A268684 -id:A268684 - cp's OEIS frontend

A002939 a(n) = 2n(2*n-1).

0, 2, 12, 30, 56, 90, 132, 182, 240, 306, 380, 462, 552, 650, 756, 870, 992, 1122, 1260, 1406, 1560, 1722, 1892, 2070, 2256, 2450, 2652, 2862, 3080, 3306, 3540, 3782, 4032, 4290, 4556, 4830, 5112, 5402, 5700, 6006, 6320, 6642, 6972, 7310, 7656, 8010, 8372

Offset: 0

N. J. A. Sloane

Write 0,1,2,... in a spiral; sequence gives numbers on one of 4 diagonals (see Example section).
For n>1 this is the Engel expansion of cosh(1), A118239. - Benoit Cloitre, Mar 03 2002
a(n) = A125199(n,n) for n>0. - Reinhard Zumkeller, Nov 24 2006
Central terms of the triangle in A195437: a(n+1) = A195437(2*n,n). - Reinhard Zumkeller, Nov 23 2011
For n>2, the terms represent the sums of those primitive Pythagorean triples with hypotenuse (H) one unit longer than the longest side (L), or H = L + 1. - Richard R. Forberg, Jun 09 2015
For n>1, a(n) is the perimeter of a Pythagorean triangle with an odd leg 2*n-1. - Agola Kisira Odero, Apr 26 2016
From Rigoberto Florez, Nov 07 2020 : (Start)
A338109(n)/a(n+1) is the Kirchhoff index of the join of the disjoint union of two complete graphs on n vertices with the empty graph on n+1 vertices.
Equivalently, the graph can be described as the graph on 3*n + 1 vertices with labels 0..3*n and with i and j adjacent iff iff  i+j> 0 mod 3.
A338588(n)/a(n+1) is the Kirchhoff index of the disjoint union of two complete graphs each on n and n+1 vertices with the empty graph on n+1 vertices.
Equivalently, the graph can be described as the graph on 3*n + 2 vertices with labels 0..3*n+1 and with i and j adjacent iff  i+j> 0 mod 3.
These graphs are cographs. (End)
a(n), n>=1, is the number of paths of minimum length (length=2) from the origin to the cross polytope of size 2 in Z^n (column 2 in A371064). - Shel Kaphan, Mar 09 2024

			G.f. = 2*x + 12*x^2 + 30*x^3 + 56*x^4 + 90*x^5 + 132*x^6 + 182*x^7 + 240*x^8 + ...
On a square lattice, place the nonnegative integers at lattice points forming a spiral as follows: place "0" at the origin; then move one step in any of the four cardinal directions and place "1" at the lattice point reached; then turn 90 degrees in either direction and place a "2" at the next lattice point; then make another 90-degree turn in the same direction and place a "3" at the lattice point; etc. The terms of the sequence will lie along one of the diagonals, as seen in the example below:
.
   99  64--65--66--67--68--69--70--71--72
    |   |                               |
   98  63  36--37--38--39--40--41--42  73
    |   |   |                       |   |
   97  62  35  16--17--18--19--20  43  74
    |   |   |   |               |   |   |
   96  61  34  15   4---5---6  21  44  75
    |   |   |   |   |       |   |   |   |
   95  60  33  14   3  *0*  7  22  45  76
    |   |   |   |   |   |   |   |   |   |
   94  59  32  13  *2*--1   8  23  46  77
    |   |   |   |           |   |   |   |
   93  58  31 *12*-11--10---9  24  47  78
    |   |   |                   |   |   |
   92  57 *30*-29--28--27--26--25  48  79
    |   |                           |   |
   91 *56*-55--54--53--52--51--50--49  80
    |                                   |
  *90*-89--88--87--86--85--84--83--82--81
.
[Edited by _Jon E. Schoenfield_, Jan 01 2017]

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
H-Y. Ching, R. Florez, and A. Mukherjee, Families of Integral Cographs within a Triangular Arrays, arXiv:2009.02770 [math.CO], 2020.
A. M. Nemirovsky et al., Marriage of exact enumeration and 1/d expansion methods: lattice model of dilute polymers, J. Statist. Phys., 67 (1992), 1083-1108.
R. Tijdeman, Some applications of Diophantine approximation, pp. 261-284 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003.
Eric Weisstein's World of Mathematics, Kirchhoff Index
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Sequences from spirals: A001107, A002939, A007742, A033951-A033953, A033954, A033989-A033991, A002943, A033996, A033988.
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. A016789, A017041, A017485, A125202.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=8). - Bruno Berselli, Jun 10 2013
Cf. A017089 (first differences), A268684 (partial sums), A010050 (partial products).
Cf. A371064.

a002939 n = (* 2) . a000384
a002939_list = scanl1 (+) a017089_list
-- Reinhard Zumkeller, Jun 08 2015

