A093644 -id:A093644 - cp's OEIS frontend

A172171 (1, 9) Pascal Triangle read by horizontal rows. Same as A093644, but mirrored and without the additional row/column (1, 9, 9, 9, 9, ...).

1, 1, 10, 1, 11, 19, 1, 12, 30, 28, 1, 13, 42, 58, 37, 1, 14, 55, 100, 95, 46, 1, 15, 69, 155, 195, 141, 55, 1, 16, 84, 224, 350, 336, 196, 64, 1, 17, 100, 308, 574, 686, 532, 260, 73, 1, 18, 117, 408, 882, 1260, 1218, 792, 333, 82

Offset: 1

Mark Dols, Jan 28 2010

Binomial transform of A017173.

			Triangle begins:
  1;
  1, 10;
  1, 11,  19;
  1, 12,  30,  28;
  1, 13,  42,  58,   37;
  1, 14,  55, 100,   95,   46;
  1, 15,  69, 155,  195,  141,   55;
  1, 16,  84, 224,  350,  336,  196,   64;
  1, 17, 100, 308,  574,  686,  532,  260,   73;
  1, 18, 117, 408,  882, 1260, 1218,  792,  333,   82;
  1, 19, 135, 525, 1290, 2142, 2478, 2010, 1125,  415,  91;
  1, 20, 154, 660, 1815, 3432, 4620, 4488, 3135, 1540, 506, 100;

G. C. Greubel, Rows n = 1..50 of the triangle, flattened

Cf. A007318, A017173, A050489 (central terms), A093644, A139634 (row sums).

T[n_, k_]:= T[n, k]= If[k<1 || k>n, 0, If[k==1, 1, If[n==2 && k==2, 10, T[n-1, k] + 2*T[n-1, k-1] - T[n-2, k-1] - T[n-2, k-2]]]];
Table[T[n, k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Apr 24 2022 *)

@CachedFunction
def T(n,k):
    if (k<1 or k>n): return 0
    elif (k==1): return 1
    elif (n==2 and k==2): return 10
    else: return T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2)
flatten([[T(n,k) for k in (1..n)] for n in (1..15)]) # G. C. Greubel, Apr 24 2022

T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(n,1) = 1, T(2,2) = 10, T(n,k) = 0 if k < 1 or if k > n.
Sum_{k=0..n} T(n, k) = A139634(n).
T(2*n-1, n) = A050489(n).

More terms from Philippe Deléham, Dec 25 2013

A020714 a(n) = 5 * 2^n.

Original entry on oeis.org

5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 655360, 1310720, 2621440, 5242880, 10485760, 20971520, 41943040, 83886080, 167772160, 335544320, 671088640, 1342177280, 2684354560, 5368709120, 10737418240

Offset: 0

David W. Wilson

Same as Pisot sequences E(5,10), L(5,10), P(5,10), T(5,10). See A008776 for definitions of Pisot sequences.
The first differences are the sequence itself. - Alexandre Wajnberg & Eric Angelini, Sep 07 2005
5 times powers of 2. - Omar E. Pol, Dec 16 2008
Subsequence of A051916. - Reinhard Zumkeller, Mar 20 2010
With the addition of "2, 3," at the beginning, this sequence gives terms (n + 3) through the first term greater than 2^n, for n odd, of the negabinary Keith sequence for 2^n, thus proving that with the exception of 2 itself, no odd-indexed power of 2 is a negabinary Keith number (see A188381). - Alonso del Arte, Feb 02 2012
Let b(0) = 5 and b(n+1) = the smallest number not in the sequence such that b(n+1) - Product_{i=0..n} b(i) divides b(n+1) - Sum_{i=0..n} b(i). Then b(n+2) = a(n) for n > 0. - Derek Orr, Jan 15 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..238
John Elias, Illustration: Alternating Tetrahedrons of Tetrahedrons
Tanya Khovanova, Recursive Sequences.
Petro Kosobutskyy, Anastasiia Yedyharova, and Taras Slobodzyan, From Newton's binomial and Pascal's triangle to Collatz's problem, Comp. Des. Sys., Theor. Practice (2023) Vol. 5, No. 1, 121-127.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1003.
Everett Sullivan, Linear chord diagrams with long chords, arXiv preprint arXiv:1611.02771 [math.CO], 2016. See Table 1.
Index entries for linear recurrences with constant coefficients, signature (2).

