A093565 -id:A093565 - cp's OEIS frontend

A001107 10-gonal (or decagonal) numbers: a(n) = n(4n-3).

0, 1, 10, 27, 52, 85, 126, 175, 232, 297, 370, 451, 540, 637, 742, 855, 976, 1105, 1242, 1387, 1540, 1701, 1870, 2047, 2232, 2425, 2626, 2835, 3052, 3277, 3510, 3751, 4000, 4257, 4522, 4795, 5076, 5365, 5662, 5967, 6280, 6601, 6930, 7267, 7612, 7965, 8326

Offset: 0

N. J. A. Sloane

Write 0, 1, 2, ... in a square spiral, with 0 at the origin and 1 immediately below it; sequence gives numbers on the negative y-axis (see Example section).
Number of divisors of 48^(n-1) for n > 0. - J. Lowell, Aug 30 2008
a(n) is the Wiener index of the graph obtained by connecting two copies of the complete graph K_n by an edge (for n = 3, approximately: |>-<|). The Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph. - Emeric Deutsch, Sep 20 2010
This sequence does not contain any squares other than 0 and 1. See A188896. - T. D. Noe, Apr 13 2011
For n > 0: right edge of the triangle A033293. - Reinhard Zumkeller, Jan 18 2012
Sequence found by reading the line from 0, in the direction 0, 10, ... and the parallel line from 1, in the direction 1, 27, ..., in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Jul 18 2012
Partial sums give A007585. - Omar E. Pol, Jan 15 2013
This is also a star pentagonal number: a(n) = A000326(n) + 5*A000217(n-1). - Luciano Ancora, Mar 28 2015
Also the number of undirected paths in the n-sunlet graph. - Eric W. Weisstein, Sep 07 2017
After 0, a(n) is the sum of 2*n consecutive integers starting from n-1. - Bruno Berselli, Jan 16 2018
Number of corona of an H0 hexagon with a T(n) triangle. - Craig Knecht, Dec 13 2024

			On a square lattice, place the nonnegative integers at lattice points forming a spiral as follows: place "0" at the origin; then move one step downward (i.e., in the negative y direction) and place "1" at the lattice point reached; then turn 90 degrees in either direction and place a "2" at the next lattice point; then make another 90-degree turn in the same direction and place a "3" at the lattice point; etc. The terms of the sequence will lie along the negative y-axis, as seen in the example below:
  99  64--65--66--67--68--69--70--71--72
   |   |                               |
  98  63  36--37--38--39--40--41--42  73
   |   |   |                       |   |
  97  62  35  16--17--18--19--20  43  74
   |   |   |   |               |   |   |
  96  61  34  15   4---5---6  21  44  75
   |   |   |   |   |       |   |   |   |
  95  60  33  14   3  *0*  7  22  45  76
   |   |   |   |   |   |   |   |   |   |
  94  59  32  13   2--*1*  8  23  46  77
   |   |   |   |           |   |   |   |
  93  58  31  12--11-*10*--9  24  47  78
   |   |   |                   |   |   |
  92  57  30--29--28-*27*-26--25  48  79
   |   |                           |   |
  91  56--55--54--53-*52*-51--50--49  80
   |                                   |
  90--89--88--87--86-*85*-84--83--82--81
[Edited by _Jon E. Schoenfield_, Jan 02 2017]

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
Bruce C. Berndt, Ramanujan's Notebooks, Part II, Springer; see p. 23.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
S. M. Ellerstein, The square spiral, J. Recreational Mathematics 29 (#3, 1998) 188; 30 (#4, 1999-2000), 246-250.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
Soren Laing Aletheia-Zomlefer, Lenny Fukshansky, and Stephan Ramon Garcia, The Bateman-Horn Conjecture: Heuristics, History, and Applications, arXiv:1807.08899 [math.NT], 2018-2019. See 6.6.3 p. 33.
Emilio Apricena, A version of the Ulam spiral.
Yin Choi Cheng, Greedy Sidon sets for linear forms, J. Num. Theor. (2024).
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 344.
Craig Knecht, Corona of the H0 hexagon with a T(n) triangle.
Minh Nguyen, 2-adic Valuations of Square Spiral Sequences, Honors Thesis, Univ. of Southern Mississippi (2021).
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Leo Tavares, Illustration: Conjoined Hexagon/Square Pairs
Eric Weisstein's World of Mathematics, Barbell Graph.
Eric Weisstein's World of Mathematics, Decagonal Number.
Eric Weisstein's World of Mathematics, Graph Path.
Eric Weisstein's World of Mathematics, Sunlet Graph.
Index to sequences related to polygonal numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A007585, A028994.
Cf. A093565 ((8, 1) Pascal, column m = 2). Partial sums of A017077.
Sequences from spirals: A001107 (this), A002939, A007742, A033951, A033952, A033953, A033954, A033989, A033990, A033991, A002943, A033996, A033988.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. A003215.

