A046092 -id:A046092 - cp's OEIS frontend

A049581 Table T(n,k) = |n-k| read by antidiagonals (n >= 0, k >= 0).

0, 1, 1, 2, 0, 2, 3, 1, 1, 3, 4, 2, 0, 2, 4, 5, 3, 1, 1, 3, 5, 6, 4, 2, 0, 2, 4, 6, 7, 5, 3, 1, 1, 3, 5, 7, 8, 6, 4, 2, 0, 2, 4, 6, 8, 9, 7, 5, 3, 1, 1, 3, 5, 7, 9, 10, 8, 6, 4, 2, 0, 2, 4, 6, 8, 10, 11, 9, 7, 5, 3, 1, 1, 3, 5, 7, 9, 11, 12, 10, 8, 6, 4, 2, 0, 2, 4, 6, 8, 10, 12

Offset: 0

N. J. A. Sloane

Commutative non-associative operator with identity 0. T(nx,kx) = x T(n,k). A multiplicative analog is A089913. - Marc LeBrun, Nov 14 2003
For the characteristic polynomial of the n X n matrix M_n with entries M_n(i, j) = |i-j| see A203993. - Wolfdieter Lang, Feb 04 2018
For the determinant of the n X n matrix M_n with entries M_n(i, j) = |i-j| see A085750. - Bernard Schott, May 13 2020
a(n) = 0 iff n = 4 times triangular number (A046092). - Bernard Schott, May 13 2020

			Displayed as a triangle t(n, k):
  n\k   0 1 2 3 4 5 6 7 8 9 10 ...
  0:    0
  1:    1 1
  2:    2 0 2
  3:    3 1 1 3
  4:    4 2 0 2 4
  5:    5 3 1 1 3 5
  6:    6 4 2 0 2 4 6
  7:    7 5 3 1 1 3 5 7
  8:    8 6 4 2 0 2 4 6 8
  9:    9 7 5 3 1 1 3 5 7 9
  10:  10 8 6 4 2 0 2 4 6 8 10
... reformatted by _Wolfdieter Lang_, Feb 04 2018
Displayed as a table:
  0 1 2 3 4 5 6 ...
  1 0 1 2 3 4 5 ...
  2 1 0 1 2 3 4 ...
  3 2 1 0 1 2 3 ...
  4 3 2 1 0 1 2 ...
  5 4 3 2 1 0 1 ...
  6 5 4 3 2 1 0 ...
  ...

Peter Kagey, Rows n = 0..125 of triangle, flattened

Cf. A003989, A003990, A003991, A003056, A004247, A002260, A004736.

Cf. A089913. Apart from signs, same as A114327. A203993.

a := Flat(List([0..12],n->List([0..n],k->Maximum(k,n-k)-Minimum(k,n-k)))); # Muniru A Asiru, Jan 26 2018

[[Abs(n-2*k): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Jun 07 2019

seq(seq(abs(n-2*k),k=0..n),n=0..12); # Robert Israel, Sep 30 2015

Table[Abs[(n-k) -k], {n,0,12}, {k,0,n}]//Flatten (* Michael De Vlieger, Sep 29 2015 *)
Table[Join[Range[n,0,-2],Range[If[EvenQ[n],2,1],n,2]],{n,0,12}]//Flatten (* Harvey P. Dale, Sep 18 2023 *)

a(n) = abs(2*(n+1)-binomial((sqrtint(8*(n+1))+1)\2, 2)-(binomial(1+floor(1/2 + sqrt(2*(n+1))), 2))-1);
vector(100, n , a(n-1)) \\ Altug Alkan, Sep 29 2015

{t(n,k) = abs(n-2*k)}; \\ G. C. Greubel, Jun 07 2019

from math import isqrt
def A049581(n): return abs((k:=n+1<<1)-((m:=isqrt(k))+(k>m*(m+1)))**2-1) # Chai Wah Wu, Nov 09 2024

