cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 23 results. Next

A014105 Second hexagonal numbers: a(n) = n*(2*n + 1).

Original entry on oeis.org

0, 3, 10, 21, 36, 55, 78, 105, 136, 171, 210, 253, 300, 351, 406, 465, 528, 595, 666, 741, 820, 903, 990, 1081, 1176, 1275, 1378, 1485, 1596, 1711, 1830, 1953, 2080, 2211, 2346, 2485, 2628, 2775, 2926, 3081, 3240, 3403, 3570, 3741, 3916, 4095, 4278
Offset: 0

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Author

N. J. A. Sloane, Jun 14 1998

Keywords

Comments

Note that when starting from a(n)^2, equality holds between series of first n+1 and next n consecutive squares: a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2*n)^2; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Henry Bottomley, Jan 22 2001; with typos fixed by Zak Seidov, Sep 10 2015
a(n) = sum of second set of n consecutive even numbers - sum of the first set of n consecutive odd numbers: a(1) = 4-1, a(3) = (8+10+12) - (1+3+5) = 21. - Amarnath Murthy, Nov 07 2002
Partial sums of odd numbers 3 mod 4, that is, 3, 3+7, 3+7+11, ... See A001107. - Jon Perry, Dec 18 2004
If Y is a fixed 3-subset of a (2n+1)-set X then a(n) is the number of (2n-1)-subsets of X intersecting Y. - Milan Janjic, Oct 28 2007
More generally (see the first comment), for n > 0, let b(n,k) = a(n) + k*(4*n + 1). Then b(n,k)^2 + (b(n,k) + 1)^2 + ... + (b(n,k) + n)^2 = (b(n,k) + n + 1 + 2*k)^2 + ... + (b(n,k) + 2*n + 2*k)^2 + k^2; e.g., if n = 3 and k = 2, then b(n,k) = 47 and 47^2 + ... + 50^2 = 55^2 + ... + 57^2 + 2^2. - Charlie Marion, Jan 01 2011
Sequence found by reading the line from 0, in the direction 0, 10, ..., and the line from 3, in the direction 3, 21, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Nov 09 2011
a(n) is the number of positions of a domino in a pyramidal board with base 2n+1. - César Eliud Lozada, Sep 26 2012
Differences of row sums of two consecutive rows of triangle A120070, i.e., first differences of A016061. - J. M. Bergot, Jun 14 2013 [In other words, the partial sums of this sequence give A016061. - Leo Tavares, Nov 23 2021]
a(n)*Pi is the total length of half circle spiral after n rotations. See illustration in links. - Kival Ngaokrajang, Nov 05 2013
For corresponding sums in first comment by Henry Bottomley, see A059255. - Zak Seidov, Sep 10 2015
a(n) also gives the dimension of the simple Lie algebras B_n (n >= 2) and C_n (n >= 3). - Wolfdieter Lang, Oct 21 2015
With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for unsigned A130757, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 13 2015
Partial sums of squares with alternating signs, ending in an even term: a(n) = 0^2 - 1^2 +- ... + (2*n)^2, cf. Example & Formula from Berselli, 2013. - M. F. Hasler, Jul 03 2018
Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central peak and the largest Dyck path has a central valley, n > 0. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
a(n) is the area of a triangle with vertices at (0,0), (2*n+1, 2*n), and ((2*n+1)^2, 4*n^2). - Art Baker, Dec 12 2018
This sequence is the largest subsequence of A000217 such that gcd(a(n), 2*n) = a(n) mod (2*n) = n, n > 0 up to a given value of n. It is the interleave of A033585 (a(n) is even) and A033567 (a(n) is odd). - Torlach Rush, Sep 09 2019
A generalization of Hasler's Comment (Jul 03 2018) follows. Let P(k,n) be the n-th k-gonal number. Then for k > 1, partial sums of {P(k,n)} with alternating signs, ending in an even term, = n*((k-2)*n + 1). - Charlie Marion, Mar 02 2021
Let U_n(H) = {A in M_n(H): A*A^H = I_n} be the group of n X n unitary matrices over the quaternions (A^H is the conjugate transpose of A. Note that over the quaternions we still have A*A^H = I_n <=> A^H*A = I_n by mapping A and A^H to (2n) X (2n) complex matrices), then a(n) is the dimension of its Lie algebra u_n(H) = {A in M_n(H): A + A^H = 0} as a real vector space. A basis is given by {(E_{st}-E_{ts}), i*(E_{st}+E_{ts}), j*(E_{st}+E_{ts}), k*(E_{st}+E_{ts}): 1 <= s < t <= n} U {i*E_{tt}, j*E_{tt}, k*E_{tt}: t = 1..n}, where E_{st} is the matrix with all entries zero except that its (st)-entry is 1. - Jianing Song, Apr 05 2021

