A123149 -id:A123149 - cp's OEIS frontend

A169623 Generalized Pascal triangle read by rows: T(n,0) = T(0,n) = 1 for n >= 0, T(n,k) = 0 for k < 0 or k > n; otherwise T(n,k) = T(n-2,k-2) + T(n-2,k-1) + T(n-2,k) for 1 <= k <= n-1.

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 1, 3, 5, 5, 3, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 9, 13, 13, 9, 4, 1, 1, 4, 10, 16, 19, 16, 10, 4, 1, 1, 5, 14, 26, 35, 35, 26, 14, 5, 1, 1, 5, 15, 30, 45, 51, 45, 30, 15, 5, 1, 1, 6, 20, 45, 75, 96, 96, 75, 45, 20, 6, 1

Offset: 0

Roger L. Bagula and Gary W. Adamson, Dec 03 2009

The borders are all 1's, with zero entries outside. To get an internal entry, use the rule that D = A+B+C here:
         A   B   C
       *   *   *   *
     *   *   D   *   *
That is, add the three terms directly above you, two rows back.
This is the triangle er(n,k) defined in the Ehrenborg and Readdy link. See Proposition 2.4 and Table 1. - Michel Marcus, Sep 14 2016
If the offset is changed from 0 to 1, this is also the table U(n,k) of the coefficients [x^k] p_n(x) of the polynomials p_n(x) = (x + 1)*p_{n-1}(x) (if n even), p_n = (x^2 + x + 1)^floor(n/2) if n odd.
May be split into two triangles by taking the even-numbered and odd-numbered rows separately: the even-numbered rows give A027907.
From Peter Bala, Aug 19 2021: (Start)
Let M denote the lower unit triangular array A070909. 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 matrix product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section below. The proof uses the hockey-stick identities from the Formula section. (End)

			Triangle begins:
                    1
                  1   1
                1   1   1
              1   2   2   1
            1   2   3   2   1
          1   3   5   5   3   1
        1   3   6   7   6   3   1
      1   4   9  13  13   9   4   1
    1   4  10  16  19  16  10   4   1
  ...
As a square array read by antidiagonals:
  1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 1, 1, 1, 1, ...
  1, 1, 2,  2,  3,  3,  4,  4,  5,  5,  6,  6, 7, 7, 8, 8, ...
  1, 2, 3,  5,  6,  9, 10, 14, 15, 20, 21, 27, ...
  1, 2, 5,  7, 13, 16, 26, 30, 45, ...
  1, 3, 6, 13, 19, 35, 45, 75, ...
  1, 3, 9, 16, 35, 51, 96, ...
  ...
From _Peter Bala_, Aug 19 2021: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins
  /1        \/1        \/1       \ /1       \        /1         \
  |1 1      ||0 1      ||0 1      ||0 1      |       |1 1       |
  |1 0 1    ||0 1 1    ||0 0 1    ||0 0 1    |...  = |1 1 1     |
  |1 0 1 1  ||0 1 0 1  ||0 0 1 1  ||0 0 0 1  |       |1 2 2 1   |
  |1 0 1 0 1||0 1 0 1 1||0 0 1 0 1||0 0 0 1 1|       |1 2 3 2 1 |
  |...      ||...       |...      ||...      |       |...       |
(End)

Rémy Sigrist, Rows 0..199, flattened
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Richard Ehrenborg and Margaret A. Readdy, The Gaussian coefficient revisited, arXiv:1609.03216 [math.CO], 2016.
Richard L. Ollerton and Anthony G. Shannon, Some properties of generalized Pascal squares and triangles, Fib. Q., 36 (No. 2, 1998), 98-109. See Table 10.
Wikipedia, Hockey-stick identity.

A123149 is essentially the same triangle, except for a diagonal of zeros.
Row sums are in A182522 (essentially A038754).
Cf. A027907, A070909.
See A295555 for the next triangle in the series A007318, A169623 (this sequence).

T:=proc(n,k) option remember;
if n >= 0 and k = 0 then 1
elif n >= 0 and k = n then 1
elif (k < 0 or k > n) then 0
else T(n-2,k-2)+T(n-2,k-1)+T(n-2,k);
fi;
end;
for n from 0 to 14 do lprint([seq(T(n,k),k=0..n)]); od: # N. J. A. Sloane, Nov 23 2017

p[x, 1] := 1;
p[x_, n_] := p[x, n] = If[Mod[n, 2] == 0, (x + 1)*p[x, n - 1], (x^2 + x + 1)^Floor[n/2]]
a = Table[CoefficientList[p[x, n], x], {n, 1, 12}]
Flatten[a] (* This is for the same sequence but with offset 1 *)

