A028347 -id:A028347 - cp's OEIS frontend

A087475 a(n) = n^2 + 4.

4, 5, 8, 13, 20, 29, 40, 53, 68, 85, 104, 125, 148, 173, 200, 229, 260, 293, 328, 365, 404, 445, 488, 533, 580, 629, 680, 733, 788, 845, 904, 965, 1028, 1093, 1160, 1229, 1300, 1373, 1448, 1525, 1604, 1685, 1768, 1853, 1940, 2029, 2120, 2213, 2308, 2405, 2504

Offset: 0

Gary W. Adamson, Sep 09 2003

Schroeder, p. 330, states "For positive n, these winding numbers are precisely those whose continued fraction expansion is periodic and has period length 1".
Positive X values of solutions to the equation X^3 - 4*X^2 = Y^2. To find Y values: b(n) = n*(n^2 + 4). - Mohamed Bouhamida, Nov 06 2007
From Artur Jasinski, Oct 03 2008: (Start)
General formula for cotangent recurrences type:
a(n+1) = a(n)^3 + 3*a(n) and a(1)=k is
a(n) = floor(((k + sqrt(k^2 + 4))/2)^(3^(n-1))). (End)
Given sequences of the form S(n) = N*S(n-1) + S(n-2) starting (1, N, ...), and having convergents with discriminant (N^2 + 4), S(p) == (a(n))^((p-1)/2) mod p, for n>0, p = odd prime. Example: with N = 2 we have the Pell series (1, 2, 5, 12, 29, 70, 169, ...) with P(7) = 169. Then 169 == 8^3 mod 7, with a(2) = 8. Cf. Schroeder, "Number Theory in Science and Communication", p. 90, for N = 1: F(p) == 5^((p-1)/2) mod p. - Gary W. Adamson, Feb 23 2009
The only two real solutions of the form f(x) = A*x^p with positive p that satisfy f^(n)(x) = f^[-1](x), x >= 0, n >= 1, with f^(n) the n-th derivative and f^[-1] the compositional inverse of f, are obtained for p = p1(n) = (n + sqrt(a(n)))/2 and p = p2(n) = (n - sqrt(a(n)))/2, n >= 1, and A = A(n) = (fallfac(p,n))^(-p/(p+1)), for p = p1(n) and p = p2(n), respectively. Here fallfac(x, k) := product(x - j, j = 0..k-1), the falling factorials. See the T. Koshy reference, pp. 263-264 (there is also a solution for negative p if n is even; see the corresponding comment in A002522). - Wolfdieter Lang, Oct 21 2010, Oct 28 2010
(n + sqrt(a(n)))/2 = [n;n,n,...], with the regular continued fraction with period length 1. For a simple proof see, e.g., the Schroeder reference, pp. 330-331. See also the first comment above.

			a(2) = 8, discriminant of algebraic representation of barover(2) = [2,2,2,...] = sqrt 2 - 1 = 0.41421356... = ((sqrt 8) - 2)/2. a(3) = 13, discriminant of barover(3) = [3,3,3,...] = 0.3027756... = ((sqrt 13) - 3)/2.

Manfred R. Schroeder, "Fractals, Chaos, Power Laws"; W.H. Freeman & Co, 1991, p. 330-331.
Manfred R. Schroeder, "Number Theory in Science and Communication", Springer Verlag, 5th ed., 2009. [From Gary W. Adamson, Feb 23 2009]
Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, New York, 2001. [From Wolfdieter Lang, Oct 21 2010]

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Near-Square Prime.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001082, A002378, A002522, A005563, A028347, A036666, A046092, A053755, A062717, A155965, A155966, A156701, A156798.

Range[0, 50]^2 + 4 (* Harvey P. Dale, Jan 05 2011 *)

