704, 2916, 3136, 3364, 3600, 3844, 4096, 4356, 4624, 4900, 5184, 5476, 5776, 6084, 6400, 6724, 7056, 7396, 7744, 8100, 8464

Offset: 0

N. J. A. Sloane

4 times the squares.
Number of edges in the complete bipartite graph of order 5n, K_{n,4n} - Roberto E. Martinez II, Jan 07 2002
It is conjectured (I think) that a regular Hadamard matrix of order n exists iff n is an even square (cf. Seberry and Yamada, Th. 10.11). A Hadamard matrix is regular if the sum of the entries in each row is the same. - N. J. A. Sloane, Nov 13 2008
Sequence arises from reading the line from 0, in the direction 0, 16, ... and the line from 4, in the direction 4, 36, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
The entries from a(1) on can be interpreted as pair sums of (2, 2), (8, 8), (18, 18), (32, 32) etc. that arise from a re-arrangement of the subshell orbitals in the periodic table of elements. 8 becomes the maximum number of electrons in the (2s,2p) or (3s,3p) orbitals, 18 the maximum number of electrons in (4s,3d,4p) or (5s,3d,5p) shells, for example. - Julio Antonio Gutiérrez Samanez, Jul 20 2008
The first two terms of the sequence (n=1, 2) give the numbers of chemical elements using only n types of atomic orbitals, i.e., there are a(1)=4 elements (H,He,Li,Be) where electrons reside only on s-orbitals, there are a(2)=16 elements (B,C,N,O,F,Ne,Na,Mg,Al,Si,P,S,Cl,Ar,K,Ca) where electrons reside only on s- and p-orbitals. However, after that, there is 37 (which is one more than a(3)=36) elements (from Sc, Scandium, atomic number 21 to La, Lanthanum, atomic number 57) where electrons reside only on s-, p- and d-orbitals. This is because Lanthanum (with the electron configuration [Xe]5d^1 6s^2) is an exception to the Aufbau principle, which would predict that its electron configuration is [Xe]4f^1 6s^2. - Antti Karttunen, Aug 14 2008.
Number of cycles of length 3 in the king's graph associated with an (n+1) X (n+1) chessboard. - Anton Voropaev (anton.n.voropaev(AT)gmail.com), Feb 01 2009
a(n+1) is the molecular topological index of the n-star graph S_n. - Eric W. Weisstein, Jul 11 2011
a(n) is the sum of two consecutives odd numbers 2*n^2-1 and 2*n^2+1 and the difference of two squares (n^2+1)^2 - (n^2-1)^2. - Pierre CAMI, Jan 02 2012
For n > 3, a(n) is the area of the irregular quadrilateral created by the points ((n-4)*(n-3)/2,(n-3)*(n-2)/2), ((n-2)*(n-1)/2,(n-1)*n/2), ((n+1)*(n+2)/2,n*(n+1)/2), and ((n+3)*(n+4)/2,(n+2)*(n+3)/2). - J. M. Bergot, May 27 2014
Number of terms less than 10^k: 1, 2, 5, 16, 50, 159, 500, 1582, 5000, 15812, 50000, 158114, 500000, ... - Muniru A Asiru, Jan 28 2018
Right-hand side of the binomial coefficient identity Sum_{k = 0..2*n} (-1)^(k+1)* binomial(2*n,k)*binomial(2*n + k,k)*(2*n - k) = a(n). - Peter Bala, Jan 12 2022

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.
Seberry, Jennifer and Yamada, Mieko; Hadamard matrices, sequences and block designs, in Dinitz and Stinson, eds., Contemporary design theory, pp. 431-560, Wiley-Intersci. Ser. Discrete Math. Optim., Wiley, New York, 1992.
W. D. Wallis, Anne Penfold Street and Jennifer Seberry Wallis, Combinatorics: Room squares, sum-free sets, Hadamard matrices, Lecture Notes in Mathematics, Vol. 292, Springer-Verlag, Berlin-New York, 1972. iv+508 pp.

Vincenzo Librandi, Table of n, a(n) for n = 0..900
R. P. Boas and N. J. A. Sloane, Correspondence, 1974.
Leo Tavares, Illustration: X Squares
Various, Electron Configuration (Discussion in Physics Forums).
Eric Weisstein's World of Mathematics, Graph Cycle.
Eric Weisstein's World of Mathematics, King Graph.
Eric Weisstein's World of Mathematics, Molecular Topological Index.
Wikipedia, Aufbau principle.
Index entries for sequences related to Hadamard matrices
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001105, A001539, A016754, A016802, A016814, A016826, A016838, A007742, A033991, A245058.
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. sequences listed in A254963.
Other n X n king graph cycle counts: A288918 (4-cycles), A288919 (5-cycles), A288920 (6-cycles).
Cf. A000384, A014105.
Cf. A016813.

