A005449 -id:A005449 - cp's OEIS frontend

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

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

A049451 Twice second pentagonal numbers.

Original entry on oeis.org

0, 4, 14, 30, 52, 80, 114, 154, 200, 252, 310, 374, 444, 520, 602, 690, 784, 884, 990, 1102, 1220, 1344, 1474, 1610, 1752, 1900, 2054, 2214, 2380, 2552, 2730, 2914, 3104, 3300, 3502, 3710, 3924, 4144, 4370, 4602, 4840, 5084, 5334, 5590, 5852, 6120, 6394, 6674, 6960, 7252, 7550, 7854

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 0,4,... . The spiral begins:
                            .
                          52
                          . \
            33--32--31--30  51
            /           . \   \
          34  16--15--14  29  50
          /   /       . \   \   \
        35  17   5---4  13  28  49
        /   /   /   . \   \   \   \
      36  18   6   0   3  12  27  48
      /   /   /   /   /   /   /   /
    37  19   7   1---2  11  26  47
      \   \   \         /   /   /
      38  20   8---9--10  25  46
        \   \             /   /
        39  21--22--23--24  45
          \                 /
          40--41--42--43--44
(End)
Number of edges in the join of the complete bipartite graph of order 2n and the cycle graph of order n, K_n,n * C_n. - Roberto E. Martinez II, Jan 07 2002
The average of the first n elements starting from a(1) is equal to (n+1)^2. - Mario Catalani (mario.catalani(AT)unito.it), Apr 10 2003
If Y is a 4-subset of an n-set X then, for n >= 4, a(n-4) is the number of (n-4)-subsets of X having either one element or two elements in common with Y. - Milan Janjic, Dec 28 2007
With offset 1: the maximum possible sum of numbers in an N x N standard Minesweeper grid. - Dmitry Kamenetsky, Dec 14 2008
a(n) = A001399(6*n-2), number of partitions of 6*n-2 into parts < 4. For example a(2)=14 where the partitions of 6*2-2=10 into parts < 4 are [1,1,1,1,1,1,1,1,1,1], [1,1,1,1,1,1,1,1,2], [1,1,1,1,1,1,1,3], [1,1,1,1,1,1,2,2], [1,1,1,1,1,2,3], [1,1,1,1,2,2,2], [1,1,1,1,3,3], [1,1,1,2,2,3], [1,1,2,2,2,2], [1,1,2,3,3], [1,2,2,2,3], [2,2,2,2,2], [1,3,3,3], [2,2,3,3]. - Adi Dani, Jun 07 2011
A003056 is the following array A read by antidiagonals:
  0,  1,  2,  3,  4,  5, ...
  1,  2,  3,  4,  5,  6, ...
  2,  3,  4,  5,  6,  7, ...
  3,  4,  5,  6,  7,  8, ...
  4,  5,  6,  7,  8,  9, ...
  5,  6,  7,  8,  9, 10, ...
and a(n) is the hook sum Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013
a(n)*Pi is the total length of 3 points circle center spiral after n rotations. The spiral length at each rotation (L(n)) is A016957. The spiral length ratio rounded down [floor(L(n)/L(1))] is A001651. See illustration in links. - Kival Ngaokrajang, Dec 27 2013
Partial sums give A114364. - Leo Tavares, Feb 25 2022
For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n+1; {2, 2n-1, 1, 4, 1, 2n-1, 2, 18n+2}]. - Magus K. Chu, Oct 13 2022

			From _Dmitry Kamenetsky_, Dec 14 2008, with slight rewording by Raymond Martineau (mart0258(AT)yahoo.com), Dec 16 2008: (Start)
For an N x N Minesweeper grid the highest sum of numbers is (N-1)(3*N-2). This is achieved by filling every second row with mines (shown as 'X'). For example, when N=5 the best grids are:
.
  X X X X X
  4 6 6 6 4
  X X X X X
  4 6 6 6 4
  X X X X X
.
  and
.
  2 3 3 3 2
  X X X X X
  4 6 6 6 4
  X X X X X
  2 3 3 3 2
.
each giving a total of 52. (End)

L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 12.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Raghavendra N. Bhat, Cristian Cobeli, and Alexandru Zaharescu, A lozenge triangulation of the plane with integers, arXiv:2403.10500 [math.NT], 2024.
Kival Ngaokrajang, Illustration of 3 points circle center spiral.
Leo Tavares, Illustration: Double Hexagonal Trapezoids.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A001105, A002378, A005449, A033580, A049450.
Similar sequences are listed in A316466.
Cf. A003215, A005408, A114364.