a(n)=n^2+4 \\ Charles R Greathouse IV, Jun 10 2011

(0 to 49).map(n => n * n + 4) // Alonso del Arte, May 29 2019

n^2 + 4 are discriminant terms in the formula for Positive Silver Mean Constants, defined as barover(n), = (sqrt (n^2 + 4) - n)/2. Such constants barover(n) = C have the property: 1/C - C = n.
a(n) = A156701(n) / A053755(n). - Reinhard Zumkeller, Feb 13 2009
a(n) = A156798(n)/A002522(n). - Reinhard Zumkeller, Feb 16 2009
a(n) = a(n-1) + 2*n-1 (with a(0)=4). - Vincenzo Librandi, Nov 22 2010
G.f.: (4 - 7*x + 5*x^2)/(1 - x)^3. - Colin Barker, Jan 06 2012
a(n)^3 = A155965(n)^2 + A155966(n)^2. - Vincenzo Librandi, Feb 22 2012
From Amiram Eldar, Jul 13 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + 2*Pi*coth(2*Pi))/8.
Sum_{n>=0} (-1)^n/a(n) = (1 + 2*Pi*cosech(2*Pi))/8 = A371803. (End)
E.g.f.: exp(x)*(4 + x + x^2). - Stefano Spezia, Jul 08 2023
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=0} (1 - 1/a(n)) = sqrt(3)*sinh(sqrt(3)*Pi)/(2*sinh(2*Pi)).
Product_{n>=0} (1 + 1/a(n)) = sqrt(5)*sinh(sqrt(5)*Pi)/(2*sinh(2*Pi)). (End)

A001614 Connell sequence: 1 odd, 2 even, 3 odd, ...

Original entry on oeis.org

1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, 19, 21, 23, 25, 26, 28, 30, 32, 34, 36, 37, 39, 41, 43, 45, 47, 49, 50, 52, 54, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 75, 77, 79, 81, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 122

Offset: 1

N. J. A. Sloane

Next (2n-1) odd numbers alternating with next 2n even numbers. Squares (A000290(n)) occur at the A000217(n)-th entry. - Lekraj Beedassy, Aug 06 2004. - Comment corrected by Daniel Forgues, Jul 18 2009
a(t_n) = a(n(n+1)/2) = n^2 relates squares to triangular numbers. - Daniel Forgues
The natural numbers not included are A118011(n) = 4n - a(n) as n=1,2,3,... - Paul D. Hanna, Apr 10 2006
As a triangle with row sums = A069778 (1, 6, 21, 52, 105, ...): /Q 1;/Q 2, 4;/Q 5, 7, 9;/Q 10, 12, 14, 16;/Q ... . - Gary W. Adamson, Sep 01 2008
The triangle sums, see A180662 for their definitions, link the Connell sequence A001614 as a triangle with six sequences, see the crossrefs. - Johannes W. Meijer, May 20 2011
a(n) = A122797(n) + n - 1. - Reinhard Zumkeller, Feb 12 2012

			From _Omar E. Pol_, Aug 13 2013: (Start)
Written as a triangle the sequence begins:
   1;
   2,  4;
   5,  7,  9;
  10, 12, 14, 16;
  17, 19, 21, 23, 25;
  26, 28, 30, 32, 34, 36;
  37, 39, 41, 43, 45, 47, 49;
  50, 52, 54, 56, 58, 60, 62, 64;
  65, 67, 69, 71, 73, 75, 77, 79, 81;
  82, 84, 86, 88, 90, 92, 94, 96, 98, 100;
  ...
Right border gives A000290, n >= 1.
(End)

C. Pickover, Computers and the Imagination, St. Martin's Press, NY, 1991, p. 276.
C. A. Pickover, The Mathematics of Oz, Chapter 39, Camb. Univ. Press UK 2002.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..1000
Ian Connell and Andrew Korsak, Problem E1382, Amer. Math. Monthly, 67 (1960), 380.
Douglas E. Iannucci and Donna Mills-Taylor, On Generalizing the Connell Sequence, J. Integer Sequences, Vol. 2, 1999, #99.1.7.
H. P. Robinson and N. J. A. Sloane, Correspondence, 1971-1972
N. J. A. Sloane, Handwritten notes on Self-Generating Sequences, 1970 (note that A1148 has now become A005282)
Gary E. Stevens, A Connell-Like Sequence, J. Integer Sequences, Vol. 1, 1998, #98.1.4.
Eric Weisstein's World of Mathematics, Connell Sequence

Cf. A117384, A118011 (complement), A118012.
Cf. A069778. - Gary W. Adamson, Sep 01 2008
From Johannes W. Meijer, May 20 2011: (Start)
Triangle columns: A002522, A117950 (n>=1), A117951 (n>=2), A117619 (n>=3), A154533 (n>=5), A000290 (n>=1), A008865 (n>=2), A028347 (n>=3), A028878 (n>=1), A028884 (n>=2), A054569 [T(2*n,n)].
Triangle sums (see the comments): A069778 (Row1), A190716 (Row2), A058187 (Related to Kn11, Kn12, Kn13, Kn21, Kn22, Kn23, Fi1, Fi2, Ze1 and Ze2), A000292 (Related to Kn3, Kn4, Ca3, Ca4, Gi3 and Gi4), A190717 (Related to Ca1, Ca2, Ze3, Ze4), A190718 (Related to Gi1 and Gi2). (End)
Cf. A014132, A057211, A023531.