[4*n^2-3*n : n in [0..50] ]; // Wesley Ivan Hurt, Jun 05 2014

A001107:=-(1+7*z)/(z-1)**3; # Simon Plouffe in his 1992 dissertation

LinearRecurrence[{3, -3, 1}, {0, 1, 10}, 60] (* Harvey P. Dale, May 08 2012 *)
Table[PolygonalNumber[RegularPolygon[10], n], {n, 0, 46}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
Table[4 n^2 - 3 n, {n, 0, 49}] (* Alonso del Arte, Jan 24 2017 *)
PolygonalNumber[10, Range[0, 20]] (* Eric W. Weisstein, Sep 07 2017 *)
LinearRecurrence[{3, -3, 1}, {1, 10, 27}, {0, 20}] (* Eric W. Weisstein, Sep 07 2017 *)

PARI
```
a(n)=4*n^2-3*n
```

a=lambda n: 4*n**2-3*n # Indranil Ghosh, Jan 01 2017
def aList(): # Intended to compute the initial segment of the sequence, not isolated terms.
     x, y = 1, 1
     yield 0
     while True:
         yield x
         x, y = x + y + 8, y + 8
A001107 = aList()
print([next(A001107) for i in range(49)]) # Peter Luschny, Aug 04 2019

a(n) = A033954(-n) = A074377(2*n-1).
a(n) = n + 8*A000217(n-1). - Floor van Lamoen, Oct 14 2005
G.f.: x*(1 + 7*x)/(1 - x)^3.
Partial sums of odd numbers 1 mod 8, i.e., 1, 1 + 9, 1 + 9 + 17, ... . - Jon Perry, Dec 18 2004
1^3 + 3^3*(n-1)/(n+1) + 5^3*((n-1)*(n-2))/((n+1)*(n+2)) + 7^3*((n-1)*(n-2)*(n-3))/((n+1)*(n+2)*(n+3)) + ... = n*(4*n-3) [Ramanujan]. - Neven Juric, Apr 15 2008
Starting (1, 10, 27, 52, ...), this is the binomial transform of [1, 9, 8, 0, 0, 0, ...]. - Gary W. Adamson, Apr 30 2008
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2, a(0)=0, a(1)=1, a(2)=10. - Jaume Oliver Lafont, Dec 02 2008
a(n) = 8*n + a(n-1) - 7 for n>0, a(0)=0. - Vincenzo Librandi, Jul 10 2010
a(n) = 8 + 2*a(n-1) - a(n-2). - Ant King, Sep 04 2011
a(n) = A118729(8*n). - Philippe Deléham, Mar 26 2013
a(8*a(n) + 29*n+1) = a(8*a(n) + 29*n) + a(8*n + 1). - Vladimir Shevelev, Jan 24 2014
Sum_{n >= 1} 1/a(n) = Pi/6 + log(2) = 1.216745956158244182494339352... = A244647. - Vaclav Kotesovec, Apr 27 2016
From Ilya Gutkovskiy, Aug 28 2016: (Start)
E.g.f.: x*(1 + 4*x)*exp(x).
Sum_{n >= 1} (-1)^(n+1)/a(n) = (sqrt(2)*Pi - 2*log(2) + 2*sqrt(2)*log(1 + sqrt(2)))/6 = 0.92491492293323294695... (End)
a(n) = A000217(3*n-2) - A000217(n-2). In general, if P(k,n) be the n-th k-gonal number and T(n) be the n-th triangular number, A000217(n), then P(T(k),n) = T((k-1)*n - (k-2)) - T(k-3)*T(n-2). - Charlie Marion, Sep 01 2020
Product_{n>=2} (1 - 1/a(n)) = 4/5. - Amiram Eldar, Jan 21 2021
a(n) = A003215(n-1) + A000290(n) - 1. - Leo Tavares, Jul 23 2022

A017077 a(n) = 8*n + 1.

