A077588 -id:A077588 - cp's OEIS frontend

A069894 Centered square numbers: a(n) = 4n^2 + 4n + 2.

2, 10, 26, 50, 82, 122, 170, 226, 290, 362, 442, 530, 626, 730, 842, 962, 1090, 1226, 1370, 1522, 1682, 1850, 2026, 2210, 2402, 2602, 2810, 3026, 3250, 3482, 3722, 3970, 4226, 4490, 4762, 5042, 5330, 5626, 5930, 6242, 6562, 6890, 7226, 7570, 7922, 8282

Offset: 0

Glenn B. Cox (igloos_r_us(AT)canada.com), Apr 10 2002

Any number may be substituted for y to yield similar sequences. The number set used determines values given (i.e., integer yields integer). All centered square integers in the set of integers may be found by this formula.
1/2 + 1/10 + 1/26 + ... = (Pi/4)*tanh(Pi/2) [Jolley]. - Gary W. Adamson, Dec 21 2006
For n > 0, a(n-1) is the number of triples (w, x, y) having all terms in {0, ..., n} and min(|w - x|, |x - y|) = 1. - Clark Kimberling, Jun 12 2012
Consider the primitive Pythagorean triples (x(n), y(n), z(n) = y(n) + 1) with n >= 0, and x(n) = 2*n + 1, y(n) = 2*n*(n + 1), z(n) = 2*n*(n + 1) + 1. The sequence, a(n), is 2*z(n). - George F. Johnson, Oct 22 2012
Ulam's spiral (SE corner). See the Wikipedia link. - Kival Ngaokrajang, Jul 25 2014
Conference matrix orders (A000952) of the form n-1 is a perfect square are all in this sequence. All values less than 1000 are conference matrices except for 226 which is still an open question (Balonin & Seberry 2014). - Colin Hall, Nov 21 2018
For n > 0, a(n-1) is the number of maximum number of regions into which the plane can be divided using n convex quadrilaterals. Related: A077588 A077591. - Keyang Li, Jun 17 2022

			If y = 3, then 81 + 144 = 225; if y = 4, then 12^2 + 16^2 = 20^2; 7^2 + 24^2 = 25^2 = 15^2 + 20^2.

L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 176.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
N. A. Balonin and Jennifer Seberry, A Review and New Symmetric Conference Matrices, Research Online, Faculty of Engineering and Information Sciences, University of Wollongong, 2014.
Keyang Li, Figure for n=1,2,3,4,5
Tintarn, n convex quadrilaterals in the plane
Wikipedia, Ulam Spiral Construction.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001105, A001844, A002522, A016754, A016825, A033996, A261377.

[4*n^2+4*n+2 : n in [0..50]]; // Wesley Ivan Hurt, Jul 26 2014

A069894:=n->4*n^2+4*n+2: seq(A069894(n), n=0..50); # Wesley Ivan Hurt, Jul 26 2014

Mathematica
```
Table[4n(n + 1) + 2, {n, 0, 45}]
```

vector(100, n, (2*n-1)^2+1); \\ Derek Orr, Jul 27 2014

[(2*n+1)^2 + 1 for n in range(50)] # G. C. Greubel, Nov 21 2018

(y*(2*x + 1))^2 + (y*(2*x^2 + 2*x))^2 = (y*(2*x^2 + 2*x + 1))^2, where y = 2. If a^2 + b^2 = c^2, then c^2 = y^2*(4*x^4 + 8*x^3 + 8*x^2 + 4*x + 1). Also 2*A001844.
a(n) = (2*n + 1)^2 + 1. - Vladimir Joseph Stephan Orlovsky, Nov 10 2008 [Corrected by R. J. Mathar, Sep 16 2009]
a(n) = 8*n + a(n-1) for n > 0, a(0)=2. - Vincenzo Librandi, Aug 08 2010
From George F. Johnson, Oct 22 2012: (Start)
G.f.: 2*(1 + x)^2/(1 - x)^3, a(0) = 2, a(1) = 10.
a(n+1) = a(n) + 4 + 4*sqrt(a(n) - 1).
a(n-1) * a(n+1) = (a(n)-4)^2 + 16.
a(n) - 1 = (2*n+1)^2 = A016754(n) for n > 0.
(a(n+1) - a(n-1))/8 = sqrt(a(n) - 1).
a(n+1) = 2*a(n) - a(n-1) + 8 for n > 2, a(0)=2, a(1)=10, a(2)=26.
a(n+1) = 3*a(n) - 3*a(n-1) + a(n-2) for n > 3; a(0)=2, a(1)=10, a(2)=26, a(3)=50.
a(n) = A033996(n) + 2 = A002522(2n + 1).
a(n)^2 = A033996(n)^2 + A016825(n)^2. (End)
a(n) = A001105(n) + A001105(n+1). - Bruno Berselli, Jul 03 2017
E.g.f.: 2*(1 + 4*x + 2*x^2)*exp(x). - G. C. Greubel, Nov 21 2018
a(n) = A261327(4*n+2). - Paul Curtz, Dec 23 2021
a(n) = 2*A001844(n) = 4*A000217(n) + 2*A002061(n+1). - Klaus Purath, Aug 13 2025

