A032765 -id:A032765 - cp's OEIS frontend

A032798 Numbers such that n(n+1)(n+2)...(n+12) / (n+(n+1)+(n+2)+...+(n+12)) is a multiple of n.

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 45, 47, 48, 49, 50, 51, 53, 54, 55, 56, 57, 58, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 79, 80, 81, 82, 83, 84

Offset: 1

Patrick De Geest, May 15 1998

nonn

Equals natural numbers minus '7,13,20,26,33,...' (= previous term +6,+7,+6,+7,...).

Cf. A032765-A032798.

From Chai Wah Wu, Dec 17 2016: (Start)
a(n) = a(n-1) + a(n-11) - a(n-12) for n > 12.
G.f.: x*(x^11 + x^10 + x^9 + x^8 + x^7 + 2*x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)/(x^12 - x^11 - x + 1). (End)

A143974 Rectangular array R by antidiagonals: label each unit square in the first quadrant lattice by its northeast vertex (x,y) and mark those having x+y=1(mod 3); then R(m,n) is the number of marked unit squares in the rectangle [0,m]x[0,n].

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 2, 3, 4, 4, 3, 2, 2, 4, 5, 5, 5, 4, 2, 2, 4, 6, 6, 6, 6, 4, 2, 3, 5, 7, 8, 8, 8, 7, 5, 3, 3, 6, 8, 9, 10, 10, 9, 8, 6, 3, 3, 6, 9, 10, 11, 12, 11, 10, 9, 6, 3, 4, 7, 10, 12, 13, 14, 14, 13, 12, 10, 7, 4, 4, 8, 11, 13, 15, 16, 16, 16, 15, 13, 11, 8, 4, 4, 8

Offset: 1

Clark Kimberling, Sep 06 2008

Diagonals: A000212, A128422, A032765, A143975, A071408.

Rows: A002264, A004523, A000027, A004773, A138235, A005843, A047349 et al.

			Northwest corner:
0 0 1 1 1 2
0 1 2 2 3 4
1 2 3 4 5 6
1 2 4 5 6 8
1 3 5 6 8 10
R(3,4) counts these marked squares: (1,3), (2,2), (3,1), (3,4).

Cf. A143976, A143977.

R(m,n)=floor(mn/3).

A032774 a(n) = floor( n(n+1)(n+2)...(n+6) / (n+(n+1)+(n+2)+...+(n+6)) ).

Original entry on oeis.org

0, 180, 1152, 4320, 12342, 29700, 63360, 123552, 224640, 386100, 633600, 1000182, 1527552, 2267460, 3283200, 4651200, 6462720, 8825652, 11866422, 15732000, 20592000, 26640900, 34100352, 43221600, 54288000, 67617642, 83566080, 102529152, 124945920, 151301700, 182131200

Offset: 0

Patrick De Geest, May 15 1998

In general, such sequences a(n) = floor((Product_{m=0..k} n+i) / (Sum_{m=0..k} n+i)) have rational generating functions. - Georg Fischer, Feb 23 2021

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1,1,-6,15,-20,15,-6,1).

Cf. A032775, A032776.

Cf. A004526 (k=2), A032765 (k=3), A032768 (k=4), A032771 (k=5), A032774 (k=6), A032777 (k=7), A032780 (k=8), A032790 (k=9).

seq(coeff(series( -(6*x^10-36*x^9 + 90*x^8 - 120*x^7 - 90*x^6 - 108*x^5 - 102*x^4 - 108*x^3 - 72*x^2 - 180*x) / (-x^13+6*x^12 - 15*x^11+20*x^10 - 15*x^9+6*x^8 - x^7+x^6 - 6*x^5+15*x^4 - 20*x^3+15*x^2 - 6*x + 1) , x, n+1), x, n), n = 0..40); # Georg Fischer, Feb 23 2021

Table[Floor[(Times @@ Range[n, n + 6])/(7 n + 21)], {n, 0, 30}] (* Harvey P. Dale, May 16 2020 *)

More terms from Georg Fischer, Feb 23 2021

A143975 a(n) = floor(n*(n+3)/3).

Original entry on oeis.org

1, 3, 6, 9, 13, 18, 23, 29, 36, 43, 51, 60, 69, 79, 90, 101, 113, 126, 139, 153, 168, 183, 199, 216, 233, 251, 270, 289, 309, 330, 351, 373, 396, 419, 443, 468, 493, 519, 546, 573, 601, 630, 659, 689, 720, 751, 783, 816, 849, 883, 918, 953, 989, 1026, 1063, 1101

Offset: 1

Clark Kimberling, Sep 06 2008

Fourth diagonal of A143974, associated with counting unit squares in a lattice.