Original entry on oeis.org

1, 9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, 97, 105, 113, 121, 129, 137, 145, 153, 161, 169, 177, 185, 193, 201, 209, 217, 225, 233, 241, 249, 257, 265, 273, 281, 289, 297, 305, 313, 321, 329, 337, 345, 353, 361, 369, 377, 385, 393, 401, 409, 417, 425, 433

Offset: 0

N. J. A. Sloane

Cf. A007519 (primes), subsequence of A047522.
a(n-1), n >= 1, gives the first column of the triangle A238475 related to the Collatz problem. - Wolfdieter Lang, Mar 12 2014
First differences of A054552. - Wesley Ivan Hurt, Jul 08 2014
An odd number is congruent to a perfect square modulo every power of 2 iff it is in this sequence. Sketch of proof: Suppose the modulus is 2^k with k at least three and note that the only odd quadratic residue (mod 8) is 1. By application of difference of squares and the fact that gcd(x-y,x+y)=2 we can show that for odd x,y, we have x^2 and y^2 congruent mod 2^k iff x is congruent to one of y, 2^(k-1)-y, 2^(k-1)+y, 2^k-y. Now when we "lift" to (mod 2^(k+1)) we see that the degeneracy between a^2 and (2^(k-1)-a)^2 "breaks" to give a^2 and a^2-2^ka+2^(2k-2). Since a is odd, the latter is congruent to a^2+2^k (mod 2^(k+1)). Hence we can form every binary number that ends with '001' by starting modulo 8 and "lifting" while adding digits as necessary. But this sequence is exactly the set of binary numbers ending in '001', so our claim is proved. - Rafay A. Ashary, Oct 23 2016
For n > 3, also the number of (not necessarily maximal) cliques in the n-antiprism graph. - Eric W. Weisstein, Nov 29 2017
Bisection of A016813. - L. Edson Jeffery, Apr 26 2022

			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
--------------------------------------------------------------
.  1       9          17              25                  33
- _Bruno Berselli_, Feb 28 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
Eric Weisstein's World of Mathematics, Antiprism Graph.
Eric Weisstein's World of Mathematics, Clique.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A002189 (subsequence), A004768, A007519, A010731 (first differences), A016813, A047522, A054552.
Column 1 of A093565. Column 5 of triangle A130154. Second leftmost column of triangle A281334.
Row 1 of the arrays A081582, A238475, A371095, and A371096.
Row 2 of A257852.
Apart from the initial term, row sums of triangle A278480.

a017077 = (+ 1) . (* 8)
a017077_list = [1, 9 ..]  -- Reinhard Zumkeller, Dec 28 2012

I:=[1,9]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..60]]; // Vincenzo Librandi, Mar 14 2014

[8*n+1 : n in [0..50]]; // Wesley Ivan Hurt, Jul 08 2014

A017077:=n->8*n+1: seq(A017077(n), n=0..50); # Wesley Ivan Hurt, Jul 08 2014

Table[8 n + 1, {n, 0, 6!}] (* Vladimir Joseph Stephan Orlovsky, Mar 10 2010 *)
CoefficientList[Series[(1 + 7 x)/(1 - x)^2, {x, 0, 60}], x] (* Vincenzo Librandi, Mar 14 2014 *)
8 Range[0, 50] + 1 (* Wesley Ivan Hurt, Jul 08 2014 *)
LinearRecurrence[{2, -1}, {9, 17}, {0, 20}] (* Eric W. Weisstein, Nov 29 2017 *)

a(n)=8*n+1 \\ Charles R Greathouse IV, Jul 10 2016

G.f.: (1+7*x)/(1-x)^2.
a(n+1) = A004768(n). - R. J. Mathar, May 28 2008
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Mar 14 2014
E.g.f.: exp(x)*(1 + 8*x). - Stefano Spezia, May 13 2021
From Elmo R. Oliveira, Apr 10 2025: (Start)
a(n) = a(n-1) + 8 with a(0)=1.
a(n) = A016813(2*n). (End)

A029653 Numbers in (2,1)-Pascal triangle (by row).

Original entry on oeis.org

1, 2, 1, 2, 3, 1, 2, 5, 4, 1, 2, 7, 9, 5, 1, 2, 9, 16, 14, 6, 1, 2, 11, 25, 30, 20, 7, 1, 2, 13, 36, 55, 50, 27, 8, 1, 2, 15, 49, 91, 105, 77, 35, 9, 1, 2, 17, 64, 140, 196, 182, 112, 44, 10, 1, 2, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 2, 21, 100, 285