Examples

			For n=6, a(6) = 0^2 - 1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2 - 7^2 + 8^2 - 9^2 + 10^2 - 11^2 + 12^2 = 78. - _Bruno Berselli_, Aug 29 2013
		

References

  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)

Crossrefs

Second column of array A094416.
Equals A033586(n) divided by 4.
See Comments of A132124.
Second n-gonal numbers: A005449, A147875, A045944, A179986, A033954, A062728, A135705.
Row sums in triangle A253580.

Programs

Formula

a(n) = 3*Sum_{k=1..n} tan^2(k*Pi/(2*(n + 1))). - Ignacio Larrosa Cañestro, Apr 17 2001
a(n)^2 = n*(a(n) + 1 + a(n) + 2 + ... + a(n) + 2*n); e.g., 10^2 = 2*(11 + 12 + 13 + 14). - Charlie Marion, Jun 15 2003
From N. J. A. Sloane, Sep 13 2003: (Start)
G.f.: x*(3 + x)/(1 - x)^3.
E.g.f.: exp(x)*(3*x + 2*x^2).
a(n) = A000217(2*n) = A000384(-n). (End)
a(n) = A084849(n) - 1; A100035(a(n) + 1) = 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = A126890(n, k) + A126890(n, n-k), 0 <= k <= n. - Reinhard Zumkeller, Dec 30 2006
a(2*n) = A033585(n); a(3*n) = A144314(n). - Reinhard Zumkeller, Sep 17 2008
a(n) = a(n-1) + 4*n - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = Sum_{k=0.2*n} (-1)^k*k^2. - Bruno Berselli, Aug 29 2013
a(n) = A242342(2*n + 1). - Reinhard Zumkeller, May 11 2014
a(n) = Sum_{k=0..2} C(n-2+k, n-2) * C(n+2-k, n), for n > 1. - J. M. Bergot, Jun 14 2014
a(n) = floor(Sum_{j=(n^2 + 1)..((n+1)^2 - 1)} sqrt(j)). Fractional portion of each sum converges to 1/6 as n -> infinity. See A247112 for a similar summation sequence on j^(3/2) and references to other such sequences. - Richard R. Forberg, Dec 02 2014
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3, with a(0) = 0, a(1) = 3, and a(2) = 10. - Harvey P. Dale, Feb 10 2015
Sum_{n >= 1} 1/a(n) = 2*(1 - log(2)) = 0.61370563888010938... (A188859). - Vaclav Kotesovec, Apr 27 2016
From Wolfdieter Lang, Apr 27 2018: (Start)
a(n) = trinomial(2*n, 2) = trinomial(2*n, 2*(2*n-1)), for n >= 1, with the trinomial irregular triangle A027907; i.e., trinomial(n,k) = A027907(n,k).
a(n) = (1/Pi) * Integral_{x=0..2} (1/sqrt(4 - x^2)) * (x^2 - 1)^(2*n) * R(4*(n-1), x), for n >= 0, with the R polynomial coefficients given in A127672, and R(-m, x) = R(m, x). [See Comtet, p. 77, the integral formula for q = 3, n -> 2*n, k = 2, rewritten with x = 2*cos(phi).] (End)
a(n) = A002943(n)/2. - Ralf Steiner, Jul 23 2019
a(n) = A000290(n) + A002378(n). - Torlach Rush, Nov 02 2020
a(n) = A003215(n) - A000290(n+1). See Squared Hexagons illustration. Leo Tavares, Nov 23 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/2 + log(2) - 2. - Amiram Eldar, Nov 28 2021

Extensions

Link added and minor errors corrected by Johannes W. Meijer, Feb 04 2010

A046092 4 times triangular numbers: a(n) = 2*n*(n+1).