Edited by Robert G. Wilson v, Apr 11 2002

Offset corrected by Charles R Greathouse IV, Jul 25 2010

A077591 Maximum number of regions into which the plane can be divided using n (concave) quadrilaterals.

Original entry on oeis.org

1, 2, 18, 50, 98, 162, 242, 338, 450, 578, 722, 882, 1058, 1250, 1458, 1682, 1922, 2178, 2450, 2738, 3042, 3362, 3698, 4050, 4418, 4802, 5202, 5618, 6050, 6498, 6962, 7442, 7938, 8450, 8978, 9522, 10082, 10658, 11250, 11858, 12482, 13122, 13778

Offset: 0

Joshua Zucker and the Castilleja School MathCounts club, Nov 07 2002

Sequence found by reading the segment (1, 2) together with the line from 2, in the direction 2, 18, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 05 2011
For a(n) > 1, a(n) are the numbers such that phi(sum of the odd divisors of a(n)) = phi(sum of even divisors of a(n)). - Michel Lagneau, Sep 14 2011
Apart from first term, subsequence of A195605. - Bruno Berselli, Sep 21 2011
Engel expansion of 1F2(1; 1/2, 1/2; 1/8). - Benedict W. J. Irwin, Jun 21 2018
Let f(n) = 4*n^2 - 5, then (x, y, z) = (a(n+1), -f(n), -f(n + 1)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 512. - XU Pingya, Apr 25 2022

			a(2) = 18 if you draw two concave quadrilaterals such that all four sides of one cross all four sides of the other.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Keyang Li, when n=1, number of regions is 2; when n=2, maximum number of regions is 18
Keyang Li, when n=3, maximum number of regions is 50
Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).

Cf. A006752, A077588, A239186.
Cf. A071974, A071975.
Cf. A016754, A069129.

Concatenation([1], List([1..2000], n->8*n^2 - 8*n + 2)); # Muniru A Asiru, Jan 29 2018

A077591:=n->`if`(n=0, 1, 8*n^2 - 8*n + 2); seq(A077591(n), n=0..50); # Wesley Ivan Hurt, Mar 12 2014

Table[2*(4*n^2 - 4*n + 1), {n,0,50}] (* G. C. Greubel, Jul 15 2017 *)

isok(n) = (sod = sumdiv(n, d, (d%2)*d)) && (sed = sumdiv(n, d, (1 - d%2)*d)) && (eulerphi(sod) == eulerphi(sed)); \\ from Michel Lagneau comment; Michel Marcus, Mar 15 2014

