A147875 -id:A147875 - cp's OEIS frontend

A306383 Number of ways to write n as x(2x+1) + y(2y+1) + z*(2z+1), where x,y,z are nonnegative integers with x <= y <= z.

1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 2, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 2, 1, 0, 2, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 2, 0, 1, 1, 0, 0, 1, 1, 1, 0

Offset: 0

Zhi-Wei Sun, Feb 11 2019

nonn

Conjecture 1: a(n) > 0 for any integer n > 138158.
We have verified this for n up to 2*10^6. Note that n*(2n+1) (n = 0,1,...) are the second hexagonal numbers (A014105).
Conjecture 2: Any integer n > 146858 can be written as the sum of three hexagonal numbers (A000384).
Conjecture 3: Any integer n > 33066 can be written as the sum of three pentagonal numbers (A000326).
Conjecture 4: Any integer n > 24036 can be written as the sum of three second pentagonal numbers (A005449).
Conjecture 5: Let N(1) = 114862, N(-1) = 166897, N(3) = 196987 and N(-3) = 273118. Then, for any r among 1, -1, 3 and -3, each integer n > N(r) can be written as x*(5x+r)/2 + y*(5y+r)/2 + z*(5z+r)/2 with x,y,z nonnegative integers.
We have verified Conjectures 2-5 for n up to 10^6.

			a(223595) = 1 with 223595 = 95*(2*95+1) + 200*(2*200+1) + 250*(2*250+1).
a(290660) = 1 with 290660 = 136*(2*136+1) + 149*(2*149+1) + 323*(2*323+1).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Universal sums of three quadratic polynomials, Sci. China Math., in press.

Cf. A000217, A000326, A000384, A000566, A005449, A005475, A005476, A008443, A014105, A147875, A306382.

QQ[n_]:=QQ[n]=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1],4]==1;
tab={};Do[r=0;Do[If[QQ[n-x(2x+1)-y(2y+1)],r=r+1],{x,0,(Sqrt[8n/3+1]-1)/4},{y,x,(Sqrt[4(n-x(2x+1))+1]-1)/4}];tab=Append[tab,r],{n,0,100}];Print[tab]

A126264 a(n) = 5n^2 + 3n.

Original entry on oeis.org

8, 26, 54, 92, 140, 198, 266, 344, 432, 530, 638, 756, 884, 1022, 1170, 1328, 1496, 1674, 1862, 2060, 2268, 2486, 2714, 2952, 3200, 3458, 3726, 4004, 4292, 4590, 4898, 5216, 5544, 5882, 6230, 6588, 6956, 7334, 7722, 8120, 8528, 8946, 9374, 9812, 10260

Offset: 1

Gary W. Adamson, Dec 22 2006

			a(24) = 5*24^2 + 3*24 = 2880 + 72 = 2952.

L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 12

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001622, A147875.

a:=n->5*n^2+3*n: seq(a(n),n=1..55); # Emeric Deutsch, Apr 17 2007

Table[n*(5*n + 3), {n,1,50}] (* G. C. Greubel, Aug 23 2017 *)

a(n)=5*n^2+3*n \\ Charles R Greathouse IV, Jun 17 2017

Sum_{i=1..n} a(i) = n*(n+1)*(5n+7)/3 = 2*A162148(n).
a(n) = 2*A147875(n+1).
From G. C. Greubel, Aug 23 2017: (Start)
G.f.: 2*x*(x + 4)/(1 - x)^3.
E.g.f.: x*(5*x + 8)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 5/9 + sqrt(1-2/sqrt(5))*Pi/6 + log(phi)*sqrt(5)/6 - 5*log(5)/12, where phi is the golden ratio (A001622). - Amiram Eldar, Aug 21 2022

More terms from Emeric Deutsch, Apr 17 2007

A225785 Numbers n such that triangular(n) + triangular(2*n) is a triangular number.

Original entry on oeis.org

0, 12, 84, 3960, 27144, 1275204, 8740380, 410611824, 2814375312, 132215732220, 906220110180, 42573055163112, 291800061102744, 13708391546789940, 93958713454973484, 4414059505011197664, 30254413932440359200, 1421313452222058857964

Offset: 1

Alex Ratushnyak, May 16 2013

Equivalently, numbers n such that oblong(n) + oblong(2*n) is an oblong number, where oblong(n) = A002378(n) = n*(n+1).
Also, x values in the equation A147875(x) = A000217(y) - see Ralf Stephan in Program lines. - Bruno Berselli, May 18 2013
Also, numbers m such that 2*m+1 and 10*m+1 are both squares. - Bruno Berselli, Mar 03 2016

			12*13/2 + 24*25/2 = 27*28/2, so 12 is in the sequence.