List([0..55], n-> n*(3*n+1)); # G. C. Greubel, Sep 01 2019

a049451 n = n * (3 * n + 1)  -- Reinhard Zumkeller, Jul 07 2012

[n*(3*n+1): n in [0..55]]; // G. C. Greubel, Sep 01 2019

Table[n(3n+1), {n,0,55}] (* or *) CoefficientList[Series[2x(2+x)/(1-x)^3, {x,0,55}], x] (* Michael De Vlieger, Apr 05 2017 *)

a(n)=n*(3*n+1) \\ Charles R Greathouse IV, Sep 24 2015

[n*(3*n+1) for n in range(60)] # Gennady Eremin, Feb 27 2022

[n*(3*n+1) for n in (0..55)] # G. C. Greubel, Sep 01 2019

a(n) = n*(3*n+1).
G.f.: 2*x*(2+x)/(1-x)^3.
Sum_{i=1..n} a(i) = A045991(n+1). - Gary W. Adamson, Dec 20 2006
a(n) = 2*A005449(n). - Omar E. Pol, Dec 18 2008
a(n) = a(n-1) + 6*n -2, n > 0. - Vincenzo Librandi, Aug 06 2010
a(n) = A100104(n+1) - A100104(n). - Reinhard Zumkeller, Jul 07 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 4, a(2) = 14. - Philippe Deléham, Mar 26 2013
a(n) = A174709(6*n+3). - Philippe Deléham, Mar 26 2013
a(n) = (24/(n+2)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+2). - Bruno Berselli, Jun 04 2013 - after the similar formula of Vladimir Kruchinin in A002411
a(n) = A002061(n+1) + A056220(n). - Bruce J. Nicholson, Sep 21 2017
a(n) = Sum_{i = 2..5} P(i,n), where P(i,m) = m*((i-2)*m-(i-4))/2. - Bruno Berselli, Jul 04 2018
E.g.f.: x*(4 + 3*x)*exp(x). - G. C. Greubel, Sep 01 2019
a(n) = A003215(n) - A005408(n). - Leo Tavares, Feb 25 2022
From Amiram Eldar, Feb 27 2022: (Start)
Sum_{n>=1} 1/a(n) = 3 - Pi/(2*sqrt(3)) - 3*log(3)/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/sqrt(3) + 2*log(2) - 3. (End)
a(n) = A001105(n) + A002378(n). - Torlach Rush, Jul 11 2022

A140091 a(n) = 3n(n + 3)/2.

Original entry on oeis.org

0, 6, 15, 27, 42, 60, 81, 105, 132, 162, 195, 231, 270, 312, 357, 405, 456, 510, 567, 627, 690, 756, 825, 897, 972, 1050, 1131, 1215, 1302, 1392, 1485, 1581, 1680, 1782, 1887, 1995, 2106, 2220, 2337, 2457, 2580, 2706, 2835, 2967

Offset: 0

Omar E. Pol, May 22 2008

a(n) is also the dimension of the irreducible representation of the Lie algebra sl(3) with the highest weight 2*L_1+n*(L_1+L_2). - Leonid Bedratyuk, Jan 04 2010
Number of edges in the hexagonal triangle, T(n) (see the He et al. reference). - Emeric Deutsch, Nov 14 2014
a(n) = twice the area of a triangle having vertices at binomials (C(n,3),C(n+3,3)), (C(n+1,3),C(n+4,3)), and (C(n+2,3),C(n+5,3)) with n>=2. - J. M. Bergot, Mar 01 2018

W. Fulton, J. Harris, Representation theory: a first course. (1991). page 224, Exercise 15.19. - Leonid Bedratyuk, Jan 04 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 5.
Q. H. He, J. Z. Gu, S. J. Xu, and W. H. Chan, Hosoya polynomials of hexagonal triangles and trapeziums, MATCH, Commun. Math. Comput. Chem., Vol. 72, No. 3 (2014), pp. 835-843. - _Emeric Deutsch_, Nov 14 2014
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A000217, A005563, A067728, A140681, A212331.