a(n) = 8*n^2 - 8*n + 2 = 2*(2*n-1)^2, n > 0, a(0)=1.
Proof from Keyang Li, Jun 18 2022: (Start)
Represent the configuration of n concave quadrilaterals by a planar graph with a node for each vertex of the quadrilaterals and for each intersection point. Let there be v_n nodes and e_n edges. By Euler's formula for planar graphs, a(n) = e_n - v_n + 2. When we go from n to n+1 quadrilaterals, each side of the new quadrilateral can meet each side of the existing quadrilaterals at most 4 times, so Dv_n := v_{n+1} - v_n <= 4*4n = 16n.
Each of these intersection points increases the number of edges in the graph by 2, so De_n := e_{n+1} - e_n = 4 + 2*Dv_n, Da_n := a(n+1) - a(n) = 4 + Dv_n <= 4+16*n.
These upper bounds can be achieved by taking n interwoven concave quadrilaterals (for n=1,2,3 see the attached Keyang Li links), and we achieve a(n) = 8n^2 - 8n + 2 (and v_n = 8n^2 - 4n, e_n = 4n*(4n-3)) for n > 0. QED (End)
For n > 0: A071974(a(n)) = 2*n+1, A071975(a(n)) = 2. - Reinhard Zumkeller, Jul 10 2011
a(n) = 1 + A069129(n), if n >= 1. - Omar E. Pol, Sep 05 2011
a(n) = 2*A016754(n-1), if n >= 1. - Omar E. Pol, Sep 05 2011
G.f.: (1 - x + 15*x^2 + x^3)/(1-x)^3. - Colin Barker, Feb 23 2012
E.g.f.: (8*x^2 + 2)*exp(x) - 1. - G. C. Greubel, Jul 15 2017
From Amiram Eldar, Jan 29 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/16.
Sum_{n>=1} (-1)^(n+1)/a(n) = G/2, where G is Catalan constant (A006752).
Product_{n>=1} (1 + 1/a(n)) = cosh(Pi/sqrt(8)).
Product_{n>=1} (1 - 1/a(n)) = cos(Pi/sqrt(8)). (End)

A096777 a(n) = a(n-1) + Sum_{k=1..n-1}(a(k) mod 2), a(1) = 1.

Original entry on oeis.org

1, 2, 3, 5, 8, 11, 15, 20, 25, 31, 38, 45, 53, 62, 71, 81, 92, 103, 115, 128, 141, 155, 170, 185, 201, 218, 235, 253, 272, 291, 311, 332, 353, 375, 398, 421, 445, 470, 495, 521, 548, 575, 603, 632, 661, 691, 722, 753, 785, 818, 851, 885, 920, 955, 991, 1028

Offset: 1

Reinhard Zumkeller, Jul 09 2004

a(n) = a(n-1) + (number of odd terms so far in the sequence). Example: 15 is 11 + 4 odd terms so far in the sequence (they are 1,3,5,11). See A007980 for the same construction with even integers. - Eric Angelini, Aug 05 2007

A016789 and A032766 give positions where even and odd terms occur; a(3*n)=A056106(n); a(3*n-1)=A077588(n); a(3*n-2)=A056108(n). - Reinhard Zumkeller, Dec 29 2007

			G.f. = x + 2*x^2 + 3*x^3 + 5*x^4 + 8*x^5 + 11*x^6 + 15*x^7 + 20*x^8 + ... - _Michael Somos_, Apr 18 2020

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J.-L. Baril, T. Mansour, A. Petrossian, Equivalence classes of permutations modulo excedances, 2014.
Eric Weisstein's World of Mathematics, Odd Number
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A004396, A007980, A016789, A032766, A056106, A056108, A061347, A077588, A097602, A131093.

a096777 n = a096777_list !! (n-1)
a096777_list = 1 : zipWith (+) a096777_list
                               (scanl1 (+) (map (`mod` 2) a096777_list))
-- Reinhard Zumkeller, Mar 11 2014

[Floor((n-2)^2/3)+n: n in [1..60]]; // Vincenzo Librandi, Dec 27 2015

A096777:=n->n + floor((n-2)^2/3); seq(A096777(n), n=1..100); # Wesley Ivan Hurt, Mar 06 2014

Table[n + Floor[(n-2)^2/3], {n, 100}] (* Wesley Ivan Hurt, Mar 06 2014 *)

a(n)=(n-2)^2\3+n \\ Charles R Greathouse IV, Mar 06 2014

a(n+1) - a(n) = A004396(n).
a(n) = floor(n/3) * (3*floor(n/3) + 2*(n mod 3) - 1) + n mod 3 + 0^(n mod 3). - Reinhard Zumkeller, Dec 29 2007
a(n) = floor((n-2)^2/3) + n. - Christopher Hunt Gribble, Mar 06 2014
G.f.: -x*(x^4+1) / ((x-1)^3*(x^2+x+1)). - Colin Barker, Mar 07 2014
Euler transform of finite sequence [2, 0, 1, 1, 0, 0, 0, -1]. - Michael Somos, Apr 18 2020
a(n) = (10 + 3*n*(n - 1) - A061347(n+1))/9. - Stefano Spezia, Sep 22 2022

A084684 Degrees of certain maps (see Comments and Formulas for more precise definitions).