List([0..100], n -> (2*n)^2); # Muniru A Asiru, Jan 28 2018

a016742 = (* 4) . (^ 2)
a016742_list = 0 : map (subtract 4) (zipWith (+) a016742_list [8, 16 ..])
-- Reinhard Zumkeller, Jun 28 2015, Apr 20 2015

[(2*n)^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

seq((2*n)^2, n=0..100); # Muniru A Asiru, Jan 28 2018

Table[(2n)^2, {n, 0, 46}] (* Alonso del Arte, Apr 26 2011 *)

makelist((2*n)^2,n,0,20); /* Martin Ettl, Jan 22 2013 */

a(n)=4*n^2 \\ Charles R Greathouse IV, Jul 28 2015

O.g.f.: 4*x*(1+x)/(1-x)^3. - R. J. Mathar, Jul 28 2008
a(n) = A000290(n)*4 = A001105(n)*2. - Omar E. Pol, May 21 2008
a(n) = A155955(n,2) for n > 1. - Reinhard Zumkeller, Jan 31 2009
Sum_{n>=1} 1/a(n) = (1/4)*Pi^2/6 = Pi^2/24. - Ant King, Nov 04 2009
a(n) = a(n-1) + 8*n - 4 (with a(0)=0). - Vincenzo Librandi, Nov 19 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 4, a(2) = 16. - Philippe Deléham, Mar 26 2013
a(n) = A118729(8n+3). - Philippe Deléham, Mar 26 2013
Pi = 2*Product_{n>=1} (1 + 1/(a(n)-1)). - Adriano Caroli, Aug 04 2013
Pi = Sum_{n>=0} 8/(a(2n+1)-1). - Adriano Caroli, Aug 06 2013
E.g.f.: exp(x)*(4x^2 + 4x). - Geoffrey Critzer, Oct 07 2013
a(n) = A000384(n) + A014105(n). - Bruce J. Nicholson, Nov 11 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/48 (A245058). - Amiram Eldar, Oct 10 2020
From Amiram Eldar, Jan 25 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/2)/(Pi/2) (A308716).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/2)/(Pi/2) = 2/Pi (A060294). (End)
a(n) = A016754(n) - A016813(n). - Leo Tavares, Feb 24 2022

More terms from Sabir Abdus-Samee (sabdulsamee(AT)prepaidlegal.com), Mar 13 2006

A049450 Pentagonal numbers multiplied by 2: a(n) = n(3n-1).

Original entry on oeis.org

0, 2, 10, 24, 44, 70, 102, 140, 184, 234, 290, 352, 420, 494, 574, 660, 752, 850, 954, 1064, 1180, 1302, 1430, 1564, 1704, 1850, 2002, 2160, 2324, 2494, 2670, 2852, 3040, 3234, 3434, 3640, 3852, 4070, 4294, 4524, 4760, 5002, 5250, 5504, 5764

Offset: 0

Joe Keane (jgk(AT)jgk.org)

From Floor van Lamoen, Jul 21 2001: (Start)
Write 1,2,3,4,... in a hexagonal spiral around 0, then a(n) is the sequence found by reading the line from 0 in the direction 0,2,.... The spiral begins:
.
                   56--55--54--53--52
                   /                 \
                 57  33--32--31--30  51
                 /   /             \   \
               58  34  16--15--14  29  50
               /   /   /         \   \   \
             59  35  17   5---4  13  28  49
             /   /   /   /     \   \   \   \
           60  36  18   6   0   3  12  27  48
           /   /   /   /   / . /   /   /   /
         61  37  19   7   1---2  11  26  47
           \   \   \   \       . /   /   /
           62  38  20   8---9--10  25  46
             \   \   \           . /   /
             63  39  21--22--23--24  45
               \   \               . /
               64  40--41--42--43--44
                 \                   .
                 65--66--67--68--69--70
(End)
Starting with offset 1 = binomial transform of [2, 8, 6, 0, 0, 0, ...]. - Gary W. Adamson, Jan 09 2009
Number of possible pawn moves on an (n+1) X (n+1) chessboard (n=>3). - Johannes W. Meijer, Feb 04 2010
a(n) = A069905(6n-1): Number of partitions of 6*n-1 into 3 parts. - Adi Dani, Jun 04 2011
Even octagonal numbers divided by 4. - Omar E. Pol, Aug 19 2011
Partial sums give A011379. - Omar E. Pol, Jan 12 2013
First differences are A016933; second differences equal 6. - Bob Selcoe, Apr 02 2015
For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n-2; {2, 2n-1, 6, 2n-1, 2, 18n-4}]. - Magus K. Chu, Oct 13 2022