			Main diagonal of A143974: (0,1,3,5,8,12,...) = A000212;
2nd diagonal: (0,2,4,6,10,14,18,...) = A128422;
3rd diagonal: (1,2,5,8,11,16,21,...) = A032765;
4th diagonal: (1,3,6,9,13,18,23,...) = A143975.

Vincenzo Librandi, Table of n, a(n) for n = 1..3000
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A099837, A143974.

[Floor(n*(n+3)/3): n in [1..60]]; // Vincenzo Librandi, May 08 2011

a[n_] := Floor[n*(n+3)/3]; Array[a, 60] (* Amiram Eldar, Oct 01 2022 *)

a(n) = floor(n*(n+3)/3).
From R. J. Mathar, Oct 05 2009: (Start)
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
G.f.: x*(-1 - x - x^2 + x^3)/( (1 + x + x^2) * (x-1)^3). (End)
9*a(n) = 3*n^2 + 9*n - 2 + A099837(n+3). - R. J. Mathar, Apr 26 2022
Sum_{n>=1} 1/a(n) = 4/3 + (tan((sqrt(13)+2)*Pi/6) - cot((sqrt(13)+1)*Pi/6)) * Pi/sqrt(13). - Amiram Eldar, Oct 01 2022
E.g.f.: (exp(x)*(3*x*(4 + x) - 2) + 2*exp(-x/2)*cos(sqrt(3)*x/2))/9. - Stefano Spezia, Oct 24 2022

A006066 Kobon triangles: maximal number of nonoverlapping triangles that can be formed from n lines drawn in the plane.

Original entry on oeis.org

0, 0, 1, 2, 5, 7, 11, 15, 21, 25, 32, 38, 47

Offset: 1

N. J. A. Sloane

The known values a = a(n) and upper bounds U (usually A032765(n)) with name of discoverer of the arrangement when known are as follows:
   n       a    U  [Found by]
------------------------------
     0    0
     0    0
     1    1
     2    2
     5    5
     7    7
    11   11
    15   16
    21   21
    25   25  [Grünbaum]
    32   33  [32-triangle solutions found by Honma and Kabanovitch; proved maximal by Savchuk 2025]
    38   38  [Kabanovitch]
    47   47  [Kabanovitch]
 >= 53   54  [Bader]
    65   65  [Suzuki]
    72   72  [Bader]
    85   85  [Bader]
 >= 93   94  [Bader]
   107  107  [Wood]
>= 116  117  [Wood]
   133  133  [Savchuk]
>= 143  144  [Savchuk]
   161  161  [Savchuk]
>= 172  173  [Savchuk]
   191  191  [Bartholdi]
     ?  205
   225  225  [Savchuk]
     ?  239
   261  261  [Bartholdi]
     ?  276
   299  299  [Wood]
     ?  316
   341  341  [Bartholdi]
Ed Pegg's web page gives the upper bound for a(6) as 8. But by considering all possible arrangements of 6 lines - the sixth term of A048872 - one can see that 8 is impossible. - N. J. A. Sloane, Nov 11 2007
Although they are somewhat similar, this sequence is strictly different from A084935, since A084935(12) = 48 exceeds the upper bound on a(12) from A032765. - Floor van Lamoen, Nov 16 2005
The name is sometimes incorrectly entered as "Kodon" triangles.
Named after the Japanese puzzle expert and mathematics teacher Kobon Fujimura (1903-1983). - Amiram Eldar, Jun 19 2021

			a(17) = 85 because the a configuration with 85 exists meeting the upper bound.