[[abs(n-2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jun 07 2019

G.f.: (x + y - 4*x*y + x^2*y + x*y^2)/((1-x)^2*(1-y)^2*(1-x*y)) = (x/(1-x)^2 + y/(1-y)^2)/(1-x*y). T(n,0) = T(0,n) = n; T(n+1,k+1) = T(n,k). - Franklin T. Adams-Watters, Feb 06 2006
a(n) = |A002260(n+1)-A004736(n+1)| or a(n) = |((n+1)-t*(t+1)/2) - ((t*t+3*t+4)/2-(n+1))| where t = floor((-1+sqrt(8*(n+1)-7))/2). - Boris Putievskiy, Dec 24 2012; corrected by Altug Alkan, Sep 30 2015
From Robert Israel, Sep 30 2015: (Start)
If b(n) = a(n+1) - 2*a(n) + a(n-1), then for n >= 3 we have
  b(n) = -1 if n = (j^2+5j+4)/2 for some integer j >= 1
  b(n) = -3 if n = (j^2+5j+6)/2 for some integer j >= 0
  b(n) = 4 if n = 2j^2 + 6j + 4 for some integer j >= 0
  b(n) = 2 if n = 2j^2 + 8j + 7 or 2j^2 + 8j + 8 for some integer j >= 0
  b(n) = 0 otherwise. (End)
Triangle t(n,k) = max(k, n-k) - min(k, n-k). - Peter Luschny, Jan 26 2018
Triangle t(n, k) = |n - 2*k| for n >= 0, k = 0..n. See the Maple and Mathematica programs. Hence t(n, k)= t(n, n-k). - Wolfdieter Lang, Feb 04 2018
a(n) = |t^2 - 2*n - 1|, where t = floor(sqrt(2*n+1) + 1/2). - Ridouane Oudra, Jun 07 2019; Dec 11 2020
As a rectangle, T(n,k) = |n-k| = max(n,k) - min(n,k). - Clark Kimberling, May 11 2020

A147973 a(n) = -2n^2 + 12n - 14.

Original entry on oeis.org

-4, 2, 4, 2, -4, -14, -28, -46, -68, -94, -124, -158, -196, -238, -284, -334, -388, -446, -508, -574, -644, -718, -796, -878, -964, -1054, -1148, -1246, -1348, -1454, -1564, -1678, -1796, -1918, -2044, -2174, -2308, -2446, -2588, -2734, -2884, -3038, -3196, -3358

Offset: 1

Vladimir Joseph Stephan Orlovsky, Nov 18 2008

-a(n+3) = 2*n^2 - 4, n >= 0, [-4,-2, 4, 14, ...] appears as the first member of the quartet for the square of [n, n+1, n+2, n+3], for n >= 0, in the Clifford algebra Cl_2. The other members are given in A046092(n), A054000(n+1) and A139570(n). The basis of Cl_2 is <1, s1, s2, s12> with s1.s1 = s2.s2 = 1, s12.s12 = -1, s1.s2 = -s2.s1 = s12. See e.g., pp. 5-6, eqs. (2.4)-(2.13) of the S. Gull et al. reference. - Wolfdieter Lang, Oct 15 2014
Related to the previous comment: if one uses the exterior (Grassmann) product with s1.s1 = s2.s2 = s12.s12 = 0 and s1.s2 = -s2.s1 = s12, then the four components of the square of [n, n+1, n+2, n+3] are [A000290(n), A046092(n), A054000(n+1), A139570(n)], n >= 0. - Wolfdieter Lang, Nov 13 2014
2 - a(n)/2 is a square. - Bruno Berselli, Apr 10 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
S. Gull, A. Lasenby and C. Doran, Imaginary Numbers are not Real - the Geometric Algebra of Spacetime, Found. Phys., Vol. 23(9) (1993), pp. 1175-1201.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A008865, A046092, A054000, A139570.

[-2*n^2+12*n-14: n in [1..50]]; // Vincenzo Librandi, Jul 10 2012

[-2*n^2+12*n-14$n=1..50]; # Muniru A Asiru, Feb 12 2019

lst={};Do[k=n^2-((n-1)^2+(n-2)^2+(n-3)^2);AppendTo[lst,k],{n,5!}];lst
Table[-2n^2+12n-14,{n,1,50}] (* Vincenzo Librandi, Jul 10 2012 *)
LinearRecurrence[{3,-3,1},{-4,2,4},50] (* Harvey P. Dale, Mar 02 2020 *)

a(n)=-2*n^2+12*n-14 \\ Charles R Greathouse IV, Sep 24 2015

Vec(-2*x*(2 - 7*x + 7*x^2) / (1 - x)^3 + O(x^40)) \\ Colin Barker, Feb 12 2019

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jul 10 2012
a(n) = -2*A008865(n-3). - J. M. Bergot, Jun 25 2018
G.f.: -2*x*(2 - 7*x + 7*x^2)/(1 - x)^3. - Colin Barker, Feb 12 2019
E.g.f.: -2*(exp(x)*(x^2 - 5*x + 7) - 7). - Elmo R. Oliveira, Nov 17 2024

A059056 Penrice Christmas gift numbers, Card-matching numbers (Dinner-Diner matching numbers): Triangle T(n,k) = number of ways to get k matches for a deck with n cards, 2 of each kind.