			On a 4 X 4 chessboard pawns at the second row have (3+4+4+3) moves and pawns at the third row have (2+3+3+2) moves so a(3) = 24. - _Johannes W. Meijer_, Feb 04 2010
From _Adi Dani_, Jun 04 2011: (Start)
a(1)=2: the partitions of 6*1-1=5 into 3 parts are [1,1,3] and[1,2,2].
a(2)=10: the partitions of 6*2-1=11 into 3 parts are [1,1,9], [1,2,8], [1,3,7], [1,4,6], [1,5,5], [2,2,7], [2,3,6], [2,4,5], [3,3,5], and [3,4,4].
(End)
.
.                                                         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 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
.    2      10         24             44                 70
- _Philippe Deléham_, Mar 30 2013

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16)
Leo Tavares, Illustration: X Hexagons
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567.
Bisection of A001859. Cf. A045944, A000326, A033579, A027599, A049451.
Cf. A033586 (King), A035005 (Queen), A035006 (Rook), A035008 (Knight) and A002492 (Bishop).
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488. [Bruno Berselli, Jun 10 2013]
Cf. sequences listed in A254963.
Cf. A003215, A016813, A000290, A000217.

List([0..50], n-> n*(3*n-1)); # G. C. Greubel, Aug 31 2019

[n*(3*n-1) : n in [0..50]]; // Wesley Ivan Hurt, Sep 24 2017

seq(n*(3*n-1),n=0..44); # Zerinvary Lajos, Jun 12 2007

Table[n(3n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,2,10},50] (* Harvey P. Dale, Jun 21 2014 *)
2*PolygonalNumber[5,Range[0,50]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 01 2018 *)

a(n)=n*(3*n-1) \\ Charles R Greathouse IV, Nov 20 2012

[n*(3*n-1) for n in (0..50)] # G. C. Greubel, Aug 31 2019

O.g.f.: A(x) = 2*x*(1+2*x)/(1-x)^3.
a(n) = A049452(n) - A033428(n). - Zerinvary Lajos, Jun 12 2007
a(n) = 2*A000326(n), twice pentagonal numbers. - Omar E. Pol, May 14 2008
a(n) = A022264(n) - A000217(n). - Reinhard Zumkeller, Oct 09 2008
a(n) = a(n-1) + 6*n - 4 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n) = A014642(n)/4 = A033579(n)/2. - Omar E. Pol, Aug 19 2011
a(n) = A000290(n) + A000384(n) = A000217(n) + A000566(n). - Omar E. Pol, Jan 11 2013
a(n+1) = A014107(n+2) + A000290(n). - Philippe Deléham, Mar 30 2013
E.g.f.: x*(2 + 3*x)*exp(x). - Vincenzo Librandi, Apr 28 2016
a(n) = (2/3)*A000217(3*n-1). - Bruno Berselli, Feb 13 2017
a(n) = A002061(n) + A056220(n). - Bruce J. Nicholson, Sep 21 2017
From Amiram Eldar, Feb 20 2022: (Start)
Sum_{n>=1} 1/a(n) = 3*log(3)/2 - Pi/(2*sqrt(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/sqrt(3) - 2*log(2). (End)
From Leo Tavares, Feb 23 2022: (Start)
a(n) = A003215(n) - A016813(n).
a(n) = 2*A000290(n) + 2*A000217(n-1). (End)

A147875 Second heptagonal numbers: a(n) = n(5n+3)/2.

Original entry on oeis.org

0, 4, 13, 27, 46, 70, 99, 133, 172, 216, 265, 319, 378, 442, 511, 585, 664, 748, 837, 931, 1030, 1134, 1243, 1357, 1476, 1600, 1729, 1863, 2002, 2146, 2295, 2449, 2608, 2772, 2941, 3115, 3294, 3478, 3667, 3861, 4060, 4264, 4473, 4687, 4906, 5130, 5359, 5593

Offset: 0

Vladimir Joseph Stephan Orlovsky, Nov 16 2008

Zero followed by partial sums of A016897.
Apparently = every 2nd term of A111710 and A085787.
Bisection of A085787. Sequence found by reading the line from 0, in the direction 0, 13, ... and the line from 4, in the direction 4, 27, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Jul 18 2012
Numbers of the form m^2 + k*m*(m+1)/2: in this case is k=3. See also A254963. - Bruno Berselli, Feb 11 2015

			G.f. = 4*x + 13*x^2 + 27*x^3 + 46*x^4 + 70*x^5 + 99*x^6 + 133*x^7 + ... - _Michael Somos_, Jan 25 2019

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Sliced Hexagons
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A016897, A111710, A000217, A085787, A224419 (positions of squares).
Second n-gonal numbers: A005449, A014105, A045944, A179986, A033954, A062728, A135705.
Cf. A000566.
Cf. A003215, A000096, A000290.

