A056000 -id:A056000 - cp's OEIS frontend

A056126 a(n) = n*(n + 17)/2.

0, 9, 19, 30, 42, 55, 69, 84, 100, 117, 135, 154, 174, 195, 217, 240, 264, 289, 315, 342, 370, 399, 429, 460, 492, 525, 559, 594, 630, 667, 705, 744, 784, 825, 867, 910, 954, 999, 1045, 1092, 1140, 1189, 1239, 1290, 1342, 1395, 1449, 1504, 1560, 1617, 1675

Offset: 0

Barry E. Williams, Jul 07 2000

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A001477, A051942, A056000, A056121.

Cf. A001008, A002805, A098849, A120071, A132760, A132761, A132765.

List([0..50], n-> n*(n+17)/2 ); # G. C. Greubel, Jan 19 2020

[n*(n+17)/2: n in [0..50]]; // G. C. Greubel, Jan 19 2020

seq( n*(n+17)/2, n=0..50); # G. C. Greubel, Jan 19 2020

Table[n(n+17)/2,{n,0,50}] (* Harvey P. Dale, Apr 25 2011 *)

a(n)=n*(n+17)/2 \\ Charles R Greathouse IV, Sep 24 2015

[n*(n+17)/2 for n in (0..50)] # G. C. Greubel, Jan 19 2020

G.f.: x*(9-8*x)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A126890(n,8) for n>7. - Reinhard Zumkeller, Dec 30 2006
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*stirling1(n-k,i)* Product_{j=0..k-1} (-a-j), then a(n) = -f(n,n-1,9), for n>=1. - Milan Janjic, Dec 20 2008
a(n) = a(n-1) + n + 8 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
a(n) = 9*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
E.g.f.: x*(18 + x)*exp(x)/2. - G. C. Greubel, Jan 19 2020
From Amiram Eldar, Jan 10 2021: (Start)
Sum_{n>=1} 1/a(n) = 2*A001008(17)/(17*A002805(17)) = 42142223/104144040.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/17 - 1768477/20828808. (End)

A126890 Triangle read by rows: T(n,k) = n(n+2k+1)/2, 0 <= k <= n.

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 6, 9, 12, 15, 10, 14, 18, 22, 26, 15, 20, 25, 30, 35, 40, 21, 27, 33, 39, 45, 51, 57, 28, 35, 42, 49, 56, 63, 70, 77, 36, 44, 52, 60, 68, 76, 84, 92, 100, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 55, 65, 75, 85, 95, 105, 115, 125, 135, 145, 155, 66, 77, 88

Offset: 0

Reinhard Zumkeller, Dec 30 2006

T(n,k) + T(n,n-k) = A014105(n);
row sums give A059270; Sum_{k=0..n-1} T(n,k) = A000578(n);
central terms give A007742; T(2*n+1,n) = A016754(n);
T(n,0)   = A000217(n);
T(n,1)   = A000096(n)   for n > 0;
T(n,2)   = A055998(n)   for n > 1;
T(n,3)   = A055999(n)   for n > 2;
T(n,4)   = A056000(n)   for n > 3;
T(n,5)   = A056115(n)   for n > 4;
T(n,6)   = A056119(n)   for n > 5;
T(n,7)   = A056121(n)   for n > 6;
T(n,8)   = A056126(n)   for n > 7;
T(n,10)  = A101859(n-1) for n > 9;
T(n,n-3) = A095794(n-1) for n > 2;
T(n,n-2) = A045943(n-1) for n > 1;
T(n,n-1) = A000326(n)   for n > 0;
T(n,n)   = A005449(n).