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 4, 0, 1, 10, 24, 27, 16, 12, 0, 1, 297, 672, 736, 480, 246, 64, 24, 0, 1, 13756, 30480, 32365, 21760, 10300, 3568, 970, 160, 40, 0, 1, 925705, 2016480, 2116836, 1418720, 677655, 243360, 67920, 14688, 2655, 320, 60, 0, 1

Offset: 0

Barbara Haas Margolius (margolius(AT)math.csuohio.edu)

This is a triangle of card matching numbers. A deck has n kinds of cards, 2 of each kind. The deck is shuffled and dealt in to n hands with 2 cards each. A match occurs for every card in the j-th hand of kind j. Triangle T(n,k) is the number of ways of achieving exactly k matches (k=0..2n). The probability of exactly k matches is T(n,k)/((2n)!/2^n).
Rows are of length 1,3,5,7,... = A005408(n). [Edited by M. F. Hasler, Sep 21 2015]
Analogous to A008290. - Zerinvary Lajos, Jun 10 2005

			There are 4 ways of matching exactly 2 cards when there are 2 different kinds of cards, 2 of each in each of the two decks so T(2,2)=4.
Triangle begins:
1
"0", 0, 1
1, '0', "4", 0, 1
10, 24, 27, '16', "12", 0, 1
297, 672, 736, 480, 246, '64', "24", 0, 1
13756, 30480, 32365, 21760, 10300, 3568, 970, '160', "40", 0, 1
925705, 2016480, 2116836, 1418720, 677655, 243360, 67920, 14688, 2655, '320', "60", 0, 1
Diagonal " ": T(n,2n-2) = 0, 4, 12, 24, 40, 60, 84, 112, 144, ... equals A046092
Diagonal ' ': T(n,2n-3) = 0, 16, 64, 160, 320, 560, 896, 1344, ... equals A102860

F. N. David and D. E. Barton, Combinatorial Chance, Hafner, NY, 1962, Ch. 7 and Ch. 12.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 174-178.
R. P. Stanley, Enumerative Combinatorics Volume I, Cambridge University Press, 1997, p. 71.

F. F. Knudsen and I. Skau, On the Asymptotic Solution of a Card-Matching Problem, Mathematics Magazine 69 (1996), 190-197.
Barbara H. Margolius, Dinner-Diner Matching Probabilities
B. H. Margolius, The Dinner-Diner Matching Problem, Mathematics Magazine, 76 (2003), 107-118.
S. G. Penrice, Derangements, permanents and Christmas presents, The American Mathematical Monthly 98(1991), 617-620.
Index entries for sequences related to card matching

Cf. A059056-A059071, A008290.

p := (x,k)->k!^2*sum(x^j/((k-j)!^2*j!),j=0..k); R := (x,n,k)->p(x,k)^n; f := (t,n,k)->sum(coeff(R(x,n,k),x,j)*(t-1)^j*(n*k-j)!,j=0..n*k);
for n from 0 to 7 do seq(coeff(f(t,n,2),t,m)/2^n,m=0..2*n); od;

p[x_, k_] := k!^2*Sum[ x^j/((k-j)!^2*j!), {j, 0, k}];
R[x_, n_, k_] := p[x, k]^n;
f[t_, n_, k_] := Sum[ Coefficient[ R[x, n, k], x, j]*(t-1)^j*(n*k-j)!, {j, 0, n*k}];
Table[ Coefficient[ f[t, n, 2]/2^n, t, m], {n, 0, 6}, {m, 0, 2*n}] // Flatten
(* Jean-François Alcover, Sep 17 2012, translated from Maple *)

G.f.: sum(coeff(R(x, n, k), x, j)*(t-1)^j*(n*k-j)!, j=0..n*k) where n is the number of kinds of cards, k is the number of cards of each kind (here k is 2) and R(x, n, k) is the rook polynomial given by R(x, n, k)=(k!^2*sum(x^j/((k-j)!^2*j!))^n (see Stanley or Riordan). coeff(R(x, n, k), x, j) indicates the j-th coefficient on x of the rook polynomial.

Additional comments from Zerinvary Lajos, Jun 18 2007

Edited by M. F. Hasler, Sep 21 2015

A085250 4 times hexagonal numbers: a(n) = 4n(2*n-1).