List([0..50], n-> n*(5*n+3)/2); # G. C. Greubel, Jul 04 2019

[n*(5*n+3)/2: n in [0..50]]; // G. C. Greubel, Jul 04 2019

Table[(n(5n+3))/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 4, 13}, 50] (* Harvey P. Dale, May 15 2013 *)

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

[n*(5*n+3)/2 for n in (0..50)] # G. C. Greubel, Jul 04 2019

G.f.: x*(4+x)/(1-x)^3.
a(n) = Sum_{k=0..n-1} A016897(k).
a(n) - a(n-1) = 5*n -1. - Vincenzo Librandi, Nov 26 2010
G.f.: U(0) where U(k) = 1 + 2*(2*k+3)/(k + 2 - x*(k+2)^2*(k+3)/(x*(k+2)*(k+3) + (2*k+2)*(2*k+3)/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 14 2012
E.g.f.: U(0) where U(k) = 1 + 2*(2*k+3)/(k + 2 - 2*x*(k+2)^2*(k+3)/(2*x*(k+2)*(k+3) + (2*k+2)^2*(2*k+3)/U(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 14 2012
a(n) = A130520(5n+3). - Philippe Deléham, Mar 26 2013
a(n) = A131242(10n+7)/2. - Philippe Deléham, Mar 27 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=4, a(2)=13. - Harvey P. Dale, May 15 2013
Sum_{n>=1} 1/a(n) = 10/9 + sqrt(1 - 2/sqrt(5))*Pi/3 - 5*log(5)/6 + sqrt(5)*log((1 + sqrt(5))/2)/3 = 0.4688420784500060750083432... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(n) + A000217(2*n). - Bruno Berselli, Jul 01 2016
From Ilya Gutkovskiy, Jul 01 2016: (Start)
E.g.f.: x*(8 + 5*x)*exp(x)/2.
Dirichlet g.f.: (5*zeta(s-2) + 3*zeta(s-1))/2. (End)
a(n) = A000566(-n) for all n in Z. - Michael Somos, Jan 25 2019
From Leo Tavares, Feb 14 2022: (Start)
a(n) = A003215(n) - A000217(n+1). See Sliced Hexagons illustration in links.
a(n) = A000096(n) + 2*A000290(n). (End)

Edited by Klaus Brockhaus and R. J. Mathar, Nov 20 2008

New name from Bruno Berselli, Jan 13 2011

A049453 Second pentagonal numbers with even index: a(n) = n(6n+1).

Original entry on oeis.org

0, 7, 26, 57, 100, 155, 222, 301, 392, 495, 610, 737, 876, 1027, 1190, 1365, 1552, 1751, 1962, 2185, 2420, 2667, 2926, 3197, 3480, 3775, 4082, 4401, 4732, 5075, 5430, 5797, 6176, 6567, 6970, 7385, 7812, 8251, 8702, 9165, 9640, 10127, 10626, 11137, 11660, 12195

Offset: 0

Joe Keane (jgk(AT)jgk.org)

Number of edges in the join of the complete tripartite graph of order 3n and the cycle graph of order n, K_n,n,n * C_n. - Roberto E. Martinez II, Jan 07 2002
Sequence found by reading the line (one of the diagonal axes) from 0, in the direction 0, 7, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Sep 08 2011
First bisection of A036498. - Bruno Berselli, Nov 25 2012

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001318, A005449, A033568, A033570, A036498, A049452, A185019, A194454.