[2*n*(2*n-1): n in [0..50]]; // Vincenzo Librandi, Jul 26 2011

A002939:=n->2*n*(2*n-1): seq(A002939(n), n=0..100); # Wesley Ivan Hurt, Jan 28 2017

Table[2*n*(2*n-1), {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2008 *)
2#(2#-1)&/@Range[0,50]  (* Harvey P. Dale, Mar 06 2011 *)

a(n)=2*binomial(2*n,2) \\ Charles R Greathouse IV, Jul 25 2011

a=lambda n: 2*n*(2*n-1) # Indranil Ghosh, Jan 01 2017

Sum_{n >= 1} 1/a(n) = log(2) (cf. Tijdeman).
Log(2) = Sum_{n >= 1} ((1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + ...) = Sum_{n >= 0} (-1)^n/(n+1). Log(2) = Integral_{x=0..1} 1/(1+x) dx. - Gary W. Adamson, Jun 22 2003
a(n) = A000384(n)*2. - Omar E. Pol, May 14 2008
From R. J. Mathar, Apr 23 2009: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 2*x*(1+3*x)/(1-x)^3. (End)
a(n) = a(n-1) + 8*n - 6 (with a(0)=0). - Vincenzo Librandi, Nov 12 2010
a(n) = A118729(8n+1). - Philippe Deléham, Mar 26 2013
Product_{k=1..n} a(k) = (2n)! = A010050(n). - Tony Foster III, Sep 06 2015
E.g.f.: 2*x*(1 + 2*x)*exp(x). - Ilya Gutkovskiy, Apr 29 2016
a(n) = A002943(-n) for all n in Z. - Michael Somos, Jan 28 2017
0 = 12 + a(n)*(-8 + a(n) - 2*a(n+1)) + a(n+1)*(-8 + a(n+1)) for all n in Z. - Michael Somos, Jan 28 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/4 - log(2)/2. - Amiram Eldar, Jul 31 2020

A333446 Table T(n,k) read by upward antidiagonals. T(n,k) = Sum_{i=1..n} Product_{j=1..k} (i-1)*k+j.

Original entry on oeis.org

1, 3, 2, 6, 14, 6, 10, 44, 126, 24, 15, 100, 630, 1704, 120, 21, 190, 1950, 13584, 30360, 720, 28, 322, 4680, 57264, 390720, 666000, 5040, 36, 504, 9576, 173544, 2251200, 14032080, 17302320, 40320, 45, 744, 17556, 428568, 8626800, 110941200, 603353520, 518958720, 362880

Offset: 1

Chai Wah Wu, Mar 23 2020

T(n,k) is the maximum value of Sum_{i=1..n} Product_{j=1..k} r[(i-1)*k+j] among all permutations r of {1..kn}. For the minimum value see A331889.

			From _Seiichi Manyama_, Jul 23 2020: (Start)
T(3,2) = Sum_{i=1..3} Product_{j=1..2} (i-1)*2+j = 1*2 + 3*4 + 5*6 = 44.
Square array begins:
   1,   2,    6,     24,      120,        720, ...
   3,  14,  126,   1704,    30360,     666000, ...
   6,  44,  630,  13584,   390720,   14032080, ...
  10, 100, 1950,  57264,  2251200,  110941200, ...
  15, 190, 4680, 173544,  8626800,  538459200, ...
  21, 322, 9576, 428568, 25727520, 1940869440, ... (End)

Seiichi Manyama, Antidiagonals n = 1..140, flattened
Chai Wah Wu, On rearrangement inequalities for multiple sequences, arXiv:2002.10514 [math.CO], 2020.

Column k=1-3 give A000217, A268684, A268685(n-1).
Main diagonal gives A336513.
Cf. A323663, A331889,

def T(n,k): # T(n,k) for A333446
    c, l = 0, list(range(1,k*n+1,k))
    lt = list(l)
    for i in range(n):
        for j in range(1,k):
            lt[i] *= l[i]+j
        c += lt[i]
    return c

T(n,k) = Sum_{i=1..n} Gamma(ik+1)/Gamma((i-1)k+1).

A300254 a(n) = 25(n + 1)(4n + 3)(5*n + 4)/3.

Original entry on oeis.org

100, 1050, 3850, 9500, 19000, 33350, 53550, 80600, 115500, 159250, 212850, 277300, 353600, 442750, 545750, 663600, 797300, 947850, 1116250, 1303500, 1510600, 1738550, 1988350, 2261000, 2557500, 2878850, 3226050, 3600100, 4002000, 4432750, 4893350, 5384800, 5908100, 6464250

Offset: 0

Bruno Berselli, Mar 12 2018

Hirschhorn has discovered that p(20*n+11,4) + p(20*n+12,4) + p(20*n+13,4) = 25*(n + 1)*(4*n + 3)*(5*n + 4)/3, where p(m,k) denote the number of partitions of m into at most k parts. Therefore, p(20*n+11,4) + p(20*n+12,4) + p(20*n+13,4) == 0 (mod 50) [see Hirschhorn's paper in References section].
a(n) == 0 (mod 3) if n is of the form 2*h + 3*floor(h/3 + 2/3) + 1.
a(n) == 0 (mod 7) if n is a member of A047278.