Original entry on oeis.org

0, 4, 24, 60, 112, 180, 264, 364, 480, 612, 760, 924, 1104, 1300, 1512, 1740, 1984, 2244, 2520, 2812, 3120, 3444, 3784, 4140, 4512, 4900, 5304, 5724, 6160, 6612, 7080, 7564, 8064, 8580, 9112, 9660, 10224, 10804, 11400, 12012, 12640, 13284

Offset: 0

Gary W. Adamson, Jun 23 2003

a(n) also can represented as n concentric squares (see example). - Omar E. Pol, Aug 21 2011

Sequence found by reading the line from 0, in the direction 0, 4, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 03 2011

			From _Omar E. Pol_, Aug 21 2011: (Start)
Illustration of initial terms as concentric squares:
.
.                           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
.
.      4          24                 60
.
(End)

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A067239, A000384, A033581, A046092, A152734, A152751, A194274.

Table[8*n^2 - 4*n, {n, 0, 50}] (* G. C. Greubel, Jul 14 2017 *)
4 PolygonalNumber[6,Range[0,50]] (* Harvey P. Dale, Oct 19 2022 *)

a(n)=4*n*(2*n-1) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = A067239(n)/2, for n>0.
Sum_{n>0} 1/a(n) = log(2)/2.
a(n) = A000384(n)*4. - Omar E. Pol, Dec 11 2008
a(n) = 16*n + a(n-1) - 12 (with a(0)=0). - Vincenzo Librandi, Aug 08 2010
G.f.: 4*x*(1 + 3*x)/(1 - 3*x + 3*x^2 - x^3). - Colin Barker, Jan 04 2012
E.g.f.: 4*x*(2*x + 1)*exp(x). - G. C. Greubel, Jul 14 2017
a(n) = A046092(2n-1), for n > 0. - Bruce J. Nicholson, Sep 04 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 - log(2)/4. - Amiram Eldar, Mar 17 2022

Edited by Don Reble, Nov 13 2005

Added zero, better definition, corrected offset and edited original formula. - Omar E. Pol, Dec 11 2008

A097080 a(n) = 2n^2 - 2n + 3.

Original entry on oeis.org

3, 7, 15, 27, 43, 63, 87, 115, 147, 183, 223, 267, 315, 367, 423, 483, 547, 615, 687, 763, 843, 927, 1015, 1107, 1203, 1303, 1407, 1515, 1627, 1743, 1863, 1987, 2115, 2247, 2383, 2523, 2667, 2815, 2967, 3123, 3283, 3447, 3615, 3787, 3963, 4143, 4327, 4515, 4707

Offset: 1

N. J. A. Sloane, Nov 02 2008

The rational numbers may be totally ordered, first by height (see A002246) and then by magnitude; every rational number of height n appears in this ordering at a position <= a(n).
This ordering of the rationals is given in A113136/A113137.
The old entry with this sequence number was a duplicate of A027356.
This is also the sum of the pairwise averages of five consecutive triangular numbers in A000217. Start with A000217(0): (0+1)/2 + (1+3)/2 + (3+6)/2 + (6+10)/2 = 15, which is the third term of this sequence. Simply continue to create this sequence. - J. M. Bergot, Jun 13 2012
2*a(n) is inverse radius (curvature) of the touching circle of the large semicircle (radius 1) and the n-th and (n-1)-st circles of the Pappus chain of the symmetric Arbelos. One can use Descartes three circle theorem to find the solution 4*n^2 - 4*n + 6, n >= 1. Note that the circle with curvature 6 is also the third circle of the clockwise Pappus chain, which is A059100(2) (by symmetry). See the illustration. - Wolfdieter Lang and Kival Ngaokrajang, Jul 01 2015
Numbers k such that 2*k - 5 is a square. - Bruno Berselli, Nov 08 2017

M. N. Huxley, Area, Lattice Points and Exponential Sums, Oxford, 1996, p. 7.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Steven Edwards and William Griffiths, Generalizations of Delannoy and cross polytope numbers, Fib. Q., Vol. 55, No. 4 (2017), pp. 356-366.
Steven Edwards and William Griffiths, On Generalized Delannoy Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.3.6.
Kival Ngaokrajang, Illustration of the Pappus chain of the symmetric Arbelos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A001845, A002246, A027356, A027575, A046092, A059100, A113136, A113137.

a097080 n = 2 * n * (n - 1) + 3  -- Reinhard Zumkeller, Dec 15 2013

Table[2n^2-2n+3,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{3,7,15},50] (* Harvey P. Dale, Aug 02 2014 *)
CoefficientList[Series[(3 - 2 x + 3 x^2)/(1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Aug 03 2014 *)

