A098181 -id:A098181 - cp's OEIS frontend

A194274 Concentric square numbers (see Comments lines for definition).

0, 1, 4, 8, 12, 17, 24, 32, 40, 49, 60, 72, 84, 97, 112, 128, 144, 161, 180, 200, 220, 241, 264, 288, 312, 337, 364, 392, 420, 449, 480, 512, 544, 577, 612, 648, 684, 721, 760, 800, 840, 881, 924, 968, 1012, 1057, 1104, 1152, 1200, 1249, 1300, 1352, 1404

Offset: 0

Omar E. Pol, Aug 20 2011

Cellular automaton on the first quadrant of the square grid. The sequence gives the number of cells "ON" in the structure after n-th stage. A098181 gives the first differences. For a definition without words see the illustration of initial terms in the example section. For other concentric polygonal numbers see A194273, A194275 and A032528.
Also, union of A046092 and A077221, the bisections of this sequence.
Also row sums of an infinite square array T(n,k) in which column k lists 4*k-1 zeros followed by the numbers A008574 (see example).

			Using the numbers A008574 we can write:
0, 1, 4, 8, 12, 16, 20, 24, 28, 32, 36, ...
0, 0, 0, 0, 0,  1,   4,  8, 12, 16, 20, ...
0, 0, 0, 0, 0,  0,   0,  0,  0,  1,  4, ...
And so on.
===========================================
The sums of the columns give this sequence:
0, 1, 4, 8, 12, 17, 24, 32, 40, 49, 60, ...
...
Illustration of initial terms:
.                                         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
.
. 1    4      8        12         17           24

Arkadiusz Wesolowski, Table of n, a(n) for n = 0..10000
Index entries for sequences related to cellular automata
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. A000290, A005563, A008574, A056594, A032528, A046092.

Cf. A077221, A085250, A098181, A194273, A194275.

[n le 2 select n-1 else (n-1)^2 - Self(n-2): n in [1..61]]; // G. C. Greubel, Jan 31 2024

Table[Floor[3*n/4] + Floor[(n*(n + 2) + 1)/2] - Floor[(3*n + 1)/4], {n, 0, 52}] (* Arkadiusz Wesolowski, Nov 08 2011 *)
RecurrenceTable[{a[0]==0,a[1]==1,a[n]==n^2-a[n-2]},a,{n,60}] (* or *) LinearRecurrence[{3,-4,4,-3,1},{0,1,4,8,12},60] (* Harvey P. Dale, Sep 11 2013 *)

prpr = 0
prev = 1
for n in range(2,777):
    print(str(prpr), end=", ")
    curr = n*n - prpr
    prpr = prev
    prev = curr
# Alex Ratushnyak, Aug 03 2012

def A194274(n): return (3*n>>2)+(n*(n+2)+1>>1)-(3*n+1>>2) # Chai Wah Wu, Jul 15 2023

def A194274(n): return n if n<2 else n^2 - A194274(n-2)
[A194274(n) for n in range(41)] # G. C. Greubel, Jan 31 2024

a(n) = n^2 - a(n-2), with a(0)=0, a(1)=1. - Alex Ratushnyak, Aug 03 2012
From R. J. Mathar, Aug 22 2011: (Start)
G.f.: x*(1 + x)/((1 + x^2)*(1 - x)^3).
a(n) = (A005563(n) - A056594(n-1))/2. (End)
a(n) = a(-n-2) = (2*n*(n+2) + (1-(-1)^n)*i^(n+1))/4, where i=sqrt(-1). - Bruno Berselli, Sep 22 2011
a(n) = floor(3*n/4) + floor((n*(n+2)+1)/2) - floor((3*n+1)/4). - Arkadiusz Wesolowski, Nov 08 2011
a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5), with a(0)=0, a(1)=1, a(2)=4, a(3)=8, a(4)=12. - Harvey P. Dale, Sep 11 2013
E.g.f.: (exp(x)*x*(3 + x) - sin(x))/2. - Stefano Spezia, Feb 26 2023

A152204 Triangle read by rows: T(n,k) = 2n-4k+5 (n >= 0, 1 <= k <= 1+floor(n/2)).