Cf. A254963 (comment).

seq(binomial(6*n+1,2)/3, n=0..42); # Zerinvary Lajos, Jan 21 2007

s=0;lst={s};Do[s+=n++ +7;AppendTo[lst, s], {n, 0, 7!, 12}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 16 2008 *)
Table[n*(6*n + 1), {n,0,50}] (* G. C. Greubel, Jun 07 2017 *)

x='x+O('x^50); concat([0], Vec(x*(7+5*x)/(1-x)^3)) \\ G. C. Greubel, Jun 07 2017

G.f.: x*(7+5*x)/(1-x)^3.
a(n) = 12*n + a(n-1) - 5 with n > 0, a(0)=0. - Vincenzo Librandi, Aug 06 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - G. C. Greubel, Jun 07 2017
From Amiram Eldar, Feb 18 2022: (Start)
Sum_{n>=1} 1/a(n) = 6 - sqrt(3)*Pi/2 - 2*log(2) - 3*log(3)/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi + log(2) + sqrt(3)*log(2 + sqrt(3)) - 6. (End)
E.g.f.: exp(x)*x*(7 + 6*x). - Elmo R. Oliveira, Dec 12 2024

A022264 a(n) = n(7n - 1)/2.

Original entry on oeis.org

0, 3, 13, 30, 54, 85, 123, 168, 220, 279, 345, 418, 498, 585, 679, 780, 888, 1003, 1125, 1254, 1390, 1533, 1683, 1840, 2004, 2175, 2353, 2538, 2730, 2929, 3135, 3348, 3568, 3795, 4029, 4270, 4518, 4773, 5035, 5304, 5580, 5863, 6153, 6450, 6754, 7065, 7383

Offset: 0

N. J. A. Sloane

Sequence found by reading the line from 0, in the direction 0, 13, ..., and the parallel line from 3, in the direction 3, 30, ..., in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. - Omar E. Pol, Sep 09 2011

G. C. Greubel, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Crysta-gons
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A022265.
Cf. sequences listed in A254963.
Cf. similar sequences listed in A022288.
Cf. A016742, A049450, A174738, A195019, A195020.
Cf. A005449, A000384.

[n*(7*n-1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Dec 04 2016

[seq(binomial(7*n,2)/7, n=0..37)]; # Zerinvary Lajos, Jan 02 2007

Table[n (7*n - 1)/2, {n, 0, 40}] (* Zerinvary Lajos, Jul 10 2009 *)

a(n)=n*(7*n-1)/2 \\ Charles R Greathouse IV, Mar 08 2013

a(n) = C(7*n,2)/7, n >= 0. - Zerinvary Lajos, Jan 02 2007
a(n) = A049450(n) + A000217(n). - Reinhard Zumkeller, Oct 09 2008
a(n) = 7*n + a(n-1) - 4 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 04 2010
a(n) = (2*n)^2 - n*(n+1)/2 = A016742(n) - A000217(n). - Philippe Deléham, Mar 08 2013
a(n) = A174738(7*n+2). - Philippe Deléham, Mar 26 2013
G.f.: x*(3 + 4*x)/(1 - x)^3. - R. J. Mathar, Aug 04 2016
a(n) = A000217(4*n-1) - A000217(3*n-1). - Bruno Berselli, Oct 17 2016
a(n) = (1/5) * Sum_{i=n..(6*n-1)} i. - Wesley Ivan Hurt, Dec 04 2016
E.g.f.: (1/2)*x*(7*x + 6)*exp(x). - G. C. Greubel, Aug 19 2017
a(n) = A005449(n) + A000384(n). See Crysta-gons illustration. - Leo Tavares, Nov 21 2021

A022267 a(n) = n(9n + 1)/2.

Original entry on oeis.org

0, 5, 19, 42, 74, 115, 165, 224, 292, 369, 455, 550, 654, 767, 889, 1020, 1160, 1309, 1467, 1634, 1810, 1995, 2189, 2392, 2604, 2825, 3055, 3294, 3542, 3799, 4065, 4340, 4624, 4917, 5219, 5530, 5850, 6179

Offset: 0

N. J. A. Sloane

From Floor van Lamoen, Jul 21 2001: (Start)
Write 0, 1, 2, 3, 4, ... in a triangular spiral; then a(n) is the sequence found by reading the line from 0 in the direction 0, 5, ... . The spiral begins:
.
            15
            / \
          16  14
          /     \
        17   3  13
        /   / \   \
      18   4   2  12
      /   /     \   \
    19   5   0---1  11
    /   /             \
  20   6---7---8---9--10
.
(End)
a(n) is the sum of n consecutive integers starting from 4*n+1: (5), (9+10), (13+14+15), ... - Klaus Purath, Jul 07 2020
a(n) with n>0 are the numbers with the periodic length 3 in the Bulgarian and Mancala solitaire. - Paul Weisenhorn, Jan 29 2022