Original entry on oeis.org

0, 4, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264, 312, 364, 420, 480, 544, 612, 684, 760, 840, 924, 1012, 1104, 1200, 1300, 1404, 1512, 1624, 1740, 1860, 1984, 2112, 2244, 2380, 2520, 2664, 2812, 2964, 3120, 3280, 3444, 3612, 3784, 3960, 4140, 4324
Offset: 0

Views

Author

Keywords

Comments

Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; sequence gives Y values. X values are 1, 3, 5, 7, 9, ... (A005408), Z values are A001844.
In the triple (X, Y, Z) we have X^2=Y+Z. Actually, the triple is given by {x, (x^2 -+ 1)/2}, where x runs over the odd numbers (A005408) and x^2 over the odd squares (A016754). - Lekraj Beedassy, Jun 11 2004
a(n) is the number of edges in n X n square grid with all horizontal and vertical segments filled in. - Asher Auel, Jan 12 2000 [Corrected by Felix Huber, Apr 09 2024]
a(n) is the only number satisfying an inequality related to zeta(2) and zeta(3): Sum_{i>a(n)+1} 1/i^2 < Sum_{i>n} 1/i^3 < Sum_{i>a(n)} 1/i^2. - Benoit Cloitre, Nov 02 2001
Number of right triangles made from vertices of a regular n-gon when n is even. - Sen-Peng Eu, Apr 05 2001
Number of ways to change two non-identical letters in the word aabbccdd..., where there are n type of letters. - Zerinvary Lajos, Feb 15 2005
a(n) is the number of (n-1)-dimensional sides of an (n+1)-dimensional hypercube (e.g., squares have 4 corners, cubes have 12 edges, etc.). - Freek van Walderveen (freek_is(AT)vanwal.nl), Nov 11 2005
From Nikolaos Diamantis (nikos7am(AT)yahoo.com), May 23 2006: (Start)
Consider a triangle, a pentagon, a heptagon, ..., a k-gon where k is odd. We label a triangle with n=1, a pentagon with n=2, ..., a k-gon with n = floor(k/2). Imagine a player standing at each vertex of the k-gon.
Initially there are 2 frisbees, one held by each of two neighboring players. Every time they throw the frisbee to one of their two nearest neighbors with equal probability. Then a(n) gives the average number of steps needed so that the frisbees meet.
I verified this by simulating the processes with a computer program. For example, a(2) = 12 because in a pentagon that's the expected number of trials we need to perform. That is an exercise in Concrete Mathematics and it can be done using generating functions. (End)
A diagonal of A059056. - Zerinvary Lajos, Jun 18 2007
If X_1,...,X_n is a partition of a 2n-set X into 2-blocks then a(n-1) is equal to the number of 2-subsets of X containing none of X_i, (i=1,...,n). - Milan Janjic, Jul 16 2007
X values of solutions to the equation 2*X^3 + X^2 = Y^2. To find Y values: b(n) = 2n(n+1)(2n+1). - Mohamed Bouhamida, Nov 06 2007
Number of (n+1)-permutations of 3 objects u,v,w, with repetition allowed, containing n-1 u's. Example: a(1)=4 because we have vv, vw, wv and ww; a(2)=12 because we can place u in each of the previous four 2-permutations either in front, or in the middle, or at the end. - Zerinvary Lajos, Dec 27 2007
Sequence found by reading the line from 0, in the direction 0, 4, ... and the same line from 0, in the direction 0, 12, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, May 03 2008
a(n) is also the least weight of self-conjugate partitions having n different even parts. - Augustine O. Munagi, Dec 18 2008
From Peter Luschny, Jul 12 2009: (Start)
The general formula for alternating sums of powers of even integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,1)-(-1)^k P(n,2k+1))/2. Here n=2, thus
a(k) = |(P(2,1) - (-1)^k*P(2,2k+1))/2|. (End)
The sum of squares of n+1 consecutive numbers between a(n)-n and a(n) inclusive equals the sum of squares of n consecutive numbers following a(n). For example, for n = 2, a(2) = 12, and the corresponding equation is 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Tanya Khovanova, Jul 20 2009
Number of roots in the root system of type D_{n+1} (for n>2). - Tom Edgar, Nov 05 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; sequence gives number of intersections of these ellipses (cf. A051890, A001844); a(n) = A051890(n+1) - 2 = A001844(n) - 1. - Jaroslav Krizek, Dec 27 2013
a(n) appears also as the second member of the quartet [p0(n), a(n), p2(n), p3(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p2(n) = A054000(n+1) and p3(n) = A139570(n). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 2014
a(n) appears also as the third and fourth member of the quartet [p0(n), p0(n), a(n), a(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p0(n) = A001105(n). - Wolfdieter Lang, Oct 16 2014
Consider two equal rectangles composed of unit squares. Then surround the 1st rectangle with 1-unit-wide layers to build larger rectangles, and surround the 2nd rectangle just to hide the previous layers. If r(n) and h(n) are the number of unit squares needed for n layers in the 1st case and the 2nd case, then for all rectangles, we have a(n) = r(n) - h(n) for n>=1. - Michel Marcus, Sep 28 2015
When greater than 4, a(n) is the perimeter of a Pythagorean triangle with an even short leg 2*n. - Agola Kisira Odero, Apr 26 2016
Also the number of minimum connected dominating sets in the (n+1)-cocktail party graph. - Eric W. Weisstein, Jun 29 2017
a(n+1) is the harmonic mean of A000384(n+2) and A014105(n+1). - Bob Andriesse, Apr 27 2019
Consider a circular cake from which wedges of equal center angle c are cut out in clockwise succession and turned around so that the bottom comes to the top. This goes on until the cake shows its initial surface again. An interesting case occurs if 360°/c is not an integer. Then, with n = floor(360°/c), the number of wedges which have to be cut out and turned equals a(n). (For the number of cutting line segments see A005408.) - According to Peter Winkler's book "Mathematical Mind-Benders", which presents the problem and its solution (see Winkler, pp. 111, 115) the problem seems to be of French origin but little is known about its history. - Manfred Boergens, Apr 05 2022
a(n-3) is the maximum irregularity over all maximal 2-degenerate graphs with n vertices. The extremal graphs are 2-stars (K_2 joined to n-2 independent vertices). (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, May 29 2023
Number of ways of placing a domino on a (n+1)X(n+1) board of squares. - R. J. Mathar, Apr 24 2024
The sequence terms are the exponents in the expansion of (1/(1 + x)) * Sum_{n >= 0} x^n * Product_{k = 1..n} (1 - x^(2*k-1))/(1 + x^(2*k+1)) = 1 - x^4 + x^12 - x^24 + x^40 - x^60 + - ... (Andrews and Berndt, Entry 9.3.3, p. 229). Cf. A153140. - Peter Bala, Feb 15 2025
Number of edges in an (n+1)-dimensional orthoplex. 2D orthoplexes (diamonds) have 4 edges, 3D orthoplexes (octahedrons) have 12 edges, 4D orthoplexes (16-cell) have 24 edges, and so on. - Aaron Franke, Mar 23 2025

