A125232 -id:A125232 - cp's OEIS frontend

A125233 Triangle T(n,k) read by rows, the (n-k)-th term of the k times repeated partial sum of the hexagonal numbers, 0 <= k < n, 0 < n.

1, 6, 1, 15, 7, 1, 28, 22, 8, 1, 45, 50, 30, 9, 1, 66, 95, 80, 39, 10, 1, 91, 161, 175, 119, 49, 11, 1, 120, 252, 336, 294, 168, 60, 12, 1, 153, 372, 588, 630, 462, 228, 72, 13, 1, 190, 525, 960, 1218, 1092, 690, 300, 85, 14, 1, 231, 715, 1485, 2178, 2310, 1782, 990, 385, 99, 15, 1

Offset: 0

Gary W. Adamson, Nov 24 2006

Left border = A000384, hexagonal numbers. The following columns are A002412, A002417, A034263, A051947, ...
Row sums = (1, 7, 23, 59, 135, 291, ...) = A126284.
A125232 is the analogous triangle for the pentagonal numbers.

			First few rows of the triangle:
   1;
   6,   1;
  15,   7,   1;
  28,  22,   8,   1;
  45,  50,  30,   9,  1;
  66,  95,  80,  39, 10,  1;
  91, 161, 175, 119, 49, 11, 1;
  ...
Example: (5,3) = 80 = 30 + 50 = (4,3) + (4,2).

Albert H. Beiler, "Recreations in the Theory of Numbers", Dover, 1964, p. 189.

Cf. A000384, A002412, A002417, A034263, A051947, A125232.

A000384Psum:= proc(n,k) coeftayl( x*(1+3*x)/(1-x)^(3+k),x=0,n) ; end: A125233 := proc(n,k) A000384Psum(n-k,k) ; end: for n from 1 to 15 do for k from 0 to n -1 do printf("%d,",A125233(n,k)) ; od: od: # R. J. Mathar, May 03 2008

T[n_, k_] := T[n, k] = Which[k == 0, n (2 n - 1), 1 <= k < n, T[n - 1, k] + T[n - 1, k - 1], True, 0];
Table[T[n, k], {n, 1, 11}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, Sep 14 2023, after R. J. Mathar *)

T(n,0)=A000384(n). T(n,k) = T(n-1,k) + T(n-1,k-1), k>1. - R. J. Mathar, May 03 2008

Edited and extended by R. J. Mathar, May 03 2008, and M. F. Hasler, Sep 29 2012

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, ...).

A125234 Triangle T(n,k) read by rows: the k-th column contains the k-fold iterated partial sum of A000566.

Original entry on oeis.org

1, 7, 1, 18, 8, 1, 34, 26, 9, 1, 55, 60, 35, 10, 1, 81, 115, 95, 45, 11, 1, 112, 196, 210, 140, 56, 12, 1, 148, 308, 406, 350, 196, 68, 13, 1, 189, 456, 714, 756, 546, 264, 81, 14, 1, 235, 645, 1170, 1470, 1302, 810, 345, 95, 15, 1, 286, 880, 1815, 2640, 2772, 2112, 1155, 440, 110, 16, 1

Offset: 1

Gary W. Adamson, Nov 24 2006

The leftmost column contains the heptagonal numbers A000566.
The adjacent columns to the right are A002413, A002418, A027800, A051946, A050484.
Row sums = 1, 8, 27, 70, 161, 348, 727, ... = 6*(2^n-1)-5*n.

			First few rows of the triangle are:
  1;
  7, 1;
  18, 8, 1;
  34, 26, 9, 1;
  55, 60, 35, 10, 1;
  81, 115, 95, 45, 11, 1;
  112, 196, 210, 140, 56, 12, 1;
Example: T(6,2) = 95 = 35 + 60 = T(5,2) + T(5,1).

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, 1966, p. 189.

Cf. A000566, A002413, A002418, A027800, A051946, A050484.

Analogous triangles for the hexagonal and pentagonal numbers are A125233 and A125232.

A000566 := proc(n) n*(5*n-3)/2 ; end: A125234 := proc(n,k) if k = 0 then A000566(n); elif k>= n then 0 ; else procname(n-1,k-1)+procname(n-1,k) ; fi; end: seq(seq(A125234(n,k),k=0..n-1),n=1..16) ; # R. J. Mathar, Sep 09 2009

A000566[n_] := n(5n-3)/2;
T[n_, k_] := Which[k == 0, A000566[n], k >= n, 0, True, T[n-1, k-1] + T[n-1, k] ];
Table[Table[T[n, k], {k, 0, n-1}], {n, 1, 11}] // Flatten (* Jean-François Alcover, Oct 26 2023, after R. J. Mathar *)

T(n,0) = A000566(n). T(n,k) = T(n-1,k) + T(n-1,k-1), k>0.

