A130518 -id:A130518 - cp's OEIS frontend

A247247 Triangular numbers that are the sum of 2 consecutive terms of A130518.

0, 1, 3, 21, 120, 300, 2080, 11781, 29403, 203841, 1154440, 2881200, 19974360, 113123361, 282328203, 1957283461, 11084934960, 27665282700, 191793804840, 1086210502741, 2710915376403, 18793835590881, 106437544333680, 265642041604800, 1841604094101520

Offset: 1

J. M. Bergot, Nov 28 2014

nonn

What will be the distribution of these triangular numbers?
Will they mostly be multiples of three?
From Hiroaki Yamanouchi, Dec 04 2014: (Start)
a(n) is some nonnegative x in the integer solutions (x, y) of
(1) (6*x + 3)^2 - 6*(6*y + 4)^2 = -15,
(2) (6*x + 3)^2 - 6*(6*y + 8)^2 = -15 or
(3) (2*x + 1)^2 - 6*(2*y + 2)^2 = 1.
(End)

			A130518(8)+A130518(9) = 9+12 = 21 = A000217(6), so 21 is in the sequence.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..1000

Cf. A000217, A130518.

f:= proc(n)
local x;
  x:= ceil((n^2+2*n)/3);
if issqr(1+8*x) then x else NULL fi
end proc:
seq(f(n),n=0..10^6); # Robert Israel, Nov 30 2014

a247247[n_Integer] := Module[{a130518, a000217, s},
  a130518[m_] := Table[i, {i, 0, m}, {3}] // Flatten // Accumulate;
  a000217[m_] := Accumulate[Range[m]];
  s[m_] :=
   a130518[m] + Most@PrependTo[a130518[m], 0] // DeleteDuplicates;
Intersection[s[n], a000217[n]]]; a247247[50000000] (* Michael De Vlieger, Nov 30 2014 after Jean-François Alcover at A130518 and Harvey P. Dale at A000217 *)

Empirical G.f.: x^2*(x+1)*(x^4+2*x^3+19*x^2+2*x+1)/((1-x)*(x^2+x+1)*(x^6-98*x^3+1)). - Robert Israel, Nov 30 2014

a(7)-a(13) from Michel Marcus, Nov 28 2014
a(14)-a(24) from Michael De Vlieger, Nov 30 2014
a(25) from Hiroaki Yamanouchi, Dec 04 2014

A005449 Second pentagonal numbers: a(n) = n(3n + 1)/2.

Original entry on oeis.org

0, 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, 260, 301, 345, 392, 442, 495, 551, 610, 672, 737, 805, 876, 950, 1027, 1107, 1190, 1276, 1365, 1457, 1552, 1650, 1751, 1855, 1962, 2072, 2185, 2301, 2420, 2542, 2667, 2795, 2926, 3060, 3197, 3337, 3480

Offset: 0

N. J. A. Sloane

Number of edges in the join of the complete graph and the cycle graph, both of order n, K_n * C_n. - Roberto E. Martinez II, Jan 07 2002
Also number of cards to build an n-tier house of cards. - Martin Wohlgemuth, Aug 11 2002
The modular form Delta(q) = q*Product_{n>=1} (1-q^n)^24 = q*(1 + Sum_{n>=1} (-1)^n*(q^(n*(3*n-1)/2)+q^(n*(3*n+1)/2)))^24 = q*(1 + Sum_{n>=1} A033999(n)*(q^A000326(n)+q^a(n)))^24. - Jonathan Vos Post, Mar 15 2006
Row sums of triangle A134403.
Bisection of A001318. - Omar E. Pol, Aug 22 2011
Sequence found by reading the line from 0 in the direction 0, 7, ... and the line from 2 in the direction 2, 15, ... in the square spiral whose vertices are the generalized pentagonal numbers, A001318. - Omar E. Pol, Sep 08 2011
A general formula for the n-th second k-gonal number is given by T(n, k) = n*((k-2)*n+k-4)/2, n>=0, k>=5. - Omar E. Pol, Aug 04 2012
Partial sums give A006002. - Denis Borris, Jan 07 2013
A002260 is the following array A read by antidiagonals:
  0,  1,  2,  3,  4,  5, ...
  0,  1,  2,  3,  4,  5, ...
  0,  1,  2,  3,  4,  5, ...
  0,  1,  2,  3,  4,  5, ...
  0,  1,  2,  3,  4,  5, ...
  0,  1,  2,  3,  4,  5, ...
and a(n) is the hook sum: Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013
From Klaus Purath, May 13 2021: (Start)
This sequence and A000326 provide all integers m such that 24*m + 1 is a square. The union of the two sequences is A001318.
If A is a sequence satisfying the recurrence t(n) = 3*t(n-1) - 2*t(n-2) with the initial values either A(0) = 1, A(1) = n + 2 or A(0) = -1, A(1) = n - 1, then a(n) = (A(i)^2 - A(i-1)*A(i+1))/2^i + n^2 for i>0. (End)
a(n+1) is the number of Dyck paths of size (3,3n+2), i.e., the number of NE lattice paths from (0,0) to (3,3n+2) which stay above the line connecting these points. - Harry Richman, Jul 13 2021
Binomial transform of [0, 2, 3, 0, 0, 0, ...], being a(n) = 2*binomial(n,1) + 3*binomial(n,2). a(3) = 15 = [0, 2, 3, 0] dot [1, 3, 3, 1] = [0 + 6 + 9 + 0]. - Gary W. Adamson, Dec 17 2022
a(n) is the sum of longest side length of all nondegenerate integer-sided triangles with shortest side length n and middle side length (n + 1), n > 0. - Torlach Rush, Feb 04 2024