The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, A005449, A045943, A115067, A140090, this sequence, A059845, A140672, A140673, A140674, A140675, A151542.

[3*n*(n+3)/2 : n in [0..50]]; // Wesley Ivan Hurt, Nov 14 2014

A140091:=n->3*n*(n+3)/2: seq(A140091(n), n=0..50); # Wesley Ivan Hurt, Nov 14 2014

Table[3 n (n + 3)/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 6, 15}, 50] (* Harvey P. Dale, Aug 15 2015 *)

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

a(n) = A000096(n)*3 = (3*n^2 + 9*n)/2 = n*(3*n+9)/2.
a(n) = a(n-1) + 3*n + 3 with n>0, a(0)=0. - Vincenzo Librandi, Nov 24 2010
G.f.: 3*x*(2 - x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 24 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2. - Harvey P. Dale, Aug 15 2015
E.g.f.: (1/2)*(3*x^2 + 12*x)*exp(x). - G. C. Greubel, Jul 17 2017
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 11/27.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/9 - 5/27. (End)

A059270 a(n) is both the sum of n+1 consecutive integers and the sum of the n immediately higher consecutive integers.

Original entry on oeis.org

0, 3, 15, 42, 90, 165, 273, 420, 612, 855, 1155, 1518, 1950, 2457, 3045, 3720, 4488, 5355, 6327, 7410, 8610, 9933, 11385, 12972, 14700, 16575, 18603, 20790, 23142, 25665, 28365, 31248, 34320, 37587, 41055, 44730, 48618, 52725, 57057, 61620

Offset: 0

Henry Bottomley, Jan 24 2001

Group the non-multiples of n as follows, e.g., for n = 4: (1,2,3), (5,6,7), (9,10,11), (13,14,15), ... Then a(n) is the sum of the members of the n-th group. Or, the sum of (n-1)successive numbers preceding n^2. - Amarnath Murthy, Jan 19 2004
Convolution of odds (A005408) and multiples of three (A008585). G.f. is the product of the g.f. of A005408 by the g.f. of A008585. - Graeme McRae, Jun 06 2006
Sums of rows of the triangle in A126890. - Reinhard Zumkeller, Dec 30 2006
Corresponds to the Wiener indices of C_{2n+1} i.e., the cycle on 2n+1 vertices (n > 0). - K.V.Iyer, Mar 16 2009
Also the product of the three numbers from A005843(n) up to A163300(n), divided by 8. - Juri-Stepan Gerasimov, Jul 26 2009
Partial sums of A033428. - Charlie Marion, Dec 08 2013
For n > 0, sum of multiples of n and (n+1) from 1 to n*(n+1). - Zak Seidov, Aug 07 2016
A generalization of Ianakiev's formula, a(n) = A005408(n)*A000217(n), follows. A005408(n+k)*A000217(n) is the sum of n+1 consecutive integers and, after skipping k integers, the sum of the n immediately higher consecutive integers. For example, for n = 3 and k = 2, 9*6 = 54 = 12+13+14+15 = 17+18+19. - Charlie Marion, Jan 25 2022

			a(5) = 25 + 26 + 27 + 28 + 29 + 30 = 31 + 32 + 33 + 34 + 35 = 165.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Roger B. Nelsen, Proof Without Words: Consecutive Sums of Consecutive Integers, Math. Mag., 63 (1990), 25.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A059255 for analog for sum of squares.
Cf. A222716 for the analogous sum of triangular numbers.
Cf. A234319 for nonexistence of analogs for sums of n-th powers, n > 2. - Jonathan Sondow, Apr 23 2014
Cf. A098737 (first subdiagonal).
Bisection of A109900.

I:=[0, 3, 15, 42]; [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..50]]; // Vincenzo Librandi, Jun 23 2012

A059270 := proc(n) n*(n+1)*(2*n+1)/2 ; end proc: # R. J. Mathar, Jul 10 2011

# (#+1)(2#+1)/2 &/@ Range[0,39] (* Ant King, Jan 03 2011 *)
CoefficientList[Series[3 x (1 + x)/(x - 1)^4, {x, 0, 39}], x]
LinearRecurrence[{4,-6,4,-1},{0,3,15,42},50] (* Vincenzo Librandi, Jun 23 2012 *)