a001614 n = a001614_list !! (n-1)
a001614_list = f 0 0 a057211_list where
   f c z (x:xs) = z' : f x z' xs where z' = z + 1 + 0 ^ abs (x - c)
-- Reinhard Zumkeller, Dec 30 2011

[2*n-Round(Sqrt(2*n)): n in [1..80]]; // Vincenzo Librandi, Apr 17 2015

A001614:=proc(n): 2*n - floor((1+sqrt(8*n-7))/2) end: seq(A001614(n),n=1..67); # Johannes W. Meijer, May 20 2011

lst={};i=0;For[j=1, j<=4!, a=i+1;b=j;k=0;For[i=a, i<=9!, k++;AppendTo[lst, i];If[k>=b, Break[]];i=i+2];j++ ];lst (* Vladimir Joseph Stephan Orlovsky, Aug 29 2008 *)
row[n_] := 2*Range[n+1]+n^2-1; Table[row[n], {n, 0, 11}] // Flatten (* Jean-François Alcover, Oct 25 2013 *)

a(n)=2*n - round(sqrt(2*n)) \\ Charles R Greathouse IV, Apr 20 2015

from math import isqrt
def A001614(n): return (m:=n<<1)-(k:=isqrt(m))-int((m<<2)>(k<<2)*(k+1)+1) # Chai Wah Wu, Jul 26 2022

a(n) = 2*n - floor( (1+ sqrt(8*n-7))/2 ).
a(n) = A005843(n) - A002024(n). - Lekraj Beedassy, Aug 06 2004
a(n) = A118012(A118011(n)). A117384( a(n) ) = n; A117384( 4*n - a(n) ) = n. - Paul D. Hanna, Apr 10 2006
a(1) = 1; then a(n) = a(n-1)+1 if a(n-1) is a square, a(n) = a(n-1)+2 otherwise. For example, a(21)=36 is a square therefore a(22)=36+1=37 which is not a square so a(23)=37+2=39 ... - Benoit Cloitre, Feb 07 2007
T(n,k) = (n-1)^2 + 2*k - 1. - Omar E. Pol, Aug 13 2013
a(n)^2 = a(n*(n+1)/2). - Ivan N. Ianakiev, Aug 15 2013
Let the sequence be written in the form of the triangle in the EXAMPLE section below and let a(n) and a(n+1) belong to the same row of the triangle. Then a(n)*a(n+1) + 1 = a(A000217(A118011(n))) = A000290(A118011(n)). - Ivan N. Ianakiev, Aug 16 2013
a(n) = 2*n-round(sqrt(2*n)). - Gerald Hillier, Apr 15 2015
From Robert Israel, Apr 20 2015 (Start):
G.f.: 2*x/(1-x)^2 - (x/(1-x))*Sum_{n>=0} x^(n*(n+1)/2) = 2*x/(1-x)^2 - (Theta2(0,x^(1/2)))*x^(7/8)/(2*(1-x)) where Theta2 is a Jacobi theta function.
a(n) = 2*n-1 - Sum_{i=0..n-2} A023531(i).  (End)
a(n) = 3*n-A014132(n). - Chai Wah Wu, Oct 19 2024

More terms from Larry Reeves (larryr(AT)acm.org), Mar 16 2001

A140090 a(n) = n(3n + 7)/2.