			From _Philippe Deléham_, Oct 03 2011: (Start)
Triangle begins:
   0;
   1,  2;
   3,  5,  7;
   6,  9, 12, 15;
  10, 14, 18, 22, 26;
  15, 20, 25, 30, 35, 40;
  21, 27, 33, 39, 45, 51, 57;
  28, 35, 42, 49, 56, 63, 70, 77; (End)

Léonard Euler, Introduction à l'analyse infinitésimale, tome premier, ACL-Editions, Paris, 1987, p. 353-354.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Émile Fourrey, Les nombres abstraits, Récreations arithmétiques, 1899 and later, Vuibert, Paris, page 86-87. Triangle without right diagonal.
Adrien-Marie Legendre, Théorie des nombres, tome 2, quatrième partie, p.131, troisième édition, Paris, 1830.

Cf. A110449.

a126890 n k = a126890_tabl !! n !! k
a126890_row n = a126890_tabl !! n
a126890_tabl = map fst $ iterate
   (\(xs@(x:_), i) -> (zipWith (+) ((x-i):xs) [2*i+1 ..], i+1)) ([0], 0)
-- Reinhard Zumkeller, Nov 10 2013

Flatten[Table[(n(n+2k+1))/2,{n,0,20},{k,0,n}]] (* Harvey P. Dale, Jun 21 2013 *)

T(n,k) = T(n,k-1) + n, for k <= n. - Philippe Deléham, Oct 03 2011

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

A111774 Numbers that can be written as a sum of at least three consecutive positive integers.

Original entry on oeis.org

6, 9, 10, 12, 14, 15, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 102

Offset: 1

Jaap Spies, Aug 15 2005

In this sequence there are no (odd) primes and there are no powers of 2.
So we have only three kinds of natural numbers: the odd primes, the powers of 2 and the numbers that can be represented as a sum of at least three consecutive integers.
Odd primes can only be written as a sum of two consecutive integers. Powers of 2 do not have a representation as a sum of k consecutive integers (other than the trivial n=n, for k=1).
Numbers of the form (x*(x+1)-y*(y+1))/2 for nonnegative integers x,y with x-y >= 3. - Bob Selcoe, Feb 21 2014
Numbers of the form (x + 1)*(x + 2*y)/2 for integers x,y with x >= 2 and y >= 1. For y = 1 only triangular numbers (A000217) >= 6 occur. - Ralf Steiner, Jun 27 2019
From Ralf Steiner, Jul 09 2019: (Start)
If k >= 1 sequences are c_k(n) = c_k(n - 1) + n + k - 1, c_k(0) = 0, means c_k(n) = n*(n + 2*k - 1)/2: A000217, A000096, A055998, A055999, A056000, ... then this sequence is the union of c_k(n), n >= 3. (End)
From Wolfdieter Lang, Oct 28 2020: (Start)
This sequence gives all positive integers that have at least one odd prime as proper divisor. The proof follows from the first two comments.
The set {a(n)}_{n>=1} equals the set {k positive integer : floor(k/2) - delta(k) >= 1}, where delta(k) = A055034(k). Proof: floor(k/2) gives the number of positive odd numbers < k, and delta(k), gives the number of positive odd numbers coprime to k. delta(1) = 1 but 1 is not < 1, therefore k = 1 is not a member of this set. Hence a member >= 2 of this set has at least one odd number > 1 and < k missing in the set of odd numbers relative prime to k. Therefore there exists at least one odd prime < k dividing k. (End)
For the multiplicity of a(n) see A338428, obtained from triangle A337940 (the array is given Bob Selcoe as example below, and in the Ralf Steiner comment above). - Wolfdieter Lang, Dec 09 2020