Row sums of (4, 1)-Pascal triangle A093561.
Row sums of (9, 1)-Pascal triangle A093644.
Row sums of (1, 4)-Pascal triangle A095666 (with leading 4).
Cf. A000079, A001045, A008776, A051916, A118416, A173786, A188381.

[5*2^n: n in [0..40]]; // Vincenzo Librandi, Apr 28 2011

Table[5*2^n, {n, 0, 31}] (* Vladimir Joseph Stephan Orlovsky, Dec 16 2008 *)
NestList[2#&,5,40] (* Harvey P. Dale, Mar 13 2022 *)

a(n)=5<Charles R Greathouse IV, Sep 24 2015

a(n) = 5*2^n. a(n) = 2*a(n-1).
G.f.: 5/(1-2*x).
If m is a term greater than 5 of this sequence then m = 5*phi(phi(m)). - Farideh Firoozbakht, Aug 16 2005
a(n) = A118416(n+1,3) for n>2. - Reinhard Zumkeller, Apr 27 2006
a(n) = A000079(n)*5. - Omar E. Pol, Dec 16 2008
a(n) = A173786(n+2,n) for n > 1. - Reinhard Zumkeller, Feb 28 2010
a(n) = A001045(n+4) - A001045(n). - Paul Curtz, Nov 08 2012
Sum_{n>=1} 1/a(n) = 2/5. - Amiram Eldar, Oct 28 2020
E.g.f.: 5*exp(2*x). - Stefano Spezia, May 15 2021

A051682 11-gonal (or hendecagonal) numbers: a(n) = n(9n-7)/2.

Original entry on oeis.org

0, 1, 11, 30, 58, 95, 141, 196, 260, 333, 415, 506, 606, 715, 833, 960, 1096, 1241, 1395, 1558, 1730, 1911, 2101, 2300, 2508, 2725, 2951, 3186, 3430, 3683, 3945, 4216, 4496, 4785, 5083, 5390, 5706, 6031, 6365, 6708, 7060, 7421, 7791, 8170

Offset: 0

Barry E. Williams

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 0,1,...
The spiral begins:
         15
         / \
       16  14
       /     \
     17   3  13
     /   / \   \
   18   4   2  12
       /     \   \
      5   0---1  11
     /             \
    6---7---8---9--10
. (End)
(1), (4+7), (7+10+13), (10+13+16+19), ... - Jon Perry, Sep 10 2004
This sequence does not contain any triangular numbers other than 0 and 1. See A188892. - T. D. Noe, Apr 13 2011
Sequence found by reading the line from 0, in the direction 0, 11, ... and the parallel line from 1, in the direction 1, 30, ..., in the square spiral whose vertices are the generalized 11-gonal numbers A195160. - Omar E. Pol, Jul 18 2012
Starting with offset 1, the sequence is the binomial transform of (1, 10, 9, 0, 0, 0, ...). - Gary W. Adamson, Aug 01 2015

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 189, 194-196.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
Murray R. Spiegel, Calculus of Finite Differences and Difference Equations, "Schaum's Outline Series", McGraw-Hill, 1971, pp. 10-20, 79-94.

T. D. Noe, Table of n, a(n) for n = 0..1000
Amelia Carolina Sparavigna, The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences, Politecnico di Torino, Italy (2019), [math.NT].
Index to sequences related to polygonal numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

First differences of A007586.
Cf. A093644 ((9, 1) Pascal, column m=2). Partial sums of A017173.
Cf. A000217, A004188, A131432, A188892, A195160, A218470.