Original entry on oeis.org

0, 5, 13, 24, 38, 55, 75, 98, 124, 153, 185, 220, 258, 299, 343, 390, 440, 493, 549, 608, 670, 735, 803, 874, 948, 1025, 1105, 1188, 1274, 1363, 1455, 1550, 1648, 1749, 1853, 1960, 2070, 2183, 2299, 2418, 2540, 2665, 2793, 2924

Offset: 0

Omar E. Pol, May 22 2008

This sequence is mentioned in the Guo-Niu Han's paper, chapter 6: Dictionary of the standard puzzle sequences, p. 19 (see link). - Omar E. Pol, Oct 28 2011

Number of cards needed to build an n-tier house of cards with a flat, one-card-wide roof. - Tyler Busby, Dec 28 2022

G. C. Greubel, Table of n, a(n) for n = 0..5000
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 5.
Julie Zhang, Noah A. Rosenberg, and Julia A. Palacios, The space of multifurcating ranked tree shapes: enumeration, lattice structure, and Markov chains, arXiv:2506.10856 [math.PR], 2025. See p. 33.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, A005449, A045943, A115067, this sequence, A140091, A059845, A140672, A140673, A140674, A140675, A151542.

Cf. numbers of the form n*(d*n + 10 - d)/2: A008587, A056000, A028347, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734, A139273.

LinearRecurrence[{3,-3,1},{0,5,13},50] (* Harvey P. Dale, Jan 17 2022 *)

a(n)=n*(3*n+7)/2 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(5 - 2*x)/(1 - x)^3. - Bruno Berselli, Feb 11 2011
a(n) = (3*n^2 + 7*n)/2.
a(n) = a(n-1) + 3*n + 2 (with a(0)=0). - Vincenzo Librandi, Nov 24 2010
E.g.f.: (1/2)*(3*x^2 + 10*x)*exp(x). - G. C. Greubel, Jul 17 2017
From Amiram Eldar, Feb 22 2022: (Start)
Sum_{n>=1} 1/a(n) = 117/98 - Pi/(7*sqrt(3)) - 3*log(3)/7.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/(7*sqrt(3)) + 4*log(2)/7 - 75/98. (End)

A067725 a(n) = 3n^2 + 6n.

Original entry on oeis.org

0, 9, 24, 45, 72, 105, 144, 189, 240, 297, 360, 429, 504, 585, 672, 765, 864, 969, 1080, 1197, 1320, 1449, 1584, 1725, 1872, 2025, 2184, 2349, 2520, 2697, 2880, 3069, 3264, 3465, 3672, 3885, 4104, 4329, 4560, 4797, 5040, 5289, 5544, 5805, 6072, 6345, 6624

Offset: 0

Robert G. Wilson v, Feb 05 2002

Numbers h such that 3*(3 + h) is a perfect square. - Alexander D. Healy, Tj Tullo, Avery Pickford, Sep 20 2004

Equivalently, numbers k such that k/3+1 is a square. - Bruno Berselli, Apr 10 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).

Cf. A005563.

Cf. numbers k such that k*(k + m) is a perfect square: A028560 (k=9), A067728 (k=8), A067727 (k=7), A067726 (k=6), A067724 (k=5), A028347 (k=4), A054000 (k=2), A005563 (k=1).

List([0..50], n-> 3*n*(n+2)); # G. C. Greubel, Sep 01 2019

[3*n*(n+2): n in [0..50]]; // G. C. Greubel, Sep 01 2019

seq(3*n*(n+2), n=0..50); # G. C. Greubel, Sep 01 2019

Select[ Range[10000], IntegerQ[ Sqrt[ 3(3 + # )]] & ]
3*(Range[50]^2 -1) (* G. C. Greubel, Sep 01 2019 *)

a(n)=3*n*(n+2) \\ Charles R Greathouse IV, Dec 07 2011

[3*n*(n+2) for n in (0..50)] # G. C. Greubel, Sep 01 2019

a(n) = 3*A005563(n). - Zerinvary Lajos, Mar 06 2007
a(n) = a(n-1) + 6*n + 3, with n>0, a(0)=0. - Vincenzo Librandi, Aug 08 2010
From Colin Barker, Apr 11 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 3*x*(3-x)/(1-x)^3. (End)
E.g.f.: 3*x*(x + 3)*exp(x). - G. C. Greubel, Jul 20 2017
From Amiram Eldar, Feb 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/4.
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/12. (End)

Edited by N. J. A. Sloane, Sep 14 2008 at the suggestion of R. J. Mathar

A062264 Coefficient triangle of certain polynomials N(4; m,x).