Harvey P. Dale, Table of n, a(n) for n = 1..798
Index entries for linear recurrences with constant coefficients, signature (1,322,-322,-1,1).

Cf. A000217, A002378, A082183.
Cf. A224419 (numbers n such that triangular(n) + triangular(2*n) is a square).
Cf. A011916 (numbers n such that triangular(2*n) - triangular(n) is a triangular number).
Cf. A225786 (numbers n such that oblong(2*n) + oblong(n) is a square).
Cf. A225839 (triangular numbers of the form triangular(x) + triangular(2*x)).

#include 
#include 
int main() {
  unsigned long long i, s, t;
  for (i = 0; i< (1ULL<<31); i++) {
    s = 2*i*(2*i+1) + i*(i+1);
    t = sqrt(s);
    if (s==t*(t+1)) printf("%llu, ", i);
  }
  return 0;
}

CoefficientList[Series[12 x (1 + 6 x + x^2)/((1 - x) (1 - 18 x + x^2) (1 + 18 x + x^2)), {x, 0, 20}], x] (* Bruno Berselli, May 18 2013 *)
LinearRecurrence[{1,322,-322,-1,1},{0,12,84,3960,27144},20] (* Harvey P. Dale, Apr 08 2021 *)

for(n=1,10^9,t=n*(5*n+3)/2;x=sqrtint(2*t);if(t==x*(x+1)/2,print(n))) /* Ralf Stephan, May 17 2013 */

G.f.: 12*x*(1+6*x+x^2)/((1-x)*(1-18*x+x^2)(1+18*x+x^2)). [Bruno Berselli, May 18 2013]
a(n) = (1/20)*((3+(-1)^n*sqrt(5))*(2-sqrt(5))^(4*floor(n/2))+(3-(-1)^n*sqrt(5))*(2+sqrt(5))^(4*floor(n/2))-6). [Bruno Berselli, May 18 2013]
a(2*n) = (Fibonacci(6*n-3)^2 + Lucas(6*n-3)*Fibonacci(6*n-1))/2. - Greg Dresden, Sep 24 2023

More terms from Bruno Berselli, May 18 2013

A255934 Number of ways to write n as the sum of four unordered generalized octagonal numbers.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 2, 3, 2, 3, 2, 1, 3, 3, 3, 3, 2, 4, 2, 1, 3, 2, 3, 4, 3, 4, 2, 2, 4, 4, 3, 4, 3, 6, 5, 2, 4, 3, 4, 5, 4, 6, 4, 1, 4, 5, 4, 5, 5, 7, 4, 1, 5, 5, 5, 6, 5, 8, 5, 3, 4, 6, 6, 6, 6, 7, 6, 3, 6, 6, 5, 6, 6, 10, 7, 1, 5, 8, 7, 7, 7, 8, 5, 3, 6, 7, 6, 8, 7, 10, 8, 3

Offset: 0

Zhi-Wei Sun, Mar 11 2015

nonn

I have proved that a(n) > 0 for all n, i.e., any nonnegative integer can be expressed as the sum of four generalized octagonal numbers. I can show that a(n) = 1 if 3*n+4 is among 7, 13, 19, 31, 43, 2^(2k), 5*2^(2k+1), 11*2^(2k+1), 23*2^(2k+1) (k = 0,1,2,...), and conjecture the converse.

I also conjecture that each nonnegative integer can be written as the sum of two heptagonal numbers, a second heptagonal number and a generalized heptagonal number.

			a(60) = 1 since 60 = 1*(3*1-2) + (-1)*(3*(-1)-2) + 3*(3*3-2) + (-3)*(3*(-3)-2).
a(1876) = 1 since 1876 = (-5)*(3*(-5)-2) + (-5)*(3*(-5)-2) + 11*(3*11-2) + (-21)*(3*(-21)-2).
a(15700) = 1 since 15700 = 11*(3*11-2) + (-21)*(3*(-21)-2) + 43*(3*43-2) + (-53)*(3*(-53)-2).
a(21844) = 1 since 21844 = 43*(3*43-2) + 43*(3*43-2) + 43*(3*43-2) + 43*(3*43-2).
a(30036) = 1 since 30036 = (-21)*(3*(-21)-2) + (-21)*(3*(-21)-2) + 43*(3*43-2) + (-85)*(3*(-85)-2).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.
Zhi-Wei Sun, A result similar to Lagrange's theorem, arXiv:1503.03743 [math.NT], 2015.