			From _Omar E. Pol_, Aug 22 2011: (Start)
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     O O O O     O O O O O
    O    O O     O O O     O O O O     O O O O O
    -    ---     -----     -------     ---------
    2     7        15         26           40
(End)

Henri Cohen, A Course in Computational Algebraic Number Theory, vol. 138 of Graduate Texts in Mathematics, Springer-Verlag, 2000.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
A. O. L. Atkin and F. Morain, Elliptic Curves and Primality Proving, Math. Comp., Vol. 61, No. 203 (1993), pp. 29-68.
Charles H. Conley and Valentin Ovsienko, Quiddities of polygon dissections and the Conway-Coxeter frieze equation, arXiv:2107.01234 [math.CO], 2021.
Leonhard Euler, De mirabilibus proprietatibus numerorum pentagonalium, Acta Academiae Scientiarum Imperialis Petropolitanae, Vol. 1780: I, pp. 56-75.
Leonhard Euler, Observatio de summis divisorum, Novi Commentarii academiae scientiarum Petropolitanae, Vol. 5, pp. 59-74.
Leonhard Euler, An observation on the sums of divisors, arXiv:math/0411587 [math.HO], 2004-2009, p. 8.
Leonhard Euler, On the remarkable properties of the pentagonal numbers, arXiv:math/0505373 [math.HO], 2005.
Alfred Hoehn, Illustration of initial terms of A000326, A005449, A045943, A115067.
D. Suprijanto and I. W. Suwarno, Observation on Sums of Powers of Integers Divisible by 3k-1, Applied Mathematical Sciences, Vol. 8, No. 45 (2014), pp. 2211-2217.
Martin Wohlgemuth, Pentagon, Kartenhaus und Summenzerlegung.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A016789 (first differences), A006002 (partial sums).
Cf. A000217, A000320, A000326, A001318, A033568, A049451, A101165, A101166.
The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, this sequence, A045943, A115067, A140090, A140091, A059845, A140672-A140675, A151542.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=3).
Cf. numbers of the form n*((2*k+1)*n+1)/2 listed in A022289 (this sequence is the case k=1).

[n*(3*n + 1) / 2: n in [0..40]]; // Vincenzo Librandi, May 02 2011

A005449:=n->n*(3*n + 1)/2; seq(A005449(k), k=0..100); # Wesley Ivan Hurt, Oct 11 2013

Table[n (3 n + 1)/2, {n, 0, 100}] (* Zak Seidov, Jan 31 2012 *)