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

[bernoulli_polynomial(n+1,3) for n in range(0, 41)] # Zerinvary Lajos, May 17 2009

a(n) = n*(n+1)*(2*n+1)/2.
a(n) = A000330(n)*3 = A006331(n)*3/2 = A055112(n)/2 = A000217(A002378(n)) - A000217(A005563(n-1)) = A000217(A005563(n)) - A000217(A002378(n)).
a(n) = A110449(n+1, n-1) for n > 1.
a(n) = Sum_{k=A000290(n) .. A002378(n)} k = Sum_{k=n^2..n^2+n} k.
a(n) = Sum_{k=n^2+n+1 .. n^2+2*n} k = Sum_{k=A002061(n+1) .. A005563(n)} k.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6 = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Ant King, Jan 03 2011
G.f.: 3*x*(1+x)/(1-x)^4. - Ant King, Jan 03 2011
a(n) = A000578(n+1) - A000326(n+1). - Ivan N. Ianakiev, Nov 29 2012
a(n) = A005408(n)*A000217(n) = a(n-1) + 3*A000290(n). -Ivan N. Ianakiev, Mar 08 2013
a(n) = n^3 + n^2 + A000217(n). - Charlie Marion, Dec 04 2013
From Ilya Gutkovskiy, Aug 08 2016: (Start)
E.g.f.: x*(6 + 9*x + 2*x^2)*exp(x)/2.
Sum_{n>=1} 1/a(n) = 2*(3 - 4*log(2)) = 0.4548225555204375246621... (End)
a(n) = Sum_{k=0..2*n} A001318(k). - Jacob Szlachetka, Dec 20 2021
a(n) = Sum_{k=0..n} A000326(k) + A005449(k). - Jacob Szlachetka, Dec 21 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(Pi-3). - Amiram Eldar, Sep 17 2022

A016922 a(n) = (6*n+1)^2.

Original entry on oeis.org

1, 49, 169, 361, 625, 961, 1369, 1849, 2401, 3025, 3721, 4489, 5329, 6241, 7225, 8281, 9409, 10609, 11881, 13225, 14641, 16129, 17689, 19321, 21025, 22801, 24649, 26569, 28561, 30625, 32761, 34969, 37249, 39601, 42025, 44521, 47089, 49729, 52441, 55225

Offset: 0

N. J. A. Sloane

Except for 2, exponents e such that x^e-x+1 is reducible.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A005449, A086727, A016778 (bisection), A016921.

[(6*n+1)^2: n in [0..60]]; // Vincenzo Librandi, May 04 2011

A016922:=n->(6*n+1)^2; seq(A016922(n), n=0..100); # Wesley Ivan Hurt, Dec 06 2013

Table[(6n+1)^2, {n,0,100}] (* or *)
CoefficientList[Series[(1 + 46*x + 25*x^2)/(1 - x)^3, {x, 0, 30}], x] (* Wesley Ivan Hurt, Oct 13 2014 *)
LinearRecurrence[{3,-3,1},{1,49,169},50] (* Harvey P. Dale, Feb 17 2023 *)

a(n)=(6*n+1)^2 \\ Charles R Greathouse IV, Jun 17 2017

G.f.: ( 1+46*x+25*x^2 ) / (1-x)^3. - R. J. Mathar, Mar 10 2011
a(n) = A016921(n)^2 = A000290(A016921(n)). - Wesley Ivan Hurt, Dec 06 2013
a(n) = 24*A005449(n)+1. - Jean-Bernard François, Oct 12 2014
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Wesley Ivan Hurt, Oct 13 2014
Sum_{n>=0} 1/a(n) = A086727. - Amiram Eldar, Nov 16 2020

A262815 Number of ordered ways to write n as x^3 + y(y+1)/2 + z(3*z+1)/2, where x, y and z are nonnegative integers.