Examples

			a(7)=112 because 112 = 2*7*(7+1).
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25), ...
The first such partitions, corresponding to a(n)=1,2,3,4, are 2+2, 4+4+2+2, 6+6+4+4+2+2, 8+8+6+6+4+4+2+2. - _Augustine O. Munagi_, Dec 18 2008
		

References

  • George E. Andrews and Bruce C. Berndt, Ramanujan's Lost Notebook, Part I, Springer, 2005.
  • Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
  • Albert H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
  • Ronald L. Graham, D. E. Knuth and Oren Patashnik, Concrete Mathematics, Reading, Massachusetts: Addison-Wesley, 1994.
  • Peter Winkler, Mathematical Mind-Benders, Wellesley, Massachusetts: A K Peters, 2007.

Crossrefs

Main diagonal of array in A001477.
Equals A033996/2. Cf. A001844. - Augustine O. Munagi, Dec 18 2008
Cf. A078371, A141530 (see Librandi's comment in A078371).
Cf. similar sequences listed in A299645.
Cf. A005408.
Cf. A016754.
Cf. A002378, A046092, A028896 (irregularities of maximal k-degenerate graphs).

Programs

Formula

a(n) = A100345(n+1, n-1) for n>0.
a(n) = 2*A002378(n) = 4*A000217(n). - Lekraj Beedassy, May 25 2004
a(n) = C(2n, 2) - n = 4*C(n, 2). - Zerinvary Lajos, Feb 15 2005
From Lekraj Beedassy, Jun 04 2006: (Start)
a(n) - a(n-1)=4*n.
Let k=a(n). Then a(n+1) = k + 2*(1 + sqrt(2k + 1)). (End)
Array read by rows: row n gives A033586(n), A085250(n+1). - Omar E. Pol, May 03 2008
O.g.f.:4*x/(1-x)^3; e.g.f.: exp(x)*(2*x^2+4*x). - Geoffrey Critzer, May 17 2009
From Stephen Crowley, Jul 26 2009: (Start)
a(n) = 1/int(-(x*n+x-1)*(step((-1+x*n)/n)-1)*n*step((x*n+x-1)/(n+1)),x=0..1) where step(x)=piecewise(x<0,0,0<=x,1) is the Heaviside step function.
Sum_{n>=1} 1/a(n) = 1/2. (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=4, a(2)=12. - Harvey P. Dale, Jul 25 2011
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} (sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011
a(n) = A001844(n) - 1. - Omar E. Pol, Oct 03 2011
(a(n) - A000217(k))^2 = A000217(2n-k)*A000217(2n+1+k) - (A002378(n) - A000217(k)), for all k. See also A001105. - Charlie Marion, May 09 2013
From Ivan N. Ianakiev, Aug 30 2013: (Start)
a(n)*(2m+1)^2 + a(m) = a(n*(2m+1)+m), for any nonnegative integers n and m.
t(k)*a(n) + t(k-1)*a(n+1) = a((n+1)*(t(k)-t(k-1)-1)), where k>=2, n>=1, t(k)=A000217(k). (End)
a(n) = A245300(n,n). - Reinhard Zumkeller, Jul 17 2014
2*a(n)+1 = A016754(n) = A005408(n)^2, the odd squares. - M. F. Hasler, Oct 02 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) - 1/2 = A187832. - Ilya Gutkovskiy, Mar 16 2017
a(n) = lcm(2*n,2*n+2). - Enrique Navarrete, Aug 30 2017
a(n)*a(n+k) + k^2 = m^2 (a perfect square), n >= 1, k >= 0. - Ezhilarasu Velayutham, May 13 2019
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(Pi/2)/(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -2*cos(sqrt(3)*Pi/2)/Pi. (End)
a(n) = A016754(n) - A001844(n). - Leo Tavares, Sep 20 2022

