A093645 -id:A093645 - cp's OEIS frontend

A017281 a(n) = 10*n + 1.

1, 11, 21, 31, 41, 51, 61, 71, 81, 91, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 201, 211, 221, 231, 241, 251, 261, 271, 281, 291, 301, 311, 321, 331, 341, 351, 361, 371, 381, 391, 401, 411, 421, 431, 441, 451, 461, 471, 481, 491, 501, 511, 521, 531

Offset: 0

N. J. A. Sloane

Equals [1, 2, 3, ...] convolved with [1, 9, 0, 0, 0, ...]. - Gary W. Adamson, May 30 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1] = -1, and A[i,j]=0 otherwise. Then, for n>=2, a(n-1) = -coeff(charpoly(A,x),x^(n-1)). - Milan Janjic, Feb 21 2010
Positive integers with last decimal digit = 1. - Wesley Ivan Hurt, Jun 17 2015
Also the number of (not necessarily maximal) cliques in the 2n-crossed prism graph. - Eric W. Weisstein, Nov 29 2017
From Martin Renner, May 28 2024: (Start)
Also number of squares in a grid cross with equally long arms and a width of two points (cf. A017113), e.g. for n = 2 there are nine squares of size 1 unit of area, four of size 2, two of size 5, four of size 8 and two of size 13, thus a total of 21 squares.
    · ·           · ·           · ·           · ·           * ·
    · ·           · ·           * ·           * ·           · ·
* * · · · ·   · · * · · ·   · · · · * ·   · · · · · ·   · · · · · *
* * · · · ·   · * · * · ·   · * · · · ·   * · · · * ·   * · · · · ·
    · ·           * ·           · *           · ·           · ·
    · ·           · ·           · ·           * ·           · *
The possible areas of the squares are given by ceiling(k^2/2) for 1 <= k <= 2*n+1, cf. A000982. In general, there are 4*n + 1 squares with one unit area to be found in the cross, cf. A016813, for n > 0 always four squares of even area and two squares of odd area > 1. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Milan Janjic, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Clique
Eric Weisstein's World of Mathematics, Crossed Prism Graph
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A093645 (column 1).
Subsequence of A034709, together with A017293, A017329, A139222, A139245, A139249, A139264, A139279 and A139280.
Cf. A048161, A154428.
Cf. A005408, A016813, A016921, A017533, A161700, A128470, A158057, A161705, A161709, A161714.
Cf. A030430 (primes).
Cf. A272914, first comment. [Bruno Berselli, May 26 2016]

List([0..60], n-> 10*n+1 ); # G. C. Greubel, Sep 18 2019

a017281 = (+ 1) . (* 10)
a017281_list = [1,11..]  -- Reinhard Zumkeller, Apr 16 2012

[10*n+1 : n in [0..60]]; // Zaki Khandaker, May 16 2015

A017281:=n->10*n + 1; seq(A017281(n), n=0..80); # Wesley Ivan Hurt, Jan 29 2014

f[n_] := FromDigits[IntegerDigits[n^2, n + 1]]; Array[f, 60] (* Robert G. Wilson v, Apr 14 2009 *)
Range[1, 1000, 10] (* Vladimir Joseph Stephan Orlovsky, May 28 2011 *)
(* From Eric W. Weisstein, Nov 29 2017: (Start) *)
Table[10n+1, {n, 0, 60}]
10*Range[0, 60] + 1
LinearRecurrence[{2, -1}, {11, 21}, {0, 60}]
CoefficientList[Series[(1+9x)/(1-x)^2, {x, 0, 60}], x] (* End *)

Vec((1+9*x)/(1-x)^2 + O(x^80)) \\ Michel Marcus, Jun 17 2015

[10*n+1 for n in (0..60)] # G. C. Greubel, Sep 18 2019

G.f.: (1+9*x)/(1-x)^2.
a(n) = 20*n - a(n-1) - 8, with a(0)=1. - Vincenzo Librandi, Nov 20 2010
a(n) = 2*a(n-1) - a(n-2), for n > 2. - Wesley Ivan Hurt, Jun 17 2015
E.g.f.: (1 + 10*x)*exp(x). - G. C. Greubel, Sep 18 2019

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

A051624 12-gonal (or dodecagonal) numbers: a(n) = n(5n-4).