Lei Zhou, Table of n, a(n) for n = 0..10000
Milan Janjic, Two Enumerative Functions
Leo Tavares, Illustration: X Triangles
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A051682, A110449, A235332.
Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A033429, A022268, A049452, A186030, A135703, A152734, A139273.
Cf. similar sequences listed in A254963.
Cf. similar sequences listed in A022289.
Cf. A060544, A016813.

seq(binomial(9*n+1,2)/9, n=0..37); # Zerinvary Lajos, Jan 21 2007

Table[ n (9 n + 1)/2, {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 19}, 40] (* Harvey P. Dale, Jul 01 2013 *)

vector(100,n,(n-1)*(9*n-8)/2) \\ Derek Orr, Feb 06 2015

a(n) = A110449(n, 4) for n>3.
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 4*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A000566(n). (End)
a(n) = 9*n + a(n-1) - 4 with n>0, a(0)=0. - Vincenzo Librandi, Aug 04 2010
a(n) = A218470(9*n+4). - Philippe Deléham, Mar 27 2013
a(n) = A000217(5*n) - A000217(4*n). - Bruno Berselli, Oct 13 2016
E.g.f.: (1/2)*(9*x^2 + 10*x)*exp(x). - G. C. Greubel, Jul 17 2017
a(n) = A060544(n+1) - A016813(n). - Leo Tavares, Mar 20 2022

A226492 a(n) = n(11n-5)/2.

Original entry on oeis.org

0, 3, 17, 42, 78, 125, 183, 252, 332, 423, 525, 638, 762, 897, 1043, 1200, 1368, 1547, 1737, 1938, 2150, 2373, 2607, 2852, 3108, 3375, 3653, 3942, 4242, 4553, 4875, 5208, 5552, 5907, 6273, 6650, 7038, 7437, 7847, 8268, 8700, 9143, 9597, 10062, 10538, 11025, 11523

Offset: 0

Bruno Berselli, Jun 11 2013

Sequences of numbers of the form n*(n*k - k + 6)/2:
. k from 0 to 10, respectively: A008585, A055998, A005563, A045943, A014105, A005475, A033428, A022264, A033991, A062741, A147874;
. k=11: a(n);
. k=12: A094159;
. k=13: 0, 3, 19, 48, 90, 145, 213, 294, 388, 495, 615, 748, 894, ...;
. k=14: 0, 3, 20, 51, 96, 155, 228, 315, 416, 531, 660, 803, 960, ...;
. k=15: A152773;
. k=16: A139272;
. k=17: 0, 3, 23, 60, 114, 185, 273, 378, 500, 639, 795, 968, ...;
. k=18: A152751;
. k=19: 0, 3, 25, 66, 126, 205, 303, 420, 556, 711, 885, 1078, ...;
. k=20: 0, 3, 26, 69, 132, 215, 318, 441, 584, 747, 930, 1133, ...;
. k=21: A152759;
. k=22: 0, 3, 28, 75, 144, 235, 348, 483, 640, 819, 1020, 1243, ...;
. k=23: 0, 3, 29, 78, 150, 245, 363, 504, 668, 855, 1065, 1298, ...;
. k=24: A152767;
. k=25: 0, 3, 31, 84, 162, 265, 393, 546, 724, 927, 1155, 1408, ...;
. k=26: 0, 3, 32, 87, 168, 275, 408, 567, 752, 963, 1200, 1463, ...;
. k=27: A153783;
. k=28: A195021;
. k=29: 0, 3, 35, 96, 186, 305, 453, 630, 836, 1071, 1335, 1628, ...;
. k=30: A153448;
. k=31: 0, 3, 37, 102, 198, 325, 483, 672, 892, 1143, 1425, 1738, ...;
. k=32: 0, 3, 38, 105, 204, 335, 498, 693, 920, 1179, 1470, 1793, ...;
. k=33: A153875.
Also:
a(n) - n         = A180223(n);
a(n) + n         = n*(11*n-3)/2 = 0, 4, 19, 45, 82, 130, 189, 259, ...;
a(n) - 2*n       = A051865(n);
a(n) + 2*n       = A022268(n);
a(n) - 3*n       = A152740(n-1);
a(n) + 3*n       = A022269(n);
a(n) - 4*n       = n*(11*n-13)/2 = 0, -1, 9, 30, 62, 105, 159, 224, ...;
a(n) + 4*n       = A254963(n);
a(n) - n*(n-1)/2 = A147874(n+1);
a(n) + n*(n-1)/2 = A094159(n) (case k=12);
a(n) - n*(n-1)   = A062741(n) (see above, this is the case k=9);
a(n) + n*(n-1)   = n*(13*n-7)/2 (case k=13);
a(n) - n*(n+1)/2 = A135706(n);
a(n) + n*(n+1)/2 = A033579(n);
a(n) - n*(n+1)   = A051682(n);
a(n) + n*(n+1)   = A186030(n);
a(n) - n^2       = A062708(n);
a(n) + n^2       = n*(13*n-5)/2 = 0, 4, 21, 51, 94, 150, 219, ..., etc.
Sum of reciprocals of a(n), for n > 0: 0.47118857003113149692081665034891...