[n*(9*n-7)/2 : n in [0..50]]; // Wesley Ivan Hurt, Aug 01 2015

Table[n (9n-7)/2,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,1,11},51] (* Harvey P. Dale, May 07 2012 *)

a(n)=(9*n-7)*n/2 \\ Charles R Greathouse IV, Jun 16 2011

a(n) = n*(9*n-7)/2.
G.f.: x*(1+8*x)/(1-x)^3.
Row sums of triangle A131432. - Gary W. Adamson, Jul 10 2007
a(n) = 9*n + a(n-1) - 8 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=1, a(2)=11. - Harvey P. Dale, May 07 2012
a(n) = A218470(9n). - Philippe Deléham, Mar 27 2013
a(9*a(n)+37*n+1) = a(9*a(n)+37*n) + a(9*n+1). - Vladimir Shevelev, Jan 24 2014
a(n+y) - a(n-y-1) = (a(n+x) - a(n-x-1))*(2*y+1)/(2*x+1), 0 <= x < n, y <= x, a(0)=0. - Gionata Neri, May 03 2015
a(n) = A000217(n-1) + A000217(3*n-2) - A000217(n-2). - Charlie Marion, Dec 21 2019
Product_{n>=2} (1 - 1/a(n)) = 9/11. - Amiram Eldar, Jan 21 2021
E.g.f.: exp(x)*x*(2 + 9*x)/2. - Stefano Spezia, Dec 25 2022

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

A017173 a(n) = 9*n + 1.

Original entry on oeis.org

1, 10, 19, 28, 37, 46, 55, 64, 73, 82, 91, 100, 109, 118, 127, 136, 145, 154, 163, 172, 181, 190, 199, 208, 217, 226, 235, 244, 253, 262, 271, 280, 289, 298, 307, 316, 325, 334, 343, 352, 361, 370, 379, 388, 397, 406, 415, 424, 433, 442, 451, 460, 469, 478

Offset: 0

N. J. A. Sloane

Also all the numbers with digital root 1; A010888(a(n)) = 1. - Rick L. Shepherd, Jan 12 2009
A116371(a(n)) = A156144(a(n)); positions where records occur in A156144: A156145(n+1) = A156144(a(n)). - Reinhard Zumkeller, Feb 05 2009
If A=[A147296] 9*n^2+2*n (n>0, 11, 40, 87, ...); Y=[A010701] 3 (3, 3, 3, ...); X=[A017173] 9*n+1 (n>0, 10, 19, 28, ...), we have, for all terms, Pell's equation X^2 - A*Y^2 = 1. Example: 10^2 - 11*3^2 = 1; 19^2 - 40*3^2 = 1; 28^2 - 87*3^2 = 1. - Vincenzo Librandi, Aug 01 2010

Mohammed Yaseen, Table of n, a(n) for n = 0..10000
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A093644 ((9,1) Pascal, column m=1).
Cf. A010701, A010888, A016777, A116371, A147296, A156144, A156145.
Numbers with digital root m: this sequence (m=1), A017185 (m=2), A017197 (m=3), A017209 (m=4), A017221 (m=5), A017233 (m=6), A017245 (m=7), A017257 (m=8), A008591 (m=9).

a017173 = (+ 1) . (* 9)
a017173_list = [1, 10 ..]  -- Reinhard Zumkeller, Feb 04 2014

Range[1, 1000, 9] (* Vladimir Joseph Stephan Orlovsky, May 28 2011 *)
LinearRecurrence[{2,-1},{1,10},60] (* Harvey P. Dale, Dec 27 2014 *)

forstep(n=1,500,9,print1(n", ")) \\ Charles R Greathouse IV, May 28 2011

[i+1 for i in range(480) if gcd(i,9) == 9] # Zerinvary Lajos, May 20 2009

G.f.: (1 + 8*x)/(1 - x)^2.
a(n) = 2*a(n-1) - a(n-2) with a(0)=1, a(1)=10. - Vincenzo Librandi, Aug 01 2010
E.g.f.: exp(x)*(1 + 9*x). - Stefano Spezia, Apr 20 2023
a(n) = A016777(3*n). - Elmo R. Oliveira, Apr 12 2025

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

A093645 (10,1) Pascal triangle.