Original entry on oeis.org

0, 1, 12, 33, 64, 105, 156, 217, 288, 369, 460, 561, 672, 793, 924, 1065, 1216, 1377, 1548, 1729, 1920, 2121, 2332, 2553, 2784, 3025, 3276, 3537, 3808, 4089, 4380, 4681, 4992, 5313, 5644, 5985, 6336, 6697, 7068, 7449, 7840, 8241, 8652

Offset: 0

Barry E. Williams

Zero followed by partial sums of A017281. - Klaus Brockhaus, Nov 20 2008
Sequence found by reading the line from 0, in the direction 0, 12, ... and the parallel line from 1, in the direction 1, 33, ..., in the square spiral whose vertices are the generalized 12-gonal numbers A195162. - Omar E. Pol, Jul 18 2012
This is also a star hexagonal number: a(n) = A000384(n) + 6*A000217(n-1). - Luciano Ancora, Mar 30 2015
Starting with offset 1, this is the binomial transform of (1, 11, 10, 0, 0, 0, ...). - Gary W. Adamson, Aug 01 2015
a(n+1) is the sum of the odd numbers from 4n+1 to 6n+1. - Wesley Ivan Hurt, Dec 14 2015
For n >= 2, a(n) is the number of intersection points of all unit circles centered on the inner lattice points of an (n+1) X (n+1) square grid. - Wesley Ivan Hurt, Dec 08 2020
The final digit of a(n) equals the final digit of n, A010879(n). - Enrique Pérez Herrero, Nov 13 2022
a(n-1) is the maximum second Zagreb index of maximal 2-degenerate graphs with n vertices.  (The second Zagreb index of a graph is the sum of the products of the degrees over all edges of the graph.) - Allan Bickle, Apr 16 2024

			The graph K_3 has 3 degree 2 vertices, so a(3-1) = 3*4 = 12.

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 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
John Elias, Illustration: compass configuration , Illustration: cross configuration.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
Allan Bickle, Zagreb Indices of Maximal k-degenerate Graphs, Australas. J. Combin. 89 1 (2024) 167-178.
J. Estes and B. Wei, Sharp bounds of the Zagreb indices of k-trees, J Comb Optim 27 (2014), 271-291.
L. Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, p. 36.
Wikipedia, Dodecagonal number
Index to sequences related to polygonal numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

First differences of A007587.
Cf. A093645 ((10, 1) Pascal, column m=2). Partial sums of A017281.
Cf. A051624, A372025, A372026 (second Zagreb indices of maximal k-degenerate graphs).
Cf. A372027 (second Zagreb index of MOPs).

[ n eq 1 select 0 else Self(n-1)+10*(n-2)+1: n in [1..43] ]; // Klaus Brockhaus, Nov 20 2008

RecurrenceTable[{a[0]==0, a[1]==1, a[2]==12, a[n]== 3*a[n-1] - 3*a[n-2] + a[n-3]}, a, {n, 30}] (* G. C. Greubel, Jul 31 2015 *)
Table[n*(5*n - 4), {n, 0, 100}] (* Robert Price, Oct 11 2018 *)

a(n)=(5*n-4)*n \\ Charles R Greathouse IV, Jun 16 2011

G.f.: x*(1+9*x)/(1-x)^3.
a(n) = Sum_{k=0..n-1} 10*k+1. - Klaus Brockhaus, Nov 20 2008
a(n) = 10*n + a(n-1) - 9 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n) = A131242(10n). - Philippe Deléham, Mar 27 2013
a(10*a(n) + 46*n + 1) = a(10*a(n) + 46*n) + a(10*n+1). - Vladimir Shevelev, Jan 24 2014
E.g.f.: x*(5*x + 1) * exp(x). - G. C. Greubel, Jul 31 2015
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=0, a(1)=1, a(2)=12. - G. C. Greubel, Jul 31 2015
Sum_{n>=1} 1/a(n) = sqrt(1 + 2/sqrt(5))*Pi/8 + 5*log(5)/16 + sqrt(5)*log((1 + sqrt(5))/2)/8 = 1.177956057922663858735173968... . - Vaclav Kotesovec, Apr 27 2016
a(n) + 4*(n-1)^2 = (3*n-2)^2. Let P(k,n) be the n-th k-gonal number. Then, in general, P(4k,n) + (k-1)^2*(n-1)^2 = (k*n-k+1)^2. - Charlie Marion, Feb 04 2020
Product_{n>=2} (1 - 1/a(n)) = 5/6. - Amiram Eldar, Jan 21 2021
a(n) = (3*n-2)^2 - (2*n-2)^2. In general, if we let P(k,n) = the n-th k-gonal number, then P(4k,n) = (k*n-(k-1))^2 - ((k-1)*n-(k-1))^2. - Charlie Marion, Nov 11 2021

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

A093644 (9,1) Pascal triangle.