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. sequences in Comments lines.
First differences are in A017425.
Cf. A033584, A180223.

Magma
```
[n*(11*n-5)/2: n in [0..50]];
```

I:=[0,3,17]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..46]]; // Vincenzo Librandi, Aug 18 2013

Table[n (11 n - 5)/2, {n, 0, 50}]
CoefficientList[Series[x (3 + 8 x) / (1 - x)^3, {x, 0, 45}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{3,-3,1},{0,3,17},50] (* Harvey P. Dale, Jan 14 2019 *)

a(n)=n*(11*n-5)/2 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(3+8*x)/(1-x)^3.
a(n) + a(-n) = A033584(n).
From Elmo R. Oliveira, Dec 27 2024: (Start)
E.g.f.: exp(x)*x*(6 + 11*x)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = n + A180223(n). (End)

A033580 Four times second pentagonal numbers: a(n) = 2n(3*n+1).

Original entry on oeis.org

0, 8, 28, 60, 104, 160, 228, 308, 400, 504, 620, 748, 888, 1040, 1204, 1380, 1568, 1768, 1980, 2204, 2440, 2688, 2948, 3220, 3504, 3800, 4108, 4428, 4760, 5104, 5460, 5828, 6208, 6600, 7004, 7420, 7848, 8288, 8740, 9204, 9680, 10168, 10668, 11180, 11704, 12240

Offset: 0

N. J. A. Sloane

Subsequence of A062717: A010052(6*a(n)+1) = 1. - Reinhard Zumkeller, Feb 21 2011
Sequence found by reading the line from 0, in the direction 0, 8,..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A139267 in the same spiral - Omar E. Pol, Sep 09 2011
a(n) is the number of edges of the octagonal network O(n,n); O(m,n) is defined by Fig. 1 of the Siddiqui et al. reference. - Emeric Deutsch May 13 2018
The partial sums of this sequence give A035006. - Leo Tavares, Oct 03 2021

Ivan Panchenko, Table of n, a(n) for n = 0..1000
M. K. Siddiqui, M. Naeem, N. A. Rahman, and M. Imran, Computing topological indices of certain networks, J. of Optoelectronics and Advanced Materials, 18, No. 9-10 (2016), pp. 884-892.
Leo Tavares, Illustration: Crossed Stars
Leo Tavares, Illustration: Four Quarter Star Crosses
Leo Tavares, Illustration: Triangulated Star Crosses
Leo Tavares, Illustration: Oblong Star Crosses
Leo Tavares, Illustration: Crossed Diamond Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A033579, A049451, A045945, A186423.
Cf. sequences listed in A254963.
Cf. A003154, A016813.
Cf. A035006.
Cf. A005449, A046092, A033996, A002378, A016742, A000290, A016754, A001105.

List([0..50], n-> 2*n*(3*n+1)); # G. C. Greubel, Oct 09 2019

[2*n*(3*n+1): n in [0..50]]; // G. C. Greubel, Oct 09 2019

seq(2*n*(3*n+1), n=0..50); # G. C. Greubel, Oct 09 2019

4*Binomial[3*Range[50]-2, 2]/3 (* G. C. Greubel, Oct 09 2019 *)

a(n)=2*n*(3*n+1) \\ Charles R Greathouse IV, Sep 28 2015

[2*n*(3*n+1) for n in (0..50)] # G. C. Greubel, Oct 09 2019

a(n) = a(n-1) +12*n -4 (with a(0)=0). - Vincenzo Librandi, Aug 05 2010
G.f.: 4*x*(2+x)/(1-x)^3. - Colin Barker, Feb 13 2012
a(-n) = A033579(n). - Michael Somos, Jun 09 2014
E.g.f.: 2*x*(4 + 3*x)*exp(x). - G. C. Greubel, Oct 09 2019
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=1} 1/a(n) = 3/2 - Pi/(4*sqrt(3)) - 3*log(3)/4.
Sum_{n>=1} (-1)^(n+1)/a(n) =  -3/2 + Pi/(2*sqrt(3)) + log(2). (End)
From Leo Tavares, Oct 12 2021: (Start)
a(n) = A003154(n+1) - A016813(n). See Crossed Stars illustration.
a(n) = 4*A005449(n). See Four Quarter Star Crosses illustration.
a(n) = 2*A049451(n).
a(n) = A046092(n-1) + A033996(n). See Triangulated Star Crosses illustration.
a(n) = 4*A000217(n-1) + 8*A000217(n).
a(n) = 4*A000217(n-1) + 4*A002378. See Oblong Star Crosses illustration.
a(n) = A016754(n) + 4*A000217(n). See Crossed Diamond Stars illustration.
a(n) = 2*A001105(n) + 4*A000217(n).
a(n) = A016742(n) + A046092(n).
a(n) = 4*A000290(n) + 4*A000217(n). (End)