Original entry on oeis.org

1, 2, 4, 8, 13, 20, 28, 38, 49, 62, 76, 92, 109, 128, 148, 170, 193, 218, 244, 272, 301, 332, 364, 398, 433, 470, 508, 548, 589, 632, 676, 722, 769, 818, 868, 920, 973, 1028, 1084, 1142, 1201, 1262, 1324, 1388, 1453, 1520, 1588, 1658, 1729, 1802, 1876, 1952, 2029, 2108, 2188, 2270, 2353, 2438, 2524, 2612, 2701, 2792, 2884, 2978, 3073, 3170, 3268, 3368, 3469, 3572

Offset: 0

N. J. A. Sloane, Jul 16 2003

Number of binary strings of length n with no substrings equal to 0001, 1001, or 1011. - R. H. Hardin, Aug 14 2009

Degree sequence d(n) of recursion x(n+1)+x(n)+x(n-1) = b + c(n)/x(n) where c(n) = c(n-1) + c(n-2) - c(n-3) and x(n) = u(n)/f(n) and x(n-1) = v(n)/f(n) in homogeneous coordinates (projectivization). Denoted by sigma_1 on page 32 of Hiertarinta and Viallet (2000). - Michael Somos, Jan 04 2022

			G.f. = 1 + 2*x + 4*x^2 + 8*x^3 + 13*x^4 + 20*x^5 + 28*x^6 + 38*x^7 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Jarmo Hietarinta and Claude Viallet, Discrete Painlevé I and singularity confinement in projective space, Chaos, Solitons and Fractals 11 (2000), pp. 29-32.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A064863, A056107 (bisection), A077588 (bisection).

Cf. also A001651, A002620, A122958.

a[ n_] := Quotient[ 3*n^2 + 6, 4]; (* Michael Somos, Feb 08 2015 *)
LinearRecurrence[{2,0,-2,1},{1,2,4,8},70] (* Harvey P. Dale, Jul 21 2021 *)

a(n)=(6*n^2 + 9 - (-1)^n)/8 \\ Charles R Greathouse IV, Sep 10 2014

{a(n) = (n^2 + 2)*3 \ 4}; /* Michael Somos, Feb 08 2015 */

a(n) = (6*n^2 + 9 - (-1)^n)/8. - Charles R Greathouse IV, Sep 10 2014
G.f.: ( 1+2*x^3 ) / ( (1+x)*(1-x)^3 ). - R. J. Mathar, Sep 11 2014
a(n) = 2*a(n-1)-2*a(n-3)+a(n-4). - Colin Barker, Sep 11 2014
a(n) = a(-n) for all n in Z. - Michael Somos, Feb 08 2015
a(n) - a(n-1) = A001651(n), a(n+1) - a(n-1) = 3*n for all n in Z. - Michael Somos, Feb 08 2015
(a(n) - a(n+1))^2 - (2*a(n) + a(n+1)) + 4 = 3*n/2 + 1 for all even n in Z. - Michael Somos, Feb 08 2015
0 = -4 + a(n)*(-a(n+1) + a(n+2)) + a(n+1)*(+3 + a(n+1) - a(n+2)) for all n in Z. - Michael Somos, Feb 08 2015
A122958(n-1) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n for all n>1. - Michael Somos, Feb 08 2015
a(n) = 2*a(n-1) - 3*A002620(n-2) for all n in Z. - Michael Somos, Dec 27 2021
a(n) = 3*(a(n-1) + a(n-4)) - 2*(a(n-2) + a(n-3)) - a(n-5) for all n in Z. - Michael Somos, Jan 04 2022

More terms from Charles R Greathouse IV, Sep 10 2014

Edited by N. J. A. Sloane, Jan 04 2022

A242658 a(n) = 3n^2 - 3n + 2.