Nicolas Bartholdi, Jérémy Blanc, and Sébastien Loisel, "On simple arrangements of lines and pseudo-lines in P^2 and R^2 with the maximum number of triangles", 2008, in Goodman, Jacob E.; Pach, János; Pollack, Richard (eds.), Surveys on Discrete and Computational Geometry: Proceedings of the 3rd AMS-IMS-SIAM Joint Summer Research Conference "Discrete and Computational Geometry—Twenty Years Later" held in Snowbird, UT, June 18-22, 2006, Contemporary Mathematics, vol. 453, Providence, Rhode Island: American Mathematical Society, pp. 105-116, doi:10.1090/conm/453/08797, ISBN 978-0-8218-4239-3, MR 2405679
Martin Gardner, Wheels, Life and Other Mathematical Amusements, Freeman, NY, 1983, pp. 170, 171, 178. Mentions that the problem was invented by Kobon Fujimura.
Branko Grünbaum, Convex Polytopes, Wiley, NY, 1967; p. 400 shows that a(10) >= 25.
Viatcheslav Kabanovitch, Kobon Triangle Solutions, Sharada (Charade, by the Russian puzzle club Diogen), pp. 1-2, June 1999.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Johannes Bader, Kobon Triangles.
Johannes Bader, Kobon Triangles. [Cached copy, with permission, pdf format]
Johannes Bader, Illustration showing a(17)=85, Nov 28 2007.
Johannes Bader, Illustration showing a(17)=85, Nov 28 2007. [Cached copy, with permission]
Nicolas Bartholdi, Jérémy Blanc, and Sébastien Loisel, On simple arrangements of lines and pseudo-lines in P^2 and R^2 with the maximum number of triangles, arXiv:0706.0723 [math.CO], 2007.
Nicolas Bartholdi, Jérémy Blanc, Sébastien Loisel, and Pavlo Savchuk, Illustration showing a(33) = 341, 2008.
Gilles Clement and Johannes Bader, Tighter Upper Bound for the Number of Kobon Triangles, Unpublished, 2007.
Gilles Clement and Johannes Bader, Tighter Upper Bound for the Number of Kobon Triangles, Unpublished, 2007. [Cached copy, with permission]
Martin Gardner, Letter to N. J. A. Sloane, Jun 20 1991.
S. Honma, 三角形の最大数
S. Honma, Illustration showing a(10)>=25
S. Honma, Illustration showing a(11)>=32
S. Honma, n本の直線でn(n-2)/3個の三角形が出来る条件についての考察: part 1, part 2, part 3.
Ed Pegg, Jr., Kobon triangles, 2006.
Ed Pegg, Jr., Kobon Triangles, 2006. [Cached copy, with permission, pdf format]
Luis Felipe Prieto-Martínez, A list of problems in Plane Geometry with simple statement that remain unsolved, arXiv:2104.09324 [math.HO], 2021.
Pavlo Savchuk, Constructing Optimal Kobon Triangle Arrangements via Table Encoding, SAT Solving, and Heuristic Straightening, arXiv:2507.07951 [math.CO], 2025. See pp. 1, 17.
Pavlo Savchuk, Illustration showing a(21) = 133
Pavlo Savchuk, Illustration showing a(23) = 161
Pavlo Savchuk, Corresponding pseudo-line arrangement for the optimal n=23 solution
Pavlo Savchuk, Illustration showing a(27) = 225
Pavlo Savchuk, Illustration showing a(22) >= 143
Pavlo Savchuk, Illustration showing a(24) >= 172
N. J. A. Sloane, Illustration for a(5) = 5 (a pentagram)
Alexandre Wajnberg, Illustration showing a(10) >= 25. [A different construction from Grünbaum's]
Eric Weisstein's World of Mathematics, Kobon Triangle.
Kyle Wood, Illustration showing a(19) = 107
Kyle Wood, Illustration showing a(20) >= 116
Kyle Wood, Illustration showing a(31) = 299

An upper bound on this sequence is given by A032765.
For any odd n > 1, if n == 1 (mod 6), a(n) <= (n^2 - (2n + 2))/3; in other odd cases, a(n) <= (n^2 - 2n)/3. For any even n  > 0, if n == 4 (mod 6), a(n) <= (n^2 - (2n + 2))/3, otherwise a(n) <= (n^2 - 2n)/3. - Sergey Pavlov, Feb 11 2017
The upper bound for even n can be improved: floor(n(n-7/3)/3), proven by Bartholdi et. al.

a(15) = 65 found by Toshitaka Suzuki on Oct 02 2005. - Eric W. Weisstein, Oct 04 2005
Grünbaum reference from Anthony Labarre, Dec 19 2005
Additional links to Japanese web sites from Alexandre Wajnberg, Dec 29 2005 and Anthony Labarre, Dec 30 2005
A perfect solution for 13 lines was found in 1999 by Kabanovitch. - Ed Pegg Jr, Feb 08 2006
Updated with results from Johannes Bader (johannes.bader(AT)tik.ee.ethz.ch), Dec 06 2007, who says "Acknowledgments and dedication to Corinne Thomet".
a(11)-a(13) from Eric W. Weisstein, Jul 26 2025

A032780 a(n) = n(n+1)(n+2)...(n+8) / (n+(n+1)+(n+2)+...+(n+8)).

