A125165 -id:A125165 - cp's OEIS frontend

A125166 Triangle R(n,k), companion to A125165, left column n^3.

1, 8, 1, 27, 9, 1, 64, 36, 10, 1, 125, 100, 46, 11, 1, 216, 225, 146, 57, 12, 1, 343, 441, 371, 203, 69, 13, 1, 512, 784, 812, 574, 272, 82, 14, 1, 729, 1296, 1596, 1386, 846, 354, 96, 15, 1, 1000, 2025, 2892, 2982, 2232, 1200, 450, 111, 16, 1

Offset: 0

Gary W. Adamson, Nov 22 2006

Riordan array ((1 + 4*x + x^2)/(1 - x)^4, x/(1 - x)). - Philippe Deléham, Dec 09 2013

			With other offset (k >= 1) from first formula: R(5,3) = 146 = R(4,3) + R(4,2) = 46 + 100.
The Riordan triangle R begins:
  n\k|     0    1    2    3    4    5   6   7   8  9
  --------------------------------------------------
   0 |     1
   1 |     8    1
   2 |    27    9    1
   3 |    64   36   10    1
   4 |   125  100   46   11    1
   5 |   216  225  146   57   12    1
   6 |   343  441  371  203   69   13   1
   7 |   512  784  812  574  272   82  14   1
   8 |   729 1296 1596 1386  846  354  96  15   1
   9 |  1000 2025 2892 2982 2232 1200 450 111  16  1
... refomatted and extended by _Wolfdieter Lang_, Mar 25 2025.

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).

Cf. A000578 (column 0), A000537 (column 1), A024166 (column 2), A101094 (column 3).
Cf. A257448 (row sums).
Cf. A008292, A028246, A097805, A125165, A050488, A382057.

A125166[n_, k_] := A125166[n, k] = Switch[k, 0, (n + 1)^3, n, 1, _, A125166[n - 1, k - 1] + A125166[n - 1, k]];
Table[A125166[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Apr 08 2025 *)

y = polygen(QQ, 'y')
x = y.parent()[['x']].gen()
gf = (1 + 4*x + x^2)/((x - 1)^3*(y*x + x - 1))
[list(u) for u in list(gf.O(11))]  # Peter Luschny, Apr 02 2025

Binomial transform of an infinite matrix M with diagonal D, subdiagonal (D-1)..., etc; as follows: (D) = (1,1,1...); (D-1) = (7,7,7...); (D-2) = (12,12,12...); (D-3) = (6,6,6...). Alternatively, given left border n^3: (1, 8, 27, 64...); for k>1, R(n,k) = R(n-1,k) + R(n-1,k-1).
From Wolfdieter Lang, Mar 27 2025: (Start)
Riordan triangle (see a comment above):
R(n, 0) = (n+1)^3, R(n, k) = R(n-1, k-1) + R(n-1, k), for k >= 1. (from the (finite) A-sequence {1, 1} with offset 0),
R(n, 0) = Sum_{k=0..n-1} Z(j)*R(n-1, k), for n >= 1, and R(0, 0) = 1, with the Riordan Z-sequence A382057. For Riordan A- and Z-sequences see the first W. Lang link in A006232.
R(n, k) = Sum{j=0..n} (j+1)^3*A097805(n-j, k) (convolution property).
R(n, k) = Sum{j=0..3} A028246(4, j+1)*binomial(n, k+j). (Proof for k=0 with a standard (n+1)^3  formula, and for k >= 1 using the Pascal type recurrence for the triangle.)
O.g.f. column k (with leading 0s): ((1 + 4*x + x^2)/(1 - x)^4)*(x/(1-x))^k. (Numerator polynomial from row 3 of the triangle A008292.)
O.g.f. row polynomials P(n, y) = Sum_{k=0..n} R(n, k)*y^k:
   G(y, x) = (1 + 4*x + x^2)/((1 - x)^3*(1 - (1+y)*x)). (End)

a(27) corrected by Georg Fischer, Feb 18 2020

A050488 a(n) = 3(2^n-1) - 2n.