Original entry on oeis.org

1, 1, 5, 1, 12, 15, 1, 21, 63, 35, 1, 32, 168, 224, 70, 1, 45, 360, 840, 630, 126, 1, 60, 675, 2400, 3150, 1512, 210, 1, 77, 1155, 5775, 11550, 9702, 3234, 330, 1, 96, 1848, 12320, 34650, 44352, 25872, 6336, 495, 1, 117, 2808, 24024, 90090, 162162, 144144, 61776, 11583, 715

Offset: 0

Wolfdieter Lang, Jun 19 2001

The e.g.f. of the m-th (unsigned) column sequence without leading zeros of the generalized (a=4) Laguerre triangle L(4; n+m,m) = A062140(n+m,m), n >= 0, is N(4; m,x)/(1-x)^(5+2*m), with the row polynomials N(4; m,x) := Sum_{k=0..m} T(m,k)*x^k.

			Triangle begins as:
  1;
  1,   5;
  1,  12,   15;
  1,  21,   63,    35;
  1,  32,  168,   224,     70;
  1,  45,  360,   840,    630,    126;
  1,  60,  675,  2400,   3150,   1512,    210;
  1,  77, 1155,  5775,  11550,   9702,   3234,    330;
  1,  96, 1848, 12320,  34650,  44352,  25872,   6336,    495;
  1, 117, 2808, 24024,  90090, 162162, 144144,  61776,  11583,   715;
  1, 140, 4095, 43680, 210210, 504504, 630630, 411840, 135135, 20020, 1001;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Family of polynomials (see A062145): A008459 (c=1), A132813 (c=2), A062196 (c=3), A062145 (c=4), this sequence (c=5), A062190 (c=6).
Columns: A028347 (k=2), A104473 (k=3), A104474 (k=4), A104475 (k=5), A027814 (k=6), A103604 (k=7), A104476 (k=8), A104478 (k=9).
Diagonals: A000332 (k=n), A027810 (k=n-1), A105249 (k=n-2), A105250 (k=n-3), A105251 (k=n-4), A105252 (k=n-5), A105253 (k=n-6), A105254 (k=n-7).
Sums: A002694 (row).

A062264:= func< n,k | Binomial(n,k)*Binomial(n+4,k) >;
[A062264(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 03 2025

A062264[n_, k_]:= Binomial[n,k]*Binomial[n+4,k];
Table[A062264[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 03 2025 *)

def A062264(n,k): return binomial(n,k)*binomial(n+4,k)
print(flatten([[A062264(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Mar 03 2025

T(m, k) = [x^k] N(4; m, x), with N(4; m, x) = ((1-x)^(2*m+5))*(d^m/dx^m)((x^m)/(m!*(1-x)^(m+5))).
N(4; m, x) = Sum_{j=0..m} (binomial(m, j)*(2*m+4-j)!/((m+4)!*(m-j)!)*(x^(m-j))*(1-x)^j).
From G. C. Greubel, Mar 03 2025: (Start)
T(n, k) = binomial(n,k)*binomial(n+4,k).
Sum_{k=0..n} (-1)^k*T(n, k) = (1/4)*( (1+(-1)^n)*(-1)^((n+2)/2)*(n^2 + 5*n - 2)*Catalan((n+2)/2)/(n+1) + 8*(1-(-1)^n)*(-1)^((n+1)/2)*Catalan((n+1)/2) ). (End)

A022267 a(n) = n(9n + 1)/2.

Original entry on oeis.org

0, 5, 19, 42, 74, 115, 165, 224, 292, 369, 455, 550, 654, 767, 889, 1020, 1160, 1309, 1467, 1634, 1810, 1995, 2189, 2392, 2604, 2825, 3055, 3294, 3542, 3799, 4065, 4340, 4624, 4917, 5219, 5530, 5850, 6179

Offset: 0

N. J. A. Sloane

From Floor van Lamoen, Jul 21 2001: (Start)
Write 0, 1, 2, 3, 4, ... in a triangular spiral; then a(n) is the sequence found by reading the line from 0 in the direction 0, 5, ... . The spiral begins:
.
            15
            / \
          16  14
          /     \
        17   3  13
        /   / \   \
      18   4   2  12
      /   /     \   \
    19   5   0---1  11
    /   /             \
  20   6---7---8---9--10
.
(End)
a(n) is the sum of n consecutive integers starting from 4*n+1: (5), (9+10), (13+14+15), ... - Klaus Purath, Jul 07 2020
a(n) with n>0 are the numbers with the periodic length 3 in the Bulgarian and Mancala solitaire. - Paul Weisenhorn, Jan 29 2022