Original entry on oeis.org

1, 10, 1, 10, 11, 1, 10, 21, 12, 1, 10, 31, 33, 13, 1, 10, 41, 64, 46, 14, 1, 10, 51, 105, 110, 60, 15, 1, 10, 61, 156, 215, 170, 75, 16, 1, 10, 71, 217, 371, 385, 245, 91, 17, 1, 10, 81, 288, 588, 756, 630, 336, 108, 18, 1, 10, 91, 369, 876, 1344, 1386, 966, 444, 126, 19, 1

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(10;n,m) gives in the columns m >= 1 the figurate numbers based on A017281, including the 12-gonal numbers A051624 (see the W. Lang link).
This is the tenth member, d=10, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-5 and A093644 for d=1..9.
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+9*z)/(1-(1+x)*z).
The SW-NE diagonals give A022100(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k), n >= 1, with n=0 value 9. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins
   1;
  10,  1;
  10, 11,  1;
  10, 21, 12,  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: 1 for n=0 and A005015(n-1), n >= 1, alternating row sums are 1 for n=0, 9 for n=2 and 0 otherwise.

The column sequences give for m=1..9: A017281, A051624 (12-gonal), A007587, A051799, A051880, A050406, A052254, A056125, A093646.

a093645 n k = a093645_tabl !! n !! k
a093645_row n = a093645_tabl !! n
a093645_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [10, 1]
-- Reinhard Zumkeller, Aug 31 2014

t[0, 0] = 1; t[n_, k_] := Binomial[n, k] + 9*Binomial[n-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 05 2013, after Philippe Deléham *)

a(n, m) = F(10;n-m, m) for 0 <= m <= n, else 0, with F(10;0, 0)=1, F(10;n, 0)=10 if n >= 1 and F(10;n, m):=(10*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)=10 if n >= 1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+9*x)/(1-x)^(m+1), m >= 0.
T(n, k) = C(n, k) + 9*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)*(10 + 21*x + 12*x^2/2! + x^3/3!) = 10 + 31*x + 64*x^2/2! + 110*x^3/3! + 170*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

A093565 (8,1) Pascal triangle.

Original entry on oeis.org

1, 8, 1, 8, 9, 1, 8, 17, 10, 1, 8, 25, 27, 11, 1, 8, 33, 52, 38, 12, 1, 8, 41, 85, 90, 50, 13, 1, 8, 49, 126, 175, 140, 63, 14, 1, 8, 57, 175, 301, 315, 203, 77, 15, 1, 8, 65, 232, 476, 616, 518, 280, 92, 16, 1, 8, 73, 297, 708, 1092, 1134, 798, 372, 108, 17, 1, 8, 81, 370, 1005

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(8;n,m) gives in the columns m>=1 the figurate numbers based on A017077, including the decagonal numbers A001107,(see the W. Lang link).
This is the eighth member, d=8, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-4, for d=1..7.
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+7*z)/(1-(1+x)*z).
The SW-NE diagonals give A022098(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 7. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins
  [1];
  [8,  1];
  [8,  9,  1];
  [8, 17, 10,  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: A005010(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 7 for n=2 and 0 else.
The column sequences give for m=1..9: A017077, A001107 (decagonal), A007585, A051797, A051878, A050404, A052226, A056001, A056122.
Cf. A093644 (d=9).

a093565 n k = a093565_tabl !! n !! k
a093565_row n = a093565_tabl !! n
a093565_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [8, 1]
-- Reinhard Zumkeller, Aug 31 2014

a(n, m)=F(8;n-m, m) for 0<= m <= n, otherwise 0, with F(8;0, 0)=1, F(8;n, 0)=8 if n>=1 and F(8;n, m):=(8*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)=8 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+7*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 7*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)*(8 + 17*x + 10*x^2/2! + x^3/3!) = 8 + 25*x + 52*x^2/2! + 90*x^3/3! + 140*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

A007586 11-gonal (or hendecagonal) pyramidal numbers: a(n) = n(n+1)(3*n-2)/2.

Original entry on oeis.org

0, 1, 12, 42, 100, 195, 336, 532, 792, 1125, 1540, 2046, 2652, 3367, 4200, 5160, 6256, 7497, 8892, 10450, 12180, 14091, 16192, 18492, 21000, 23725, 26676, 29862, 33292, 36975, 40920, 45136, 49632, 54417, 59500, 64890, 70596, 76627, 82992, 89700, 96760, 104181

Offset: 0

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

Starting with 1 equals binomial transform of [1, 11, 19, 9, 0, 0, 0, ...]. - Gary W. Adamson, Nov 02 2007

			From _Vincenzo Librandi_, Feb 12 2014: (Start)
After 0, the sequence is provided by the row sums of the triangle (see above, third formula):
  1;
  2, 10;
  3, 20, 19;
  4, 30, 38, 28;
  5, 40, 57, 56, 37;
  6, 50, 76, 84, 74, 46; etc. (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).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A051682.
Cf. A093644 ((9,1) Pascal, column m=3).
Cf. similar sequences listed in A237616.