Edited and extended by R. J. Mathar, Sep 09 2009

A095264 a(n) = 2^(n+2) - 3*n - 4.

Original entry on oeis.org

1, 6, 19, 48, 109, 234, 487, 996, 2017, 4062, 8155, 16344, 32725, 65490, 131023, 262092, 524233, 1048518, 2097091, 4194240, 8388541, 16777146, 33554359, 67108788, 134217649, 268435374, 536870827, 1073741736, 2147483557, 4294967202, 8589934495, 17179869084, 34359738265

Offset: 1

Gary W. Adamson, May 31 2004

A sequence derived from a 3rd-order matrix generator.
The number of positive 3-strand braids of degree at most n. - R. J. Mathar, May 04 2006
Define a triangle T by T(n,n) = n*(n+1)/2, T(n,1) = n*(n-1) + 1, and T(r,c) = T(r-1,c-1) + T(r-1,c). Its rows are 1; 3,3; 7,6,6; 13,13,12,10; 21,26,25,22,15; etc. The sum of the terms in the n-th row is a(n). - J. M. Bergot, May 03 2013

			a(5) = 109 = 2^7 - 3*5 - 4.
a(5) = 109 since M^5 * [1 0 0] = [1 5 109].
a(7) = 487 = 4*234 - 5*109 + 2*48.

P. Dehornoy, Combinatorics of normal sequences of braids, arXiv:math/0511114 [math.CO], 2005.
Shishuo Fu, Zhicong Lin, and Yaling Wang, Refined Wilf-equivalences by Comtet statistics, arXiv:2009.04269 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A036563, A125232, A126277.

a[n_] := (MatrixPower[{{1, 0, 0}, {1, 1, 0}, {1, 3, 2}}, n].{{1}, {0}, {0}})[[3, 1]]; Table[ a[n], {n, 30}] (* Robert G. Wilson v, Jun 05 2004 *)
Table[2^(n+2)-3n-4,{n,40}] (* or *) LinearRecurrence[{4,-5,2},{1,6,19},40] (* Harvey P. Dale, Sep 24 2021 *)

Let M = [1 0 0 / 1 1 0 / 1 3 2], then M^n * [1 0 0] = [1 n a(n)]. The characteristic polynomial of M is x^3 - 4*x^2 + 5*x - 2.
a(n+3) = 4*a(n+2) - 5*a(n+1) + 2*a(n).
a(n) = Sum_{i=2..n+1} A036563(i) [A036563 is 2^i-3]. - Gerald McGarvey, Jun 28 2004
Row sums of A125232; 5th diagonal from the right of A126277; binomial transform of [1, 5, 8, 8, 8, ...]. - Gary W. Adamson, Dec 23 2006
a(n) = 2*a(n-1) + (3n-2). - Gary W. Adamson, Sep 30 2007
G.f.: -x*(1+2*x)/((2*x-1)*(x-1)^2). - R. J. Mathar, Nov 18 2007
E.g.f.: exp(x)*(4*exp(x) - 3*x - 4). - Elmo R. Oliveira, Apr 01 2025

Edited, corrected and extended by Robert G. Wilson v, Jun 05 2004

More terms from Elmo R. Oliveira, Apr 01 2025

A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

Original entry on oeis.org

1, 4, 1, 7, 5, 1, 10, 12, 6, 1, 13, 22, 18, 7, 1, 16, 35, 40, 25, 8, 1, 19, 51, 75, 65, 33, 9, 1, 22, 70, 126, 140, 98, 42, 10, 1, 25, 92, 196, 266, 238, 140, 52, 11, 1, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1

Offset: 0

Gary W. Adamson, Dec 01 2007

A007318 * a bidiagonal matrix with all 1's in the main diagonal and all 3's in the subdiagonal.
Row sums give A036563(n+2), n >= 0.
From Wolfdieter Lang, Mar 23 2015: (Start)
This is the triangle of iterated partial sums of A016777. Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given).
This is therefore the Riordan triangle ((1+2*x)/(1-x)^2, x/(1-x)) with o.g.f. of the columns ((1+2*x)/(1-x)^2)*(x/(1-x))^k, k >= 0.
The column sequences are A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310, for k = 0..8.
The alternating row sums are A122553(n) = {1, repeat(3)}.
The Riordan A-sequence is A(y) = 1 + y (implying the Pascal triangle recurrence for k >= 1).
The Riordan Z-sequence is A256096, leading to a recurrence for T(n,0) given in the formula section. See the link "Sheffer a- and z-sequences" under A006232 also for Riordan A- and Z-sequences with references. (End)
When the first column (k = 0) is removed from this triangle, the result is A125232. - Georg Fischer, Jul 26 2023