From Peter Bala, Aug 19 2021: (Start)
T(2*n,k) = T(2*n-1,k-1) + T(2*n-2,k).
T(2*n,k) = T(2*n-1,k) + T(2*n-2,k-2).
T(2*n+1,k) = T(2*n,k) + T(2*n,k-1).
Hockey stick identities (relate row k entries to entries in row k-1):
T(2*n,k) = T(2*n-1,k-1) + T(2*n-3,k-1) + T(2*n-5,k-1) + ....
T(2*n+1,k) = T(2*n,k-1) + ( T(2*n-1,k-1) + T(2*n-3,k-1) + T(2*n-5,k-1) + ... ). (End)

Keyword:tabl added, notation standardized, formula added by the Assoc. Editors of the OEIS, Feb 02 2010

Entry revised by N. J. A. Sloane, Nov 23 2017

A182522 a(0) = 1; thereafter a(2n + 1) = 3^n, a(2n + 2) = 2 * 3^n.

Original entry on oeis.org

1, 1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, 729, 1458, 2187, 4374, 6561, 13122, 19683, 39366, 59049, 118098, 177147, 354294, 531441, 1062882, 1594323, 3188646, 4782969, 9565938, 14348907, 28697814, 43046721, 86093442, 129140163, 258280326, 387420489

Offset: 0

Michael Somos, May 03 2012

Row sums of triangle in A123149. - Philippe Deléham, May 04 2012
This is simply the classic sequence A038754 prefixed by a 1. - N. J. A. Sloane, Nov 23 2017
Binomial transform is A057960.
Range of row n of the circular Pascal array of order 6. - Shaun V. Ault, May 30 2014
a(n) is also the number of achiral color patterns in a row or cycle of length n using three or fewer colors. Two color patterns are the same if we permute the colors, so ABCAB=BACBA. For a cycle, we can rotate the colors, so ABCAB=CABAB. A row is achiral if it is the same as some color permutation of its reverse. Thus the reversal of ABCAB is BACBA, which is equivalent to ABCAB when we permute A and B. A cycle is achiral if it is the same as some rotation of some color permutation of its reverse. Thus CABAB reversed is BABAC. We can permute A and B to get ABABC and then rotate to get CABAB, so CABAB is achiral. It is interesting that the number of achiral color patterns is the same for rows and cycles. - Robert A. Russell, Mar 10 2018
Also, the number of walks of length n on the graph 0--1--2--3--4 starting at vertex 0. - Sean A. Irvine, Jun 03 2025

			G.f. = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 9*x^5 + 18*x^6 + 27*x^7 + 54*x^8 + ...
From _Robert A. Russell_, Mar 10 2018: (Start)
For a(4) = 6, the achiral color patterns for rows are AAAA, AABB, ABAB, ABBA, ABBC, and ABCA.  Note that for cycles AABB=ABBA and ABBC=ABCA.  The achiral patterns for cycles are AAAA, AAAB, AABB, ABAB, ABAC, and ABBC.  Note that AAAB and ABAC are not achiral rows.
For a(5) = 9, the achiral color patterns (for both rows and cycles) are AAAAA, AABAA, ABABA, ABBBA, AABCC, ABACA, ABBBC, ABCAB, and ABCBA. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Shaun V. Ault and Charles Kicey, Counting paths in corridors using circular Pascal arrays, arXiv:1407.2197 [math.CO], 2014.
Shaun V. Ault and Charles Kicey, Counting paths in corridors using circular Pascal arrays, Discrete Mathematics, Volume 332, October 2014, Pages 45-54.
Daniel Birmajer, Juan B. Gil, Jordan O. Tirrell, and Michael D. Weiner, Pattern-avoiding stabilized-interval-free permutations, arXiv:2306.03155 [math.CO], 2023.
Johann Cigler, Some remarks and conjectures related to lattice paths in strips along the x-axis, arXiv:1501.04750 [math.CO], 2015-2016.
Johann Cigler, Recurrences for certain sequences of binomial sums in terms of (generalized) Fibonacci and Lucas polynomials, arXiv:2212.02118 [math.NT], 2022.
Sean A. Irvine, Walks on Graphs.
Index entries for linear recurrences with constant coefficients, signature (0,3).

Cf. A038754 (essentially the same sequence).
Cf. A008277, A052551, A057427, A057960.
Cf. A123149, A140735, A293181, A304973.
Also row sums of triangle in A169623.
Column 3 of A305749.
Cf. A124302 (oriented), A001998 (unoriented), A107767 (chiral), for rows, varying offsets.
Cf. A002076 (oriented), A056353 (unoriented), A320743 (chiral), for cycles.