{a(n) = n * (3*n + 1) / 2} /* Michael Somos, Jul 18 2003 */

a(n) = A110449(n, 1) for n>0.
G.f.: x*(2+x)/(1-x)^3. E.g.f.: exp(x)*(2*x + 3*x^2/2). a(n) = n*(3*n + 1)/2. a(-n) = A000326(n). - Michael Somos, Jul 18 2003
a(n) = A001844(n) - A000217(n+1) = A101164(n+2,2) for n>0. - Reinhard Zumkeller, Dec 03 2004
a(n) = Sum_{j=1..n} n+j. - Zerinvary Lajos, Sep 12 2006
a(n) = A126890(n,n). - Reinhard Zumkeller, Dec 30 2006
a(n) = 2*C(3*n,4)/C(3*n,2), n>=1. - Zerinvary Lajos, Jan 02 2007
a(n) = A000217(n) + A000290(n). - Zak Seidov, Apr 06 2008
a(n) = a(n-1) + 3*n - 1 for n>0, a(0)=0. - Vincenzo Librandi, Nov 18 2010
a(n) = A129267(n+5,n). - Philippe Deléham, Dec 21 2011
a(n) = 2*A000217(n) + A000217(n-1). - Philippe Deléham, Mar 25 2013
a(n) = A130518(3*n+1). - Philippe Deléham, Mar 26 2013
a(n) = (12/(n+2)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+2). - Vladimir Kruchinin, Jun 04 2013
a(n) = floor(n/(1-exp(-2/(3*n)))) for n>0. - Richard R. Forberg, Jun 22 2013
a(n) = Sum_{i=1..n} (3*i - 1) for n >= 1. - Wesley Ivan Hurt, Oct 11 2013 [Corrected by Rémi Guillaume, Oct 24 2024]
a(n) = (A000292(6*n+k+1)-A000292(k))/(6*n+1) - A000217(3*n+k+1), for any k >= 0. - Manfred Arens, Apr 26 2015
Sum_{n>=1} 1/a(n) = 6 - Pi/sqrt(3) - 3*log(3) = 0.89036376976145307522... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(2*n) - A000217(n). - Bruno Berselli, Sep 21 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/sqrt(3) + 4*log(2) - 6. - Amiram Eldar, Jan 18 2021
From Klaus Purath, May 13 2021: (Start)
Partial sums of A016789 for n > 0.
a(n) = 3*n^2 - A000326(n).
a(n) = A000326(n) + n.
a(n) = A002378(n) + A000217(n-1) for n >= 1. [Corrected by Rémi Guillaume, Aug 14 2024] (End)
From Klaus Purath, Jul 14 2021: (Start)
b^2 = 24*a(n) + 1 we get by b^2 = (a(n+1) - a(n-1))^2 = (a(2*n)/n)^2.
a(2*n) = n*(a(n+1) - a(n-1)), n > 0.
a(2*n+1) = n*(a(n+1) - a(n)). (End)
A generalization of Lajos' formula, dated Sep 12 2006, follows. Let SP(k,n) = the n-th second k-gonal number. Then SP(2k+1,n) = Sum_{j=1..n} (k-1)*n+j+k-2. - Charlie Marion, Jul 13 2024
a(n) = Sum_{k = 0..3*n} (-1)^(n+k+1) * binomial(k, 2) * binomial(3*n+k, 2*k). - Peter Bala, Nov 03 2024
For integer m, (6*m + 1)^2*a(n) + a(m) = a((6*m+1)*n + m). - Peter Bala, Jan 09 2025

A002264 Nonnegative integers repeated 3 times.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24, 25

Offset: 0

N. J. A. Sloane

Complement of A010872, since A010872(n) + 3*a(n) = n. - Hieronymus Fischer, Jun 01 2007
Chvátal proved that, given an arbitrary n-gon, there exist a(n) points such that all points in the interior are visible from at least one of those points; further, for all n >= 3, there exists an n-gon which cannot be covered in this fashion with fewer than a(n) points. This is known as the "art gallery problem". - Charles R Greathouse IV, Aug 29 2012
The inverse binomial transform is 0, 0, 0, 1, -3, 6, -9, 9, 0, -27, 81, -162, 243, -243, 0, 729,.. (see A000748). - R. J. Mathar, Feb 25 2023

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Václav Chvátal, A combinatorial theorem in plane geometry, Journal of Combinatorial Theory, Series B 18 (1975), pp. 39-41, doi:10.1016/0095-8956(75)90061-1.
Clark Kimberling, A Combinatorial Classification of Triangle Centers on the Line at Infinity, J. Int. Seq., Vol. 22 (2019), Article 19.5.4.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A001477, A002265, A002266, A004526, A008615, A008620, A010761, A010762, A010872, A010873, A010874, A022003, A110532, A110533, A137221 (binomial transform).
Partial sums give A130518.
Cf. A004523 interlaced with A004396.
Apart from the zeros, this is column 3 of A235791.

a002264 n = a002264_list !! n
a002264_list = 0 : 0 : 0 : map (+ 1) a002264_list
-- Reinhard Zumkeller, Nov 06 2012, Apr 16 2012

[Floor(n/3): n in [0..100]]; // Vincenzo Librandi, Apr 29 2015

&cat [[n,n,n]: n in [0..30]]; // Bruno Berselli, Apr 29 2015

seq(i$3,i=0..100); # Robert Israel, Aug 04 2014

Flatten[Table[{n, n, n}, {n, 0, 25}]] (* Harvey P. Dale, Jun 09 2013 *)
Floor[Range[0, 20]/3] (* Eric W. Weisstein, Aug 12 2023 *)
Table[Floor[n/3], {n, 0, 20}] (* Eric W. Weisstein, Aug 12 2023 *)
Table[(n - Cos[2 (n - 2) Pi/3] + Sin[2 (n - 2) Pi/3]/Sqrt[3] - 1)/3, {n, 0, 20}] (* Eric W. Weisstein, Aug 12 2023 *)
Table[(n - ChebyshevU[n - 2, -1/2] - 1)/3, {n, 0, 20}] (* Eric W. Weisstein, Aug 12 2023 *)
LinearRecurrence[{1, 0, 1, -1}, {0, 0, 0, 1}, 20] (* Eric W. Weisstein, Aug 12 2023 *)
CoefficientList[Series[x^3/((-1 + x)^2 (1 + x + x^2)), {x, 0, 20}], x] (* Eric W. Weisstein, Aug 12 2023 *)