Cf. A000566, A001082, A085787, A147875, A255350.

T[n_]:=Union[Table[x(3x-2),{x,-Floor[(Sqrt[3n+1]-1)/3],Floor[(Sqrt[3n+1]+1)/3]}]]
Do[r=0;Do[If[n-Part[T[n],x]-Part[T[n],y]-Part[T[n],z]

A174814 a(n) = n(n+1)(5*n+1)/3.

Original entry on oeis.org

0, 4, 22, 64, 140, 260, 434, 672, 984, 1380, 1870, 2464, 3172, 4004, 4970, 6080, 7344, 8772, 10374, 12160, 14140, 16324, 18722, 21344, 24200, 27300, 30654, 34272, 38164, 42340, 46810, 51584, 56672, 62084, 67830, 73920, 80364, 87172, 94354, 101920, 109880

Offset: 0

Bruno Berselli, Dec 01 2010 - Dec 02 2010

Also zero followed by bisection (even part) of A088003.

Numbers ending in 0, 2 or 4 (cf. 2*A053796(n)). Therefore we can easily see that a(m)^(2*k+1)==-1 (mod 5) only for m in A047219, while a(m)^(2*k)==-1 (mod 5) only for m in A016873 and k odd.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

[n*(n+1)*(5*n+1)/3: n in [0..50]]; // Vincenzo Librandi, May 30 2011

Table[1/3 n (1+n) (1+5 n), {n,0,50}]  (* Harvey P. Dale, Feb 25 2011 *)

a(n) = n*(n+1)*(5*n+1)/3 \\ Charles R Greathouse IV, May 30 2011

G.f.: 2*x*(2+3*x)/(1-x)^4.
a(n) = 2*A033994(n) for n>0.
a(n) = n*A147875(n+1)-sum(k=1..n, A147875(k)) for n>0.
a(-n) = -A144945(n).

A253187 Number of ordered ways to write n as the sum of a pentagonal number, a second pentagonal number and a generalized decagonal number.

Original entry on oeis.org

1, 2, 2, 2, 1, 1, 1, 3, 4, 2, 2, 1, 4, 3, 3, 4, 2, 3, 1, 3, 2, 2, 5, 3, 3, 3, 3, 6, 3, 6, 4, 2, 3, 1, 7, 2, 4, 5, 5, 4, 1, 5, 5, 2, 3, 4, 4, 5, 5, 5, 3, 5, 7, 6, 4, 3, 1, 6, 6, 8, 5, 3, 6, 4, 7, 4, 2, 6, 5, 5, 3, 4, 8, 3, 3, 3, 6, 6, 7, 9, 6, 2, 5, 6, 7, 7, 4, 6, 6, 7, 5, 3, 10, 6, 3, 4, 5, 7, 3, 10, 7

Offset: 0

Zhi-Wei Sun, Apr 07 2015

nonn

Conjecture: a(n) > 0 for all n. Also, for any ordered pair (k,m) among (5,7), (5,9), (5,13), (6,5), (6,7), (7,5), each nonnegative integer n can be written as the sum of a k-gonal number, a second k-gonal number and a generalized m-gonal number.

See also the author's similar conjectures in A254574, A254631, A255916 and the two linked papers.

			a(33) = 1 since 33 = 0*(3*0-1)/2 + 4*(3*4+1)/2 + 1*(4*1+3).
a(56) = 1 since 56 = 4*(3*4-1)/2 + 2*(3*2+1)/2 + 3*(4*3+3).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.
Zhi-Wei Sun, On universal sums a*x^2+b*y^2+f(z), a*T_x+b*T_y+f(z) and a*T_x+b*y^2+f(z), arXiv:1502.03056 [math.NT], 2015.

Cf. A000326, A000384, A000566, A005449, A014105, A074377, A085787, A147875, A254574, A254631.