Offset: 0

Mohammad K. Azarian

Reverse of A029635. Row sums are A003945. Diagonal sums are Fibonacci(n+2) = Sum_{k=0..floor(n/2)} (2n-3k)*C(n-k,n-2k)/(n-k). - Paul Barry, Jan 30 2005
Riordan array ((1+x)/(1-x), x/(1-x)). The signed triangle (-1)^(n-k)T(n,k) or ((1-x)/(1+x), x/(1+x)) is the inverse of A055248. Row sums are A003945. Diagonal sums are F(n+2). - Paul Barry, Feb 03 2005
Row sums = A003945: (1, 3, 6, 12, 24, 48, 96, ...) = (1, 3, 7, 15, 31, 63, 127, ...) - (0, 0, 1, 3, 7, 15, 31, ...); where (1, 3, 7, 15, ...) = A000225. - Gary W. Adamson, Apr 22 2007
Triangle T(n,k), read by rows, given by (2,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 17 2011
A029653 is jointly generated with A208510 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+x*v(n-1)x and v(n,x)=u(n-1,x)+x*v(n-1,x)+1. See the Mathematica section. - Clark Kimberling, Feb 28 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle, see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle, see A228576. - Boris Putievskiy, Sep 04 2013
The n-th row polynomial is (2 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Feb 25 2018

			The triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7  8  9 10 ...
0:  1
1:  2  1
2:  2  3  1
3:  2  5  4   1
4:  2  7  9   5   1
5:  2  9 16  14   6   1
6:  2 11 25  30  20   7   1
7:  2 13 36  55  50  27   8   1
8:  2 15 49  91 105  77  35   9  1
9:  2 17 64 140 196 182 112  44 10  1
10: 2 19 81 204 336 378 294 156 54 11  1
... Reformatted. - _Wolfdieter Lang_, Jan 09 2015
With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
/1        \/1         \/1        \      /1        \
|2 1      ||0 1       ||0 1      |      |2 1      |
|2 1 1    ||0 2 1     ||0 0 1    |... = |2 3 1    |
|2 1 1 1  ||0 2 1 1   ||0 0 2 1  |      |2 5 4 1  |
|2 1 1 1 1||0 2 1 1 1 ||0 0 2 1 1|      |2 7 9 5 1|
|...      ||...       ||...      |      |...      |
- _Peter Bala_, Dec 27 2014

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Paul Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.5.4.
Hacene Belbachir and Athmane Benmezai, Expansion of Fibonacci and Lucas Polynomials: An Answer to Prodinger's Question, Journal of Integer Sequences, Vol. 15 (2012), #12.7.6.
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 39.
H. Hosoya, Pascal's triangle, non-adjacent numbers and D-dimensional atomic orbitals, J. Math. Chemistry, vol. 23, 1998, 169-178.
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
M. Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5
Huyile Liang, Yanni Pei, and Yi Wang, Analytic combinatorics of coordination numbers of cubic lattices, arXiv:2302.11856 [math.CO], 2023. See p. 8.
Mark C. Wilson, Asymptotics for generalized Riordan arrays. International Conference on Analysis of Algorithms DMTCS proc. AD. Vol. 323. 2005.

(d, 1) Pascal triangles: A007318(d=1), A093560(3), A093561(4), A093562(5), A093563(6), A093564(7), A093565(8), A093644(9), A093645(10).
Cf. A003945, A208510, A228196, A228576.
Cf. A078812, A106195.

a029653 n k = a029653_tabl !! n !! k
a029653_row n = a029653_tabl !! n
a029653_tabl = [1] : iterate
               (\xs -> zipWith (+) ([0] ++ xs) (xs ++ [0])) [2, 1]
-- Reinhard Zumkeller, Dec 16 2013

A029653 :=  proc(n,k)
if n = 0 then
  1;
else
  binomial(n-1, k)+binomial(n, k)
fi
end proc: # R. J. Mathar, Jun 30 2013

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A208510 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A029653 *)
(* Clark Kimberling, Feb 28 2012 *)

from sympy import Poly
from sympy.abc import x
def u(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x)
def v(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x) + 1
def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 27 2017

from math import comb, isqrt
def A029653(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*((r<<1)-a)//r if n else 1 # Chai Wah Wu, Nov 12 2024

T(n, k) = C(n-2, k-1) + C(n-2, k) + C(n-1, k-1) + C(n-1, k) except for n=0.
G.f.: (1 + x + y + xy)/(1 - y - xy). - Ralf Stephan, May 17 2004
T(n, k) = (2n-k)*binomial(n, n-k)/n, n, k > 0. - Paul Barry, Jan 30 2005
Sum_{k=0..n} T(n, k)*x^k gives A003945-A003954 for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, Jul 10 2005
T(n, k) = C(n-1, k) + C(n, k). - Philippe Deléham, Jul 10 2005
Equals A097806 * A007318, i.e., the pairwise operator * Pascal's Triangle as infinite lower triangular matrices. - Gary W. Adamson, Apr 22 2007
From Peter Bala, Dec 27 2014: (Start)
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(2 + 5*x + 4*x^2/2! + x^3/3!) = 2 + 7*x + 16*x^2/2! + 30*x^3/3! + 50*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the lower unit triangular array with 1's on the main diagonal and 1's everywhere else below the main diagonal except for the first column which consists of the sequence [1,2,2,2,...]. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)

More terms from James Sellers

A005010 a(n) = 9*2^n.

