A093353 -id:A093353 - cp's OEIS frontend

A294274 Sum of the seventh powers of the parts in the partitions of n into two parts.

0, 2, 129, 2444, 18700, 99012, 376761, 1216688, 3297456, 8158550, 18080425, 37847532, 73399404, 136971464, 241561425, 414517952, 680856256, 1095977898, 1703414961, 2607286700, 3877286700, 5697862412, 8172733129, 11613390384, 16164030000, 22330294142

Offset: 1

Wesley Ivan Hurt, Oct 26 2017

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for sequences related to partitions
Index entries for linear recurrences with constant coefficients, signature (1,8,-8,-28,28,56,-56,-70,70,56,-56,-28,28,8,-8,-1,1).

Sum of k-th powers of the parts in the partitions of n into two parts for k=0..10: A052928 (k=0), A093353 (k=1), A226141 (k=2), A294270 (k=3), A294271 (k=4), A294272 (k=5), A294273 (k=6), this sequence (k=7), A294275 (k=8), A294276 (k=9), A294279 (k=10).

[n^2*(64 - 224*n^2 + 448*n^4 - 381*n^5 + 96*n^6 + 3*n^5*(-1)^n)/768 : n in [1..50]]; // Wesley Ivan Hurt, Jul 12 2025

Table[Sum[i^7 + (n - i)^7, {i, Floor[n/2]}], {n, 40}]

concat(0, Vec(x^2*(2 + 127*x + 2299*x^2 + 15240*x^3 + 61848*x^4 + 151257*x^5 + 262139*x^6 + 306832*x^7 + 260914*x^8 + 151257*x^9 + 60777*x^10 + 15240*x^11 + 2180*x^12 + 127*x^13+ x^14) / ((1 - x)^9*(1 + x)^8) + O(x^40))) \\ Colin Barker, Nov 20 2017

a(n) = Sum_{i=1..floor(n/2)} i^7 + (n-i)^7.
From Colin Barker, Nov 20 2017: (Start)
G.f.: x^2*(2 + 127*x + 2299*x^2 + 15240*x^3 + 61848*x^4 + 151257*x^5 + 262139*x^6 + 306832*x^7 + 260914*x^8 + 151257*x^9 + 60777*x^10 + 15240*x^11 + 2180*x^12 + 127*x^13+ x^14) / ((1 - x)^9*(1 + x)^8).
a(n) = a(n-1) + 8*a(n-2) - 8*a(n-3) - 28*a(n-4) + 28*a(n-5) + 56*a(n-6) - 56*a(n-7) - 70*a(n-8) + 70*a(n-9) + 56*a(n-10) - 56*a(n-11) - 28*a(n-12) + 28*a(n-13) + 8*a(n-14) - 8*a(n-15) - a(n-16) + a(n-17) for n>17.
(End)
a(n) = n^2*(64 - 224*n^2 + 448*n^4 - 381*n^5 + 96*n^6 + 3*n^5*(-1)^n)/768. - Wesley Ivan Hurt, Jul 12 2025

A294275 Sum of the eighth powers of the parts in the partitions of n into two parts.

Original entry on oeis.org

0, 2, 257, 7074, 72354, 469540, 2142595, 7972932, 24684612, 68121958, 167731333, 383769830, 812071910, 1633567432, 3103591687, 5683259528, 9961449608, 16980253770, 27957167625, 45040730666, 70540730666, 108577948908, 163239463563, 241980430540, 351625763020

Offset: 1

Wesley Ivan Hurt, Oct 26 2017

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for sequences related to partitions
Index entries for linear recurrences with constant coefficients, signature (1,9,-9,-36,36,84,-84,-126,126,126,-126,-84,84,36,-36,-9,9,1,-1).

Sum of k-th powers of the parts in the partitions of n into two parts for k=0..10: A052928 (k=0), A093353 (k=1), A226141 (k=2), A294270 (k=3), A294271 (k=4), A294272 (k=5), A294273 (k=6), A294274 (k=7), this sequence (k=8), A294276 (k=9), A294279 (k=10).