Original entry on oeis.org

0, 1, 5, 15, 37, 83, 177, 367, 749, 1515, 3049, 6119, 12261, 24547, 49121, 98271, 196573, 393179, 786393, 1572823, 3145685, 6291411, 12582865, 25165775, 50331597, 100663243, 201326537, 402653127, 805306309, 1610612675, 3221225409, 6442450879, 12884901821, 25769803707

Offset: 0

James Sellers, Dec 26 1999

Number of words of length n+1 where first element is from {0,1,2}, other elements are from {0,1} and sequence does not decrease (for n=2 there are 3*2^2 sequences, but 000, 100, 110, 111, 200, 210, 211 decrease, so a(2) = 12-7 = 5).
Number of subgroups of C_(2^n) X C_(2^n) (see A060724).
Starting with 1 = row sums of triangle A054582. - Gary W. Adamson, Jun 23 2008
Starting with "1" equals the eigensequence of a triangle with integer squares (1, 4, 9, 16, ...) as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010
(1 + 2x + 2x^2 + 2x^3 + ...)*(1 + 3x + 7x^2 + 15x^3 + ...) = (1 + 5x + 15x^2 + 37x^3 + ...). - Gary W. Adamson, Mar 14 2012
The partial sums of A033484. - J. M. Bergot, Oct 03 2012
Binomial transform is 0, 1, 7, 33, ... (shifted A066810); inverse binomial transform is 0, 1, 3, 3, ... (3 repeated). - R. J. Mathar, Oct 05 2012
Define a triangle by T(n,0) = n*(n+1) + 1, T(n,n) = n + 1, and T(r,c) = T(r-1,c-1) + T(r-1,c) otherwise; then a(n+1) is the sum of the terms of row n. - J. M. Bergot, Mar 30 2013
Starting with "1" are also the antidiagonal sums of the array formed by partial sums of integer squares (1, 4, 9, 16, ...). - Luciano Ancora, Apr 24 2015
Sums of 2 adjacent terms in diagonal k=2 of Eulerian triangle A008292. I.e., T(n,2)+T(n-1,2) for n > 0. Also, 4th NW-SE diagonal of A126277. In other words, a(n) = A000295(n) + A000295(n+1). - Gregory Gerard Wojnar, Sep 30 2018

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Adam Buck, Jennifer Elder, Azia A. Figueroa, Pamela E. Harris, Kimberly Harry, and Anthony Simpson, Flattened Stirling Permutations, arXiv:2306.13034 [math.CO], 2023. See p. 14.
Tamas Lengyel, On p-adic properties of the Stirling numbers of the first kind, Journal of Number Theory, 148 (2015) 73-94.
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

A050487(2^m-1).
Equals (1/2) A051667.
Cf. A000079, A000225, A000295, A005408, A008292, A033484, A054852, A060724, A066810, A125165, A126277, A142963, A156919, A156920, A156925.

List([0..30],n->3*(2^n-1)-2*n); # Muniru A Asiru, Oct 26 2018

a050488 n = sum $ zipWith (*) a000079_list (reverse $ take n a005408_list)
-- Reinhard Zumkeller, Jul 24 2015

[3*(2^n-1) - 2*n: n in [0..30]]; // G. C. Greubel, Oct 23 2018

seq(coeff(series(x*(x+1)/((1-x)^2*(1-2*x)),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Oct 26 2018

Table[3(2^n-1)-2n,{n,0,30}] (* or *) LinearRecurrence[{4,-5,2}, {0,1,5}, 40] (* Harvey P. Dale, Apr 09 2018 *)

a(n)=3*(2^n-1)-2*n \\ Charles R Greathouse IV, Sep 24 2015

for n in range(0, 30): print(3*(2**n-1) - 2*n, end=', ') # Stefano Spezia, Oct 27 2018

Row sums of A125165: (1, 5, 15, 37, ...). Binomial transform of [1, 4, 6, 6, 6, ...] = [1, 5, 15, 37, ...]. 4th diagonal from the right of A126777 = (1, 5, 15, ...). - Gary W. Adamson, Dec 23 2006
a(n) = 2*a(n-1) + (2n-1). - Gary W. Adamson, Sep 30 2007
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n+1) = A156920(n+1,1).
a(n+1) = A156919(n+1,1)/2^n.
a(n+1) = A142963(n+2,1)/2.
a(n) = 4a(n-1) - 5a(n-2) + 2a(n-3) for n>2 with a(0) = 0, a(1) = 1, a(2) = 5.
G.f.: z*(1+z)/((1-z)^2*(1-2*z)).
(End)
a(n) = 2*n + 2*a(n-1) - 1 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n+1) = Sum_{k=0..n} A000079(k) * A005408(n-k), convolution of the powers of 2 with the odd numbers. - Reinhard Zumkeller, Mar 08 2012
E.g.f.: exp(x)*(3*exp(x) - 2*x - 3). - Stefano Spezia, May 15 2023