I:=[1,1,2]; [n le 3 select I[n] else 3*Self(n-2): n in [1..40]]; // Bruno Berselli, Mar 19 2013

Join[{1}, RecurrenceTable[{a[1]==1, a[2]==2, a[n]==3 a[n-2]}, a, {n, 40}]] (* Bruno Berselli, Mar 19 2013 *)
CoefficientList[Series[(1+x-x^2)/(1-3*x^2), {x,0,50}], x] (* G. C. Greubel, Apr 14 2017 *)
Table[If[EvenQ[n], StirlingS2[(n+6)/2,3] - 4 StirlingS2[(n+4)/2,3] + 5 StirlingS2[(n+2)/2,3] - 2 StirlingS2[n/2,3], StirlingS2[(n+5)/2,3] - 3 StirlingS2[(n+3)/2,3] + 2 StirlingS2[(n+1)/2,3]], {n,0,40}] (* Robert A. Russell, Oct 21 2018 *)
Join[{1},Table[If[EvenQ[n], 2 3^((n-2)/2), 3^((n-1)/2)],{n,40}]] (* Robert A. Russell, Oct 28 2018 *)

makelist(if n=0 then 1 else (1+mod(n-1,2))*3^floor((n-1)/2), n, 0, 40); /* Bruno Berselli, Mar 19 2013 */

{a(n) = if( n<1, n==0, n--; (n%2 + 1) * 3^(n \ 2))}

my(x='x+O('x^50)); Vec((1+x-x^2)/(1-3*x^2)) \\ G. C. Greubel, Apr 14 2017

def A182522(n): return (3 -(3-2*sqrt(3))*((n+1)%2))*3^((n-3)/2) + int(n==0)/3
[A182522(n) for n in range(41)] # G. C. Greubel, Jul 17 2023

G.f.: (1 + x - x^2) / (1 - 3*x^2).
Expansion of 1 / (1 - x / (1 - x / (1 + x / (1 + x)))) in powers of x.
a(n+1) = A038754(n).
a(n) = Sum_{k=0..n} A123149(n,k). - Philippe Deléham, May 04 2012
a(n) = (3-(1+(-1)^n)*(3-2*sqrt(3))/2)*sqrt(3)^(n-3) for n>0, a(0)=1. - Bruno Berselli, Mar 19 2013
a(0) = 1, a(1) = 1, a(n) = a(n-1) + a(n-2) if n is odd, and a(n) = a(n-1) + a(n-2) + a(n-3) if n is even. - Jon Perry, Mar 19 2013
For odd n = 2m-1, a(2m-1) = T(m,1)+T(m,2)+T(m,3) for triangle T(m,k) of A140735; for even n = 2m, a(2m) = T(m,1)+T(m,2)+T(m,3) for triangle T(m,k) of A293181. - Robert A. Russell, Mar 10 2018
From Robert A. Russell, Oct 21 2018: (Start)
a(2m) = S2(m+3,3) - 4*S2(m+2,3) + 5*S2(m+1,3) - 2*S2(m,3).
a(2m-1) = S2(m+2,3) - 3*S2(m+1,3) + 2*S2(m,3), where S2(n,k) is the Stirling subset number A008277.
a(n) = 2*A001998(n-1) - A124302(n) = A124302(n) - 2*A107767(n-1) = A001998(n-1) - A107767(n-1).
a(n) = 2*A056353(n) - A002076(n) = A002076(n) - 2*A320743(n) = A056353(n) - A320743(n).
a(n) = A057427(n) + A052551(n-2) + A304973(n). (End)

Edited by Bruno Berselli, Mar 19 2013

Definition simplified by N. J. A. Sloane, Nov 23 2017

cp's OEIS Frontend

A169623 Generalized Pascal triangle read by rows: T(n,0) = T(0,n) = 1 for n >= 0, T(n,k) = 0 for k < 0 or k > n; otherwise T(n,k) = T(n-2,k-2) + T(n-2,k-1) + T(n-2,k) for 1 <= k <= n-1.

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A182522 a(0) = 1; thereafter a(2n + 1) = 3^n, a(2n + 2) = 2 * 3^n.

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cp's OEIS Frontend

A169623 Generalized Pascal triangle read by rows: T(n,0) = T(0,n) = 1 for n >= 0, T(n,k) = 0 for k < 0 or k > n; otherwise T(n,k) = T(n-2,k-2) + T(n-2,k-1) + T(n-2,k) for 1 <= k <= n-1.

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Examples

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Crossrefs

Programs

Maple

Mathematica

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A182522 a(0) = 1; thereafter a(2*n + 1) = 3^n, a(2*n + 2) = 2 * 3^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

PARI

SageMath

Formula

Extensions

A182522 a(0) = 1; thereafter a(2n + 1) = 3^n, a(2n + 2) = 2 * 3^n.