			a(1)=6 because 6 is the first number that can be written as a sum of three consecutive positive integers: 6 = 1+2+3.
From _Bob Selcoe_, Feb 23 2014: (Start)
Let the top row of an array be A000217(n). Let the diagonals (reading down and left) be A000217(n)-A000217(1),  A000217(n)-A000217(2), A000217(n)-A000217(3)..., A000217(n)-A000217(n-3).  This is A049777 read as a square array, starting with the third column. The array begins as follows:
   6 10 15 21 28 36 45 55 66
   9 14 20 27 35 44 54 65
  12 18 25 33 42 52 63
  15 22 30 39 49 60
  18 26 35 45 56
  21 30 40 51
  24 34 45
  27 38
  30
This is (x*(x+1)-y*(y+1))/2 for nonnegative integers x,y with x-y >= 3, because it is equivalent to 1+2+3/+4/+5/...+x/-0/-1/-2/-3/-4/-5/...-(x+3)/ for all possible strings of consecutive integers, which represents every possible way to sum three or more consecutive positive integers. So for example, 4+5+6+7 = 1+2+3+4+5+6+7-1-2-3 = 22, which is (x*(x+1)-y*(y+1))/2 when x=7, y=3. Notice that values can appear more than once in the array because some numbers can be represented as sums of more than one string of three or more consecutive positive integers. For example, 30 = (x*(x+1)-y*(y+1))/2 when (a) x=11, y=8: 9+10+11; (b) x=9, y=5: 6+7+8+9; and (c) x=8, y=3: 4+5+6+7+8. By definition, x-y is the number of integers in the string. (End)

Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991, Solution to problem 3G p. 179.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC, Problem C, Jun 2005, p. 181-182.
Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC, Problem C: Solution of this Problem
J. Spies, Sage program for computing A111774

Cf. A000040, A000079, A066542, A174090 (complement), A000217, A049777, A055034.

Cf. A337940, A338428.

ispoweroftwo := proc(n) local a, t; t := 1; while (n > t) do t := 2*t end do; if (n = t) then a := true else a := false end if; return a; end proc; f:= proc(n) if (not isprime(n)) and (not ispoweroftwo(n)) then return n end if; end proc; seq(f(i),i = 1..150);

max=6!;lst={};Do[z=n+(n+1);Do[z+=(n+x);If[z>max,Break[]];AppendTo[lst,z],{x,2,max}],{n,max}];Union[lst] (* Vladimir Joseph Stephan Orlovsky, Mar 06 2010 *)

isok(n) = !(n == 1) && !isprime(n) && !(isprimepower(n, &p) && (p == 2)); \\ Michel Marcus, Jul 02 2019

from sympy import primepi
def A111774(n):
    def f(x): return int(n+(0 if x<=1 else primepi(x)-1)+x.bit_length())
    m, k = n, f(n)
    while m != k: m, k = k, f(k)
    return m # Chai Wah Wu, Sep 19 2024

A056115 a(n) = n*(n+11)/2.

Original entry on oeis.org

0, 6, 13, 21, 30, 40, 51, 63, 76, 90, 105, 121, 138, 156, 175, 195, 216, 238, 261, 285, 310, 336, 363, 391, 420, 450, 481, 513, 546, 580, 615, 651, 688, 726, 765, 805, 846, 888, 931, 975, 1020, 1066, 1113, 1161, 1210, 1260, 1311, 1363, 1416, 1470, 1525

Offset: 0

Barry E. Williams, Jul 04 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A001477, A055999, A056000, A056119.

Third column of Pascal (1, 6) triangle A096956.

List([0..50], n-> n*(n+11)/2 ); # G. C. Greubel, Jan 18 2020

[n*(n+11)/2: n in [0..50]]; // G. C. Greubel, Jan 18 2020

((2*Range[0,50]+11)^2 -11^2)/8 (* G. C. Greubel, Jan 18 2020 *)

a(n)=n*(n+11)/2; \\ Joerg Arndt, Oct 25 2014

[n*(n+11)/2 for n in (0..50)] # G. C. Greubel, Jan 18 2020

G.f.: x*(6-5*x)/(1-x)^3.
a(n) = A000096(n) + 4*A001477(n) = A056000(n) + A001477(n) = A056119(n) - A001477(n). - Zerinvary Lajos, Oct 01 2006
a(n) = A126890(n,5) for n>4. - Reinhard Zumkeller, Dec 30 2006
Equals A119412/2. - Zerinvary Lajos, Feb 12 2007
If we define f(n,i,a) = Sum_{k=0..n-i} ( binomial(n,k)*stirling1(n-k,i) *Product_{j=0..k-1} (-a-j) ), then a(n) = -f(n,n-1,6), for n>=1. - Milan Janjic, Dec 20 2008
a(n) = a(n-1) + n + 5 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
Sum_{n>=1} 1/a(n) = 83711/152460. - R. J. Mathar, Jul 14 2012
a(n) = 6*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
E.g.f.: x*(12 + x)*exp(x)/2. - G. C. Greubel, Jan 18 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/11 - 20417/152460. - Amiram Eldar, Jan 10 2021

A139273 a(n) = n(8n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734.

[ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011

Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)

a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 3*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

A056121 a(n) = n*(n + 15)/2.

Original entry on oeis.org

0, 8, 17, 27, 38, 50, 63, 77, 92, 108, 125, 143, 162, 182, 203, 225, 248, 272, 297, 323, 350, 378, 407, 437, 468, 500, 533, 567, 602, 638, 675, 713, 752, 792, 833, 875, 918, 962, 1007, 1053, 1100, 1148, 1197, 1247, 1298, 1350, 1403, 1457, 1512, 1568, 1625

Offset: 0

Barry E. Williams, Jul 06 2000

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A001477, A056000, A056119, A056126.

List([0..60], n-> n*(n+15)/2 ); # G. C. Greubel, Jan 18 2020

[n*(n+15)/2: n in [0..60]]; // G. C. Greubel, Jan 18 2020

Maple
```
a:=n->n*(n+15)/2: seq(a(n),n=0..60);
```

Table[n*(n + 15)/2, {n, 0, 100}] (* Paolo Xausa, Aug 02 2024 *)

a(n)=n*(n+15)/2 \\ Charles R Greathouse IV, Sep 24 2015

[n*(n+15)/2 for n in (0..60)] # G. C. Greubel, Jan 18 2020

G.f.: x*(8-7*x)/(1-x)^3.
a(n) = A000096(n) + 6*n = A056119(n) + n = A056126(n) - n. - Zerinvary Lajos, Oct 01 2006
a(n-15) = binomial(n,2) - 7*n. - Zerinvary Lajos, Nov 26 2006
a(n) = A126890(n,7) for n>6. - Reinhard Zumkeller, Dec 30 2006
Let f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then a(n) = -f(n,n-1,8), for n>=1. - Milan Janjic, Dec 20 2008
a(n) = a(n-1)+ n + 7 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
Sum_{n>=1} 1/a(n) = 1195757/2702700 via A132760. - R. J. Mathar, Jul 14 2012
a(n) = 8*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
E.g.f.: x*(16 + x)*exp(x)/2. - G. C. Greubel, Jan 18 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/15 - 52279/540540. - Amiram Eldar, Jan 10 2021

More terms from James Sellers, Jul 07 2000

A096940 Pascal (1,5) triangle.

Original entry on oeis.org

5, 1, 5, 1, 6, 5, 1, 7, 11, 5, 1, 8, 18, 16, 5, 1, 9, 26, 34, 21, 5, 1, 10, 35, 60, 55, 26, 5, 1, 11, 45, 95, 115, 81, 31, 5, 1, 12, 56, 140, 210, 196, 112, 36, 5, 1, 13, 68, 196, 350, 406, 308, 148, 41, 5, 1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5, 1, 15, 95, 345, 810, 1302

Offset: 0

Wolfdieter Lang, Jul 16 2004

This is the fifth member, q=5, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A095666 (q=4), A096956 (q=6).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(5-4*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(5-4*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022096(n-2), n>=2, with n=1 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins:
  5;
  1,  5;
  1,  6,  5;
  1,  7, 11,   5;
  1,  8, 18,  16,   5;
  1,  9, 26,  34,  21,   5;
  1, 10, 35,  60,  55,  26,   5;
  1, 11, 45,  95, 115,  81,  31,   5;
  1, 12, 56, 140, 210, 196, 112,  36,   5;
  1, 13, 68, 196, 350, 406, 308, 148,  41,  5;
  1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5; etc.