Original entry on oeis.org

9, 18, 36, 72, 144, 288, 576, 1152, 2304, 4608, 9216, 18432, 36864, 73728, 147456, 294912, 589824, 1179648, 2359296, 4718592, 9437184, 18874368, 37748736, 75497472, 150994944, 301989888, 603979776, 1207959552, 2415919104, 4831838208, 9663676416, 19327352832

Offset: 0

N. J. A. Sloane, Jun 14 1998

Row sums of (8, 1)-Pascal triangle A093565. - N. J. A. Sloane, Sep 22 2004
The first differences are the sequence itself. - Alexandre Wajnberg & Eric Angelini, Sep 07 2005
For n>=1, a(n) is equal to the number of functions f:{1,2,...,n+2}->{1,2,3} such that for fixed, different x_1, x_2,...,x_n in {1,2,...,n+2} and fixed y_1, y_2,...,y_n in {1,2,3} we have f(x_i)<>y_i, (i=1,2,...,n). - Milan Janjic, May 10 2007
9 times powers of 2. - Omar E. Pol, Dec 16 2008
a(n) = A173786(n+3,n) for n>2. - Reinhard Zumkeller, Feb 28 2010
Let D(m) = {d(m,i)}, i = 1..q, denote the set of the q divisors of a number m, and consider s0(m) and s1(m) the sums of the divisors that are congruent to 2 and 3 (mod 4) respectively. For n>0, the sequence a(n) lists the numbers m such that s0(m) = 26 and s1(m) = 3. - Michel Lagneau, Feb 10 2017

Mia Boudreau, Table of n, a(n) for n = 0..3000 (first 236 terms from Vincenzo Librandi)
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A000079, A093565, A173786, A118416, A007283.

[9*2^n: n in [0..40]]; // Vincenzo Librandi, Apr 28 2011

9*2^Range[0, 60] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)

a(n)=9<Charles R Greathouse IV, Apr 17 2012

a(n) = 9*2^n.
G.f.: 9/(1-2*x).
a(n) = A118416(n+1,5) for n>4. - Reinhard Zumkeller, Apr 27 2006
a(n) = 2*a(n-1), n>0; a(0)=9. - Philippe Deléham, Nov 23 2008
a(n) = 9*A000079(n). - Omar E. Pol, Dec 16 2008
a(n) = 3*A007283(n). - Omar E. Pol, Jul 14 2015
E.g.f.: 9*exp(2*x). - Elmo R. Oliveira, Aug 16 2024

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

Original entry on oeis.org

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024

Offset: 1

Boris Putievskiy, Aug 15 2013

The third row is (n^4 - n^2 + 24*n + 24)/12.

For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Rely Pellicer and David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4
Index entries for triangles and arrays related to Pascal's triangle

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).

Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

T:= function(n,k)
    if k=0 then return n^2;
    elif k=n then return 2^n;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019

T:= proc(n, k) option remember;
      if k=0 then n^2
    elif k=n then 2^k
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019

T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)

T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019

def funcL(n):
   q = n**2
   return q
def funcR(n):
   q = 2**n
   return q
for n in range (1,9871):
   t=int((math.sqrt(8*n-7) - 1)/ 2)
   i=n-t*(t+1)/2-1
   j=(t*t+3*t+4)/2-n-1
   sum1=0
   sum2=0
   for m1 in range (1,i+1):
      sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
   for m2 in range (1,j+1):
      sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
   sum=sum1+sum2

@CachedFunction
def T(n, k):
    if (k==0): return n^2
    elif (k==n): return 2^n
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A007585 10-gonal (or decagonal) pyramidal numbers: a(n) = n(n + 1)(8*n - 5)/6.