Original entry on oeis.org

0, 8064, 67200, 316800, 1108800, 3203200, 8072064, 18345600, 38438400, 75398400, 140025600, 248312064, 423259200, 697132800, 1114220800, 1734163200, 2635928064, 3922512000, 5726448000, 8216208000, 11603592000, 16152200064, 22187088000, 30105712000

Offset: 0

Patrick De Geest, May 15 1998

nonn

a(5n+1) == 4 modulo 10.

The product of any k consecutive integers is divisible by the sum of the same k integers for odd nonprime k's: 1 (trivial case), 9 (this sequence), 15, etc. - Zak Seidov, Mar 18 2014

T. D. Noe, Table of n, a(n) for n = 0..1000

Cf. A032765-A032798.

Cf. A104678.

nn = 9; Table[c = Range[n, n + nn - 1]; Times @@ c/Total[c], {n, 0, 25}] (* T. D. Noe, Mar 18 2014 *)

a(n) = prod(i=0, 8, n+i)/sum(i=0, 8, n+i); \\ Michel Marcus, Mar 18 2014

a(-n) = a(n-8) for all n in Z. - Michael Somos, Mar 18 2014
a(n) = 64 * A104678(n-1) = 64 * binomial(n+3, 4) * binomial(n+8, 4). - Michael Somos, Mar 18 2014
From Chai Wah Wu, Dec 17 2016: (Start)
a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n > 8.
G.f.: 64*x*(-x^4 + 9*x^3 - 36*x^2 + 84*x - 126)/(x - 1)^9. (End)

Typo in name fixed by Zak Seidov, Mar 18 2014

More terms from Michel Marcus, Mar 18 2014

A032796 Numbers that are congruent to {1, 2, 3, 5, 6} mod 7.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 9, 10, 12, 13, 15, 16, 17, 19, 20, 22, 23, 24, 26, 27, 29, 30, 31, 33, 34, 36, 37, 38, 40, 41, 43, 44, 45, 47, 48, 50, 51, 52, 54, 55, 57, 58, 59, 61, 62, 64, 65, 66, 68, 69, 71, 72, 73, 75, 76, 78, 79, 80, 82, 83, 85, 86, 87, 89, 90, 92, 93, 94, 96, 97, 99

Offset: 1

Patrick De Geest, May 15 1998

If k is a term, then k*(k+1)*(k+2)*...*(k+6)/(k+(k+1)+(k+2)+...+(k+6)) is a multiple of k.

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Cf. A032765..A032798.

Cf. A010874.

[n: n in [0..120] | n mod 7 in {1, 2, 3, 5, 6}]; // Vincenzo Librandi, Dec 29 2010

#+{1,2,3,5,6}&/@(7*Range[0,15])//Flatten (* or *) LinearRecurrence[ {1,0,0,0,1,-1},{1,2,3,5,6,8},100] (* Harvey P. Dale, Oct 07 2018 *)

Equals natural numbers minus '4, 7, 11, 14, 18, ...' (= previous term +3, +4, +3, +4, ...).
G.f.: x*(x^5 + x^4 + 2*x^3 + x^2 + x + 1)/((1-x)*(1-x^5)).
a(n) = (m^3 - 6*m^2 + 17*m + 6*(7*floor(n/5)-1))/6, where m = n mod 5. - Luce ETIENNE,Oct 17 2018

A032767 a(n) = floor ( n(n+1)(n+2)(n+3) / (n+(n+1)+(n+2)+(n+3)) ).

Original entry on oeis.org

0, 2, 8, 20, 38, 64, 100, 148, 208, 282, 373, 480, 606, 753, 921, 1112, 1328, 1571, 1841, 2140, 2471, 2833, 3229, 3661, 4129, 4635, 5182, 5769, 6399, 7074, 7794, 8561, 9377, 10244, 11162, 12133, 13160, 14242, 15382, 16582, 17842, 19164

Offset: 0

Patrick De Geest, May 15 1998

nonn

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A032765-A032798.