DQ[n_]:=IntegerQ[Sqrt[16n+9]]
Do[r=0;Do[If[DQ[n-x(3x-1)/2-y(3y+1)/2],r=r+1],{x,0,(Sqrt[24n+1]+1)/6},{y,0,(Sqrt[24(n-x(3x-1)/2)+1]-1)/6}];
Print[n," ",r];Continue,{n,0,100}]

A254623 Number of ways to write n as x^2 + y(3y+1)/2 + z(5z+3)/2 with x,y,z nonnegative integers.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 3, 1, 2, 1, 4, 4, 1, 1, 3, 4, 1, 2, 2, 3, 1, 1, 4, 3, 5, 3, 5, 2, 1, 2, 3, 4, 1, 4, 2, 5, 1, 3, 5, 4, 3, 3, 2, 3, 4, 2, 5, 2, 6, 4, 5, 3, 5, 2, 1, 2, 3, 8, 1, 6, 4, 3, 2, 3, 5, 6, 5, 2, 4, 2, 3, 5, 6, 7, 5, 1, 6, 3, 4, 3, 4, 8, 2, 5, 5, 4, 3, 3, 6, 4, 4, 3, 7, 1, 2, 6

Offset: 0

Zhi-Wei Sun, Feb 03 2015

nonn

Conjecture: (i) a(n) > 0 for all n, and a(n) > 1 for all n > 118.
(ii) For each m = 5,7,8, any nonnegative integer n can be written as the sum of two triangular numbers and a second m-gonal number, where the second m-gonal numbers are given by (m-2)*k*(k+1)/2-k (k = 0,1,...).
(iii) For every m = 5,6,7,9,11, any nonnegative integer n can be written as the sum of a triangular number, a square and a second m-gonal number.
Note that k*(3*k+1)/2 (k = 0,1,...) are second pentagonal numbers and k*(5*k+3)/2 (k = 0,1,...) are second heptagonal numbers. The conjecture has been verified for all n = 0.. 2*10^6.

			a(41) = 1 since 41 = 1^2 + 5*(3*5+1)/2 + 0*(5*0+3)/2.
a(98) = 1 since 98 = 8^2 + 2*(3*2+1)/2 + 3*(5*3+3)/2.
a(118) = 1 since 118 = 2^2 + 3*(3*3+1)/2 + 6*(5*6+3)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:1405.0635 [math.NT], 2009-2015.

Cf. A000217, A000290, A005449, A147875.

SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-y(3y+1)/2-z(5z+3)/2],r=r+1],{y,0,(Sqrt[24n+1]-1)/6},{z,0,(Sqrt[40(n-y(3y+1)/2)+9]-3)/10}];
Print[n," ",r];Continue,{n,0,100}]

A367964 Triangle of 2-parameter triangular numbers, read by rows. T(n, k) = (n(n + 1) + k(k + 1)) / 2.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 6, 7, 9, 12, 10, 11, 13, 16, 20, 15, 16, 18, 21, 25, 30, 21, 22, 24, 27, 31, 36, 42, 28, 29, 31, 34, 38, 43, 49, 56, 36, 37, 39, 42, 46, 51, 57, 64, 72, 45, 46, 48, 51, 55, 60, 66, 73, 81, 90, 55, 56, 58, 61, 65, 70, 76, 83, 91, 100, 110

Offset: 0

Peter Luschny, Dec 07 2023

If the rows of the triangle are extended for k > n, the array A144216 is created, which is symmetrical to the main diagonal and therefore contains no new information compared to this triangle.