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

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

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

Examples

			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
		

Crossrefs

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.

Programs

  • GAP
    List([0..50], n-> n*(3*n-1)); # G. C. Greubel, Aug 31 2019
  • Magma
    [n*(3*n-1) : n in [0..50]]; // Wesley Ivan Hurt, Sep 24 2017
    
  • Maple
    seq(n*(3*n-1),n=0..44); # Zerinvary Lajos, Jun 12 2007
  • Mathematica
    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 *)
  • PARI
    a(n)=n*(3*n-1) \\ Charles R Greathouse IV, Nov 20 2012
    
  • Sage
    [n*(3*n-1) for n in (0..50)] # G. C. Greubel, Aug 31 2019
    

Formula

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)

A035008 Total number of possible knight moves on an (n+2) X (n+2) chessboard, if the knight is placed anywhere.

Original entry on oeis.org

0, 16, 48, 96, 160, 240, 336, 448, 576, 720, 880, 1056, 1248, 1456, 1680, 1920, 2176, 2448, 2736, 3040, 3360, 3696, 4048, 4416, 4800, 5200, 5616, 6048, 6496, 6960, 7440, 7936, 8448, 8976, 9520, 10080, 10656, 11248, 11856, 12480, 13120, 13776
Offset: 0

Views

Author

Ulrich Schimke (ulrschimke(AT)aol.com), Dec 11 1999

Keywords

Comments

16 times the triangular numbers A000217.
Centered 16-gonal numbers A069129, minus 1. Also, sequence found by reading the segment (0, 16) together with the line from 16, in the direction 16, 48, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Apr 26 2008, Nov 20 2008
For n >= 1, number of permutations of n+1 objects selected from 5 objects v, w, x, y, z with repetition allowed, containing n-1 v's. Examples: at n=1, n-1=0 (i.e., zero v's), and a(1)=16 because we have ww, wx, wy, wz, xw, xx, xy, xz, yw, yx, yy, yz, zw, zx, zy, zz; at n=2, n-1=1 (i.e., one v), and there are 3 permutations corresponding to each one in the n=1 case (e.g., the single v can be inserted in any of three places in the 2-object permutation xy, yielding vxy, xvy, and xyv), so a(2) = 3*a(1) = 3*16 = 48; at n=3, n-1=2 (i.e., two v's), and a(3) = C(4,2)*a(1) = 6*16 = 96; etc. - Zerinvary Lajos, Aug 07 2008 (this needs clarification, Joerg Arndt, Feb 23 2014)
Sequence found by reading the line from 0, in the direction 0, 16, ... and the same line from 0, in the direction 0, 48, ..., in the square spiral whose vertices are the generalized 18-gonal numbers. - Omar E. Pol, Oct 03 2011
For n > 0, a(n) is the area of the triangle with vertices at ((n-1)^2, n^2), ((n+1)^2, (n+2)^2), and ((n+3)^2, (n+2)^2). - J. M. Bergot, May 22 2014
For n > 0, a(n) is the number of self-intersecting points in star polygon {4*(n+1)/(2*n+1)}. - Bui Quang Tuan, Mar 28 2015
Equivalently: integers k such that k$ / (k/2)! and k$ / (k/2+1)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information). - Bernard Schott, Dec 02 2021

Examples

			3 X 3-Board: knight can be placed in 8 positions with 2 moves from each, so a(1) = 16.
		

Crossrefs

Cf. A033586 (King), A035005 (Queen), A035006 (Rook), A002492 (Bishop) and A049450 (Pawn).
Cf. A348692.
Subsequence of A008586 and of A349081.