A032767 := proc(n)
        n*(n+1)*(n+2)*(n+3) /(4*n+6) ;
        floor(%) ;
end proc: # R. J. Mathar, May 20 2013

Table[Floor[Times@@(n+Range[0,3])/Total[n+Range[0,3]]],{n,0,50}] (* Harvey P. Dale, Oct 18 2024 *)

The g.f. has denominator (1-x)^4(1-x^16). - Ralf Stephan, May 16 2005

A032784 Numbers k such that k(k+1)(k+2)...(k+11) / (k+(k+1)+(k+2)+...+(k+11)) is an integer.

Original entry on oeis.org

0, 2, 5, 7, 8, 11, 12, 17, 19, 22, 26, 32, 33, 35, 44, 47, 55, 62, 68, 77, 82, 89, 107, 110, 116, 117, 132, 143, 152, 176, 187, 197, 215, 242, 257, 264, 278, 297, 332, 341, 362, 407, 418, 440, 467, 539, 572, 602, 607, 656, 737, 782, 803, 845, 902, 957, 1007, 1034

Offset: 1

Patrick De Geest, May 15 1998

(d-11)/2 where d>=7 is a divisor of 36018675. In particular, the sequence is finite. - Robert Israel, Jul 12 2018

Robert Israel, Table of n, a(n) for n = 1..157

Cf. A032765-A032798.

sort(convert(select(type,map(t -> (t-11)/2,numtheory:-divisors(36018675)),nonnegint),list));

Select[Range[0,1100],IntegerQ[Times@@Range[#,#+11]/Total[Range[#,#+11]]]&] (* Harvey P. Dale, Sep 02 2016 *)

Definition corrected by Harvey P. Dale, Sep 02 2016

Offset changed by Robert Israel, Jul 12 2018

A032785 Numbers k such that k(k+1)(k+2) ... (k+13) / (k+(k+1)+(k+2)+ ... +(k+13)) is an integer.

Original entry on oeis.org

0, 1, 4, 6, 7, 10, 11, 13, 16, 21, 25, 26, 31, 32, 34, 39, 43, 46, 52, 54, 61, 65, 76, 78, 81, 88, 91, 106, 109, 115, 130, 131, 142, 151, 156, 169, 175, 186, 196, 208, 221, 241, 247, 256, 277, 286, 296, 331, 340, 351, 358, 403, 406, 416, 417, 439, 466, 481, 494

Offset: 1

Patrick De Geest, May 15 1998

(d-13)/2 for divisors d>=13 of 2608781175. In particular, the sequence is finite. - Robert Israel, Jul 13 2018

Robert Israel, Table of n, a(n) for n = 1..372

Cf. A032765-A032798.

seq((t-13)/2, t=select(`>=`,numtheory:-divisors(2608781175),13)); # Robert Israel, Jul 13 2018

Select[Range[0,500],IntegerQ[Times@@Range[#,#+13]/Total[Range[ #, #+13]]]&] (* Harvey P. Dale, Sep 02 2016 *)

Definition corrected by Harvey P. Dale, Sep 02 2016

cp's OEIS Frontend

A032798 Numbers such that n(n+1)(n+2)...(n+12) / (n+(n+1)+(n+2)+...+(n+12)) is a multiple of n.

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A143974 Rectangular array R by antidiagonals: label each unit square in the first quadrant lattice by its northeast vertex (x,y) and mark those having x+y=1(mod 3); then R(m,n) is the number of marked unit squares in the rectangle [0,m]x[0,n].

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A032774 a(n) = floor( n*(n+1)*(n+2)*...*(n+6) / (n+(n+1)+(n+2)+...+(n+6)) ).

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A143975 a(n) = floor(n*(n+3)/3).

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A006066 Kobon triangles: maximal number of nonoverlapping triangles that can be formed from n lines drawn in the plane.

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A032780 a(n) = n(n+1)(n+2)...(n+8) / (n+(n+1)+(n+2)+...+(n+8)).

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A032796 Numbers that are congruent to {1, 2, 3, 5, 6} mod 7.

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Magma

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A032767 a(n) = floor ( n(n+1)(n+2)(n+3) / (n+(n+1)+(n+2)+(n+3)) ).

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A032774 a(n) = floor( n(n+1)(n+2)...(n+6) / (n+(n+1)+(n+2)+...+(n+6)) ).

A032785 Numbers k such that k(k+1)(k+2) ... (k+13) / (k+(k+1)+(k+2)+ ... +(k+13)) is an integer.