			Triangle T(n, k) starts:
  0 |  0;
  1 |  1,  2;
  2 |  3,  4,  6;
  3 |  6,  7,  9, 12;
  4 | 10, 11, 13, 16, 20;
  5 | 15, 16, 18, 21, 25, 30;
  6 | 21, 22, 24, 27, 31, 36, 42;
  7 | 28, 29, 31, 34, 38, 43, 49, 56;
  8 | 36, 37, 39, 42, 46, 51, 57, 64, 72;
  9 | 45, 46, 48, 51, 55, 60, 66, 73, 81,  90;
 10 | 55, 56, 58, 61, 65, 70, 76, 83, 91, 100, 110;
.
Start at row 0, column 0 with 0. Go down by adding the column index in step n. At row n, restart the counting and go n steps right by adding the row index in step n, then change direction and go down again by adding the column index. After 3*n steps on this path you are at T(2*n, n) which is 2*triangular(n) + (triangular(2*n) - triangular(n)) = (5*n^2 + 3*n)/2. These are the sliced heptagonal numbers A147875 (see the illustration of Leo Tavares).
.
The equation T(n, k) = (n*(n + 1) + k*(k + 1))/2 can be extended to all n, k in ZZ.
  [n\k] ... -6  -5  -4  -3  -2  -1   0   1   2   3   4   5  ...
  -------------------------------------------------------------
  [-5] ..., 25, 20, 16, 13, 11, 10, 10, 11, 13, 16, 20, 25, ...
  [-4] ..., 21, 16, 12,  9,  7,  6,  6,  7,  9, 12, 16, 21, ...
  [-3] ..., 18, 13,  9,  6,  4,  3,  3,  4,  6,  9, 13, 18, ...
  [-2] ..., 16, 11,  7,  4,  2,  1,  1,  2,  4,  7, 11, 16, ...
  [-1] ..., 15, 10,  6,  3,  1,  0,  0,  1,  3,  6, 10, 15, ...
  [ 0] ..., 15, 10,  6,  3,  1,  0,  0,  1,  3,  6, 10, 15, ...
  [ 1] ..., 16, 11,  7,  4,  2,  1,  1,  2,  4,  7, 11, 16, ...
  [ 2] ..., 18, 13,  9,  6,  4,  3,  3,  4,  6,  9, 13, 18, ...
  [ 3] ..., 21, 16, 12,  9,  7,  6,  6,  7,  9, 12, 16, 21, ...
  [ 4] ..., 25, 20, 16, 13, 11, 10, 10, 11, 13, 16, 20, 25, ...

Columns: A000217, A000124, A152950, A166136, A121263.
Diagonals: A002378, A000290, A002061, A059100, A027689.
Cf. A147875 (T(2*n, n)), A016061 (row sums), A367965 (alternating row sums), A143216 (the multiplicative equivalent), A144216 (extended array).

T := (n, k) -> (n*(n + 1) + k*(k + 1)) / 2:
for n from 0 to 10 do seq(T(n, k), k = 0..n) od;

Module[{n=1},NestList[Append[#+n,n*++n]&,{0},10]] (* or *)
Table[(n(n+1)+k(k+1))/2,{n,0,10},{k,0,n}] (* Paolo Xausa, Dec 07 2023 *)

# A purely additive construction:
from functools import cache
@cache
def a_row(n: int) -> list[int]:
    if n == 0: return [0]
    row = a_row(n - 1) + [0]
    for k in range(n): row[k] += n
    row[n] = row[n - 1] + n
    return row

Recurrence: T(n, n) = n + T(n, n-1) starting with T(0, 0) = 0.
For k <> n: T(n, k) = n + T(n-1, k).
T(n, k) = t(n) + t(k), where t(n) are the triangular numbers A000217.
G.f.: (x + x*(2 - 5*x + x^2)*y + x^4*y^2)/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Dec 07 2023

A225839 Triangular numbers representable as triangular(m) + triangular(2m).

Original entry on oeis.org

0, 378, 17766, 39209940, 1842032556, 4065365016846, 190985619471570, 421505175637435176, 19801770996209306328, 43702499616375188919330, 2053087220237987679246270, 4531162564803507161896556028, 212868189148913267563402477956, 469799997000254729943383533193910

Offset: 1

Alex Ratushnyak, May 17 2013

Triangular numbers of the sequence A147875: a(n) = A147875(A225785(n)) - see also Ralf Stephan in Program lines. [Bruno Berselli, May 20 2013]

Index entries for linear recurrences with constant coefficients, signature (1,103682,-103682,-1,1).

Cf. A000217, A147875.
Cf. A108281 (triangular numbers representable as triangular(m) + m^2).
Cf. A225785 (numbers n such that triangular(n) + triangular(2n) is a triangular number).