Original entry on oeis.org

1, 9, 1, 9, 10, 1, 9, 19, 11, 1, 9, 28, 30, 12, 1, 9, 37, 58, 42, 13, 1, 9, 46, 95, 100, 55, 14, 1, 9, 55, 141, 195, 155, 69, 15, 1, 9, 64, 196, 336, 350, 224, 84, 16, 1, 9, 73, 260, 532, 686, 574, 308, 100, 17, 1, 9, 82, 333, 792, 1218, 1260, 882, 408, 117, 18, 1, 9, 91, 415

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(9;n,m) gives in the columns m>=1 the figurate numbers based on A017173, including the 11-gonal numbers A051682 (see the W. Lang link).
This is the ninth member, d=9, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-5, for d=1..8.
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+8*z)/(1-(1+x)*z).
The SW-NE diagonals give A022099(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 8. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Triangle T(n,k), read by rows, given by (9,-8,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 10 2011

			Triangle begins
  [1];
  [9,  1];
  [9, 10,  1];
  [9, 19, 11,  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
Wolfdieter Lang, First 10 rows and array of figurate numbers .

Row sums: A020714(n-1), n >= 1, 1 for n=0, alternating row sums are 1 for n=0, 8 for n=2 and 0 otherwise.
The column sequences give for m=1..9: A017173, A051682 (11-gonal), A007586, A051798, A051879, A050405, A052206, A056117, A056003.
Cf. A093645 (d=10).

a093644 n k = a093644_tabl !! n !! k
a093644_row n = a093644_tabl !! n
a093644_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [9, 1]
-- Reinhard Zumkeller, Aug 31 2014

Join[{1},Table[Binomial[n,k]+8Binomial[n-1,k],{n,20},{k,0,n}]//Flatten] (* Harvey P. Dale, Aug 17 2024 *)

a(n, m) = F(9;n-m, m) for 0 <= m <= n, otherwise 0, with F(9;0, 0)=1, F(9;n, 0)=9 if n >= 1 and F(9;n, m):=(9*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)=9 if n >= 1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+8*x)/(1-x)^(m+1), m >= 0.
T(n, k) = C(n, k) + 8*C(n-1, k). - Philippe Deléham, Aug 28 2005
Row n: Expansion of (9+x)*(1+x)^(n-1), n > 0. - Philippe Deléham, Oct 10 2011
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(9 + 19*x + 11*x^2/2! + x^3/3!) = 9 + 28*x + 58*x^2/2! + 100*x^3/3! + 155*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
G.f.: (-1-8*x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015

A005015 a(n) = 11*2^n.

Original entry on oeis.org

11, 22, 44, 88, 176, 352, 704, 1408, 2816, 5632, 11264, 22528, 45056, 90112, 180224, 360448, 720896, 1441792, 2883584, 5767168, 11534336, 23068672, 46137344, 92274688, 184549376, 369098752, 738197504, 1476395008, 2952790016, 5905580032, 11811160064

Offset: 0

N. J. A. Sloane

The first differences are the sequence itself. - Alexandre Wajnberg & Eric Angelini, Sep 07 2005
11 times powers of 2. - Omar E. Pol, Dec 16 2008
A144472 = -1,2,9,13,31,57,.... a(n) = A144472(n+1)+A144472(n+2). Also a(n) = A144472(n+3)-A144472(n+1). A144472(n+1) is a Jacobsthal sequence from 2 and 9: A144472(n+3) = A144472(n+2)+2*A144472(n+1). Note a(n) mod 9 = period 6: repeat 2,4,8,7,5,1 = A153130(n+1). - Paul Curtz, Jan 06 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2).

Row sums of (10, 1)-Pascal triangle A093645.

Cf. A000079, A144472, A153130.

[11*2^n: n in [0..40]]; // Vincenzo Librandi, Aug 14 2011

11*2^Range[0, 60] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
NestList[2#&,11,30] (* Harvey P. Dale, Jun 11 2021 *)

a(n)=11<Charles R Greathouse IV, Oct 07 2015

G.f.: 11/(1-2*x).
a(n) = 2*a(n-1), n>0; a(0)=11. - Philippe Deléham, Nov 23 2008
a(n) = A000079(n)*11. - Omar E. Pol, Dec 16 2008
E.g.f.: 11*exp(2*x). - Elmo R. Oliveira, Aug 16 2024

A164844 Generalized Pascal Triangle - satisfying the same recurrence as Pascal's triangle, but with a(n,0)=1 and a(n,n)=10^n (instead of both being 1).