Lei Zhou, Table of n, a(n) for n = 0..10000
Milan Janjic, Two Enumerative Functions
Leo Tavares, Illustration: X Triangles
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A051682, A110449, A235332.
Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A033429, A022268, A049452, A186030, A135703, A152734, A139273.
Cf. similar sequences listed in A254963.
Cf. similar sequences listed in A022289.
Cf. A060544, A016813.

seq(binomial(9*n+1,2)/9, n=0..37); # Zerinvary Lajos, Jan 21 2007

Table[ n (9 n + 1)/2, {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 19}, 40] (* Harvey P. Dale, Jul 01 2013 *)

vector(100,n,(n-1)*(9*n-8)/2) \\ Derek Orr, Feb 06 2015

a(n) = A110449(n, 4) for n>3.
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 4*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A000566(n). (End)
a(n) = 9*n + a(n-1) - 4 with n>0, a(0)=0. - Vincenzo Librandi, Aug 04 2010
a(n) = A218470(9*n+4). - Philippe Deléham, Mar 27 2013
a(n) = A000217(5*n) - A000217(4*n). - Bruno Berselli, Oct 13 2016
E.g.f.: (1/2)*(9*x^2 + 10*x)*exp(x). - G. C. Greubel, Jul 17 2017
a(n) = A060544(n+1) - A016813(n). - Leo Tavares, Mar 20 2022

A158405 Triangle T(n,m) = 1+2*m of odd numbers read along rows, 0<=m

Original entry on oeis.org

1, 1, 3, 1, 3, 5, 1, 3, 5, 7, 1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 11, 1, 3, 5, 7, 9, 11, 13, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, 7, 9, 11, 13, 15, 17, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23

Offset: 1

Paul Curtz, Mar 18 2009

Row sums are n^2 = A000290(n).
The triangle sums, see A180662 for their definitions, link this triangle of odd numbers with seventeen different sequences, see the crossrefs. The knight sums Kn14 - Kn110 have been added. - Johannes W. Meijer, Sep 22 2010
A208057 is the eigentriangle of A158405 such that as infinite lower triangular matrices, A158405 * A208057 shifts the latter, deleting the right border of 1's. - Gary W. Adamson, Feb 22 2012
T(n,k) = A099375(n-1,n-k), 1<=k<=n. [Reinhard Zumkeller, Mar 31 2012]

			The triangle contains the first n odd numbers in row n:
  1;
  1,3;
  1,3,5;
  1,3,5,7;
From _Seiichi Manyama_, Dec 02 2017: (Start)
    |       a(n)        |                               | A000290(n)
   -----------------------------------------------------------------
   0|                                                      (=  0)
   1|                 1 = 1/3 * ( 3)                       (=  1)
   2|             1 + 3 = 1/3 * ( 5 +  7)                  (=  4)
   3|         1 + 3 + 5 = 1/3 * ( 7 +  9 + 11)             (=  9)
   4|     1 + 3 + 5 + 7 = 1/3 * ( 9 + 11 + 13 + 15)        (= 16)
   5| 1 + 3 + 5 + 7 + 9 = 1/3 * (11 + 13 + 15 + 17 + 19)   (= 25)
(End)

Seiichi Manyama, Rows n = 1..140, flattened
Daniel Erman, The Josephus Problem, Numberphile video (2016)
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Index entries for sequences related to the Josephus Problem

Cf. A129326, A099375, A005408.
Triangle sums (see the comments): A000290 (Row1; Kn11 & Kn4 & Ca1 & Ca4 & Gi1 & Gi4); A000027 (Row2); A005563 (Kn12); A028347 (Kn13); A028560 (Kn14); A028566 (Kn15); A098603 (Kn16); A098847 (Kn17); A098848 (Kn18); A098849 (Kn19); A098850 (Kn110); A000217 (Kn21. Kn22, Kn23, Fi2, Ze2); A000384 (Kn3, Fi1, Ze3); A000212 (Ca2 & Ze4); A000567 (Ca3, Ze1); A011848 (Gi2); A001107 (Gi3). - Johannes W. Meijer, Sep 22 2010
Cf. A063656, A063657, A208057, A292610, A292611.

a158405 n k = a158405_row n !! (k-1)
a158405_row n = a158405_tabl !! (n-1)
a158405_tabl = map reverse a099375_tabl
-- Reinhard Zumkeller, Mar 31 2012

Table[2 Range[1, n] - 1, {n, 12}] // Flatten (* Michael De Vlieger, Oct 01 2015 *)

a(n) = 2*(n-floor((-1+sqrt(8*n-7))/2)*(floor((-1+sqrt(8*n-7))/2)+1)/2)-1;
vector(100, n, a(n)) \\ Altug Alkan, Oct 01 2015

a(n) = 2*i-1, where i = n-t(t+1)/2, t = floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Feb 03 2013

a(n) = 2*A002262(n-1) + 1. - Eric Werley, Sep 30 2015

Edited by R. J. Mathar, Oct 06 2009

A120071 a(n) = n*(n+20).