A126277 Triangle generated from Eulerian numbers.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 7, 4, 1, 5, 11, 15, 5, 1, 6, 15, 26, 31, 6, 1, 7, 19, 37, 57, 63, 7, 1, 8, 23, 48, 83, 120, 127, 8, 1, 9, 27, 59, 109, 177, 247, 255, 9, 1, 10, 31, 70, 135, 234, 367, 502, 511, 10

Offset: 1

Gary w. Adamson, Dec 23 2006

N-th diagonal starting from the right = binomial transform of [1, N, q, q, q, ...) where q = 2*N - 2. Given the infinite set of triangles "T" composed of partial column sums of the polygonal numbers, the N-th diagonal starting from the right = row sums of triangle "T": (T=3 = A104712; T=4 = A125165; T=5 = A125232; T=6 = A125233; T=7 = A125234, T=8 = A125235; and so on). For example, 3rd diagonal from the right = the offset Eulerian numbers, (1, 4, 11, 26, 57, 120, ...) = row sums of Triangle A104712 having partial column sums of the triangular numbers: 1; 3, 1; 6, 4, 1; 10, 10, 5, 1; 15, 20, 15, 6, 1; ... Row sums = A124671: (1, 3, 7, 16, 37, 85, 191, ...).

			First few rows of the triangle:
  1;
  1,  2;
  1,  3,  3;
  1,  4,  7,  4;
  1,  5, 11, 15,   5;
  1,  6, 15, 26,  31,   6;
  1,  7, 19, 37,  57,  63,   7;
  1,  8, 23, 48,  83, 120, 127,   8;
  1,  9, 27, 59, 109, 177, 247, 255,   9;
  1, 10, 31, 70, 135, 234, 367, 502, 511, 10;
  ...
T(7,4) = 37 = A000295(4) + T(6,4) = 11 + 26.

G. C. Greubel, Rows n=1..100 of triangle, flattened

Cf. A126268, A000295, A104712, A125165, A104712, A125232 - A125235.

T[n_,1]:=1; T[n_,n_]:=n; T[n_,k_]:= T[n-1,k] + 2^k - k - 1; Table[T[n,k], {n,1,15}, {k,1,n}]//Flatten (* G. C. Greubel, Oct 23 2018 *)

{T(n,k) = if(k==1, 1, if(k==n, n, 2^k - k - 1 + T(n-1,k)))};
for(n=1,10, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 23 2018

Given right border = (1,2,3,...), T(n,k) = A000295(k) + T(n-1,k); where A000295 = the Eulerian numbers starting (0, 1, 4, 11, 26, 57, ...).

A183153 T(n,k) is the number of order-preserving partial isometries of an n-chain of height k (height of alpha = |Im(alpha)|).

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 9, 5, 1, 1, 16, 14, 6, 1, 1, 25, 30, 20, 7, 1, 1, 36, 55, 50, 27, 8, 1, 1, 49, 91, 105, 77, 35, 9, 1, 1, 64, 140, 196, 182, 112, 44, 10, 1, 1, 81, 204, 336, 378, 294, 156, 54, 11, 1, 1, 100, 285, 540, 714, 672, 450, 210, 65, 12, 1, 1, 121, 385, 825, 1254, 1386, 1122, 660

Offset: 0

Abdullahi Umar, Dec 27 2010

The matrix inverse starts
1;
-1,1;
3,-4,1;
-7,11,-5,1;
15,-26,16,-6,1;
-31,57,-42,22,-7,1;
63,-120,99,-64,29,-8,1;
-127,247,-219,163,-93,37,-9,1;
255,-502,466,-382,256,-130,46,-10,1;
...perhaps related to A054143. - R. J. Mathar, Mar 29 2013