[-n*(768-5120*n^2+10752*n^4-15360*n^6+11475*n^7-2560*n^8-45*n^7*(-1)^n)/23040 : n in [1..50]]; // Wesley Ivan Hurt, Jul 12 2025

Table[Sum[i^8 + (n - i)^8, {i, Floor[n/2]}], {n, 40}]

concat(0, Vec( x^2*(2 + 255*x + 6799*x^2 + 62985*x^3 + 335905*x^4 + 1094715*x^5 + 2500907*x^6 + 3982845*x^7 + 4690633*x^8 + 3982845*x^9 + 2489581*x^10 + 1094715*x^11 + 331859*x^12 + 62985*x^13 + 6553*x^14 + 255*x^15 + x^16) / ((1 - x)^10*(1 + x)^9) + O(x^40))) \\ Colin Barker, Nov 20 2017

a(n) = Sum_{i=1..floor(n/2)} i^8 + (n-i)^8.
From Colin Barker, Nov 20 2017: (Start)
G.f.: x^2*(2 + 255*x + 6799*x^2 + 62985*x^3 + 335905*x^4 + 1094715*x^5 + 2500907*x^6 + 3982845*x^7 + 4690633*x^8 + 3982845*x^9 + 2489581*x^10 + 1094715*x^11 + 331859*x^12 + 62985*x^13 + 6553*x^14 + 255*x^15 + x^16) / ((1 - x)^10*(1 + x)^9).
a(n) = a(n-1) + 9*a(n-2) - 9*a(n-3) - 36*a(n-4) + 36*a(n-5) + 84*a(n-6) - 84*a(n-7) - 126*a(n-8) + 126*a(n-9) + 126*a(n-10) - 126*a(n-11) - 84*a(n-12) + 84*a(n-13) + 36*a(n-14) - 36*a(n-15) - 9*a(n-16) + 9*a(n-17) + a(n-18) - a(n-19) for n>19.
(End)
a(n) = -n*(768-5120*n^2+10752*n^4-15360*n^6+11475*n^7-2560*n^8-45*n^7*(-1)^n)/23040. - Wesley Ivan Hurt, Jul 12 2025

A294276 Sum of the ninth powers of the parts in the partitions of n into two parts.

Original entry on oeis.org

0, 2, 513, 20708, 282340, 2255148, 12313161, 52928912, 186884496, 576258110, 1574304985, 3942330372, 9092033028, 19736886008, 40357579185, 78935156288, 147520415296, 266495712282, 464467582161, 788155279940, 1299155279940, 2095793274212, 3300704544313

Offset: 1

Wesley Ivan Hurt, Oct 26 2017

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for sequences related to partitions
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-45,45,120,-120,-210,210,252,-252,-210,210,120,-120,-45,45,10,-10,-1,1).

Sum of k-th powers of the parts in the partitions of n into two parts for k=0..10: A052928 (k=0), A093353 (k=1), A226141 (k=2), A294270 (k=3), A294271 (k=4), A294272 (k=5), A294273 (k=6), A294274 (k=7), A294275 (k=8), this sequence (k=9), A294279 (k=10).

[-n^2*(768-2560*n^2+3584*n^4-3840*n^6+2555*n^7-512*n^8-5*n^7*(-1)^n)/5120 : n in [1..50]]; // Wesley Ivan Hurt, Jul 12 2025

Table[Sum[i^9 + (n - i)^9, {i, Floor[n/2]}], {n, 40}]

concat(0, Vec(x^2*(2 + 511*x + 20175*x^2 + 256522*x^3 + 1770948*x^4 + 7464688*x^5 + 21796206*x^6 + 45087574*x^7 + 69569484*x^8 + 79813090*x^9 + 69501528*x^10 + 45087574*x^11 + 21722580*x^12 + 7464688*x^13 + 1756842*x^14 + 256522*x^15 + 19674*x^16 + 511*x^17+ x^18) / ((1 - x)^11*(1 + x)^10) + O(x^40))) \\ Colin Barker, Nov 21 2017