Original entry on oeis.org

0, 21, 44, 69, 96, 125, 156, 189, 224, 261, 300, 341, 384, 429, 476, 525, 576, 629, 684, 741, 800, 861, 924, 989, 1056, 1125, 1196, 1269, 1344, 1421, 1500, 1581, 1664, 1749, 1836, 1925, 2016, 2109, 2204, 2301, 2400, 2501, 2604, 2709, 2816, 2925, 3036, 3149, 3264

Offset: 0

Wolfdieter Lang, Jul 20 2006

Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

a(n-10), n >= 11, tenth column (used for the n=10 series of the hydrogen atom) of triangle A120070.
For n*(n+18) see A098850.
Cf. A001008, A001477, A005563, A028347, A028552, A028557, A028560, A028563, A028566, A028569, A098603, A098847, A098848, A098849, A102928, A120061, A132759, A132760, A132761, A132762, A132765.

seq(sum(4*k-n, k=7..n), n=6..51); # Zerinvary Lajos, Feb 17 2008

s=0;lst={};Do[s+=n;AppendTo[lst,s],{n,21,6!,2}];lst (* Vladimir Joseph Stephan Orlovsky, Feb 26 2009 *)

a(n)=n*(n+20) \\ Charles R Greathouse IV, Jan 21 2015

a(n) = (n+10)^2 - 10^2 = n*(n+20), n >= 0.
G.f.: x*(21-19*x)/(1-x)^3.
a(n) = 2*n + a(n-1) + 19 (with a(0)=0). - Vincenzo Librandi, Nov 13 2010
From Amiram Eldar, Jan 16 2021: (Start)
Sum_{n>=1} 1/a(n) = H(20)/20 = A001008(20)/A102928(20) = 11167027/62078016, where H(k) is the k-th harmonic number.
Sum_{n>=1} (-1)^(n+1)/a(n) = 155685007/4655851200. (End)
From Elmo R. Oliveira, Jan 12 2025: (Start)
E.g.f.: exp(x)*x*(21 + x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

A132765 a(n) = n*(n + 23).

Original entry on oeis.org

0, 24, 50, 78, 108, 140, 174, 210, 248, 288, 330, 374, 420, 468, 518, 570, 624, 680, 738, 798, 860, 924, 990, 1058, 1128, 1200, 1274, 1350, 1428, 1508, 1590, 1674, 1760, 1848, 1938, 2030, 2124, 2220, 2318, 2418, 2520, 2624, 2730, 2838, 2948, 3060, 3174, 3290, 3408

Offset: 0

Omar E. Pol, Aug 28 2007

G. C. Greubel, Table of n, a(n) for n = 0..5000
Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, ResearchGate, 2014.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001008, A001477, A002378, A005563, A028347, A028552, A028557, A028560, A028563, A028566, A028569, A056126, A098603, A098847, A098848, A098849, A098850, A102928, A120071, A132759, A132760, A132761, A132762, A132763, A132764, A132765, A132766.

Table[n (n + 23), {n, 0, 50}] (* Bruno Berselli, Sep 03 2018 *)

a(n)=n*(n+23) \\ Charles R Greathouse IV, Jun 17 2017

[n*(n+23) for n in (0..50)] # G. C. Greubel, Mar 14 2022

a(n) = n*(n + 23).
a(n) = 2*n + a(n-1) + 22 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Chai Wah Wu, Dec 17 2016: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
G.f.: 2*x*(12 - 11*x)/(1-x)^3. (End)
From Amiram Eldar, Jan 16 2021: (Start)
Sum_{n>=1} 1/a(n) = H(23)/23 = A001008(23)/A102928(23) = 444316699/2736605872, where H(k) is the k-th harmonic number.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/23 - 3825136961/123147264240. (End)
E.g.f.: x*(24 + x)*exp(x). - G. C. Greubel, Mar 14 2022

A094728 Triangle read by rows: T(n,k) = n^2 - k^2, 0 <= k < n.