Original entry on oeis.org

2, 2, 8, 20, 38, 62, 92, 128, 170, 218, 272, 332, 398, 470, 548, 632, 722, 818, 920, 1028, 1142, 1262, 1388, 1520, 1658, 1802, 1952, 2108, 2270, 2438, 2612, 2792, 2978, 3170, 3368, 3572, 3782, 3998, 4220, 4448, 4682, 4922, 5168, 5420, 5678, 5942, 6212, 6488, 6770, 7058, 7352

Offset: 0

N. J. A. Sloane, May 30 2014

An exercise in Smith (1950), my secondary school algebra book.

For n > 0, also the number of (not necessarily maximal) cliques in the (n-1)-triangular grid graph. - Eric W. Weisstein, Nov 29 2017

C. Smith, A Treatise on Algebra, Macmillan, London, 5th ed., 1950, p. 429, Example 2(i).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Clique
Eric Weisstein's World of Mathematics, Triangular Grid Graph
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

A077588 is the same except for the initial term.

Cf. A002378, A242659.

[3*n^2 - 3*n + 2: n in [0..70]]; // Vincenzo Librandi, Sep 05 2016

Table[3 n^2 - 3 n + 2, {n, 0, 100}] (* Vincenzo Librandi, Sep 05 2016 *)
LinearRecurrence[{3, -3, 1}, {2, 8, 20}, {0, 20}] (* Eric W. Weisstein, Nov 29 2017 *)
CoefficientList[Series[-2 (1 - 2 x + 4 x^2)/(-1 + x)^3, {x, 0, 20}], x] (* Eric W. Weisstein, Nov 29 2017 *)

a(n) = 3*n^2-3*n+2 \\ Altug Alkan, Sep 05 2016

From Chai Wah Wu, May 30 2016: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
G.f.: 2*(-4*x^2 + 2*x - 1)/(x - 1)^3. (End)
E.g.f.: exp(x)*(2 + 3*x^2). - Stefano Spezia, Dec 27 2021
a(n) = A002378(n) + A002378(n-1) + A002378(n-2). - Peter Bala, Jun 11 2024

A004538 a(n) = 3n^2 + 3n - 1.

Original entry on oeis.org

-1, 5, 17, 35, 59, 89, 125, 167, 215, 269, 329, 395, 467, 545, 629, 719, 815, 917, 1025, 1139, 1259, 1385, 1517, 1655, 1799, 1949, 2105, 2267, 2435, 2609, 2789, 2975, 3167, 3365, 3569, 3779, 3995, 4217, 4445

Offset: 0

N. J. A. Sloane, Eric T. Lane (ERICLANE(AT)UTCVM.UTC.EDU)

Numbers k such that (4*k + 7)/3 is a square. - Bruno Berselli, Sep 11 2018

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

First differences of A033445.

Cf. A028896, A003215, A077588, A000538, A000330.

[3*n^2 + 3*n -1: n in [0..50]]; // G. C. Greubel, Sep 10 2018

Table[5*Sum[k^4,{k,1,n}]/Sum[k^2,{k,1,n}], {n,1,20}] (* Alexander Adamchuk, Apr 12 2006 *)
Table[3n^2+3n-1,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{-1,5,17},40] (* Harvey P. Dale, Jan 18 2019 *)

a(n)=3*n^2+3*n-1 \\ Charles R Greathouse IV, Jun 17 2017

From Alexander Adamchuk, Apr 12 2006: (Start)
a(n) = 5 * Sum_{k=1..n} k^4 / Sum_{k=1..n} k^2, n > 0.
a(n) = 5 * A000538(n) / A000330(n), n > 0. (End)
a(n) = a(n-1) + 6*n with a(0)=-1. - Vincenzo Librandi, Nov 18 2010
From G. C. Greubel, Sep 10 2018: (Start)
G.f.: (-1 + 8*x - x^2)/(1 - x)^3.
E.g.f.: (-1 + 6*x + 3*x^2)*exp(x). (End)
Sum_{n>=0} 1/a(n) = ( psi(1/2+sqrt(21)/6) - psi(1/2-sqrt(21)/6)) /sqrt(21) = -0.6286929... R. J. Mathar, Apr 24 2024

A081576 Square array of binomial transforms of Fibonacci numbers, read by antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 3, 2, 0, 1, 5, 8, 3, 0, 1, 7, 20, 21, 5, 0, 1, 9, 38, 75, 55, 8, 0, 1, 11, 62, 189, 275, 144, 13, 0, 1, 13, 92, 387, 905, 1000, 377, 21, 0, 1, 15, 128, 693, 2305, 4256, 3625, 987, 34, 0, 1, 17, 170, 1131, 4955, 13392, 19837, 13125, 2584, 55

Offset: 0

Paul Barry, Mar 22 2003

Array rows are solutions of the recurrence a(n) = (2*k+1)*a(n-1) - A028387(k-1)*a(n-2) where a(0) = 0 and a(1) = 1.