Programs

Formula

a(n) = 8*n*(n+1).
G.f.: 16*x/(1-x)^3.
a(n) = A069129(n+1) - 1. - Omar E. Pol, Apr 26 2008
a(n) = binomial(n+1,2)*4^2, n >= 0. - Zerinvary Lajos, Aug 07 2008
a(n) = 8*n^2 + 8*n = 16*A000217(n) = 8*A002378(n) = 4*A046092(n) = 2*A033996(n). - Omar E. Pol, Dec 12 2008
a(n) = a(n-1) + 16*n, with a(0)=0. - Vincenzo Librandi, Nov 17 2010
E.g.f.: 8*exp(x)*x*(2 + x). - Stefano Spezia, May 19 2021
From Amiram Eldar, Feb 22 2023: (Start)
Sum_{n>=1} 1/a(n) = 1/8.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2*log(2) - 1)/8.
Product_{n>=1} (1 - 1/a(n)) = -(8/Pi)*cos(sqrt(3/2)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (8/Pi)*cos(Pi/(2*sqrt(2))). (End)

Extensions

More terms from Erich Friedman
Minor errors corrected and edited by Johannes W. Meijer, Feb 04 2010

A002492 Sum of the first n even squares: 2*n*(n+1)*(2*n+1)/3.

Original entry on oeis.org

0, 4, 20, 56, 120, 220, 364, 560, 816, 1140, 1540, 2024, 2600, 3276, 4060, 4960, 5984, 7140, 8436, 9880, 11480, 13244, 15180, 17296, 19600, 22100, 24804, 27720, 30856, 34220, 37820, 41664, 45760, 50116, 54740, 59640, 64824, 70300, 76076, 82160
Offset: 0

Views

Author

Keywords

Comments

Total number of possible bishop moves on an n+1 X n+1 chessboard, if the bishop is placed anywhere. E.g., on a 3 X 3-Board: bishop has 8 X 2 moves and 1 X 4 moves, so a(2)=20. - Ulrich Schimke (ulrschimke(AT)aol.com)
Let M_n denote the n X n matrix M_n(i,j)=(i+j)^2; then the characteristic polynomial of M_n is x^n - a(n)x^(n-1) - .... - Michael Somos, Nov 14 2002
Partial sums of A016742. - Lekraj Beedassy, Jun 19 2004
0,4,20,56,120 gives the number of electrons in closed shells in the double shell periodic system of elements. This is a new interpretation of the periodic system of the elements. The factor 4 in the formula 4*n(n+1)(2n+1)/6 plays a significant role, since it designates the degeneracy of electronic states in this system. Closed shells with more than 120 electrons are not expected to exist. - Karl-Dietrich Neubert (kdn(AT)neubert.net)
Inverse binomial transform of A240434. - Wesley Ivan Hurt, Apr 13 2014
Atomic number of alkaline-earth metals of period 2n. - Natan Arie Consigli, Jul 03 2016
a(n) are the negative cubic coefficients in the expansion of sin(kx) into powers of sin(x) for the odd k: sin(kx) = k sin(x) - c(k) sin^3(x) + O(sin^5(x)); a(n) = c(2n+1) = A000292(2n). - Mathias Zechmeister, Jul 24 2022
Also the number of distinct series-parallel networks under series-parallel reduction on three unlabeled edges of n element kinds. - Michael R. Hayashi, Aug 02 2023

References

  • A. O. Barut, Group Structure of the Periodic System, in Wybourne, Ed., The Structure of Matter, University of Canterbury Press, Christchurch, 1972, p. 126.
  • Edward G. Mazur, Graphic Representation of the Periodic System during One Hundred Years, University of Alabama Press, Alabama, 1974.
  • W. Permans and J. Kemperman, "Nummeringspribleem van S. Dockx, Mathematisch Centrum. Amsterdam," Rapport ZW; 1949-005, 4 leaves, 19.8 X 34 cm.
  • 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).

Crossrefs

Cf. A033586 (King), A035005 (Queen), A035006 (Rook), A035008 (Knight) and A049450 (Pawn).