a(n)=n\3  /* Jaume Oliver Lafont, Mar 25 2009 */

v=[0,0];for(n=2,50,v=concat(v,n-2-v[#v]-v[#v-1]));v \\ Derek Orr, Apr 28 2015

[floor(n/3) for n in range(0,79)] # Zerinvary Lajos, Dec 01 2009

a(n) = floor(n/3).
a(n) = (3*n-3-sqrt(3)*(1-2*cos(2*Pi*(n-1)/3))*sin(2*Pi*(n-1)/3))/9. - Hieronymus Fischer, Sep 18 2007
a(n) = (n - A010872(n))/3. - Hieronymus Fischer, Sep 18 2007
Complex representation: a(n) = (n - (1 - r^n)*(1 + r^n/(1 - r)))/3 where r = exp(2*Pi/3*i) = (-1 + sqrt(3)*i)/2 and i = sqrt(-1). - Hieronymus Fischer, Sep 18 2007; - corrected by Guenther Schrack, Sep 26 2019
a(n) = Sum_{k=0..n-1} A022003(k). - Hieronymus Fischer, Sep 18 2007
G.f.: x^3/((1-x)*(1-x^3)). - Hieronymus Fischer, Sep 18 2007
a(n) = (n - 1 + 2*sin(4*(n+2)*Pi/3)/sqrt(3))/3. - Jaume Oliver Lafont, Dec 05 2008
For n >= 3, a(n) = floor(log_3(3^a(n-1) + 3^a(n-2) + 3^a(n-3))). - Vladimir Shevelev, Jun 22 2010
a(n) = (n - 3 + A010872(n-1) + A010872(n-2))/3 using Zumkeller's 2008 formula in A010872. - Adriano Caroli, Nov 23 2010
a(n) = A004526(n) - A008615(n). - Reinhard Zumkeller, Apr 28 2014
a(2*n) = A004523(n) and a(2*n+1) = A004396(n). - L. Edson Jeffery, Jul 30 2014
a(n) = n - 2 - a(n-1) - a(n-2) for n > 1 with a(0) = a(1) = 0. - Derek Orr, Apr 28 2015
From Wesley Ivan Hurt, May 27 2015: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4), n > 4.
a(n) = (n - 1 + 0^((-1)^(n/3) - (-1)^n) - 0^((-1)^(n/3)*(-1)^(1/3) + (-1)^n))/3. (End)
a(n) = (3*n - 3 + r^n*(1 - r) + r^(2*n)*(r + 2))/9 where r = (-1 + sqrt(-3))/2. - Guenther Schrack, Sep 26 2019
E.g.f.: exp(x)*(x - 1)/3 + exp(-x/2)*(3*cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2))/9. - Stefano Spezia, Oct 17 2022

A131242 Partial sums of A059995: a(n) = sum_{k=0..n} floor(k/10).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 156, 162, 168, 174, 180, 186, 192, 198

Offset: 0

Hieronymus Fischer, Jun 21 2007

Complementary with A130488 regarding triangular numbers, in that A130488(n)+10*a(n)=n(n+1)/2=A000217(n).

			As square array :
    0,   0,   0,   0,   0,   0,   0,   0,   0,    0
    1,   2,   3,   4,   5,   6,   7,   8,   9,   10
   12,  14,  16,  18,  20,  22,  24,  26,  28,   30
   33,  36,  39,  42,  45,  48,  51,  54,  57,   60
   64,  68,  72,  76,  80,  84,  88,  92,  96,  100
  105, 110, 115, 120, 125, 130, 135, 140, 145,  150
  156, 162, 168, 174, 180, 186, 192, 198, 204,  210
... - _Philippe Deléham_, Mar 27 2013

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

Cf. A008728, A059995, A010879, A002266, A130488, A000217, A002620, A130518, A130519, A130520, A174709, A174738, A118729, A218470.

