

A014105


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


192



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
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OFFSET

0,2


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) = 41, 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 3subset of a (2n+1)set X then a(n) is the number of (2n1)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 nth kgonal number. Then for k > 1, partial sums of {P(k,n)} with alternating signs, ending in an even term, = n*((k2)*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


REFERENCES

Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 7778. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)


LINKS



FORMULA

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
G.f.: x*(3 + x)/(1  x)^3.
E.g.f.: exp(x)*(3*x + 2*x^2).
a(n) = Sum_{k=0..2} C(n2+k, n2) * C(n+2k, 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(n1)  3*a(n2) + a(n3) for n >= 3, with a(0) = 0, a(1) = 3, and a(2) = 10.  Harvey P. Dale, Feb 10 2015
a(n) = trinomial(2*n, 2) = trinomial(2*n, 2*(2*n1)), 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*(n1), 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)
Sum_{n>=1} (1)^(n+1)/a(n) = Pi/2 + log(2)  2.  Amiram Eldar, Nov 28 2021


EXAMPLE

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


MAPLE



MATHEMATICA

LinearRecurrence[{3, 3, 1}, {0, 3, 10}, 50] (* Harvey P. Dale, Feb 10 2015 *)
CoefficientList[Series[x*(3 + x)/(1  x)^3, {x, 0, 50}], x] (* Stefano Spezia, Sep 02 2018 *)


PROG

(PARI) a(n)=n*(2*n+1)
(Haskell)
a014105 n = n * (2 * n + 1)
(Sage) [n*(2*n+1) for n in range(50)] # G. C. Greubel, Dec 16 2018


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