Original entry on oeis.org

0, 1, 11, 38, 90, 175, 301, 476, 708, 1005, 1375, 1826, 2366, 3003, 3745, 4600, 5576, 6681, 7923, 9310, 10850, 12551, 14421, 16468, 18700, 21125, 23751, 26586, 29638, 32915, 36425, 40176, 44176, 48433, 52955, 57750

Offset: 0

N. J. A. Sloane, R. K. Guy

Binomial transform of [1, 10, 17, 8, 0, 0, 0, ...] = (1, 11, 38, 90, ...). - Gary W. Adamson, Mar 18 2009
This sequence is related to A000384 by a(n) = n*A000384(n) - Sum_{i=0..n-1} A000384(i) and this is the case d=4 in the identity n*(n*(d*n-d+2)/2) - Sum_{k=0..n-1} k*(d*k-d+2)/2 = n*(n+1)*(2*d*n - 2*d + 3)/6. - Bruno Berselli, Apr 21 2010
For n>0, (a(n)) is the principal diagonal of the convolution array A213750. - Clark Kimberling, Jun 20 2012
From Ant King, Oct 30 2012: (Start)
The partial sums of the figurate decagonal numbers A001107.
For n>1, the digital roots of this sequence A010888(A007585(n)) form the purely periodic 27-cycle {1,2,2,9,4,4,8,6,6,7,8,8,6,1,1,5,3,3,4,5,5,3,7,7,2,9,9}.
For n>1, the units’ digits of this sequence A010879(A007585(n)) form the purely periodic 20-cycle {1,1,8,0,5,1,6,8,5,5,6,6,3,5,0,6,1,3,0,0}.
(End)

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Harvey P. Dale, Table of n, a(n) for n = 0..1000
B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1)

Cf. A000384.
Cf. A093565 ((8, 1) Pascal, column m=3). Partial sums of A001107.
Cf. similar sequences listed in A237616.

List([0..40], n-> n*(n+1)*(8*n-5)/6); # G. C. Greubel, Aug 30 2019

[n*(n+1)*(8*n-5)/6: n in [0..40]]; // G. C. Greubel, Aug 30 2019

seq(n*(n+1)*(8*n-5)/6, n=0..40); # G. C. Greubel, Aug 30 2019

Table[n(n+1)(8n-5)/6, {n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{4,-6,4,-1},{0,1,11,38},40] (* Harvey P. Dale, Dec 20 2011 *)

a(n)=(8*n^3+3*n^2-5*n)/6 \\ Charles R Greathouse IV, Sep 24 2015

[n*(n+1)*(8*n-5)/6 for n in (0..40)] # G. C. Greubel, Aug 30 2019

a(n) = (8*n-5)*binomial(n+1, 2)/3.
G.f.: x*(1+7*x)/(1-x)^4.
a(n) = (8*n^3 + 3*n^2 - 5*n)/6. - Vincenzo Librandi, Aug 01 2010
a(0)=0, a(1)=1, a(2)=11, a(3)=38, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Dec 20 2011
From Ant King, Oct 30 2012: (Start)
a(n) = a(n-1) + n*(4*n-3).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 8.
a(n) = (n+1)*(2*A001107(n) + n)/6.
a(n) = A000292(n) + 7*A000292(n-1).
a(n) = A007584(n) + A000292(n-1).
a(n) = A000217(n) + 8*A000292(n-1).
a(n) = binomial(n+2,3) + 7*binomial(n+1,3).
Sum_{n>=1} 1/a(n) = 6*(4*pi*(sqrt(2)-1) + 4*(8-sqrt(2))*log(2) + 8*sqrt(2)*log(2-sqrt(2))-5)/65 =  1.145932345...
(End)
a(n) = Sum_{i=0..n-1} (n-i)*(8*i+1), with a(0)=0. - Bruno Berselli, Feb 10 2014
E.g.f.: x*(6 + 27*x + 8*x^2)*exp(x)/6. - Ilya Gutkovskiy, May 12 2017

A022098 Fibonacci sequence beginning 1, 8.

Original entry on oeis.org

1, 8, 9, 17, 26, 43, 69, 112, 181, 293, 474, 767, 1241, 2008, 3249, 5257, 8506, 13763, 22269, 36032, 58301, 94333, 152634, 246967, 399601, 646568, 1046169, 1692737, 2738906, 4431643, 7170549, 11602192, 18772741, 30374933, 49147674, 79522607, 128670281

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(8; n-1-k, k) with n >= 1, a(-1) = 7. These are the SW-NE diagonals in P(8; n, k), the (8, 1) Pascal triangle A093565. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Pisano period lengths: 1, 3, 8, 6, 4, 24, 16, 12, 24, 12, 10, 24, 28, 48, 8, 24, 36, 24, 18, 12, ... (is this the same as A106291?). - R. J. Mathar, Aug 10 2012
Also the sum of five consecutive Lucas numbers starting with L(-3). - Alonso del Arte, Sep 26 2013
The Pisano period lengths of this sequence are exactly the same as those of the Lucas sequence (A000032), given in A106291. - Klaus Purath, Apr 20 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000032, A000045.