a(n) = Sum_{i=1..floor(n/2)} i^9 + (n-i)^9.
From Colin Barker, Nov 21 2017: (Start)
G.f.: x^2*(2 + 511*x + 20175*x^2 + 256522*x^3 + 1770948*x^4 + 7464688*x^5 + 21796206*x^6 + 45087574*x^7 + 69569484*x^8 + 79813090*x^9 + 69501528*x^10 + 45087574*x^11 + 21722580*x^12 + 7464688*x^13 + 1756842*x^14 + 256522*x^15 + 19674*x^16 + 511*x^17+ x^18) / ((1 - x)^11*(1 + x)^10).
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - 45*a(n-4) + 45*a(n-5) + 120*a(n-6) - 120*a(n-7) - 210*a(n-8) + 210*a(n-9) + 252*a(n-10) - 252*a(n-11) - 210*a(n-12) + 210*a(n-13) + 120*a(n-14) - 120*a(n-15) - 45*a(n-16) + 45*a(n-17) + 10*a(n-18) - 10*a(n-19) - a(n-20) + a(n-21) for n>21.
(End)
a(n) = -n^2*(768-2560*n^2+3584*n^4-3840*n^6+2555*n^7-512*n^8-5*n^7*(-1)^n)/5120. - Wesley Ivan Hurt, Jul 12 2025

A294279 Sum of the tenth powers of the parts in the partitions of n into two parts.

Original entry on oeis.org

0, 2, 1025, 61098, 1108650, 10933324, 71340451, 354864276, 1427557524, 4924107550, 14914341925, 40912232702, 102769130750, 240910097848, 529882277575, 1107606410024, 2206044295976, 4225524980826, 7792505423049, 13933571680850, 24163571680850, 40869390083652

Offset: 1

Wesley Ivan Hurt, Oct 26 2017

Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for sequences related to partitions
Index entries for linear recurrences with constant coefficients, signature (1,11,-11,-55,55,165,-165,-330,330,462,-462,-462,462,330,-330,-165,165,55,-55,-11,11,1,-1).

Sum of k-th powers of the parts in the partitions of n into two parts for k=0..10: A052928 (k=0), A093353 (k=1), A226141 (k=2), A294270 (k=3), A294271 (k=4), A294272 (k=5), A294273 (k=6), A294274 (k=7), A294275 (k=8), A294276 (k=9), this sequence (k=10).

[n*(5120-33792*n^2+67584*n^4-67584*n^6+56320*n^8-33759*n^9+6144*n^10+33*n^9*(-1)^n)/67584 : n in [1..50]]; // Wesley Ivan Hurt, Jul 13 2025

f:= proc(n)
if n::even then (1/66)*n*(6*n^10-(16863/512)*n^9+55*n^8-66*n^6+66*n^4-33*n^2+5)
  else (1/66*(n-1))*n*(2*n-1)*(n^2-n-1)*(3*n^6-9*n^5+2*n^4+11*n^3+3*n^2-10*n-5)
fi end proc:
map(f, [$1..50]); # Robert Israel, Oct 27 2017

Table[Sum[i^10 + (n - i)^10, {i, Floor[n/2]}], {n, 30}]

a(n) = Sum_{i=1..floor(n/2)} i^10 + (n-i)^10.
From Robert Israel, Oct 27 2017: (Start)
a(2*k) = (6144*k^10-16863*k^9+14080*k^8-4224*k^6+1056*k^4-132*k^2+5)*k/33.
a(2*k+1) = (6144*k^10+16896*k^9+14080*k^8-4224*k^6+1056*k^4-132*k^2+5)*k/33.
G.f.: x^2*(x^20+1023*x^19+59039*x^18+1036299*x^17+9117154*x^16+48940320*x^15
+178348744*x^14+465661416*x^13+907378474*x^12+1340492142*x^11+1528402822*x^10
+1340492142*x^9+908233636*x^8+465661416*x^7+178756096*x^6+48940320*x^5
+9163981*x^4+1036299*x^3+60051*x^2+1023*x+2)/((x^2-1)^11*(x-1)). (End)
a(n) = n*(5120-33792*n^2+67584*n^4-67584*n^6+56320*n^8-33759*n^9+6144*n^10+33*n^9*(-1)^n)/67584. - Wesley Ivan Hurt, Jul 13 2025
a(n) = a(n-1) + 11*a(n-2) - 11*a(n-3) - 55*a(n-4) + 55*a(n-5) + 165*a(n-6) - 165*a(n-7) - 330*a(n-8) + 330*a(n-9) + 462*a(n-10) - 462*a(n-11) - 462*a(n-12) + 462*a(n-13) + 330*a(n-14) - 330*a(n-15) - 165*a(n-16) + 165*a(n-17) + 55*a(n-18) - 55*a(n-19) - 11*a(n-20) + 11*a(n-21) + a(n-22) - a(n-23). - Wesley Ivan Hurt, Jul 13 2025