a(n)=2*n^2-2*n+3 \\ Charles R Greathouse IV, Jun 13 2012

Vec(x*(3-2*x+3*x^2)/(1-x)^3 + O(x^50)) \\ Altug Alkan, Nov 11 2015

a(n) = 4*(n-1) + a(n-1) for n > 1, a(1)=3. - Vincenzo Librandi, Nov 16 2010
a(n) = A046092(n) + 3. - Reinhard Zumkeller, Dec 15 2013
G.f.: x*(3 - 2*x + 3*x^2)/(1 - x)^3. - Vincenzo Librandi, Aug 03 2014
a(n) = A027575(n-2)/2. - Michel Marcus, Nov 11 2015
Sum_{n>=1} 1/a(n) = Pi*tanh(sqrt(5)*Pi/2)/(2*sqrt(5)). - Amiram Eldar, Dec 23 2022
From Elmo R. Oliveira, Nov 16 2024: (Start)
E.g.f.: exp(x)*(2*x^2 + 3) - 3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A024966 7 times triangular numbers: 7n(n+1)/2.

Original entry on oeis.org

0, 7, 21, 42, 70, 105, 147, 196, 252, 315, 385, 462, 546, 637, 735, 840, 952, 1071, 1197, 1330, 1470, 1617, 1771, 1932, 2100, 2275, 2457, 2646, 2842, 3045, 3255, 3472, 3696, 3927, 4165, 4410, 4662, 4921, 5187, 5460, 5740, 6027, 6321, 6622

Offset: 0

Joe Keane (jgk(AT)jgk.org), Dec 11 1999

Sequence found by reading the line from 0, in the direction 0, 7, ... and the same line from 0, in the direction 1, 21, ..., in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the main diagonal in the spiral. - Omar E. Pol, Sep 09 2011
Also sequence found by reading the same line mentioned above in the square spiral whose vertices are the generalized enneagonal numbers A118277. Axis perpendicular to A195145 in the same spiral. - Omar E. Pol, Sep 18 2011
Sequence provides all integers m such that 56*m + 49 is a square. - Bruno Berselli, Oct 07 2015
Sum of the numbers from 3*n to 4*n. - Wesley Ivan Hurt, Dec 22 2015

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A001106, A002378, A022264, A022265, A028895, A028896, A033996.
Cf. A045943, A046092, A069099, A118277, A174738, A179986, A186029, A193053.
Cf. A195019, A195020, A195145, A218471.

[ (7*n^2 + 7*n)/2 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

[seq(7*binomial(n,2), n=1..44)]; # Zerinvary Lajos, Nov 24 2006

7 Table[n (n + 1)/2, {n, 0, 43}] (* or *)
Table[Sum[i, {i, 3 n, 4 n}], {n, 0, 43}] (* or *)
Table[SeriesCoefficient[7 x/(1 - x)^3, {x, 0, n}], {n, 0, 43}] (* Michael De Vlieger, Dec 22 2015 *)
7*Accumulate[Range[0,50]] (* or *) LinearRecurrence[{3,-3,1},{0,7,21},50] (* Harvey P. Dale, Jul 20 2025 *)

x='x+O('x^100); concat(0, Vec(7*x/(1-x)^3)) \\ Altug Alkan, Dec 23 2015

a(n) = (7/2)*n*(n+1).
G.f.: 7*x/(1-x)^3.
a(n) = (7*n^2 + 7*n)/2 = 7*A000217(n). - Omar E. Pol, Dec 12 2008
a(n) = a(n-1) + 7*n with n > 0, a(0)=0. - Vincenzo Librandi, Nov 19 2010
a(n) = A069099(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(n) = a(-n-1), a(n+2) = A193053(n+2) + 2*A193053(n+1) + A193053(n). - Bruno Berselli, Oct 21 2011
From Philippe Deléham, Mar 26 2013: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 7, a(2) = 21.
a(n) = A174738(7*n+6).
a(n) = A179986(n) + n = A186029(n) + 2*n = A022265(n) + 3*n = A022264(n) + 4*n = A218471(n) + 5*n = A001106(n) + 6*n. (End)
a(n) = Sum_{i=3*n..4*n} i. - Wesley Ivan Hurt, Dec 22 2015
E.g.f.: (7/2)*x*(x+2)*exp(x). - G. C. Greubel, Aug 19 2017
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/7.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2/7)*(2*log(2) - 1). (End)
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(7/(2*Pi))*cos(sqrt(15/7)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (7/(2*Pi))*cosh(Pi/(2*sqrt(7))). (End)

A059993 Pinwheel numbers: a(n) = 2n^2 + 6n + 1.

