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A014105 Second hexagonal numbers: a(n) = n*(2*n + 1). 170

%I

%S 0,3,10,21,36,55,78,105,136,171,210,253,300,351,406,465,528,595,666,

%T 741,820,903,990,1081,1176,1275,1378,1485,1596,1711,1830,1953,2080,

%U 2211,2346,2485,2628,2775,2926,3081,3240,3403,3570,3741,3916,4095,4278

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

%C 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

%C 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

%C Partial sums of odd numbers 3 mod 4, that is, 3, 3+7, 3+7+11, ... See A001107. - _Jon Perry_, Dec 18 2004

%C 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

%C 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

%C 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

%C a(n) is the number of positions of a domino in a pyramidal board with base 2n+1. - _César Eliud Lozada_, Sep 26 2012

%C Differences of row sums of two consecutive rows of triangle A120070, i.e., first differences of A016061. - _J. M. Bergot_, Jun 14 2013

%C a(n)*Pi is the total length of half circle spiral after n rotations. See illustration in links. - _Kival Ngaokrajang_, Nov 05 2013

%C For corresponding sums in first comment by Henry Bottomley, see A059255. - _Zak Seidov_, Sep 10 2015

%C 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

%C 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

%C 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

%C 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

%C 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

%C 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

%C 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

%C 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

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

%H Reinhard Zumkeller, <a href="/A014105/b014105.txt">Table of n, a(n) for n = 0..10000</a>

%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]

%H Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.

%H Milan Janjic, <a href="https://pmf.unibl.org/janjic/">Two Enumerative Functions</a>, University of Banja Luka (Bosnia and Herzegovina, 2017).

%H Ângela Mestre and José Agapito, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Mestre/mestre2.html">Square Matrices Generated by Sequences of Riordan Arrays</a>, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.

%H Kival Ngaokrajang, <a href="/A014105/a014105.pdf">Illustration of half circle spiral</a>.

%H Amelia Carolina Sparavigna, <a href="https://doi.org/10.5281/zenodo.3471358">The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences</a>, Politecnico di Torino, Italy (2019).

%H Amelia Carolina Sparavigna, <a href="https://doi.org/10.5281/zenodo.3470205">The groupoid of the Triangular Numbers and the generation of related integer sequences</a>, Politecnico di Torino, Italy (2019).

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = 3*Sum_{k=1..n} tan^2(k*Pi/(2*(n + 1))). - _Ignacio Larrosa Cañestro_, Apr 17 2001

%F 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

%F From _N. J. A. Sloane_, Sep 13 2003: (Start)

%F G.f.: x*(3 + x)/(1 - x)^3.

%F E.g.f.: exp(x)*(3*x + 2*x^2).

%F a(n) = A000217(2*n) = A000384(-n). (End)

%F a(n) = A084849(n) - 1; A100035(a(n) + 1) = 1. - _Reinhard Zumkeller_, Oct 31 2004

%F a(n) = A126890(n, k) + A126890(n, n-k), 0 <= k <= n. - _Reinhard Zumkeller_, Dec 30 2006

%F a(2*n) = A033585(n); a(3*n) = A144314(n). - _Reinhard Zumkeller_, Sep 17 2008

%F a(n) = a(n-1) + 4*n - 1 (with a(0) = 0). - _Vincenzo Librandi_, Dec 24 2010

%F a(n) = Sum_{k=0.2*n} (-1)^k*k^2. - _Bruno Berselli_, Aug 29 2013

%F a(n) = A242342(2*n + 1). - _Reinhard Zumkeller_, May 11 2014

%F 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

%F 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

%F 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

%F Sum_{n >= 1} 1/a(n) = 2*(1 - log(2)) = 0.61370563888010938... . - _Vaclav Kotesovec_, Apr 27 2016

%F From _Wolfdieter Lang_, Apr 27 2018: (Start)

%F 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).

%F 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)

%F a(n) = A002943(n)/2. - _Ralf Steiner_, Jul 23 2019

%F a(n) = A000290(n) + A002378(n). - _Torlach Rush_, Nov 02 2020

%e 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

%p seq(binomial(2*n+1,2), n=0..46); # _Zerinvary Lajos_, Jan 21 2007

%t Table[n*(2*n+1), {n,0,100}] (* _Vladimir Joseph Stephan Orlovsky_, Nov 16 2008 *)

%t LinearRecurrence[{3,-3,1},{0,3,10},50] (* _Harvey P. Dale_, Feb 10 2015 *)

%t CoefficientList[Series[x*(3 + x)/(1 - x)^3,{x, 0, 50}], x] (* _Stefano Spezia_, Sep 02 2018 *)

%o (PARI) a(n)=n*(2*n+1)

%o (Haskell)

%o a014105 n = n * (2 * n + 1)

%o a014105_list = scanl (+) 0 a004767_list -- _Reinhard Zumkeller_, Oct 03 2012

%o (MAGMA) [ n*(2*n+1) : n in [0..50] ]; // _Wesley Ivan Hurt_, Jun 14 2014

%o (GAP) List([0..50],n->n*(2*n+1)); # _Muniru A Asiru_, Oct 31 2018

%o (Sage) [n*(2*n+1) for n in range(50)] # _G. C. Greubel_, Dec 16 2018

%Y Cf. A000217, A000290, A000384, A002378, A002943, A100040, A100041, A081266, A144312.

%Y Cf. A033567, A033585, A059255, A130757, A132440, A027907.

%Y Second column of array A094416.

%Y Equals A033586(n) divided by 4.

%Y See Comments of A132124.

%Y Second n-gonal numbers: A005449, A147875, A045944, A179986, A033954, A062728, A135705.

%Y Row sums in triangle A253580.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Jun 14 1998

%E Link added and minor errors corrected by _Johannes W. Meijer_, Feb 04 2010

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Last modified April 22 07:26 EDT 2021. Contains 343163 sequences. (Running on oeis4.)