Original entry on oeis.org

1, 2, 2, 3, 2, 1, 2, 2, 4, 3, 3, 4, 1, 3, 2, 3, 5, 3, 5, 1, 1, 3, 3, 4, 2, 3, 3, 3, 4, 6, 6, 3, 2, 3, 2, 4, 6, 6, 3, 2, 3, 3, 4, 5, 8, 2, 3, 3, 5, 3, 2, 5, 3, 3, 3, 7, 3, 4, 4, 3, 3, 3, 5, 8, 2, 4, 3, 5, 7, 4, 7, 4, 5, 3, 6, 1, 4, 4, 6, 7, 4, 8, 5, 1, 4, 7, 7, 4, 4, 5, 2, 3, 5, 10, 6, 4, 2, 1, 3, 5, 7

Offset: 0

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 5, 12, 19, 20, 75, 83, 97, 117.
Conjecture verified up to 10^11. - Mauro Fiorentini, Jul 20 2023
See also A262813 and A262816 for similar conjectures.
By Theorem 1.7(i) in the linked paper, each natural number can be written as the sum of a triangular number, an even square and a generalized pentagonal number.

			a(0) = 1 since 0 = 0^3 + 0*1/2 + 0*(3*0+1)/2.
a(5) = 1 since 5 = 0^3 + 2*3/2 + 1*(3*1+1)/2.
a(12) = 1 since 12 = 0^3 + 4*5/2 + 1*(3*1+1)/2.
a(19) = 1 since 19 = 1^3 + 2*3/2 + 3*(3*3+1)/2.
a(20) = 1 since 20 = 2^3 + 4*5/2 + 1*(3*1+1)/2.
a(75) = 1 since 75 = 2^3 + 4*5/2 + 6*(3*6+1)/2.
a(83) = 1 since 83 = 0^3 + 3*4/2 + 7*(3*7+1)/2.
a(97) = 1 since 97 = 3^3 + 10*11/2 + 3*(3*3+1)/2.
a(117) = 1 since 117 = 0^3 + 13*14/2 + 4*(3*4+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000217, A000578, A005449, A160325, A262813, A262816.

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^3-z(3z+1)/2],r=r+1],{x,0,n^(1/3)},{z,0,(Sqrt[24(n-x^3)+1]-1)/6}];Print[n," ",r];Continue,{n,0,100}]

A029618 Numbers in (3,2)-Pascal triangle (by row).

Original entry on oeis.org

1, 3, 2, 3, 5, 2, 3, 8, 7, 2, 3, 11, 15, 9, 2, 3, 14, 26, 24, 11, 2, 3, 17, 40, 50, 35, 13, 2, 3, 20, 57, 90, 85, 48, 15, 2, 3, 23, 77, 147, 175, 133, 63, 17, 2, 3, 26, 100, 224, 322, 308, 196, 80, 19, 2, 3, 29, 126, 324, 546, 630, 504, 276, 99, 21, 2, 3, 32, 155, 450, 870

Offset: 0

Mohammad K. Azarian

Reverse of A029600. - Philippe Deléham, Nov 21 2006
Triangle T(n,k), read by rows, given by (3,-2,0,0,0,0,0,0,0,...) DELTA (2,-1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 10 2011
Row n: expansion of (3+2x)*(1+x)^(n-1), n>0. - Philippe Deléham, Oct 10 2011
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

			Triangle begins as:
  1;
  3,  2;
  3,  5,  2;
  3,  8,  7,  2;
  3, 11, 15,  9,  2;
  ...

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A007318, A029600, A084938, A228196, A228576, A016789 (2nd column), A005449 (3rd column), A006002 (4th column).

T:= function(n,k)
    if n=0 and k=0 then return 1;
    elif k=0 then return 3;
    elif k=n then return 2;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019

A029618 := proc(n,k)
    if k < 0 or k > n then
        0;
    elif  n = 0 then
        1;
    elif k=0 then
        3;
    elif k = n then
        2;
    else
        procname(n-1,k-1)+procname(n-1,k) ;
    end if;
end proc: # R. J. Mathar, Jul 08 2015

T[n_, k_]:= T[n, k]= If[n==0 && k==0, 1, If[k==0, 3, If[k==n, 2, T[n-1, k-1] + T[n-1, k] ]]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Nov 13 2019 *)

T(n,k) = if(n==0 && k==0, 1, if(k==0, 3, if(k==n, 2, T(n-1, k-1) + T(n-1, k) ))); \\ G. C. Greubel, Nov 12 2019

@CachedFunction
def T(n, k):
    if (n==0 and k==0): return 1
    elif (k==0): return 3
    elif (k==n): return 2
    else: return T(n-1,k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

T(n,k) = T(n-1,k-1) + T(n-1,k) with T(0,0)=1, T(n,0)=3, T(n,n)=2; n, k > 0. - Boris Putievskiy, Sep 04 2013

G.f.: (-1-x*y-2*x)/(-1+x*y+x). - R. J. Mathar, Aug 11 2015

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

A081266 Staggered diagonal of triangular spiral in A051682.