Original entry on oeis.org

1, 9, 21, 37, 57, 81, 109, 141, 177, 217, 261, 309, 361, 417, 477, 541, 609, 681, 757, 837, 921, 1009, 1101, 1197, 1297, 1401, 1509, 1621, 1737, 1857, 1981, 2109, 2241, 2377, 2517, 2661, 2809, 2961, 3117, 3277, 3441, 3609, 3781, 3957, 4137, 4321, 4509, 4701, 4897

Offset: 0

Naohiro Nomoto, Mar 14 2001

Nonnegative integers m such that 2*m + 7 is a square. - Vincenzo Librandi, Mar 01 2013
Numbers of the form 4*(h+1)*(2*h-1) + 1, where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ... . - Bruno Berselli, Feb 03 2017
a(n) is also the number of vertices of the Aztec diamond AZ(n) (see Lemma 2.1 of the Imran et al. paper). - Emeric Deutsch, Sep 23 2017

M. Imran and S. Hayat, On computation of topological indices of Aztec diamonds, Sci. Int. (Lahore), Vol. 26(4), 2014, pp. 1407-1412. - Emeric Deutsch, Sep 23 2017

Harry J. Smith, Table of n, a(n) for n = 0..1000
Author?, figure. [Wayback Machine link]
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. numbers n such that 2*n + 2*k + 1 is a square: A046092 (k=0), A142463 (k=1), A090288 (k=2), this sequence (k=3), A139570 (k=4), A222182 (k=5), A181510 (k=6).

[2*n^2+6*n+1: n in [0..50]]; // Vincenzo Librandi, Mar 01 2013

I:=[1,9]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2)+4: n in [1..50]]; // Vincenzo Librandi, Mar 01 2013

Table[2 n^2 + 6 n + 1, {n, 0, 46}] (* Zerinvary Lajos, Jul 10 2009 *)
LinearRecurrence[{3,-3,1},{1,9,21},50] (* Harvey P. Dale, Oct 01 2018 *)

a(n) = { 2*n^2 + 6*n + 1 } \\ Harry J. Smith, Jul 01 2009

a(n) = 4*n + a(n-1) + 4 for n > 0, a(0)=1. - Vincenzo Librandi, Aug 07 2010
G.f.: (1 + 6*x - 3*x^2)/(1-x)^3. - Arkadiusz Wesolowski, Dec 24 2011
a(n) = 2*a(n-1) - a(n-2) + 4. - Vincenzo Librandi, Mar 01 2013
a(n) = Hyper2F1([-2, n], [1], -2). - Peter Luschny, Aug 02 2014
Sum_{n>=0} 1/a(n) = 1/3 + Pi*tan(sqrt(7)*Pi/2)/(2*sqrt(7)). - Amiram Eldar, Dec 13 2022
From Elmo R. Oliveira, Nov 16 2024: (Start)
E.g.f.: exp(x)*(1 + 8*x + 2*x^2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A062717 Numbers m such that 6*m+1 is a perfect square.

Original entry on oeis.org

0, 4, 8, 20, 28, 48, 60, 88, 104, 140, 160, 204, 228, 280, 308, 368, 400, 468, 504, 580, 620, 704, 748, 840, 888, 988, 1040, 1148, 1204, 1320, 1380, 1504, 1568, 1700, 1768, 1908, 1980, 2128, 2204, 2360, 2440, 2604, 2688, 2860, 2948, 3128, 3220, 3408, 3504

Offset: 1

Jason Earls, Jul 14 2001

X values of solutions to the equation 6*X^3 + X^2 = Y^2. - Mohamed Bouhamida, Nov 06 2007
Arithmetic averages of the k triangular numbers 0, 1, 3, 6, ..., (k-1)*k/2 that take integer values. - Vladimir Joseph Stephan Orlovsky, Aug 05 2009 [edited by Jon E. Schoenfield, Jan 10 2015]
Even terms in A186423; union of A033579 and A033580, A010052(6*a(n)+1) = 1. - Reinhard Zumkeller, Feb 21 2011
a(n) are integers produced by Sum_{i = 1..k-1} i*(k-i)/k, for some k > 0. Values for k are given by A007310 = sqrt(6*a(n)+1), the square roots of those perfect squares. - Richard R. Forberg, Feb 16 2015
Equivalently, numbers of the form 2*h*(3*h+1), where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ... (see also the sixth comment of A152749). - Bruno Berselli, Feb 02 2017

Harry J. Smith, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Equals 4 * A001318.
Cf. A005563, A046092, A001082, A002378, A036666.
Cf. A160757, A000217. - Vladimir Joseph Stephan Orlovsky, Aug 05 2009
Cf. A007310.
Diagonal of array A323674. - Sally Myers Moite, Feb 03 2019