			Square array begins as:
  0, 1,  1,   2,    3,    5,     8, ... A000045;
  0, 1,  3,   8,   21,   55,   144, ... A001906;
  0, 1,  5,  20,   75,  275,  1000, ... A030191;
  0, 1,  7,  38,  189,  905,  4256, ... A099453;
  0, 1,  9,  62,  387, 2305, 13392, ... A081574;
  0, 1, 11,  92,  693, 4955, 34408, ... A081575;
  0, 1, 13, 128, 1131, 9455, 76544, ...
The antidiagonal triangle begins as:
  0;
  0, 1;
  0, 1,  1;
  0, 1,  3,  2;
  0, 1,  5,  8,   3;
  0, 1,  7, 20,  21,   5;
  0, 1,  9, 38,  75,  55,   8;
  0, 1, 11, 62, 189, 275, 144, 13;

G. C. Greubel, Antidiagonal rows n = 0..50, flattened

Array row n: A000045 (n=0), A001906 (n=1), A030191 (n=2), A099453 (n=3), A081574 (n=4), A081575 (n=5).
Array columns k: A005408 (k=3), A077588 (k=4).
Cf. A028387, A081573, A081574, A081575.

A081576:= func< n,k | (&+[Binomial(k,j)*Fibonacci(j)*(n-k)^(k-j): j in [0..k]]) >;
[A081576(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021

T[n_, k_]:= If[n==0, Fibonacci[k], Sum[Binomial[k, j]*Fibonacci[j]*n^(k-j), {j, 0, k}]]; Table[T[n-k, k], {n,0,12}, {k,0,n}] //Flatten (* G. C. Greubel, May 26 2021 *)

def A081576(n,k): return sum( binomial(k,j)*fibonacci(j)*(n-k)^(k-j) for j in (0..k) )
flatten([[A081576(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021

Rows are successive binomial transforms of F(n).
T(n, k) = ( ( (2*n + 1 + sqrt(5))/2 )^k - ( (2*n + 1 - sqrt(5))/2 )^k )/sqrt(5).
From G. C. Greubel, May 26 2021: (Start)
T(n, k) = Sum_{j=0..k} binomial(k,j)*Fibonacci(j)*n^(k-j) with T(0, k) = Fibonacci(k) (square array).
T(n, k) = Sum_{j=0..k} binomial(k,j)*Fibonacci(j)*(n-k)^(k-j) (antidiagonal triangle). (End)

A295288 Binomial transform of the centered triangular numbers A005448.

Original entry on oeis.org

1, 5, 19, 62, 184, 512, 1360, 3488, 8704, 21248, 50944, 120320, 280576, 647168, 1478656, 3350528, 7536640, 16842752, 37421056, 82706432, 181927936, 398458880, 869269504, 1889533952, 4093640704, 8841592832, 19042140160, 40902852608

Offset: 0

Franck Maminirina Ramaharo, Nov 19 2017

The sequence is column 3 of triangle in A207630.

First difference is given by A055818(n+3,3) for n > 0.

			a(0) = (3*0^2 + 9*0 + 8)*2^(-3) = 8/8 = 1.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 1994.

G. C. Greubel, Table of n, a(n) for n = 0..1000
C. Corsani, D. Merlini, and R. Sprugnoli, Left-inversion of combinatorial sums, Discrete Mathematics, 180 (1998) 107-122.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv:1406.3081 [math.CO], 2014.
Index entries for linear recurrences with constant coefficients, signature (6,-12,8).

Cf. A000079, A001787, A001792, A005448, A055818, A077588, A207629, A207630.