			T(3,2)=5 because there are exactly 5 order-preserving partial isometries (on a 3-chain) of height 2, namely: (1,2)-->(1,2); (1,2)-->(2,3); (2,3)-->(1,2); (2,3)-->(2,3); (1,3)-->(1,3), the mappings are coordinate-wise.
Triangle begins as:
1;
1, 1;
1, 4, 1;
1, 9, 5, 1;
1, 16, 14, 6, 1;
1, 25, 30, 20, 7, 1;
1, 36, 55, 50, 27, 8, 1;
1, 49, 91, 105, 77, 35, 9, 1;

R. Kehinde, A. Umar, On the semigroup of partial isometries of a finite chain, arXiv:1101.0049

Cf. A097813 (row sums), A125165, A121306, A029653.

T(n,k)=if(k==0,1, (2*n-k+1)*binomial(n,k)/(k+1));
for(n=0,17,for(k=0,n,print1(T(n,k),", ")))

T(n,0)=1. T(n,k)=(2*n-k+1)*C(n,k)/(k+1) if k>0.

A233295 Riordan array ((1+x)/(1-x)^3, 2*x/(1-x)).

Original entry on oeis.org

1, 4, 2, 9, 10, 4, 16, 28, 24, 8, 25, 60, 80, 56, 16, 36, 110, 200, 216, 128, 32, 49, 182, 420, 616, 560, 288, 64, 64, 280, 784, 1456, 1792, 1408, 640, 128, 81, 408, 1344, 3024, 4704, 4992, 3456, 1408, 256, 100, 570, 2160, 5712, 10752, 14400, 13440, 8320, 3072, 512

Offset: 0

Philippe Deléham, Dec 07 2013

Subtriangle of the triangle in A208532.
Row sums are A060188(n+2).
Diagonal sums are A000295(n+2)=A125128(n+1)=A130103(n+2).

			Triangle begins :
1
4, 2
9, 10, 4
16, 28, 24, 8
25, 60, 80, 56, 16
36, 110, 200, 216, 128, 32
49, 182, 420, 616, 560, 288, 64
64, 280, 784, 1456, 1792, 1408, 640, 128
81, 408, 1344, 3024, 4704, 4992, 3456, 1408, 256
100, 570, 2160, 5712, 10752, 14400, 13440, 8320, 3072, 512

Cf. Columns: A000290, A006331, A112742.

Cf. Diagonal: A000079.

G.f. for the column k: 2^k*(1+x)/(1-x)^(k+3).
T(n,k) = 2^k*(binomial(n,k)+3*binomial(n,k+1)+2*binomial(n,k+2)), 0<=k<=n.
T(n,0) = 2*T(n-1,0)-T(n-2,0)+2, T(n,k)=2*T(n-1,k)+2*T(n-1,k-1)-2*T(n-2,k-1)-T(n-2,k) for k>=1, T(0,0)=1, T(1,0)=4, T(1,1)=2, T(n,k)=0 if k<0 or if k>n.
Sum_{k=0..n} T(n,k) = A060188(n+2).
Sum_{k=0..n} T(n,k)*(-1)^k = n+1.
T(n,k) = 2*sum_{j=1..n-k+1} T(n-j,k-1).
T(n,k) = 2^k*A125165(n,k).
T(n,n) = 2^n=A000079(n).
T(n,0) = (n+1)^2=A000290(n+1).
exp(2*x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(2*x)*(16 + 28*x + 24*x^2/2! + 8*x^3/3!) = 16 + 60*x + 200*x^2/2! + 616*x^3/3! + 1792*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), 2*x/(1 - x) ). Cf. A125165. - Peter Bala, Dec 21 2014

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A125166 Triangle R(n,k), companion to A125165, left column n^3.

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A050488 a(n) = 3*(2^n-1) - 2*n.

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A126277 Triangle generated from Eulerian numbers.

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A183153 T(n,k) is the number of order-preserving partial isometries of an n-chain of height k (height of alpha = |Im(alpha)|).

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A233295 Riordan array ((1+x)/(1-x)^3, 2*x/(1-x)).

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A050488 a(n) = 3(2^n-1) - 2n.