[(6*n*(n-1) + (2*n-1)*(-1)^n + 1)/4: n in [1..70]]; // Wesley Ivan Hurt, Apr 21 2021

seq(n^2+n+2*ceil(n/2)^2,n=0..48); # Gary Detlefs, Feb 23 2010

Select[Range[0, 3999], IntegerQ[Sqrt[6# + 1]] &] (* Harvey P. Dale, Mar 10 2013 *)

je=[]; for(n=0,7000, if(issquare(6*n+1),je=concat(je,n))); je

{ n=0; for (m=0, 10^9, if (issquare(6*m + 1), write("b062717.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 09 2009

def A062717(n): return (n*(3*n + 4) + 1 if n&1 else n*(3*n + 2))>>1 # Chai Wah Wu, Jan 31 2023

G.f.: 4*x^2*(1 + x + x^2) / ( (1+x)^2*(1-x)^3 ).
a(2*k) = k*(6*k+2), a(2*k+1) = 6*k^2 + 10*k + 4. - Mohamed Bouhamida, Nov 06 2007
a(n) = n^2 - n + 2*ceiling((n-1)/2)^2. - Gary Detlefs, Feb 23 2010
From Bruno Berselli, Nov 28 2010: (Start)
a(n) = (6*n*(n-1) + (2*n-1)*(-1)^n + 1)/4.
6*a(n) + 1 = A007310(n)^2. (End)
E.g.f.: (3*x^2*exp(x) - x*exp(-x) + sinh(x))/2. - Ilya Gutkovskiy, Jun 18 2016
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Apr 21 2021
From Amiram Eldar, Mar 11 2022: (Start)
Sum_{n>=2} 1/a(n) = (9-sqrt(3)*Pi)/6.
Sum_{n>=2} (-1)^n/a(n) = 3*(log(3)-1)/2. (End)

A078371 a(n) = (2n+5)(2*n+1).

Original entry on oeis.org

5, 21, 45, 77, 117, 165, 221, 285, 357, 437, 525, 621, 725, 837, 957, 1085, 1221, 1365, 1517, 1677, 1845, 2021, 2205, 2397, 2597, 2805, 3021, 3245, 3477, 3717, 3965, 4221, 4485, 4757, 5037, 5325, 5621, 5925, 6237, 6557, 6885, 7221, 7565, 7917, 8277, 8645

Offset: 0

Wolfdieter Lang, Nov 29 2002

This is the generic form of D in the (nontrivially) solvable Pell equation x^2 - D*y^2 = +4. See A077428 and A078355.
Consider all primitive Pythagorean triples (a,b,c) with c-a=8, sequence gives values of a. (Corresponding values for b are A017113(n), while c follows A078370(n).) - Lambert Klasen (Lambert.Klasen(AT)gmx.net), Nov 19 2004
From Vincenzo Librandi, Aug 08 2010: (Start)
The identity (4*n^3 + 18*n^2 + 24*n + 9)^2 - (4*n^2 + 12*n + 5)*(2*n^2 + 6*n + 4)^2 = 1 (see Ramasamy's paper in link) can be written as A141530(n+2)^2 - a(n)*A046092(n+1)^2 = 1.
a(n)^3 + 6*a(n)^2 + 9*a(n) + 4 is a square: in fact, a(n)^3 + 6*a(n)^2 + 9*a(n) + 4 = (a(n) + 1)^2*(a(n) + 4), where a(n) + 4 = (2*n+3)^2. (End)
Products of two positive odd integers with difference 4 (i.e., 1*5, 3*7, 5*9, 7*11, 9*13, ...). - Wesley Ivan Hurt, Nov 19 2013
Starting with stage 1, the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 675", based on the 5-celled von Neumann neighborhood. - Robert Price, May 21 2016
The continued fraction expansion of (sqrt(a(n))-1)/2 is [n; {1,2*n+1}] with periodic part of length 2: repeat {1,2*n+1}. - Ron Knott, May 11 2017
a(n) is the sum of 2*n+5 consecutive integers starting from n-1. - Bruno Berselli, Jan 16 2018
The continued fraction expansion of sqrt(a(n)) is [2n+2; {1, n, 2, n, 1, 4n+4}]. For n=0, this collapses to [2; {4}]. - Magus K. Chu, Aug 26 2022

Bruno Berselli, Table of n, a(n) for n = 0..1000
Soren Laing Aletheia-Zomlefer, Lenny Fukshansky, and Stephan Ramon Garcia, The Bateman-Horn Conjecture: Heuristics, History, and Applications, arXiv:1807.08899 [math.NT], 2018-2019. See Example 6.6.5 p. 34.
A. M. S. Ramasamy, Polynomial solutions for the Pell's equation, Indian Journal of Pure and Applied Mathematics 25 (1994), p. 578.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Subsequence of A077425 (D values (not a square) for which Pell x^2 - D*y^2 = +4 is solvable in positive integers).
Cf. A017113, A046092, A061037, A077428, A078355, A078370.
Supersequence of A143206.