Original entry on oeis.org

1, 3, 5, 1, 7, 3, 9, 5, 1, 11, 7, 3, 13, 9, 5, 1, 15, 11, 7, 3, 17, 13, 9, 5, 1, 19, 15, 11, 7, 3, 21, 17, 13, 9, 5, 1, 23, 19, 15, 11, 7, 3, 25, 21, 17, 13, 9, 5, 1, 27, 23, 19, 15, 11, 7, 3, 29, 25, 21, 17, 13, 9, 5, 1, 31, 27, 23, 19, 15, 11, 7, 3, 33

Offset: 0

Gary W. Adamson, Nov 29 2008

All terms are odd, decreasing across rows. Row sums = A000217, the triangular numbers.
From Johannes W. Meijer, Sep 08 2013: (Start)
Triangle read by rows formed from the antidiagonals of triangle A099375.
The alternating row sums equal A098181(n). (End)

			First few rows of the triangle:
  1
  3
  5  1
  7  3
  9  5  1
  11 7  3
  13 9  5  1
  15 11 7  3
  17 13 9  5 1
  19 15 11 7 3
  21 17 13 9 5 1
  ...

Nathaniel Johnston, Rows n = 0..200 of irregular triangle, flattened

Cf. A000217.

T := proc(n,k) return 2*n-4*k+5: end: seq(seq(T(n,k), k=1..1+floor(n/2)), n=0..20); # Nathaniel Johnston, May 01 2011

By columns, odd terms in every column, n-th column starts at row (2*n).
From Johannes W. Meijer, Sep 08 2013: (Start)
T(n, k) = A099375(n-k+1, k-1), n >= 0 and 1 <= k <= 1+floor(n/2).
T(n, k) = A158405(n+1, n-2*k+2). (End)

Edited by N. J. A. Sloane, Sep 25 2010, following a suggestion from Emeric Deutsch

Offset corrected by Johannes W. Meijer, Sep 07 2013

A174882 A (3/2,-1) Somos-4 sequence.

Original entry on oeis.org

1, 1, -2, -8, -16, -16, 32, 128, 256, 256, -512, -2048, -4096, -4096, 8192, 32768, 65536, 65536, -131072, -524288, -1048576, -1048576, 2097152, 8388608, 16777216, 16777216, -33554432, -134217728, -268435456, -268435456

Offset: 0

Paul Barry, Mar 31 2010

Hankel transform of A051286. a(n+2) = -(-1)^floor(n/4) * 2^A098181(n).

			G.f. = 1 + x - 2*x^2 - 8*x^3 - 16*x^4 - 16*x^5 + 32*x^6 + 128*x^7 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,-16).

Cf. A051286, A098181.

Q:=Rationals(); R:=PowerSeriesRing(Q, 40); Coefficients(R!((1-2*x)*(4*x^2+3*x+1)/(1+16*x^4))) // G. C. Greubel, Feb 21 2018

a[ n_] := (-1)^Quotient[n + 2, 4] 2^(n - Mod[ Quotient[n + 1, 2], 2]); (* Michael Somos, Sep 18 2014 *)
CoefficientList[Series[(1-2*x)*(4*x^2+3*x+1)/(1+16*x^4), {x,0,50}], x] (* G. C. Greubel, Feb 21 2018 *)

{a(n) = (-1)^((n+2) \ 4) * 2^(n - ((n+1) \ 2 % 2))}; /* Michael Somos, Jan 06 2011 */

x='x+O('x^30); Vec((1-2*x)*(4*x^2+3*x+1)/(1+16*x^4)) \\ G. C. Greubel, Feb 21 2018

a(n) = ((3/2)*a(n-1)*a(n-3) - a(n-2)^2)/a(n-4), n>3.
a(-n) = a(n-1) / 2^(2*n - 1) for all n in Z. - Michael Somos, Jan 06 2011
0 = a(n)*(+2*a(n+4)) + a(n+1)*(-3*a(n+3)) + a(n+2)*(+2*a(n+2)) for all n in Z. - Michael Somos, Sep 18 2014
a(n+4) = -16 * a(n) for all n in Z. - Michael Somos, Sep 02 2015
G.f.: -(2*x-1)*(4*x^2+3*x+1)/(1+16*x^4) . - R. J. Mathar, Aug 18 2017

A386820 a(n) is the size of largest subset of {1, 1/2, ..., 1/n} that can be partitioned into two parts, the sum of elements of which are equal.