David A. Corneth, Table of n, a(n) for n = 0..9999
Wolfdieter Lang, First 10 rows.

Row sums: A007283(n-1), n>=1, 5 if n=0; g.f.: (5-4*x)/(1-2*x). Alternating row sums are [5, -4, followed by 0's].

Column sequences (without leading zeros) give for m=1..9, with n>=0: A000027(n+5), A056000(n-1), A096941-7.

a(n,k):=piecewise(n=0,5,0Mircea Merca, Apr 08 2012

a(n) = {if(n <= 1, return(5 - 4*(n==1))); my(m = (sqrtint(8*n + 1) - 1)\2, t = n - binomial(m + 1, 2)); (1+4*t/m)*binomial(m,t)} \\ David A. Corneth, Aug 28 2019

Recursion: a(n, m)=0 if m>n, a(0, 0)= 5; a(n, 0)=1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (5-4*x)/(1-x)^(m+1), m>=0.
a(n,k) = (1+4*k/n)*binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012

A056119 a(n) = n*(n+13)/2.

Original entry on oeis.org

0, 7, 15, 24, 34, 45, 57, 70, 84, 99, 115, 132, 150, 169, 189, 210, 232, 255, 279, 304, 330, 357, 385, 414, 444, 475, 507, 540, 574, 609, 645, 682, 720, 759, 799, 840, 882, 925, 969, 1014, 1060, 1107, 1155, 1204, 1254, 1305, 1357, 1410, 1464, 1519, 1575

Offset: 0

Barry E. Williams, Jul 04 2000

G. C. Greubel, Table of n, a(n) for n = 0..1000
P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A001477, A056000, A056115, A056121.

List([0..50], n-> n*(n+13)/2 ); # G. C. Greubel, Jan 18 2020

[n*(n+13)/2: n in [0..50]]; // G. C. Greubel, Jan 18 2020

Table[n*(n+13)/2, {n, 0, 50}] (* Paolo Xausa, Jun 26 2024 *)

a(n)=n*(n+13)/2 \\ Charles R Greathouse IV, Oct 07 2015

[n*(n+13)/2 for n in (0..50)] # G. C. Greubel, Jan 18 2020

G.f.: x*(7-6*x)/(1-x)^3.
a(n) = A126890(n,6) for n > 5. - Reinhard Zumkeller, Dec 30 2006
a(n) = A000096(n) + 5*A001477(n) = A056115(n) + A001477(n) = A056121(n) - A001477(n). - Zerinvary Lajos, Feb 22 2008
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then a(n) = -f(n,n-1,7), for n >= 1. - Milan Janjic, Dec 20 2008
a(n) = n + a(n-1) + 6 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
Sum_{n>=1} 1/a(n) = 1145993/2342340 via A132759. - R. J. Mathar, Jul 14 2012
a(n) = 7*n - floor(n/2) + floor(n/2). - Wesley Ivan Hurt, Jun 15 2013
E.g.f.: x*(14 + x)*exp(x)/2. - G. C. Greubel, Jan 18 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/13 - 263111/2342340. - Amiram Eldar, Jan 10 2021

More terms from James Sellers, Jul 05 2000

A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

Original entry on oeis.org

0, 1, 1, 9, 4, 25, 9, 49, 16, 81, 25, 121, 36, 169, 49, 225, 64, 289, 81, 361, 100, 441, 121, 529, 144, 625, 169, 729, 196, 841, 225, 961, 256, 1089, 289, 1225, 324, 1369, 361, 1521, 400, 1681, 441, 1849, 484, 2025, 529, 2209, 576, 2401, 625, 2601