I:=[1,5,19]; [n le 3 select I[n] else 6*Self(n-1) -12*Self(n-2) +8*Self(n-3): n in [1..40]]; // G. C. Greubel, Oct 17 2018

A:=n->(3*n^2+9*n+8)*2^(n-3); seq(A(n), n=0..70);

Table[(3 n^2 + 9 n + 8) 2^(n-3), {n, 0, 70}]
LinearRecurrence[{6,-12,8}, {1,5,19}, 50] (* G. C. Greubel, Oct 17 2018 *)

makelist((3*n^2 + 9*n + 8)*2^(n - 3), n, 0, 70);

a(n) = (3*n^2 + 9*n + 8)*2^(n - 3) \\ Felix Fröhlich, Nov 19 2017

G.f.: (1 - x + x^2)/(1 - 2*x)^3.
a(n+3) = 8*a(n) - 12*a(n+1) + 6*a(n+2).
a(n+1) = 2*a(n) + 3*(n + 2)*2^(n-1).
a(n+1) = 2*a(n) + 3*A001792(n) = 2*a(n) + A001787(n+2) - A001792(n).
a(n) = (3*n^2 + 9*n + 8)*2^(n - 3).
a(n) = (1/8)*A077588(n+2)*A000079(n).

A383466 a(0) = 1; thereafter a(n) = 10n^2 - 5n + 2.

Original entry on oeis.org

1, 7, 32, 77, 142, 227, 332, 457, 602, 767, 952, 1157, 1382, 1627, 1892, 2177, 2482, 2807, 3152, 3517, 3902, 4307, 4732, 5177, 5642, 6127, 6632, 7157, 7702, 8267, 8852, 9457, 10082, 10727, 11392, 12077, 12782, 13507, 14252, 15017, 15802, 16607, 17432, 18277, 19142, 20027, 20932, 21857, 22802, 23767, 24752, 25757, 26782, 27827

Offset: 0

Scott R. Shannon and N. J. A. Sloane, Jul 22 2025

Definition: A regular pentagram of radius R is formed by placing five equally-spaced points P_0 .. P_4 around the boundary of a circle of radius R, and drawing line segments P_0 - P_2 - P_4 - P_1 - P_3 - P_0.
Theorem 1: a(n) is the maximum number of regions that can be formed in the plane by drawing n regular pentagrams with the same radius and the same center.
Conjecture 2: a(n) is the maximum number of regions that can be formed in the plane by drawing n regular pentagrams with any radii and any centers.
The following construction works for any n >= 1. Take 5*n equally-spaced points P_i around a circle, and draw a pentagram through P_i, P_{i+n}, P_{i+2*n}, P_{i+3*n}, P_{i+4*n} for i = 0, ..., n-1.
The resulting planar graph decomposes into 5*n triangular regions each with 2*n-1 cells (see the red triangle in "Illustration for a(n)..."), plus the interior and exterior regions, for a total of 10*n^2 - 5*n + 2 regions. There are 10*n^2 vertices (10 for n=1, 40 for n=2, and so on).

Paolo Xausa, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Illustration for a(1) = 7. [Note that the cell counts shown on these four figures do not include the black exterior region, so the totals are off by 1]
Scott R. Shannon, Illustration for a(2) = 32.
Scott R. Shannon, Illustration for a(3) = 77.
Scott R. Shannon, Illustration for a(8) = 602.
N. J. A. Sloane, Illustration for a(1) = 7.
N. J. A. Sloane, Illustration for a(2) = 32.
N. J. A. Sloane, Illustration for a(n), n >= 1, showing a(3) = 77.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

See A077588, A069894, and A386477 for analogous sequences based on triangles, squares, and hexagrams.

Without the "+2" in the definition, the sequence is A152745.

A383466[n_] := If[n == 0, 1, 5*n*(2*n - 1) + 2]; Array[A383466, 50, 0] (* or *)
Join[{1}, 5*PolygonalNumber[6, Range[49]] + 2] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 7, 32, 77}, 50] (* Paolo Xausa, Jul 22 2025 *)

From Elmo R. Oliveira, Sep 03 2025: (Start)
G.f.: (1 + 4*x + 14*x^2 + x^3)/(1 - x)^3.
E.g.f.: exp(x)*(2 + 5*x + 10*x^2) - 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A386477 a(0) = 1; thereafter a(n) = 2(6n^2 - 3*n + 1).