A265225 Total number of ON (black) cells after n iterations of the "Rule 54" elementary cellular automaton starting with a single ON (black) cell.

Original entry on oeis.org

1, 4, 6, 12, 15, 24, 28, 40, 45, 60, 66, 84, 91, 112, 120, 144, 153, 180, 190, 220, 231, 264, 276, 312, 325, 364, 378, 420, 435, 480, 496, 544, 561, 612, 630, 684, 703, 760, 780, 840, 861, 924, 946, 1012, 1035, 1104, 1128, 1200, 1225, 1300, 1326, 1404, 1431

Offset: 0

Robert Price, Dec 05 2015

Take the first 2n positive integers and choose n of them such that their sum: a) is divisible by n, and b) is minimal. It seems their sum equals a(n). - Ivan N. Ianakiev, Feb 16 2019

			From _Michael De Vlieger_, Dec 14 2015: (Start)
First 12 rows, replacing "0" with "." for better visibility of ON cells, followed by the total number of 1's per row, and the running total up to that row:
                        1                          =  1 ->  1
                      1 1 1                        =  3 ->  4
                    1 . . . 1                      =  2 ->  6
                  1 1 1 . 1 1 1                    =  6 -> 12
                1 . . . 1 . . . 1                  =  3 -> 15
              1 1 1 . 1 1 1 . 1 1 1                =  9 -> 24
            1 . . . 1 . . . 1 . . . 1              =  4 -> 28
          1 1 1 . 1 1 1 . 1 1 1 . 1 1 1            = 12 -> 40
        1 . . . 1 . . . 1 . . . 1 . . . 1          =  5 -> 45
      1 1 1 . 1 1 1 . 1 1 1 . 1 1 1 . 1 1 1        = 15 -> 60
    1 . . . 1 . . . 1 . . . 1 . . . 1 . . . 1      =  6 -> 66
  1 1 1 . 1 1 1 . 1 1 1 . 1 1 1 . 1 1 1 . 1 1 1    = 18 -> 84
1 . . . 1 . . . 1 . . . 1 . . . 1 . . . 1 . . . 1  =  7 -> 91
(End)

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Robert Price, Table of n, a(n) for n = 0..999
Emanuele Munarini, Topological indices for the antiregular graphs, Le Mathematiche (2021) Vol. 76, No. 1, see p. 301.
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
Index entries for sequences related to cellular automata
Index to Elementary Cellular Automata

Cf. A071030, A118108, A118109, A133872.

A265225:=n->1/4*(n+1)*(2*n-(-1)^n+5): seq(A265225(n), n=0..60); # Wesley Ivan Hurt, Dec 25 2016

rule = 54; rows = 30; Table[Total[Take[Table[Total[Table[Take[CellularAutomaton[rule,{{1},0},rows-1,{All,All}][[k]],{rows-k+1,rows+k-1}],{k,1,rows}][[k]]],{k,1,rows}],k]],{k,1,rows}]
Accumulate[Total /@ CellularAutomaton[54, {{1}, 0}, 52]]

Conjectures from Colin Barker, Dec 08 2015 and Apr 20 2019: (Start)
a(n) = (n+1)*(2*n -(-1)^n +5)/4.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>4.
G.f.: (1+3*x) / ((1-x)^3*(1+x)^2).
(End)
a(n) = n + 1 + (n+1) * floor((n+1)/2), conjectured. - Wesley Ivan Hurt, Dec 25 2016
a(n) = A093353(n) + n + 1, conjectured. - Matej Veselovac, Jan 21 2020

A211374 Product of all the parts in the partitions of n into exactly 2 parts.