Original entry on oeis.org

0, 0, 0, 0, 0, 4, 4, 4, 4, 4, 4, 6, 6, 6, 9, 9, 9, 11, 11, 11, 14, 14, 14

Offset: 1

Yifan Xie, Aug 03 2025

			For a(6) = 4, the set {1, 1/2, 1/3, 1/6} is chosen because 1 = 1/2 + 1/3 + 1/6.
The two parts for a(18) = 11 are 1 + 1/5 + 1/6 + 1/15 = 1/2 + 1/3 + 1/4 + 1/9 + 1/10 + 1/12 + 1/18.
The two parts for a(20) = 11 are 1 + 1/9 + 1/10 + 1/15 + 1/18 = 1/2 + 1/3 + 1/5 + 1/6 + 1/12 + 1/20.

Art of Problem Solving, 2020 GQMO Problem 4, which shows that a(n) >= 2*n/5 for sufficiently large n.
Vincent Jugé, Proof that a(n) > c*n for all real numbers c < 1 and sufficiently large n

Cf. A001008, A098181.

a(n) = a(n-1) if n/k = p is a prime and p > A001008(k).

A116188 Triangle T(n,k), 0<=k<=n : T(n,k)is smallest number such that T(n,k)>=T(n-1,k-1), T(n,k)>=T(n-1,k), T(n,k)and T(n-1,k-1)+T(n-1,k) have the same parity, T(0,0)=1 .

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 4, 4, 1, 1, 5, 4, 4, 5, 1, 1, 6, 5, 4, 5, 6, 1, 1, 7, 7, 5, 5, 7, 7, 1, 1, 8, 8, 8, 6, 8, 8, 8, 1, 1, 9, 8, 8, 8, 8, 8, 8, 9, 1, 1, 10, 9, 8, 8, 8, 8, 8, 9, 10, 1, 1, 11, 11, 9, 8, 8, 8, 8, 9, 11, 11, 1, 1, 12, 12, 12, 9, 8, 8, 8, 9, 12, 12, 12, 1, 1, 13, 12, 12, 13, 9

Offset: 0

Philippe Deléham, Apr 08 2007

Sequence read mod 2 gives A047999 .

			Triangle begins:
1;
1, 1;
1, 2, 1;
1, 3, 3, 1;
1, 4, 4, 4, 1;
1, 5, 4, 4, 5, 1;
1, 6, 5, 4, 5, 6, 1;
1, 7, 7, 5, 5, 7, 7, 1;
1, 8, 8, 8, 6, 8, 8, 8, 1;
1, 9, 8, 8, 8, 8, 8, 8, 9, 1;

Diagonals : A000012, A000027, A098181, A127932, A127934.

cp's OEIS Frontend

A194274 Concentric square numbers (see Comments lines for definition).

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A152204 Triangle read by rows: T(n,k) = 2*n-4*k+5 (n >= 0, 1 <= k <= 1+floor(n/2)).

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A174882 A (3/2,-1) Somos-4 sequence.

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A386820 a(n) is the size of largest subset of {1, 1/2, ..., 1/n} that can be partitioned into two parts, the sum of elements of which are equal.

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A116188 Triangle T(n,k), 0<=k<=n : T(n,k)is smallest number such that T(n,k)>=T(n-1,k-1), T(n,k)>=T(n-1,k), T(n,k)and T(n-1,k-1)+T(n-1,k) have the same parity, T(0,0)=1 .

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A152204 Triangle read by rows: T(n,k) = 2n-4k+5 (n >= 0, 1 <= k <= 1+floor(n/2)).