Michael D. Hirschhorn, Congruences modulo 5 for partitions into at most four parts, The Fibonacci Quarterly, Vol. 56, Number 1, 2018, pages 32-37 [the equation 1.7 contains a typo].

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Subsequence of A014112, A212964, A228958, A268684.

Cf. A300522, A300523.

List([0..40], n -> 25*(n+1)*(4*n+3)*(5*n+4)/3);

[div(25*(n+1)*(4*n+3)*(5*n+4), 3) for n in 0:40] |> println

[25*(n+1)*(4*n+3)*(5*n+4)/3: n in [0..40]];

Table[25 (n + 1) (4 n + 3) (5 n + 4)/3, {n, 0, 40}]

makelist(25*(n+1)*(4*n+3)*(5*n+4)/3, n, 0, 40);

vector(40, n, n--; 25*(n+1)*(4*n+3)*(5*n+4)/3)

Vec(50*(2 + 13*x + 5*x^2) / (1 - x)^4 + O(x^60)) \\ Colin Barker, Mar 13 2018

[25*(n+1)*(4*n+3)*(5*n+4)/3 for n in range(40)]

[25*(n+1)*(4*n+3)*(5*n+4)/3 for n in (0..40)]

O.g.f.: 50*(2 + 13*x + 5*x^2)/(1 - x)^4 [formula 4.3 in Hirschhorn's paper].
E.g.f.: 25*(12 + 114*x + 111*x^2 + 20*x^3)*exp(x)/3.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4)
a(n) = A014112(10*n+8) = A212964(10*n+9) = A228958(10*n+8) = A268684(5*n+4).

A267522 a(n) = 4(n + 1)(n + 2)(4n + 3)/3.

Original entry on oeis.org

8, 56, 176, 400, 760, 1288, 2016, 2976, 4200, 5720, 7568, 9776, 12376, 15400, 18880, 22848, 27336, 32376, 38000, 44240, 51128, 58696, 66976, 76000, 85800, 96408, 107856, 120176, 133400, 147560, 162688, 178816, 195976, 214200, 233520, 253968, 275576, 298376, 322400

Offset: 0

Ilya Gutkovskiy, Apr 09 2016

Partial sums of A152750.

			a(0) = (0 + 2)*(1 + 3) = 8;
a(1) = (0 + 2)*(1 + 3) + (2 + 4)*(3 + 5) = 56;
a(2) = (0 + 2)*(1 + 3) + (2 + 4)*(3 + 5) + (4 + 6)*(5 + 7) = 176;
a(3) = (0 + 2)*(1 + 3) + (2 + 4)*(3 + 5) + (4 + 6)*(5 + 7) + (6 + 8)*(7 + 9) = 400, etc

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002412, A152750, A268684.

Table[(4 (n + 1)) (n + 2) ((4 n + 3)/3), {n, 0, 38}]
LinearRecurrence[{4, -6, 4, -1}, {8, 56, 176, 400}, 39]

a(n) = 4*(n + 1)*(n + 2)*(4*n + 3)/3; \\ Michel Marcus, Apr 10 2016

x='x+O('x^99); Vec(8*(1+3*x)/(1-x)^4) \\ Altug Alkan, Apr 10 2016

G.f.: 8*(1 + 3*x)/(1 - x)^4.
E.g.f.: (4/3)*exp(x)*(6 + 36*x + 27*x^2 + 4*x^3).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 4*A268684(n + 1).
Sum_{n>=0} 1/a(n) = -3*(2*Pi - 12*log(2) + 1)/20 = 0.15518712893...
a(n) = 8*A002412(n+1). - Yasser Arath Chavez Reyes, Feb 23 2024

cp's OEIS Frontend

A002939 a(n) = 2*n*(2*n-1).

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A333446 Table T(n,k) read by upward antidiagonals. T(n,k) = Sum_{i=1..n} Product_{j=1..k} (i-1)*k+j.

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A300254 a(n) = 25*(n + 1)*(4*n + 3)*(5*n + 4)/3.

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A267522 a(n) = 4*(n + 1)*(n + 2)*(4*n + 3)/3.

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A002939 a(n) = 2n(2*n-1).

A300254 a(n) = 25(n + 1)(4n + 3)(5*n + 4)/3.

A267522 a(n) = 4(n + 1)(n + 2)(4n + 3)/3.