Table[(1/2)*Floor[n/10]*(2*n - 8 - 10*Floor[n/10]), {n,0,50}] (* G. C. Greubel, Dec 13 2016 *)
Accumulate[Table[FromDigits[Most[IntegerDigits[n]]],{n,0,110}]] (* or *) LinearRecurrence[{2,-1,0,0,0,0,0,0,0,1,-2,1},{0,0,0,0,0,0,0,0,0,0,1,2},120] (* Harvey P. Dale, Apr 06 2017 *)

for(n=0,50, print1((1/2)*floor(n/10)*(2n-8-10*floor(n/10)), ", ")) \\ G. C. Greubel, Dec 13 2016

a(n)=my(k=n\10); k*(n-5*k-4) \\ Charles R Greathouse IV, Dec 13 2016

a(n) = (1/2)*floor(n/10)*(2n-8-10*floor(n/10)).
a(n) = A059995(n)*(2n-8-10*A059995(n))/2.
a(n) = (1/2)*A059995(n)*(n-8+A010879(n)).
a(n) = (n-A010879(n))*(n+A010879(n)-8)/20.
G.f.: x^10/((1-x^10)(1-x)^2).
From Philippe Deléham, Mar 27 2013: (Start)
a(10n)   = A051624(n).
a(10n+1) = A135706(n).
a(10n+2) = A147874(n+1).
a(10n+3) = 2*A005476(n).
a(10n+4) = A033429(n).
a(10n+5) = A202803(n).
a(10n+6) = A168668(n).
a(10n+7) = 2*A147875(n).
a(10n+8) = A135705(n).
a(10n+9) = A124080(n). (End)
a(n) = A008728(n-10) for n>= 10. - Georg Fischer, Nov 03 2018

A174738 Partial sums of floor(n/7).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 15, 17, 19, 21, 24, 27, 30, 33, 36, 39, 42, 46, 50, 54, 58, 62, 66, 70, 75, 80, 85, 90, 95, 100, 105, 111, 117, 123, 129, 135, 141, 147, 154, 161, 168, 175, 182, 189, 196, 204, 212, 220, 228, 236

Offset: 0

Mircea Merca, Nov 30 2010

Apart from the initial zeros, the same as A011867.

			a(9) = floor(0/7) + floor(1/7) + floor(2/7) + floor(3/7) + floor(4/7) + floor(5/7) + floor(6/7) + floor(7/7) + floor(8/7) + floor(9/7) = 3.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions, J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,1,-2,1).

Cf. A001106, A022264, A022265, A024966, A179986, A186029, A218471.

Cf. Similar sequences: A000217, A002620, A130518, A130519, A130520, A174709, A118729, A218470, A131242.

List([0..60], n-> Int((n-2)*(n-3)/14)); # G. C. Greubel, Aug 31 2019

[Round(n*(n-5)/14): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011

A174738 := proc(n) round(n*(n-5)/14) ; end proc:
seq(A174738(n),n=0..30) ;

Table[Floor[(n - 2)*(n - 3)/14], {n,0,60}] (* G. C. Greubel, Dec 13 2016 *)

a(n)=(n-2)*(n-3)\14 \\ Charles R Greathouse IV, Sep 24 2015

[floor((n-2)*(n-3)/14) for n in (0..60)] # G. C. Greubel, Aug 31 2019

a(n) = round(n*(n-5)/14).
a(n) = floor((n-2)*(n-3)/14).
a(n) = ceiling((n+1)*(n-6)/14).
a(n) = a(n-7) + n - 6, n > 6.
a(n) = +2*a(n-1) - a(n-2) + a(n-7) - 2*a(n-8) + a(n-9). - R. J. Mathar, Nov 30 2010
G.f.: x^7/( (1 + x + x^2 + x^3 + x^4 + x^5 + x^6)*(1-x)^3 ). - R. J. Mathar, Nov 30 2010
a(7n) = A001106(n), a(7n+1) = A218471(n), a(7n+2) = A022264(n), a(7n+3) = A022265(n), a(7n+4) = A186029(n), a(7n+5) = A179986(n), a(7n+6) = A024966(n). - Philippe Deléham, Mar 26 2013

A118729 Rectangular array where row r contains the 8 numbers 4r^2 - 3r, 4r^2 - 2r, ..., 4r^2 + 4r.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 42, 45, 48, 52, 56, 60, 64, 68, 72, 76, 80, 85, 90, 95, 100, 105, 110, 115, 120, 126, 132, 138, 144, 150, 156, 162, 168

Offset: 0

Stuart M. Ellerstein (ellerstein(AT)aol.com), May 21 2006

The numbers in row r span the interval ]8*A000217(r-1), 8*A000217(r)].
The first difference between the entries in row r is r.
Partial sums of floor(n/8). - Philippe Deléham, Mar 26 2013
Apart from the initial zeros, the same as A008726. - Philippe Deléham, Mar 28 2013
a(n+7) is the number of key presses required to type a word of n letters, all different, on a keypad with 8 keys where 1 press of a key is some letter, 2 presses is some other letter, etc., and under an optimal mapping of letters to keys and presses (answering LeetCode problem 3014). - Christopher J. Thomas, Feb 16 2024

			The array starts, with row r=0, as
  r=0:   0  0  0  0  0  0  0  0;
  r=1:   1  2  3  4  5  6  7  8;
  r=2:  10 12 14 16 18 20 22 24;
  r=3:  27 30 33 36 39 42 45 48;

G. C. Greubel, Table of n, a(n) for n = 0..1000
LeetCode, 3014. Minimum Number of Pushes to Type Word I.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,1,-2,1).