a(n) = A109754(7, n+1) = A101220(7, 0, n+1).

a0:=1; a1:=8; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1, 1}, {1, 8}, 40] (* Alonso del Arte, Sep 26 2013 *)
CoefficientList[Series[(1 + 7 x)/(1 - x - x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 27 2013 *)
Table[LucasL[n + 3] + LucasL[n - 3] - 3 LucasL[n], {n, 2, 40}] (* Bruno Berselli, Dec 30 2016 *)

a(n)=([0,1; 1,1]^n*[1;8])[1,1] \\ Charles R Greathouse IV, Oct 07 2016

a(n) = a(n-1) + a(n-2) for n>1, a(0)=1, a(1)=8, and a(-1):=7.
G.f.: (1 + 7*x)/(1 - x - x^2).
a(n) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5)) + 3.5*((1 + sqrt(5))^(n-1) - (1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)) for n>0. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 2^(-1-n)*((1 - sqrt(5))^n*(-15 + sqrt(5)) + (1 + sqrt(5))^n*(15 + sqrt(5)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
a(n) = 7*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 9*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = Lucas(n+5) + Lucas(n-1) - 3*Lucas(n+2). - Bruno Berselli, Dec 29 2016, corrected by Greg Dresden, Mar 02 2022
a(n) = Lucas(n+3) - Lucas(n-2). - Greg Dresden and Michael Nyc, Mar 02 2022

A093564 (7,1) Pascal triangle.

Original entry on oeis.org

1, 7, 1, 7, 8, 1, 7, 15, 9, 1, 7, 22, 24, 10, 1, 7, 29, 46, 34, 11, 1, 7, 36, 75, 80, 45, 12, 1, 7, 43, 111, 155, 125, 57, 13, 1, 7, 50, 154, 266, 280, 182, 70, 14, 1, 7, 57, 204, 420, 546, 462, 252, 84, 15, 1, 7, 64, 261, 624, 966, 1008, 714, 336, 99, 16, 1, 7, 71, 325, 885

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(7;n,m) gives in the columns m>=1 the figurate numbers based on A016993, including the 9-gonal numbers A001106, (see the W. Lang link).
This is the seventh member, d=7, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-3, for d=1..6.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). 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)=(1+6*z)/(1-(1+x)*z).
The SW-NE diagonals give A022097(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 6. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins
  [1];
  [7,  1];
  [7,  8,  1];
  [7, 15,  9,  1];
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch. 5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
W. Lang, First 10 rows and array of figurate numbers .

Row sums: A000079(n+2), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 6 for n=2 and 0 otherwise.
The column sequences give for m=1..9: A016993, A001106 (9-gonal), A007584, A051740, A051877, A050403, A027818, A034266, A055994.
Cf. A093565 (d=8).

a093564 n k = a093564_tabl !! n !! k
a093564_row n = a093564_tabl !! n
a093564_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [7, 1]
-- Reinhard Zumkeller, Sep 01 2014

N:= 20: # to get the first N rows
T:=Matrix(N,N):
T[1,1]:= 1:
for m from 2 to N do
T[m,1]:= 7:
T[m,2..m]:= T[m-1,1..m-1] + T[m-1,2..m];
od:
for m from 1 to N do
convert(T[m,1..m],list)
od; # Robert Israel, Dec 28 2014

a(n, m)=F(7;n-m, m) for 0<= m <= n, otherwise 0, with F(7;0, 0)=1, F(7;n, 0)=7 if n>=1 and F(7;n, m):=(7*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=7 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+6*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 6*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 15*x + 9*x^2/2! + x^3/3!) = 7 + 22*x + 46*x^2/2! + 80*x^3/3! + 125*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A051797 Partial sums of A007585.

Original entry on oeis.org

1, 12, 50, 140, 315, 616, 1092, 1800, 2805, 4180, 6006, 8372, 11375, 15120, 19720, 25296, 31977, 39900, 49210, 60060, 72611, 87032, 103500, 122200, 143325, 167076, 193662, 223300, 256215, 292640, 332816, 376992, 425425, 478380, 536130

Offset: 0

Barry E. Williams, Dec 11 1999

a(n-1) is the n-th antidiagonal sum of the convolution array A213835. - Clark Kimberling, Jul 04 2012

Convolution of A000027 with A001107 (excluding 0). - Bruno Berselli, Dec 07 2012

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
Murray R. Spiegel, Calculus of Finite Differences and Difference Equations, "Schaum's Outline Series", McGraw-Hill, 1971, pp. 10-20, 79-94.
Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-8.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Index to sequences related to pyramidal numbers.