CoefficientList[Series[378 x (1 + 46 x + x^2)/((1 - x) (1 - 322 x + x^2) (1 + 322 x + x^2)), {x, 0, 20}], x] (* Bruno Berselli, May 20 2013 *)
LinearRecurrence[{1,103682,-103682,-1,1},{0,378,17766,39209940,1842032556},20] (* Harvey P. Dale, Jan 16 2019 *)

for(n=1,10^9,t=n*(5*n+3)/2;x=sqrtint(2*t);if(t==x*(x+1)/2,print(t))) \\ Ralf Stephan, May 17 2013

G.f.: 378*x*(1+46*x+x^2)/((1-x)*(1-322*x+x^2)*(1+322*x+x^2)). [Bruno Berselli, May 20 2013]

More terms from Bruno Berselli, May 20 2013

A332495 a(n-2) = a(n-6) + 5(1+2n) with a(0)=0, a(1)=2, a(2)=7, a(3)=15 for n>=4.

Original entry on oeis.org

0, 2, 7, 15, 25, 37, 52, 70, 90, 112, 137, 165, 195, 227, 262, 300, 340, 382, 427, 475, 525, 577, 632, 690, 750, 812, 877, 945, 1015, 1087, 1162, 1240, 1320, 1402, 1487, 1575, 1665, 1757, 1852, 1950, 2050, 2152, 2257

Offset: 0

Paul Curtz, Feb 14 2020

a(-2)=2, a(-1)=0. 4 evens followed by 4 odds.
Last digit is only 0, 2, 5, 7.
The vertical spoke S-N of the pentagonal spiral for A004526.
                          37
                      37  25  25
                  36  24  15  15  26
              36  24  14   7   8  16  26
          35  23  14   7   2   3   8  16  27
      35  23  13   6   2   0   0   3   9  17  27
        34  22  13   6   1   1   4   9  17  28
          34  22  12   5   5   4  10  18  28
            33  21  12  11  11  10  18  29
              33  21  20  20  19  19  29
                32  32  31  31  30  30
Rank of multiples of 10: 0, 7, 8, 15, 16, ... = A047521. Compare to A154260 in the formula.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. A004526, A033429, A062786, A168668, A135706, A147874, 2*A147875 (all in the spiral).

Cf. A005408, A047521, A154260

CoefficientList[Series[x (2 + x + 2 x^2)/((1 - x)^3*(1 + x^2)), {x, 0, 42}], x] (* Michael De Vlieger, Feb 14 2020 *)

concat(0, Vec(x*(2 + x + 2*x^2) / ((1 - x)^3*(1 + x^2)) + O(x^40))) \\ Colin Barker, Feb 14 2020

a(-1-n) = a(n).
a(2*n) + a(1+2*n) = 2, 22, 62, ... = A273366(n).
Second differences give the sequence of period 4: repeat [3, 3, 2, 2].
From Colin Barker, Feb 14 2020: (Start)
G.f.: x*(2 + x + 2*x^2) / ((1 - x)^3*(1 + x^2)).
a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) for n>4.
(End)
Multiples of 10: 10*(0, 7, 9, 30, 34, ... = A154260).
4*a(n) = A087960(n) +5*n -1 +5*n^2. - R. J. Mathar, Feb 28 2020

cp's OEIS Frontend

A306383 Number of ways to write n as x*(2x+1) + y*(2y+1) + z*(2z+1), where x,y,z are nonnegative integers with x <= y <= z.

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A126264 a(n) = 5*n^2 + 3*n.

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Maple

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A225785 Numbers n such that triangular(n) + triangular(2*n) is a triangular number.

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C

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A255934 Number of ways to write n as the sum of four unordered generalized octagonal numbers.

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A174814 a(n) = n*(n+1)*(5*n+1)/3.

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Magma

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A253187 Number of ordered ways to write n as the sum of a pentagonal number, a second pentagonal number and a generalized decagonal number.

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A254623 Number of ways to write n as x^2 + y*(3*y+1)/2 + z*(5*z+3)/2 with x,y,z nonnegative integers.

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A306383 Number of ways to write n as x(2x+1) + y(2y+1) + z*(2z+1), where x,y,z are nonnegative integers with x <= y <= z.

A126264 a(n) = 5n^2 + 3n.

A174814 a(n) = n(n+1)(5*n+1)/3.

A254623 Number of ways to write n as x^2 + y(3y+1)/2 + z(5z+3)/2 with x,y,z nonnegative integers.

A367964 Triangle of 2-parameter triangular numbers, read by rows. T(n, k) = (n(n + 1) + k(k + 1)) / 2.

A332495 a(n-2) = a(n-6) + 5(1+2n) with a(0)=0, a(1)=2, a(2)=7, a(3)=15 for n>=4.