Original entry on oeis.org

1, 1, 10, 1, 11, 100, 1, 12, 111, 1000, 1, 13, 123, 1111, 10000, 1, 14, 136, 1234, 11111, 100000, 1, 15, 150, 1370, 12345, 111111, 1000000, 1, 16, 165, 1520, 13715, 123456, 1111111, 10000000, 1, 17, 181, 1685, 15235, 137171, 1234567, 11111111, 100000000, 1, 18, 198, 1866, 16920, 152406, 1371738, 12345678, 111111111, 1000000000, 1, 19, 216, 2064, 18786, 169326, 1524144, 13717416, 123456789, 1111111111, 10000000000

Offset: 0

Mark Dols, Aug 28 2009

Like with Pascal's triangle, the columns grown polynomially. For example, a(n,1)=10+n, a(n,2)=(1/2)*(180+19n+n^2), a(n,3)=(1/6)*(5400 + 569n + 30n^2 + n^3). Likewise, diagonals grow exponentially: a(n,n)=10^n, a(n,n-1) = (10^n-1) / 9. - Kellen Myers, Jan 24 2010

			Triangle begins:
  1
  1,10
  1,11,100
  1,12,111,1000
  1,13,123,1111,10000
  1,14,136,1234,11111,100000

Robert Israel, Table of n, a(n) for n = 0..10152 (rows 0 to 141, flattened)

Cf. A007318, A093645, A011557, A228196.

f:= proc(n,k) option remember;
if k=n then 10^n elif k=0 then 1 else procname(n-1,k-1)+procname(n-1,k) fi
end proc:
seq(seq(f(n,k),k=0..n),n=0..10); # Robert Israel, Jul 01 2016

f[r_, k_] := Sum[10^i*Binomial[r - i - 1, r - k - 1], {i, 0, k}]; TableForm[Table[f[n, k], {n, 0, 15}, {k, 0, n}]] (* Alex Meiburg, Aug 21 2010 *)
a[n_, k_] := a[n, k] = Piecewise[{{0, k > n || k < 0}, {1, k == 0}, {10^n, k == n}}, a[n - 1, k - 1] + a[n - 1, k]]; TableForm[Table[a[n, k], {n, 0, 10}, {k, 0, n}]] (* Kellen Myers, Jan 24 2010 *)

From  Kellen Myers, Jan 24 2010: (Start)
a(n,k) = Sum_{i = 0..k} 10^i * binomial(n-i-1, n-k-1), for 0<=k<=n.
a(n,0) = 1, a(n,n) = 10^n, a(n,k) = a(n-1,k-1)+a(n-1,k). (End)
T(n,k) = T(n-1,k)+11*T(n-1,k-1)-10*T(n-2,k-1)-10*T(n-2,k-2), T(0,0)=1, T(1,0)=1, T(1,1)=10, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
G.f. of triangle: g(x,y) = (1-xy)/((1-10xy)(1-x-xy)). - Robert Israel, Jul 01 2016

Definition clarified, more terms, and revision of Meiburg's Mathematica code by Kellen Myers, Jan 24 2010

A022100 Fibonacci sequence beginning 1, 10.

Original entry on oeis.org

1, 10, 11, 21, 32, 53, 85, 138, 223, 361, 584, 945, 1529, 2474, 4003, 6477, 10480, 16957, 27437, 44394, 71831, 116225, 188056, 304281, 492337, 796618, 1288955, 2085573, 3374528, 5460101, 8834629, 14294730, 23129359, 37424089, 60553448, 97977537, 158530985

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(10; n-1-k, k), n >= 1, with a(-1)=9. These are the SW-NE diagonals in P(10; n, k), the (10,1) Pascal triangle A093645. 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

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

Cf. A000045.

a(n) = A109754(9, n+1) = A101220(9, 0, n+1).

a0:=1; a1:=10; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1,1},{1,10},40] (* Harvey P. Dale, May 17 2017 *)

a(n) = a(n-1) + a(n-2) for n >= 2, a(0)=1, a(1)=10, a(-1):=9.
G.f.: (1 + 9*x)/(1 - x - x^2).
a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*(9*binomial(1, k) - 8*binomial(0, k)). - Paul Barry, May 05 2005
a(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + (9/2)*((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
From Bruno Berselli, Feb 20 2017: (Start)
a(n) =  9*A000045(n) + A000045(n+1).
a(n) = 11*A000045(n) - A000045(n-2). (End)

A007587 12-gonal (or dodecagonal) pyramidal numbers: a(n) = n(n+1)(10*n-7)/6.