Original entry on oeis.org

1, 1, 2, 12, 24, 360, 720, 20160, 40320, 1814400, 3628800, 239500800, 479001600, 43589145600, 87178291200, 10461394944000, 20922789888000, 3201186852864000, 6402373705728000, 1216451004088320000, 2432902008176640000, 562000363888803840000

Offset: 1

Wesley Ivan Hurt, Feb 06 2013

			Define a(1):=1; a(2) = 1 since 2 = 1+1 and (1)*(1) = 1; a(3) = 2 since 3 = 2+1 and (2)*(1) = 2; a(4) = 12 since 4 = 3+1 = 2+2 and (3)*(1)*(2)*(2) = 12; a(5) = 24 since 5 = 4+1 = 3+2 and (4)*(1)*(3)*(2) = 24.

G. C. Greubel, Table of n, a(n) for n = 1..449
Index entries for sequences related to partitions.

Cf. A001318, A002620, A081123, A081125, A093353.

Bisections: A002674, A010050.

[(Factorial(n-1) * Factorial(Floor(n/2)))/Factorial(n-1-Floor(n/2)) : n in [1..25]]; // Wesley Ivan Hurt, Oct 16 2014

A211374:=n->( (n-1)! * floor(n/2)! )/( (n-1) - floor(n/2) )!: seq(A211374(k), k=1..25);
with(combinat, numbperm): seq(numbperm(k-1, floor(k/2))*floor(k/2)!, k = 1..25); # Wesley Ivan Hurt, Jun 07 2013

Table[Times @@ Flatten[Select[Partitions[n], Length[#] == 2 &]], {n, 25}] (* T. D. Noe, Feb 11 2013 *)
Table[((n - 1)!*Floor[n/2]!)/(n - 1 - Floor[n/2])!, {n, 25}] (* Wesley Ivan Hurt, Oct 16 2014 *)

a(n) = prod(i=1, n\2, i*(n-i)); \\ Michel Marcus, Nov 14 2017

a(n) = ( (n-1)! * floor(n/2)! )/( n-1-floor(n/2) )!.
a(n) = P(n-1, floor(n/2)) * floor(n/2)!, where P(n,k) are the k-permutations of n objects. - Wesley Ivan Hurt, Jun 07 2013
a(2n) = A002674(n); a(2n+1) = A010050(n). - Wesley Ivan Hurt, Oct 16 2014
a(n) = Product_{i=1..floor(n/2)} i * (n-i). - Wesley Ivan Hurt, Nov 14 2017
From Amiram Eldar, Mar 10 2022: (Start)
Sum_{n>=1} 1/a(n) = 3*cosh(1) - 2.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2 - cosh(1). (End)

A242669 a(n) = n*floor(n/3).

Original entry on oeis.org

0, 0, 0, 3, 4, 5, 12, 14, 16, 27, 30, 33, 48, 52, 56, 75, 80, 85, 108, 114, 120, 147, 154, 161, 192, 200, 208, 243, 252, 261, 300, 310, 320, 363, 374, 385, 432, 444, 456, 507, 520, 533, 588, 602, 616, 675, 690, 705, 768, 784, 800, 867, 884, 901, 972, 990

Offset: 0

Bruno Berselli, Jul 01 2014

For n = 0, 1, 2, 4, 8, 49, 98, 676, 1352, 9409, 18818, 131044, 262088, 1825201, 3650402, ... a(n) is a square.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,2,-2,0,-1,1).

Cf. A000290 (n^2), A010762 (floor(n/2)*floor(n/3)), A093353 (n*floor(n/2)), A213033 (n*floor(n/2)*floor(n/3)), A233035 (n*floor(n/4)).
Cf. A033428, A045944, A049451.
Cf. A002264 (floor(n/3)).