Cf. similar sequences: A000217, A002620, A130518, A130519, A130520, A174709, A174738, A218470, A131242.

Flatten[Table[4r^2+r(Range[-3,4]),{r,0,6}]] (* or *) LinearRecurrence[ {2,-1,0,0,0,0,0,1,-2,1},{0,0,0,0,0,0,0,0,1,2},60] (* Harvey P. Dale, Nov 26 2015 *)

From Philippe Deléham, Mar 26 2013: (Start)
a(8k)   = A001107(k).
a(8k+1) = A002939(k).
a(8k+2) = A033991(k).
a(8k+3) = A016742(k).
a(8k+4) = A007742(k).
a(8k+5) = A002943(k).
a(8k+6) = A033954(k).
a(8k+7) = A033996(k). (End)
G.f.: x^8/((1-x)^2*(1-x^8)). - Philippe Deléham, Mar 28 2013
a(n) = floor(n/8)*(n-3-4*floor(n/8)). - Ridouane Oudra, Jun 04 2019
a(n+7) = (1/2)*(n+(n mod 8))*(floor(n/8)+1). - Christopher J. Thomas, Feb 13 2024

Redefined as a rectangular tabf array and description simplified by R. J. Mathar, Oct 20 2010

A218530 Partial sums of floor(n/11).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 70, 74, 78, 82, 86, 90, 94, 98, 102, 106, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 171

Offset: 0

Philippe Deléham, Mar 27 2013

Apart from the initial zeros, the same as A008729.

			As square array:
..0....0....0....0....0....0....0....0....0....0....0
..1....2....3....4....5....6....7....8....9...10...11
.13...15...17...19...21...23...25...27...29...31...33
.36...39...42...45...48...51...54...57...60...63...66
.70...74...78...82...86...90...94...98..102..106..110
115..120..125..130..135..140..145..150..155..160..165
171..177..183..189..195..201..207..213..219..225..231
238..245..252..259..266..273..280..287..294..301..308
316..324..332..340..348..356..364..372..380..388..396
405..414..423..432..441..450..459..468..477..486..495
505..515..525..535..545..555..565..575..585..595..605
...

Cf. similar sequences: A000217, A002820, A130518, A130519, A130520, A174709, A174738, A118729, A218470, A131242.

a(11n)    = A051865(n).
a(11n+1)  = A180223(n).
a(11n+4)  = A022268(n).
a(11n+5)  = A022269(n).
a(11n+6)  = A254963(n)
a(11n+9)  = A211013(n).
a(11n+10) = A152740(n).
G.f.: x^11/((1-x)^2*(1-x^11)).

A134546 Triangle read by rows: T(n, k) = Sum_{j=0..n} floor(j / k).

Original entry on oeis.org

1, 3, 1, 6, 2, 1, 10, 4, 2, 1, 15, 6, 3, 2, 1, 21, 9, 5, 3, 2, 1, 28, 12, 7, 4, 3, 2, 1, 36, 16, 9, 6, 4, 3, 2, 1, 45, 20, 12, 8, 5, 4, 3, 2, 1, 55, 25, 15, 10, 7, 5, 4, 3, 2, 1, 66, 30, 18, 12, 9, 6, 5, 4, 3, 2, 1, 78, 36, 22, 15, 11, 8, 6, 5, 4, 3, 2, 1, 91, 42, 26, 18, 13, 10, 7, 6, 5, 4, 3, 2, 1

Offset: 1

Gary W. Adamson, Oct 31 2007

From Bob Selcoe, Aug 08 2016: (Start)
Columns are partial sums of k-repeating increasing positive integers:
  Column 1 is {1+2+3+4+5+...}         = A000217 (triangular numbers);
  Column 2 is {1+1+2+2+3+3+4+4+...}   = A002620 (quarter-squares);
  Column 3 is {1+1+1+2+2+2+3+3+3+...} = A130518.
  Columns k = 4..7 are A130519, A130520, A174709 and A174738, respectively.
T(n, k) is the number of positive multiples of k which can be expressed as i-j, {i=1..n; j=0..n-1}. So for example, T(5, 2) = 6 because there are 6 ways to express i-j {i<=5} as a multiple of 2: {5-3, 4-2, 3-1, 2-0, 5-1 and 4-0}. (End)
Conjecture: For T(n, k) n >= k^(3/2), there is at least one prime in the interval [T(n-1, k+1), T(n, k)]. - Bob Selcoe, Aug 21 2016
Theorem: For n >= 3*k, T(n, k) is composite. - Daniel Hoying, Jul 08 2020