Original entry on oeis.org

0, 1, 13, 46, 110, 215, 371, 588, 876, 1245, 1705, 2266, 2938, 3731, 4655, 5720, 6936, 8313, 9861, 11590, 13510, 15631, 17963, 20516, 23300, 26325, 29601, 33138, 36946, 41035, 45415, 50096, 55088, 60401, 66045, 72030, 78366, 85063, 92131, 99580, 107420, 115661

Offset: 0

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

Binomial transform of [1, 12, 21, 10, 0, 0, 0, ...] = (1, 13, 46, 110, ...). - Gary W. Adamson, Nov 28 2007

This sequence is related to A000566 by a(n) = n*A000566(n) - Sum_{i=0..n-1} A000566(i) and this is the case d=5 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, Oct 18 2010

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
B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian), 2008.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A093645 ((10, 1) Pascal, column m=3). Partial sums of A051624.
Cf. A000566.
See similar sequences listed in A237616.

List([0..45], n-> n*(n+1)*(10*n-7)/6); # G. C. Greubel, Aug 30 2019

[ n eq 1 select 0 else Self(n-1)+(n-1)*(5*n-9): n in [1..45] ]; // Klaus Brockhaus, Nov 20 2008

A007587:=n->n*(n+1)*(10*n-7)/6: seq(A007587(n), n=0..50); # Wesley Ivan Hurt, Oct 23 2014

CoefficientList[Series[x(1+9x)/(1-x)^4, {x,0,45}], x] (* Vincenzo Librandi, Jun 20 2013 *)
Table[n(n+1)(10n-7)/6,{n,0,50}] (* Harvey P. Dale, Nov 13 2013 *)

a(n)=if(n,([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,4,-6,4]^n*[0;1;13;46])[1,1],0) \\ Charles R Greathouse IV, Oct 07 2015

vector(45, n, n*(n-1)*(10*n-17)/6) \\ G. C. Greubel, Aug 30 2019

[n*(n+1)*(10*n-7)/6 for n in (0..45)] # G. C. Greubel, Aug 30 2019

a(n) = (10*n-7)*binomial(n+1, 2)/3.
G.f.: x*(1+9*x)/(1-x)^4.
a(n) = Sum_{k=0..n} k*(5*k-4). - Klaus Brockhaus, Nov 20 2008
a(n) = Sum_{i=0..n-1} (n-i)*(10*i+1), with a(0)=0. - Bruno Berselli, Feb 10 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Wesley Ivan Hurt, Oct 23 2014
E.g.f.: exp(x)*x*(6 + 33*x + 10*x^2)/6. - Elmo R. Oliveira, Aug 04 2025

A022103 Fibonacci sequence beginning 1, 13.

Original entry on oeis.org

1, 13, 14, 27, 41, 68, 109, 177, 286, 463, 749, 1212, 1961, 3173, 5134, 8307, 13441, 21748, 35189, 56937, 92126, 149063, 241189, 390252, 631441, 1021693, 1653134, 2674827, 4327961, 7002788, 11330749, 18333537, 29664286, 47997823, 77662109, 125659932, 203322041, 328981973

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(13;n-1-k,k) for n>=1, a(-1)=12. These are the SW-NE diagonals in P(13;n,k), the (13,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. 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, 4, 24, 16, 12, 24, 12, 10, 24, 28, 48, 8, 24, 36, 24, 18, 12, ... (is this A106291?). - R. J. Mathar, Aug 10 2012

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

a(n) = A109754(12, n+1) = A101220(12, 0, n+1).

Cf. A000032, A000045.

a0:=1; a1:=13; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1, 1}, {1, 13}, 40] (* or *) Table[LucasL[n + 5] - 5 LucasL[n], {n, 0, 40}] (* Bruno Berselli, Dec 30 2016 *)

a(n) = a(n-1) + a(n-2) for n>=2, a(0)=1, a(1)=13, and a(-1):=12.
G.f.: (1 + 12*x)/(1 - x - x^2).
a(n) = ((1 + sqrt(5))^n-(1 - sqrt(5))^n)/(2^n*sqrt(5))+ 6*((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) = 12*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 14*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = Lucas(n+5) - 5*Lucas(n). - Bruno Berselli, Dec 30 2016

cp's OEIS Frontend

A017281 a(n) = 10*n + 1.

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

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A051624 12-gonal (or dodecagonal) numbers: a(n) = n*(5*n-4).

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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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A093644 (9,1) Pascal triangle.

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A005015 a(n) = 11*2^n.

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A051624 12-gonal (or dodecagonal) numbers: a(n) = n(5n-4).

A007587 12-gonal (or dodecagonal) pyramidal numbers: a(n) = n(n+1)(10*n-7)/6.