Offset: 0

Paul Curtz, Nov 18 2009

From Paul Curtz, Mar 26 2011: (Start)
Successive A026741(n) * A026741(n+p):
  p=0:  0, 1,  1,  9,  4, 25,  9,     a(n),
  p=1:  0, 1,  3,  6, 10, 15, 21,  A000217,
  p=2:  0, 3,  2, 15,  6, 35, 12,  A142705,
  p=3:  0, 2,  5,  9, 14, 20, 27,  A000096,
  p=4:  0, 5,  3, 21,  8, 45, 15,  A171621,
  p=5:  0, 3,  7, 12, 18, 25, 33,  A055998,
  p=6:  0, 7,  4, 27, 10, 55, 18,
  p=7:  0, 4,  9, 15, 22, 30, 39,  A055999,
  p=8:  0, 9,  5, 33, 12, 65, 21,  (see A061041),
  p=9:  0, 5, 11, 18, 26, 35, 45,  A056000. (End)
The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016
Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A040001, A168068, A181318.

I:=[0,1,1,9,4,25]; [n le 6 select I[n] else 3*Self(n-2)-3*Self(n-4)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 10 2016

a := proc(n): n^2*(5-3*(-1)^n)/8 end: seq(a(n), n=0..46); # Johannes W. Meijer, Jul 03 2016

LinearRecurrence[{0,3,0,-3,0,1},{0,1,1,9,4,25},60] (* Harvey P. Dale, May 14 2011 *)
f[n_] := Numerator[(n/2)^2]; Array[f, 60, 0] (* Robert G. Wilson v, Dec 18 2012 *)
CoefficientList[Series[x(1+x+6x^2+x^3+x^4)/((1-x)^3(1+x)^3), {x,0,60}], x] (* Vincenzo Librandi, Jul 10 2016 *)

concat(0, Vec(x*(1+x+6*x^2+x^3+x^4)/((1-x)^3*(1+x)^3) + O(x^60))) \\ Altug Alkan, Jul 04 2016

a(n) = lcm(4, n^2)/4; \\ Andrew Howroyd, Jul 26 2018

(x*(1+x+6*x^2+x^3+x^4)/(1-x^2)^3).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 20 2019

From R. J. Mathar, Jan 22 2011: (Start)
G.f.: x*(1 + x + 6*x^2 + x^3 + x^4) / ((1-x)^3*(1+x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(n) = n^2*(5 - 3*(-1)^n)/8. (End)
a(n) = A026741(n)^2.
a(2*n) = A000290(n); a(2*n+1) = A016754(n).
a(n) - a(n-4) = 4*A064680(n+2). - Paul Curtz, Mar 27 2011
4*a(n) = A061038(n) * A010121(n+2) = A109043(n)^2, n >= 2. - Paul Curtz, Apr 07 2011
a(n) = A129194(n) / A040001(n). - Andrew Howroyd, Jul 26 2018
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 4)) = n^2/(gcd(n^2,4)) = (n/gcd(n,2))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 4, ...] is a purely periodic sequence of period 2. Thus a(n) is a quasi-polynomial in n.
O.g.f.: x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3.
Cf. A181318. (End)
From Werner Schulte, Aug 30 2020: (Start)
Multiplicative with a(2^e) = 2^(2*e-2) for e > 0, and a(p^e) = p^(2*e) for prime p > 2.
Dirichlet g.f.: zeta(s-2) * (1 - 3/2^s).
Dirichlet convolution with A259346 equals A000290.
Sum_{n>0} 1/a(n) = Pi^2 * 7 / 24. (End)
Sum_{k=1..n} a(k) ~ (5/24) * n^3. - Amiram Eldar, Nov 28 2022

cp's OEIS Frontend

A056126 a(n) = n*(n + 17)/2.

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A126890 Triangle read by rows: T(n,k) = n*(n+2*k+1)/2, 0 <= k <= n.

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

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A111774 Numbers that can be written as a sum of at least three consecutive positive integers.

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A056115 a(n) = n*(n+11)/2.

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A139273 a(n) = n*(8*n - 3).

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A126890 Triangle read by rows: T(n,k) = n(n+2k+1)/2, 0 <= k <= n.

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

A139273 a(n) = n(8n - 3).