			The triangle T(n, k) begins:
   n\k  1   2   3   4  5  6  7  8  9  10 ...
   1:   1
   2:   3   1
   3:   6   2   1
   4:  10   4   2   1
   5:  15   6   3   2  1
   6:  21   9   5   3  2  1
   7:  28  12   7   4  3  2  1
   8:  36  16   9   6  4  3  2  1
   9:  45  20  12   8  5  4  3  2  1
  10:  55  25  15  10  7  5  4  3  2   1
... Reformatted. - _Wolfdieter Lang_, Feb 04 2015
T(10,3) = 15: 3*floor(10/3)*floor(13/3)/2 - floor(10/3)*(3-1 - 13 mod 3) = 3*3*4/2 - 3*(3-1-1) = 18 - 3 = 15. - _Bob Selcoe_, Aug 21 2016

Michael De Vlieger, Table of n, a(n) for n = 1..11325 (1 <= n <= 150)

Cf. A078567 (row sums), A000217 (column 1).

Cf. A004736, A051731, A002620, A130518, A130519, A130520, A174709, A174738.

T := proc(n, k) option remember: `if`(n = k, 1, T(n-1, k) + iquo(n,k)) end:
seq(seq(T(n,k), k=1..n),n=1..16); # Peter Luschny, May 26 2020

nn = 12; f[w_] := Map[PadRight[#, nn] &, w]; MapIndexed[Take[#1, First@ #2] &, f@ Table[Reverse@ Range@ n, {n, nn}].f@ Table[Boole@ Divisible[n, #] & /@ Range@ n, {n, nn}]] // Flatten (* Michael De Vlieger, Aug 10 2016 *)

t(n, k) = if (k>n, 0, if (n==1, 1, t(n-1, k) + n\k));
tabl(nn) = {m = matrix(nn, nn, n , k, t(n,k)); for (n=1, nn, for (k=1, n, print1(m[n, k], ", ");); print(););} \\ Michel Marcus, Jan 18 2015

trg(nn) = {ma = matrix(nn, nn, n, k, if (k<=n, n-k+1, 0)); mb = matrix(nn, nn, n, k, if (k<=n, !(n%k), 0)); ma*mb;} \\ Michel Marcus, Jan 20 2015

Original definition: T = A004736 * A051731 as infinite lower triangular matrices.
In other words: T(n, k) = Sum_{m=k..n} A004736(n, m)*A051731(m, k).
T(n, k) = 0 if n < k, T(1, 1) = 1, and T(n, k) = T(n-1, k) + floor(n/k), for n >= 2. - Richard R. Forberg, Jan 17 2015
T(n, k) = k*floor(n/k)*floor((n+k)/k)/2 - floor(n/k)*(k-1-(n mod k)). - Bob Selcoe, Aug 21 2016
T(n, k) = k*A000217(b) + (b+1)*[(n +1)-(b + 1)*k] for 1 <= k <= floor[(n + 1) / 2] where b = floor[(n - k + 1) / k], T(n, k) = n-k+1 for floor[(n + 1) / 2] < k <= n and T(n, k) = 0 for k > n. - Henri Gonin, May 12 2020
T(n, k) = (-k/2)*floor(n/k)^2+(n-k/2+1)*floor(n/k). - Daniel Hoying, May 25 2020
From Daniel Hoying, Jul 06 2020: (Start)
T(m + 2*n - 1, m + n) = n for n > 0, m >= 0.
T(3*m + 3*ceiling((n-3)/6) + (n+1)/2, 2*m + 2*ceiling((n-3)/6) + 1) = n for n > 0, n odd, 0 <= m <= floor(n/3).
T(3*m + 3*ceiling(n/6) + n/2 - 1, 2*m + 2*ceiling(n/6)) = n for n > 0, n even, 0 <= m <= floor(n/3). (End)

Edited. Name clarified. Formulas rewritten. - Wolfdieter Lang, Feb 04 2015
Corrected and extended by Michael De Vlieger, Aug 10 2016
Edited and new name from Peter Luschny, Apr 02 2025

A368343 Square array T(n,k), n >= 3, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} k^(n-j) * floor(j/3).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 1, 4, 7, 5, 2, 1, 5, 13, 16, 7, 2, 1, 6, 21, 41, 34, 9, 3, 1, 7, 31, 86, 125, 70, 12, 3, 1, 8, 43, 157, 346, 377, 143, 15, 3, 1, 9, 57, 260, 787, 1386, 1134, 289, 18, 4, 1, 10, 73, 401, 1562, 3937, 5547, 3405, 581, 22, 4

Offset: 3

Seiichi Manyama, Dec 22 2023

			Square array begins:
  1,  1,   1,    1,    1,     1,     1, ...
  1,  2,   3,    4,    5,     6,     7, ...
  1,  3,   7,   13,   21,    31,    43, ...
  2,  5,  16,   41,   86,   157,   260, ...
  2,  7,  34,  125,  346,   787,  1562, ...
  2,  9,  70,  377, 1386,  3937,  9374, ...
  3, 12, 143, 1134, 5547, 19688, 56247, ...