Original entry on oeis.org

0, 6, 21, 45, 78, 120, 171, 231, 300, 378, 465, 561, 666, 780, 903, 1035, 1176, 1326, 1485, 1653, 1830, 2016, 2211, 2415, 2628, 2850, 3081, 3321, 3570, 3828, 4095, 4371, 4656, 4950, 5253, 5565, 5886, 6216, 6555, 6903, 7260, 7626, 8001, 8385, 8778, 9180

Offset: 0

Paul Barry, Mar 15 2003

Staggered diagonal of triangular spiral in A051682, between (0,4,17) spoke and (0,7,23) spoke.
Binomial transform of (0, 6, 9, 0, 0, 0, ...).
If Y is a fixed 3-subset of a (3n+1)-set X then a(n) is the number of (3n-1)-subsets of X intersecting Y. - Milan Janjic, Oct 28 2007
Partial sums give A085788. - Leo Tavares, Nov 23 2023

			a(1)=9*1+0-3=6, a(2)=9*2+6-3=21, a(3)=9*3+21-3=45.
For n=3, a(3) = -0^2+1^2-2^2+3^2-4^2+5^2-6^2+7^2-8^2+9^2 = 45.

Muniru A Asiru, Table of n, a(n) for n = 0..10000
Tomislav Došlić and Luka Podrug, Sweet division problems: from chocolate bars to honeycomb strips and back, arXiv:2304.12121 [math.CO], 2023.
Milan Janjic, Two Enumerative Functions
Milan Janjic and B. Petkovic, A Counting Function, arXiv:1301.4550 [math.CO], 2013.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Leo Tavares, Star illustration
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000290, A005449, A014105, A022266, A033585, A062725, A144312, A144314, A218470.

Cf. A003154, A133694.

List([0..50],n->Binomial(3*n+1,2)); # Muniru A Asiru, Feb 28 2019

seq(binomial(3*n+1,2), n=0..45); # Zerinvary Lajos, Jan 21 2007

LinearRecurrence[{3,-3,1},{0,6,21},50] (* Harvey P. Dale, Aug 29 2015 *)

a(n)=3*n*(3*n+1)/2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 6*C(n,1) + 9*C(n,2).
a(n) = 3*n*(3*n+1)/2.
G.f.: (6*x+3*x^2)/(1-x)^3.
a(n) = A000217(3*n); a(2*n) = A144314(n). - Reinhard Zumkeller, Sep 17 2008
a(n) = 3*A005449(n). - R. J. Mathar, Mar 27 2009
a(n) = 9*n+a(n-1)-3 for n>0, a(0)=0. - Vincenzo Librandi, Aug 08 2010
a(n) = A218470(9n+5). - Philippe Deléham, Mar 27 2013
a(n) = Sum_{k=0..3n} (-1)^(n+k)*k^2. - Bruno Berselli, Aug 29 2013
E.g.f.: 3*exp(x)*x*(4 + 3*x)/2. - Stefano Spezia, Jun 06 2021
From Amiram Eldar, Aug 11 2022: (Start)
Sum_{n>=1} 1/a(n) = 2 - Pi/(3*sqrt(3)) - log(3).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/(3*sqrt(3)) + 4*log(2)/3 - 2. (End)
From Leo Tavares, Nov 23 2023: (Start)
a(n) = 3*A000217(n) + 3*A000290(n).
a(n) = A003154(n+1) - A133694(n+1). (End)

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Sage

Formula

Extensions

A049451 Twice second pentagonal numbers.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Haskell

Magma

Mathematica

PARI

Python

Sage

Formula

A140091 a(n) = 3*n*(n + 3)/2.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A059270 a(n) is both the sum of n+1 consecutive integers and the sum of the n immediately higher consecutive integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

A016922 a(n) = (6*n+1)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A262815 Number of ordered ways to write n as x^3 + y*(y+1)/2 + z*(3*z+1)/2, where x, y and z are nonnegative integers.

Views

Author

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

A140091 a(n) = 3n(n + 3)/2.

A262815 Number of ordered ways to write n as x^3 + y(y+1)/2 + z(3*z+1)/2, where x, y and z are nonnegative integers.

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

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