Original entry on oeis.org

1, 8, 38, 92, 170, 272, 398, 548, 722, 920, 1142, 1388, 1658, 1952, 2270, 2612, 2978, 3368, 3782, 4220, 4682, 5168, 5678, 6212, 6770, 7352, 7958, 8588, 9242, 9920, 10622, 11348, 12098, 12872, 13670, 14492, 15338, 16208, 17102, 18020, 18962, 19928, 20918, 21932, 22970, 24032, 25118, 26228, 27362, 28520, 29702, 30908, 32138, 33392, 34670

Offset: 0

Scott R. Shannon and N. J. A. Sloane, Jul 22 2025

Definition: A regular hexagram of radius R is formed by placing six equally-spaced points P_0 .. P_5 around the boundary of a circle of radius R, and drawing line segments P_0 - P_2 - P_4 - P_0 and P_1 - P_3 - P_5 - P_1.
Theorem 1: a(n) is the maximum number of regions that can be formed in the plane by drawing n regular hexagrams with the same radius and the same center.
Conjecture 2: a(n) is the maximum number of regions that can be formed in the plane by drawing n regular hexagrams with any radii and any centers.
The following construction works for any n >= 1. Take 6*n equally-spaced points P_i around a circle, and draw hexagrams starting at P_i for i = 0, ..., n-1.
The resulting planar graph decomposes into 6*n triangular cells each with 2*n-1 cells (see the red triangle in the "Three pentagons" illustration), plus the interior and exterior regions, for a total of 12*n^2 - 6*n + 2 regions. There are 12*n^2 vertices, for n>0.

Paolo Xausa, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Illustration for a(1) = 8 [Note that the cell counts shown on these five figures do not include the black exterior region, so the totals are off by 1]
Scott R. Shannon, Illustration for a(2) = 38
Scott R. Shannon, Illustration for a(3) = 92
Scott R. Shannon, Illustration for a(8) = 722
Scott R. Shannon, Illustration for a(10) = 1142
N. J. A. Sloane, Sketch to illustrate a(2) = 38. The two hexagrams are colored red and black, respectively.
N. J. A. Sloane, Sketch to illustrate a(3) = 92. The three hexagrams are colored red, blue, and black, respectively.
N. J. A. Sloane, Analogous illustration for three pentagrams
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A094159, A152746.

See A077588, A069894, and A383466 for analogous sequences based on triangles, squares, and pentagrams.

A386477[n_] := If[n == 0, 1, 6*n*(2*n - 1) + 2]; Array[A386477, 50, 0] (* or *)
Join[{1}, 6*PolygonalNumber[6, Range[49]] + 2] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 8, 38, 92}, 50] (* Paolo Xausa, Jul 24 2025 *)

From Stefano Spezia, Jul 23 2025: (Start)
G.f.: (1 + 5*x + 17*x^2 + x^3)/(1 - x)^3.
E.g.f.: 2*exp(x)*(1 + 3*x + 6*x^2) - 1. (End)
a(n) = A152746(n) + 2, for n >= 1. - Paolo Xausa, Jul 24 2025

cp's OEIS Frontend

A069894 Centered square numbers: a(n) = 4*n^2 + 4*n + 2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A077591 Maximum number of regions into which the plane can be divided using n (concave) quadrilaterals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Maple

Mathematica

PARI

Formula

A096777 a(n) = a(n-1) + Sum_{k=1..n-1}(a(k) mod 2), a(1) = 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

Formula

A084684 Degrees of certain maps (see Comments and Formulas for more precise definitions).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

Extensions

A242658 a(n) = 3*n^2 - 3*n + 2.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A004538 a(n) = 3*n^2 + 3*n - 1.

Views

Author

Keywords

Comments

Links

A069894 Centered square numbers: a(n) = 4n^2 + 4n + 2.

A242658 a(n) = 3n^2 - 3n + 2.

A004538 a(n) = 3n^2 + 3n - 1.

A383466 a(0) = 1; thereafter a(n) = 10n^2 - 5n + 2.

A386477 a(0) = 1; thereafter a(n) = 2(6n^2 - 3*n + 1).