Magma
```
[n*Floor(n/3): n in [0..60]];
```
Mathematica
```
Table[n Floor[n/3], {n, 0, 60}]
```

a(n)=n\3*n \\ Charles R Greathouse IV, Oct 07 2015

Sage
```
[n*floor(n/3) for n in (0..60)];
```

G.f.: x^3*(3 + x + x^2 + x^3)/((1 - x)^3*(1 + x + x^2)^2).
a(3m) = A033428(m), a(3m+1) = A049451(m), a(3m+2) = A045944(m).
Sum_{n>=3} (-1)^(n+1)/a(n) = 9/4 + Pi^2/36 - Pi/(2*sqrt(3)) - 2*log(2). - Amiram Eldar, Mar 30 2023

A128623 Triangle read by rows: A128621 * A000012 as infinite lower triangular matrices.

Original entry on oeis.org

1, 2, 2, 6, 3, 3, 8, 8, 4, 4, 15, 10, 10, 5, 5, 18, 18, 12, 12, 6, 6, 28, 21, 21, 14, 14, 7, 7, 32, 32, 24, 24, 16, 16, 8, 8, 45, 36, 36, 27, 27, 18, 18, 9, 9, 50, 50, 40, 40, 30, 30, 20, 20, 10, 10, 66, 55, 55, 44, 44, 33, 33, 22, 22, 11, 11, 72, 72, 60, 60, 48, 48, 36, 36, 24, 24, 12, 12, 91, 78, 78, 65, 65, 52, 52, 39, 39, 26, 26, 13, 13

Offset: 1

Gary W. Adamson, Mar 14 2007

			First few rows of the triangle are:
   1;
   2,  2;
   6,  3,  3;
   8,  8,  4,  4;
  15, 10, 10,  5,  5;
  18, 18, 12, 12,  6, 6;
  28, 21, 21, 14, 14, 7, 7;
  ...

G. C. Greubel, Rows n = 1..100 of the triangle, flattened

Cf. A000027, A093005, A093353, A115514, A128621, A245524.

Cf. A128624 (row sums).

[n*Floor((n-k+2)/2): k in [1..n], n in [1..15]]; // G. C. Greubel, Mar 13 2024

Table[n*Floor[(n-k+2)/2], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Mar 13 2024 *)

flatten([[n*((n-k+2)//2) for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Mar 13 2024

Sum_{k=1..n} T(n, k) = A128624(n) (row sums).
T(n,k) = n*(1+floor((n-k)/2)), 1 <= k <= n. - R. J. Mathar, Jun 27 2012
From G. C. Greubel, Mar 13 2024: (Start)
T(n, k) = n*A115514(n, k).
T(n, k) = Sum_{j=k..n} A128621(n, j).
T(n, 1) = A093005(n).
T(n, 2) = A093353(n-1), n >= 2.
T(n, n) = A000027(n).
T(2*n-1, n) = A245524(n).
Sum_{k=1..n} (-1)^k*T(n, k) = (1/2)*(1-(-1)^n)*A000384(floor((n+1)/2)). (End)

a(41) = 27 inserted and more terms from Georg Fischer, Jun 05 2023

A171769 Partial sums of A042964 (numbers congruent to 2 or 3 mod 4).

Original entry on oeis.org

2, 5, 11, 18, 28, 39, 53, 68, 86, 105, 127, 150, 176, 203, 233, 264, 298, 333, 371, 410, 452, 495, 541, 588, 638, 689, 743, 798, 856, 915, 977, 1040, 1106, 1173, 1243, 1314, 1388, 1463, 1541, 1620, 1702, 1785, 1871, 1958, 2048, 2139, 2233, 2328, 2426, 2525

Offset: 1

Jaroslav Krizek, Dec 18 2009

If we insert an initial 0, and alternate the signs: 0,2,-5,11,-18,28,..., we get a sequence where the average of the first n terms is an integer, with no repeats: specifically A001057(n-1). The sum of the first n terms is (-1)^(n-1)*A093353(n-1). - Franklin T. Adams-Watters, May 20 2010

Suppose that n cards have the numbers 1..2n written on them randomly, one number to a side, and are set out on a table randomly. You have the task of maximizing the sum of the visible numbers by flipping cards. If you have no information other than the numbers on the upper faces, and may not flip any particular card more than once, a(n) is the largest sum you can guarantee in the worst case. - Andrew Woods, Jun 06 2013