Programs

  • Magma
    [2*n*(n+1)*(2*n+1)/3: n in [0..40]]; // Vincenzo Librandi, Jun 16 2011
  • Maple
    A002492:=n->2*n*(n+1)*(2*n+1)/3; seq(A002492(n), n=0..50); # Wesley Ivan Hurt, Apr 04 2014
  • Mathematica
    Table[2n(n+1)(2n+1)/3, {n,0,40}] (* or *) Binomial[2*Range[0,40]+2,3] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,4,20,56},40] (* Harvey P. Dale, Aug 15 2012 *)
    Accumulate[(2*Range[0,40])^2] (* Harvey P. Dale, Jun 04 2019 *)
  • PARI
    a(n)=2*n*(n+1)*(2*n+1)/3
    

Formula

G.f.: 4*x*(1+x)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
a(-1-n) = -a(n).
a(n) = 4*A000330(n) = 2*A006331(n) = A000292(2*n).
a(n) = (-1)^(n+1)*A053120(2*n+1,3) (fourth unsigned column of Chebyshev T-triangle, zeros omitted).
a(n) = binomial(2*n+2, 3). - Lekraj Beedassy, Jun 19 2004
A035005(n+1) = a(n) + A035006(n+1) since Queen = Bishop + Rook. - Johannes W. Meijer, Feb 04 2010
a(n) - a(n-1) = 4*n^2. - Joerg Arndt, Jun 16 2011
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3. - Harvey P. Dale, Aug 15 2012
a(n) = Sum_{k=0..3} C(n-2+k,n-2)*C(n+3-k,n), for n>2. - J. M. Bergot, Jun 14 2014
a(n) = 2*A006331(n). - R. J. Mathar, May 28 2016
From Natan Arie Consigli Jul 03 2016: (Start)
a(n) = A166464(n) - 1.
a(n) = A168380(2*n). (End)
a(n) = Sum_{i=0..n} A005408(i)*A005408(i-1)+1 with A005408(-1):=-1. - Bruno Berselli, Jan 09 2017
a(n) = A002412(n) + A016061(n). - Bruce J. Nicholson, Nov 12 2017
From Amiram Eldar, Jan 04 2022: (Start)
Sum_{n>=1} 1/a(n) = 9/2 - 6*log(2).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*Pi/2 - 9/2. (End)
a(n) = A081277(3, n-1) = (1+2*n)*binomial(n+1, n-2)*2^2/(n-1) for n > 0. - Mathias Zechmeister, Jul 26 2022
E.g.f.: 2*exp(x)*x*(6 + 9*x + 2*x^2)/3. - Stefano Spezia, Jul 31 2022

Extensions

Minor errors corrected and edited by Johannes W. Meijer, Feb 04 2010
Title modified by Charles R Greathouse IV at the suggestion of J. M. Bergot, Apr 05 2014

A051870 18-gonal (or octadecagonal) numbers: a(n) = n*(8*n-7).

Original entry on oeis.org

0, 1, 18, 51, 100, 165, 246, 343, 456, 585, 730, 891, 1068, 1261, 1470, 1695, 1936, 2193, 2466, 2755, 3060, 3381, 3718, 4071, 4440, 4825, 5226, 5643, 6076, 6525, 6990, 7471, 7968, 8481, 9010, 9555, 10116, 10693, 11286, 11895, 12520
Offset: 0

Views

Author

N. J. A. Sloane, Dec 15 1999

Keywords

Comments

Also, sequence found by reading the segment (0, 1) together with the line from 1, in the direction 1, 18, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Apr 26 2008
This sequence does not contain any triangular numbers other than 0 and 1. See A188892. - T. D. Noe, Apr 13 2011
Also sequence found by reading the line from 0, in the direction 0, 18, ... and the parallel line from 1, in the direction 1, 51, ..., in the square spiral whose vertices are the generalized 18-gonal numbers. - Omar E. Pol, Jul 18 2012
Partial sums of 16n + 1 (with offset 0), compare A005570. - Jeremy Gardiner, Aug 04 2012
All x values for Diophantine equation 32*x + 49 = y^2 are given by this sequence and A139278. - Bruno Berselli, Nov 11 2014
This is also a star enneagonal number: a(n) = A001106(n) + 9*A000217(n-1). - Luciano Ancora, Mar 30 2015

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 189.
  • Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing, 2012, page 6.

Crossrefs

Programs

Formula

G.f.: x*(1+15*x)/(1-x)^3. - Bruno Berselli, Feb 04 2011
a(n) = 16*n + a(n-1) - 15, with n > 0, a(0) = 0. - Vincenzo Librandi, Aug 06 2010
a(16*a(n)+121*n+1) = a(16*a(n)+121*n) + a(16*n+1). - Vladimir Shevelev, Jan 24 2014
E.g.f.: (8*x^2 + x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = ((1+sqrt(2))*Pi + 2*sqrt(2)*arccoth(sqrt(2)) + 8*log(2))/14. - Amiram Eldar, Oct 20 2020
Product_{n>=2} (1 - 1/a(n)) = 8/9. - Amiram Eldar, Jan 22 2021

A139275 a(n) = n*(8*n+1).