[(2*n+5)*(2*n+1): n in [0..100]]; // G. C. Greubel, Sep 19 2018

seq((2*n+5)*(2*n+1), n=0..48); # Emeric Deutsch, Feb 24 2005

Table[(2 n + 5) (2 n + 1), {n, 0, 100}] (* Wesley Ivan Hurt, Nov 19 2013 *)
LinearRecurrence[{3,-3,1},{5,21,45},50] (* Harvey P. Dale, Oct 18 2020 *)

lista(nn) = {for (n=0, nn, print1((2*n+1)*(2*n+5), ", "));} \\ Michel Marcus, Nov 21 2013

a(n) = 8*(binomial(n+2, 2)-1)+5, hence subsequence of A004770 (5 (mod 8) numbers).
G.f.: (5 + 6*x - 3*x^2)/(1-x)^3.
a(n) = A061037(2*n+1) = (2*n+3)^2 - 4. For A061037: a(2*n+1) = (2*n+1)*(2*n+5) = (2*n+3)^2-4. - Paul Curtz, Sep 24 2008
a(n) = 8*(n+1) + a(n-1) for n > 0, a(0)=5. - Vincenzo Librandi, Aug 08 2010
From Ilya Gutkovskiy, May 22 2016: (Start)
E.g.f.: (5 + 4*x*(4 + x))*exp(x).
Sum_{n>=0} 1/a(n) = 1/3. (End)
Sum_{n>=0} (-1)^n/a(n) = 1/6. - Amiram Eldar, Oct 08 2023

More terms from Emeric Deutsch, Feb 24 2005

A144396 The odd numbers greater than 1.

Original entry on oeis.org

3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133

Offset: 1

Paul Curtz, Oct 03 2008

Last number of the n-th row of the triangle described in A142717.
If negated, these are also the values at local minima of the sequence A141620.
a(n) is the shortest leg of the n-th Pythagorean triple with consecutive longer leg and hypotenuse. The n-th such triple is given by (2n+1,2n^2+2n, 2n^2+2n+1), so that the longer legs are A046092(n) and the hypotenuses are A099776(n). - Ant King, Feb 10 2011
Numbers k such that the symmetric representation of sigma(k) has a pair of bars as its ends (cf. A237593). - Omar E. Pol, Sep 28 2018
Numbers k such that there is a prime knot with k crossings and braid index 2. (IS this true with "prime" removed?) - Charles R Greathouse IV, Feb 14 2023

Muniru A Asiru, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1)

Complement of A004275 and of A004277.
Cf. A000217, A002378, A005408, A007395, A046092, A099776, A237593, A254858.
Essentially the same as A140139, A130773, A062545, A020735, A005818.

List([3,5..200],n->n); # Muniru A Asiru, Sep 28 2018

a144396 = (+ 1) . (* 2)
a144396_list = [3, 5 ..] -- Reinhard Zumkeller, Feb 09 2015

seq(n,n=3..200,2); # Muniru A Asiru, Sep 28 2018

Range[3, 200, 2] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2012 *)

a(n)=2*n+1 \\ Charles R Greathouse IV, Feb 14 2023

a(n) = A005408(n+1) = A000290(n+1) - A000290(n).
G.f.: x*(3-x)/(1-x)^2. - Jaume Oliver Lafont, Aug 30 2009
a(n) = A254858(n-1,2). - Reinhard Zumkeller, Feb 09 2015

Edited by R. J. Mathar, May 21 2009

cp's OEIS Frontend

A049581 Table T(n,k) = |n-k| read by antidiagonals (n >= 0, k >= 0).

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A147973 a(n) = -2*n^2 + 12*n - 14.

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A059056 Penrice Christmas gift numbers, Card-matching numbers (Dinner-Diner matching numbers): Triangle T(n,k) = number of ways to get k matches for a deck with n cards, 2 of each kind.

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A085250 4 times hexagonal numbers: a(n) = 4*n*(2*n-1).

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A097080 a(n) = 2*n^2 - 2*n + 3.

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A024966 7 times triangular numbers: 7*n*(n+1)/2.

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A147973 a(n) = -2n^2 + 12n - 14.

A085250 4 times hexagonal numbers: a(n) = 4n(2*n-1).

A097080 a(n) = 2n^2 - 2n + 3.

A024966 7 times triangular numbers: 7n(n+1)/2.

A059993 Pinwheel numbers: a(n) = 2n^2 + 6n + 1.

A078371 a(n) = (2n+5)(2*n+1).