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A042964, A001057, A093353. - Franklin T. Adams-Watters, May 20 2010

a:=[2,5,11,18];; for n in [5..60] do a[n]:=2*a[n-1]-2*a[n-3] + a[n-4]; od; a; # G. C. Greubel, Jul 02 2019

[Ceiling((2*n+1)*n/2): n in [1..60]]; // Vincenzo Librandi, Jul 02 2019

a[n_]:=Ceiling[((2n+1)n/2)]; Array[a, 60] (* Vincenzo Librandi, Jul 02 2019 *)
LinearRecurrence[{2,0,-2,1}, {2,5,11,18}, 60] (* G. C. Greubel, Jul 02 2019 *)

Vec(x*(x^2+x+2)/((1-x)^3*(x+1)) + O(x^60)) \\ Colin Barker, Jun 04 2014

[ceiling(n*(1+2*n)/2) for n in (1..60)] # G. C. Greubel, Jul 02 2019

a(n) = Sum_{i=1..n} A042964(i).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - R. H. Hardin, Nov 13 2011
a(n) = ceiling((2*n+1)*n/2). - Andrew Woods, Jun 06 2013
G.f.: x*(2+x+x^2) / ((1-x)^3*(x+1)). - Colin Barker, Jun 04 2014
a(n) = round(n/(1-exp(-1/n))). - Richard R. Forberg, Jan 28 2015

A379726 Minimum number of kings that must be placed on an n X n chessboard such that each square is attacked or occupied by at least two kings.

Original entry on oeis.org

2, 3, 8, 8, 10, 18, 18, 21, 32, 32, 36, 50, 50, 55, 72, 72, 78, 98, 98, 105, 128, 128, 136, 162, 162, 171, 200, 200, 210, 242, 242, 253, 288, 288, 300, 338, 338, 351, 392, 392, 406, 450, 450, 465, 512, 512, 528, 578, 578, 595, 648, 648, 666, 722, 722, 741, 800, 800, 820, 882, 882, 903, 968, 968, 990, 1058, 1058, 1081, 1152, 1152, 1176, 1250, 1250, 1275, 1352, 1352, 1378, 1458, 1458, 1485, 1568, 1568, 1596, 1682, 1682, 1711, 1800, 1800, 1830, 1922, 1922, 1953, 2048, 2048, 2080, 2178, 2178, 2211, 2312

Offset: 2

Matthew Scroggs, Dec 31 2024

nonn

At most one king can be placed on each square.
Every third term is conjectured to be A014105. Other terms are A001105. A093353 is conjectured to be this sequence with repeated terms removed.
The above conjectures are true (see Beveridge link). - Colin Beveridge, Jan 13 2025

			For a 3 by 3 chessboard, the three kings could be placed like this (where o is an empty square and k is a king):
   ooo
   kkk
   ooo
For a 4 by 4 chessboard, the kings could be placed like this:
   oooo
   kkkk
   okko
   okko

Rob Pratt, Table of n, a(n) for n = 2..100
Colin Beveridge, Proof of the case where n is a multiple of 3
Matthew Scroggs, December 23
Matthew Scroggs, Friendly squares
Matthew Scroggs, Python code to compute A379726
Puzzling StackExchange, Minimum Number of Squares to Color
Dominic McCarty, Illustration of a(n) for n = 2..100

Cf. A001105, A014105, A075561, A093353, A379759, A379766.

If n is not a multiple of 3, a(n) = 2*floor((n+2)/3)^2.
If n is a multiple of 3, it is conjectured that a(n)=2*(n/3)^2+n/3.
The above conjectures are true (see Beveridge link). - Colin Beveridge, Jan 13 2025

a(15)-a(100) via integer linear programming by Rob Pratt, Jan 02 2025

cp's OEIS Frontend

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A265225 Total number of ON (black) cells after n iterations of the "Rule 54" elementary cellular automaton starting with a single ON (black) cell.

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A211374 Product of all the parts in the partitions of n into exactly 2 parts.

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