Cf. A000027, A001107, A007585, A080851.
Cf. A093565 ((8, 1) Pascal, column m=4).
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.

List([0..40], n-> (2*n+1)*Binomial(n+3,3)); # G. C. Greubel, Aug 30 2019

/* A000027 convolved with A001107 (excluding 0): */
A001107:=func; [&+[(n-i+1)*A001107(i): i in [1..n]]: n in [1..35]]; // Bruno Berselli, Dec 07 2012

[(2*n+1)*Binomial(n+3,3): n in [0..40]]; // G. C. Greubel, Aug 30 2019

seq((2*n+1)*binomial(n+3,3), n=0..40); # G. C. Greubel, Aug 30 2019

Table[(2*n+1)*Binomial[n+3,3], {n,0,40}] (*  Vladimir Joseph Stephan Orlovsky, Apr 19 2011, modified by G. C. Greubel, Aug 30 2019 *)

vector(40, n, (2*n-1)*binomial(n+2,3)) \\ G. C. Greubel, Aug 30 2019

[(2*n+1)*binomial(n+3,3) for n in (0..40)] # G. C. Greubel, Aug 30 2019

a(n) = binomial(n+3,3)*(2*n+1) = (n+1)*(n+2)*(n+3)*(2*n+1)/6.
G.f.: (1+7*x)/(1-x)^5.
a(n) = A080851(8,n). - R. J. Mathar, Jul 28 2016
E.g.f.: (6 + 66*x + 81*x^2 + 25*x^3 + 2*x^4)*exp(x)/6. - G. C. Greubel, Aug 30 2019
From Amiram Eldar, Feb 11 2022: (Start)
Sum_{n>=0} 1/a(n) = (32*log(2) - 11)/10.
Sum_{n>=0} (-1)^n/a(n) = (8*Pi - 56*log(2) + 23)/10. (End)

A051878 Partial sums of A051797.

Original entry on oeis.org

1, 13, 63, 203, 518, 1134, 2226, 4026, 6831, 11011, 17017, 25389, 36764, 51884, 71604, 96900, 128877, 168777, 217987, 278047, 350658, 437690, 541190, 663390, 806715, 973791, 1167453, 1390753, 1646968, 1939608, 2272424, 2649416, 3074841, 3553221, 4089351, 4688307, 5355454, 6096454

Offset: 0

Barry E. Williams, Dec 14 1999

Convolution of triangular numbers (A000217) and decagonal numbers (A001107). [Bruno Berselli, Jul 21 2015]

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-16.

Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000217, A001107, A051797.

Cf. A093565 ((8, 1) Pascal, column m=5).

List([0..40], n-> (8*n+5)*Binomial(n+4,4)/5); # G. C. Greubel, Aug 30 2019

[(8*n+5)*Binomial(n+4,4)/5: n in [0..40]]; // G. C. Greubel, Aug 30 2019

seq((8*n+5)*binomial(n+4,4)/5, n=0..40); # G. C. Greubel, Aug 30 2019

Table[(8*n+5)*Binomial[n+4,4]/5, {n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Apr 19 2011, modified by G. C. Greubel, Aug 30 2019 *)

vector(40, n, (8*n-3)*binomial(n+3,4)/5) \\ G. C. Greubel, Aug 30 2019

[(8*n+5)*binomial(n+4,4)/5 for n in (0..30)] # G. C. Greubel, Aug 30 2019

a(n) = binomial(n+4, 4)*(8*n+5)/5.
G.f.: (1+7*x)/(1-x)^6.
E.g.f.: (120 +*1440*x +2280*x^2 +1040*x^3 +165*x^4 +8*x^5)*exp(x)/120. - G. C. Greubel, Aug 30 2019

Terms a(28) onward added by G. C. Greubel, Aug 30 2019

cp's OEIS Frontend

A001107 10-gonal (or decagonal) numbers: a(n) = n*(4*n-3).

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A017077 a(n) = 8*n + 1.

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A029653 Numbers in (2,1)-Pascal triangle (by row).

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A005010 a(n) = 9*2^n.

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A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

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A007585 10-gonal (or decagonal) pyramidal numbers: a(n) = n*(n + 1)*(8*n - 5)/6.

A001107 10-gonal (or decagonal) numbers: a(n) = n(4n-3).

A007585 10-gonal (or decagonal) pyramidal numbers: a(n) = n(n + 1)(8*n - 5)/6.