Original entry on oeis.org

0, 9, 34, 75, 132, 205, 294, 399, 520, 657, 810, 979, 1164, 1365, 1582, 1815, 2064, 2329, 2610, 2907, 3220, 3549, 3894, 4255, 4632, 5025, 5434, 5859, 6300, 6757, 7230, 7719, 8224, 8745, 9282, 9835, 10404, 10989, 11590, 12207, 12840
Offset: 0

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 9,..., in the square spiral whose vertices are the triangular numbers A000217.

Crossrefs

Programs

  • Mathematica
    Table[n (8 n + 1), {n, 0, 40}] (* Bruno Berselli, Sep 21 2016 *)
    LinearRecurrence[{3,-3,1},{0,9,34},50] (* Harvey P. Dale, Apr 21 2020 *)
  • PARI
    a(n) = n*(8*n+1); \\ Altug Alkan, Sep 21 2016

Formula

a(n) = 8*n^2 + n.
Sequences of the form a(n) = 8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 7 with n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
a(n) = A000217(5*n) - A000217(3*n). - Bruno Berselli, Sep 21 2016
Sum_{n>=1} 1/a(n) = 8 - (1+sqrt(2))*Pi/2 - 4*log(2) - sqrt(2) * log(1+sqrt(2)) = 0.1887230016056779928... . - Vaclav Kotesovec, Sep 21 2016
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(7*x + 9)/(1-x)^3.
E.g.f.: (8*x^2 + 9*x)*exp(x). (End)

A139271 a(n) = 2*n*(4*n-3).

Original entry on oeis.org

0, 2, 20, 54, 104, 170, 252, 350, 464, 594, 740, 902, 1080, 1274, 1484, 1710, 1952, 2210, 2484, 2774, 3080, 3402, 3740, 4094, 4464, 4850, 5252, 5670, 6104, 6554, 7020, 7502, 8000, 8514, 9044, 9590, 10152, 10730, 11324, 11934, 12560, 13202, 13860, 14534, 15224
Offset: 0

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 2, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A033585 in the same spiral.
Twice decagonal numbers (or twice 10-gonal numbers). - Omar E. Pol, May 15 2008
a(n) is the number of walks in a cubic lattice of n dimensions that reach the point of origin for the first time after 4 steps. - Shel Kaphan, Mar 20 2023

Crossrefs

Cf. A001107.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=16). - Bruno Berselli, Jun 10 2013
Row n=2 of A361397.

Programs

Formula

a(n) = 8*n^2 - 6*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = A001107(n)*2. - Omar E. Pol, May 15 2008
a(n) = 16*n + a(n-1) - 14 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: (2*x)*(7*x+1)/(1-x)^3.
E.g.f.: (8*x^2 + 2*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = Pi/12 + log(2)/2. - Amiram Eldar, Mar 28 2023

Extensions

Corrected by Harvey P. Dale, Sep 26 2016

A139273 a(n) = n*(8*n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680
Offset: 0

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

Crossrefs

Programs

  • Magma
    [ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011
    
  • Mathematica
    Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)
  • PARI
    a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = 8*n^2 - 3*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

A139278 a(n) = n*(8*n+7).

Original entry on oeis.org

0, 15, 46, 93, 156, 235, 330, 441, 568, 711, 870, 1045, 1236, 1443, 1666, 1905, 2160, 2431, 2718, 3021, 3340, 3675, 4026, 4393, 4776, 5175, 5590, 6021, 6468, 6931, 7410, 7905, 8416, 8943, 9486, 10045, 10620, 11211, 11818, 12441, 13080
Offset: 0

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

Sequence found by reading the segment (0, 15) together with the line from 15, in the direction 15, 46, ..., in the square spiral whose vertices are the triangular numbers A000217.

Crossrefs

Programs

  • Mathematica
    Table[n (8 n + 7), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 15, 46}, 50] (* Harvey P. Dale, Oct 07 2015 *)
  • PARI
    a(n)=n*(8*n+7) \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 8*n^2 + 7*n.
Sequences of the form a(n)=8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n)= 3a(n-1)-3a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n+a(n-1)-1 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(x+15)/(1-x)^3.
E.g.f.: (8*x^2 + 15*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/49 + (sqrt(2)+1)*Pi/14 - 4*log(2)/7 - sqrt(2)*log(sqrt(2)+1)/7. - Amiram Eldar, Mar 17 2022
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