Original entry on oeis.org

1, 4, 3, 9, 8, 5, 16, 15, 12, 7, 25, 24, 21, 16, 9, 36, 35, 32, 27, 20, 11, 49, 48, 45, 40, 33, 24, 13, 64, 63, 60, 55, 48, 39, 28, 15, 81, 80, 77, 72, 65, 56, 45, 32, 17, 100, 99, 96, 91, 84, 75, 64, 51, 36, 19, 121, 120, 117, 112, 105, 96, 85, 72, 57, 40, 21

Offset: 1

Reinhard Zumkeller, May 24 2004

(T(n,k) mod 4) <> 2, see A042965, A016825.
All numbers m occur A034178(m) times.
The row polynomials T(n,x) appear in the calculation of the column g.f.s of triangle A120070 (used to find the frequencies of the spectral lines of the hydrogen atom).

			n=3: T(3,x) = 9+8*x+5*x^2.
Triangle begins:
   1;
   4,  3;
   9,  8,  5;
  16, 15, 12,  7;
  25, 24, 21, 16,  9;
  36, 35, 32, 27, 20, 11;
  49, 48, 45, 40, 33, 24, 13;
  64, 63, 60, 55, 48, 39, 28, 15;
  81, 80, 77, 72, 65, 56, 45, 32, 17;
  ... etc. - _Philippe Deléham_, Mar 07 2013

G. C. Greubel, Table of n, a(n) for n = 1..5050

Cf. A000290, A000567, A004736, A005408, A005563, A008586, A008590.
Cf. A008594, A008598, A016825, A016945, A017329, A028347, A028560.
Cf. A028566, A034178, A042965, A094727, A120070, A128624.
Cf. A000384 (alternating row sums), A002412 (row sums).

[n^2-k^2: k in [0..n-1], n in [1..15]]; // G. C. Greubel, Mar 12 2024

Table[n^2 - k^2, {n,12}, {k,0,n-1}]//Flatten (* Michael De Vlieger, Nov 25 2015 *)

flatten([[n^2-k^2 for k in range(n)] for n in range(1,16)]) # G. C. Greubel, Mar 12 2024

Row polynomials: T(n,x) = n^2*Sum_{m=0..n} x^m - Sum_{m=0..n} m^2*x^m = Sum_{k=0..n-1} T(n,k)*x^k, n >= 1.
T(n, k) = A004736(n,k)*A094727(n,k).
T(n, 0) = A000290(n).
T(n, 1) = A005563(n-1) for n>1.
T(n, 2) = A028347(n) for n>2.
T(n, 3) = A028560(n-3) for n>3.
T(n, 4) = A028566(n-4) for n>4.
T(n, n-1) = A005408(n).
T(n, n-2) = A008586(n-1) for n>1.
T(n, n-3) = A016945(n-2) for n>2.
T(n, n-4) = A008590(n-2) for n>3.
T(n, n-5) = A017329(n-3) for n>4.
T(n, n-6) = A008594(n-3) for n>5.
T(n, n-8) = A008598(n-2) for n>7.
T(A005408(k), k) = A000567(k).
Sum_{k=0..n} T(n, k) = A002412(n) (row sums).
From G. C. Greubel, Mar 12 2024: (Start)
Sum_{k=0..n-1} (-1)^k * T(n, k) = A000384(floor((n+1)/2)).
Sum_{k=0..floor((n-1)/2)} T(n-k, k) = A128624(n).
Sum_{k=0..floor((n-1)/2)} (-1)^k*T(n-k, k) = (1/2)*n*(n+1 - (-1)^n*cos(n*Pi/2)). (End)
G.f.: x*(1 - 3*x^2*y + x*(1 + y))/((1 - x)^3*(1 - x*y)^2). - Stefano Spezia, Aug 04 2025

cp's OEIS Frontend

A087475 a(n) = n^2 + 4.

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A001614 Connell sequence: 1 odd, 2 even, 3 odd, ...

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A140090 a(n) = n*(3*n + 7)/2.

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A067725 a(n) = 3*n^2 + 6*n.

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A140090 a(n) = n(3n + 7)/2.

A067725 a(n) = 3n^2 + 6n.

A022267 a(n) = n(9n + 1)/2.