List([0..45], n-> n*(n+1)*(3*n-2)/2); # G. C. Greubel, Aug 30 2019

I:=[0,1,12,42]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4) : n in [1..50]]; // Vincenzo Librandi, Feb 12 2014

seq(n*(n+1)*(3*n-2)/2, n=0..45); # G. C. Greubel, Aug 30 2019

Table[n(n+1)(3n-2)/2,{n,0,45}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,1,12,42}, 45] (* Harvey P. Dale, Apr 09 2012 *)
CoefficientList[Series[x(1+8x)/(1-x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Feb 12 2014 *)

a(n)=n*(n+1)*(3*n-2)/2 \\ Charles R Greathouse IV, Oct 07 2015

[n*(n+1)*(3*n-2)/2 for n in (0..45)] # G. C. Greubel, Aug 30 2019

G.f.: x*(1+8*x)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3, a(0)=0, a(1)=1, a(2)=12, a(3)=42. - Harvey P. Dale, Apr 09 2012
a(n) = Sum_{i=0..n-1} (n-i)*(9*i+1), with a(0)=0. - Bruno Berselli, Feb 10 2014
From Amiram Eldar, Jun 28 2020: (Start)
Sum_{n>=1} 1/a(n) = (9*log(3) + sqrt(3)*Pi - 4)/10.
Sum_{n>=1} (-1)^(n+1)/a(n) = (sqrt(3)*Pi + 2 - 4*log(2))/5. (End)
E.g.f.: exp(x)*x*(2 + 10*x + 3*x^2)/2. - Elmo R. Oliveira, Aug 03 2025

More terms from Vincenzo Librandi, Feb 12 2014

A022099 Fibonacci sequence beginning 1, 9.

Original entry on oeis.org

1, 9, 10, 19, 29, 48, 77, 125, 202, 327, 529, 856, 1385, 2241, 3626, 5867, 9493, 15360, 24853, 40213, 65066, 105279, 170345, 275624, 445969, 721593, 1167562, 1889155, 3056717, 4945872, 8002589, 12948461, 20951050, 33899511, 54850561, 88750072, 143600633, 232350705

Offset: 0

N. J. A. Sloane, Jun 14 1998

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(9;n-1-k,k) with n>=1, a(-1)=8. These are the SW-NE diagonals in P(9;n,k), the (9,1) Pascal triangle A093644. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have: b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths:  1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (perhaps the same as A001175). - R. J. Mathar, Aug 10 2012
No, it is not the same as in A001175. The Pisano periods are different for moduli 71 and 142, where they are 35 and 105 instead of 70 and 210. Otherwise they coincide with those of the Fibonacci sequence. - Klaus Purath, Jun 26 2022

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000045, A001175, A093644, A101220, A109754.

a0:=1; a1:=9; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1, 1}, {1, 9}, 36] (* Robert G. Wilson v, Apr 11 2014 *)

a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=9. a(-1):=8.
G.f.: (1+8*x)/(1-x-x^2).
a(n) = A109754(8, n+1) = A101220(8, 0, n+1).
a(n+1) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5))+ 4*((1 + sqrt(5))^(n-1) - (1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) =  8*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 10*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = Lucas(n+3) + Fibonacci(n-4). - Greg Dresden and Mary Beth Pittman, Mar 12 2022

cp's OEIS Frontend

A172171 (1, 9) Pascal Triangle read by horizontal rows. Same as A093644, but mirrored and without the additional row/column (1, 9, 9, 9, 9, ...).

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A020714 a(n) = 5 * 2^n.

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A051682 11-gonal (or hendecagonal) numbers: a(n) = n*(9*n-7)/2.

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

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A017173 a(n) = 9*n + 1.

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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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A051682 11-gonal (or hendecagonal) numbers: a(n) = n(9n-7)/2.

A007586 11-gonal (or hendecagonal) pyramidal numbers: a(n) = n(n+1)(3*n-2)/2.