A069125 a(n) = (11n^2 - 11n + 2)/2.

Original entry on oeis.org

1, 12, 34, 67, 111, 166, 232, 309, 397, 496, 606, 727, 859, 1002, 1156, 1321, 1497, 1684, 1882, 2091, 2311, 2542, 2784, 3037, 3301, 3576, 3862, 4159, 4467, 4786, 5116, 5457, 5809, 6172, 6546, 6931, 7327, 7734, 8152, 8581, 9021, 9472, 9934, 10407, 10891, 11386, 11892

Offset: 1

Terrel Trotter, Jr., Apr 07 2002

Centered hendecagonal (11-gonal) numbers. - Omar E. Pol, Oct 03 2011

Numbers of the form (2*m+1)^2 + k*m*(m+1)/2: in this case is k=3. See also A254963. - Bruno Berselli, Feb 11 2015

			a(5)=111 because 111 = (11*5^2 - 11*5 + 2)/2 = (275 - 55 + 2)/2 = 222/2.

T. D. Noe, Table of n, a(n) for n = 1..1000
X. Acloque, Polynexus Numbers and other mathematical wonders. [broken link]
Leo Tavares, Illustration: Clipped Stars.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001263, A001844, A003215, A005448, A005891, A069099.

Cf. A000217, A003154, A152740, A254963.

FoldList[#1 + #2 &, 1, 11 Range@ 45] (* Robert G. Wilson v *)
Table[(11n^2-11n+2)/2,{n,60}] (* or *) LinearRecurrence[{3,-3,1},{1,12,34},60] (* Harvey P. Dale, Jun 25 2011 *)

a(n)=(11*n^2-11*n+2)/2 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 1 + Sum_{j=0..n-1} (11*j). - Xavier Acloque, Oct 22 2003
Binomial transform of [1, 11, 11, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 11, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
a(n) = 11*n + a(n-1) - 11 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 08 2010
G.f.: -x*(1+9*x+x^2)/(x-1)^3. - R. J. Mathar, Jun 05 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=1, a(1)=12, a(2)=34. - Harvey P. Dale, Jun 25 2011
a(n) = A152740(n-1) + 1. - Omar E. Pol, Oct 03 2011
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi*tan(sqrt(3/11)*Pi/2)/sqrt(33).
Sum_{n>=1} a(n)/n! = 13*e/2 - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 13/(2*e) - 1. (End)
a(n) = A003154(n) - A000217(n-1). - Leo Tavares, Mar 29 2022
E.g.f.: exp(x)*(1 + 11*x^2/2) - 1. - Elmo R. Oliveira, Oct 18 2024

More terms from Harvey P. Dale, Jun 25 2011

Name rewritten by Bruno Berselli, Feb 11 2015

cp's OEIS Frontend

A000566 Heptagonal numbers (or 7-gonal numbers): n*(5*n-3)/2.

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A016742 Even squares: a(n) = (2*n)^2.

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A049450 Pentagonal numbers multiplied by 2: a(n) = n*(3*n-1).

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A147875 Second heptagonal numbers: a(n) = n*(5*n+3)/2.

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A049453 Second pentagonal numbers with even index: a(n) = n*(6*n+1).

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Maple

A000566 Heptagonal numbers (or 7-gonal numbers): n(5n-3)/2.

A049450 Pentagonal numbers multiplied by 2: a(n) = n(3n-1).

A147875 Second heptagonal numbers: a(n) = n(5n+3)/2.

A049453 Second pentagonal numbers with even index: a(n) = n(6n+1).

A022264 a(n) = n(7n - 1)/2.

A022267 a(n) = n(9n + 1)/2.

A226492 a(n) = n(11n-5)/2.

A033580 Four times second pentagonal numbers: a(n) = 2n(3*n+1).

A069125 a(n) = (11n^2 - 11n + 2)/2.