Seiichi Manyama, Antidiagonals n = 3..142, flattened

Columns k=0..4 give A002264, A130518, A178455, A368344, A368345.

Cf. A055129, A368296.

PARI
```
T(n, k) = sum(j=0, n, k^(n-j)*(j\3));
```

T(n,k) = T(n-3,k) + Sum_{j=0..n-3} k^j.
T(n,k) = 1/(k-1) * Sum_{j=0..n} floor(k^j/(k^2+k+1)) = Sum_{j=0..n} floor(k^j/(k^3-1)) for k > 1.
T(n,k) = (k+1)*T(n-1,k) - k*T(n-2,k) + T(n-3,k) - (k+1)*T(n-4,k) + k*T(n-5,k).
G.f. of column k: x^3/((1-x) * (1-k*x) * (1-x^3)).
T(n,k) = 1/(k-1) * (floor(k^(n+1)/(k^3-1)) - floor((n+1)/3)) for k > 1.

A062781 Number of arithmetic progressions of four terms and any mean which can be extracted from the set of the first n positive integers.

Original entry on oeis.org

0, 0, 0, 1, 2, 3, 5, 7, 9, 12, 15, 18, 22, 26, 30, 35, 40, 45, 51, 57, 63, 70, 77, 84, 92, 100, 108, 117, 126, 135, 145, 155, 165, 176, 187, 198, 210, 222, 234, 247, 260, 273, 287, 301, 315, 330, 345, 360, 376, 392

Offset: 1

Santi Spadaro, Jul 18 2001

This sequence seems to be a shifted version of the Somos sequence A058937.
Equal to the partial sums of A002264 (cf. A130518) but with initial index 1 instead of 0. - Hieronymus Fischer, Jun 01 2007
Apart from offset, the same as A130518. - R. J. Mathar, Jun 13 2008
Apart from offset, the same as A001840. - Michael Somos, Sep 18 2010

Muniru A Asiru, Table of n, a(n) for n = 1..750
Michael Somos, Somos Polynomials
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A058937, A001840.

Cf. A002620, A130519, A130520.

seq(coeff(series(x^4/((1-x^3)*(1-x)^2),x,n+1), x, n), n = 1 .. 50); # Muniru A Asiru, Nov 25 2018

RecurrenceTable[{a[0]==0, a[n]==Floor[n/3] + a[n-1]}, a, {n, 49}] (* Jon Maiga, Nov 25 2018 *)

[floor(binomial(n,2)/3) for n in range(0,50)] # Zerinvary Lajos, Dec 01 2009

a(n) = P(n,4), where P(n,k) = n*floor(n/(k - 1)) - (1/2)(k - 1)(floor(n/(k - 1))*(floor(n/(k - 1)) + 1)); recursion: a(n) = a(n-3) + n - 3; a(1) = a(2) = a(3) = 0.
From Hieronymus Fischer, Jun 01 2007: (Start)
a(n) = (1/2)*floor((n-1)/3)*(2*n - 3 - 3*floor((n-1)/3)).
G.f.: x^4/((1 - x^3)*(1 - x)^2). (End)
a(n) = floor((n-1)/3) + a(n-1). - Jon Maiga, Nov 25 2018
E.g.f.: ((4 - 6*x + 3*x^2)*exp(x) - 4*exp(-x/2)*cos(sqrt(3)*x/2))/18. - Franck Maminirina Ramaharo, Nov 25 2018

cp's OEIS Frontend

A247247 Triangular numbers that are the sum of 2 consecutive terms of A130518.

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A005449 Second pentagonal numbers: a(n) = n*(3*n + 1)/2.

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A002264 Nonnegative integers repeated 3 times.

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A131242 Partial sums of A059995: a(n) = sum_{k=0..n} floor(k/10).

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A174738 Partial sums of floor(n/7).

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A118729 Rectangular array where row r contains the 8 numbers 4*r^2 - 3*r, 4*r^2 - 2*r, ..., 4*r^2 + 4*r.

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A005449 Second pentagonal numbers: a(n) = n(3n + 1)/2.

A118729 Rectangular array where row r contains the 8 numbers 4r^2 - 3r, 4r^2 - 2r, ..., 4r^2 + 4r.