			The triangle T(n, k) begins:
n\k  0   1   2    3    4    5    6   7   8  9 10 11
0:   1
1:   4   1
2:   7   5   1
3:  10  12   6    1
4:  13  22  18    7    1
5:  16  35  40   25    8    1
6:  19  51  75   65   33    9    1
7:  22  70 126  140   98   42   10   1
8:  25  92 196  266  238  140   52  11   1
9:  28 117 288  462  504  378  192  63  12  1
10: 31 145 405  750  966  882  570 255  75 13  1
11: 34 176 550 1155 1716 1848 1452 825 330 88 14  1
... reformatted and extended by _Wolfdieter Lang_, Mar 23 2015
From _Wolfdieter Lang_, Mar 23 2015: (Start)
T(3, 1) = T(2, 0) + T(2, 1) = 7 + 5 = 12 (Pascal, from the A-sequence given above).
T(4, 0) = 4*T(3, 0) - 9*T(3, 1) + 27*T(3, 2) - 81* T(3, 3) = 4*10 - 9*12 + 27*6 - 81*1 = 13, from the Z-sequence given above and in A256096.
T(4, 0) = 2*T(3, 0) - T(2, 0) = 2*10 - 7 = 13.
(End)

Columns k=0-8 give: A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310.
Cf. A036563, A125232.
Cf. A110813. - Philippe Deléham, Feb 08 2009

Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal and all 3's in the subdiagonal; i.e., by columns - every column = (1, 3, 0, 0, 0, ...).
T(n,k) = (3n-2k+1)*binomial(n+1,k+1)/(n+1). - Philippe Deléham, Feb 08 2009
From Wolfdieter Lang, Mar 23 2015: (Start)
O.g.f. for row polynomials: (1 + 2*z)/((1- z*(1 + x))*(1 - z)) (see the Riordan property from the comment).
O.g.f. for column k (without leading zeros): (1 + 2*x)/(1-x)^(2+k), k >= 0, (Riordan property).
T(n, k) = T(n-1, k-1) + T(n-1, k) for k >= 1. From the Riordan A-sequence given above in a comment.
T(n, 0) = Sum_{j=0..n} Z(j)*T(n-1, j), for n >= 1, from the Riordan Z-sequence A256096 mentioned above in a comment. Of course, T(n, 0) = 2*T(n-1, 0) - T(n-2, 0) for n >= 2 (see A016777).
(End)

Edited. Offset is 0 from the old name and the Philippe Deléham formula. New name, old name as first comment. - Wolfdieter Lang, Mar 23 2015

A125235 Triangle with the partial column sums of the octagonal numbers.

Original entry on oeis.org

1, 8, 1, 21, 9, 1, 40, 30, 10, 1, 65, 70, 40, 11, 1, 96, 135, 110, 51, 12, 1, 133, 231, 245, 161, 63, 13, 1, 176, 364, 476, 406, 224, 76, 14, 1, 225, 540, 840, 882, 630, 300, 90, 15, 1, 280, 765, 1380, 1722, 1512, 930, 390, 105, 16, 1

Offset: 1

Gary W. Adamson, Nov 24 2006

"Partial column sums" means the octagonal numbers are the 1st column, the 2nd column are the partial sums of the 1st column, the 3rd column are the partial sums of the 2nd, etc.

Row sums are 1, 9, 31, 81, 187, 405, 847 = 7*(2^n-1) - 6*n. - R. J. Mathar, Sep 06 2011

			First few rows of the triangle:
   1;
   8,   1;
  21,   9,   1;
  40,  30,  10,   1;
  65,  70,  40,  11,   1;
  96, 135, 110,  51,  12,   1;
  ...

Albert H. Beiler, Recreations in the Theory of Numbers, Dover (1966), p. 189.

Cf. A000567, A002414, A002419, A051843, A027810, A125232 - A125234.

t(n, k) = if (n <0, 0, if (k==1, n*(3*n-2), if (k > 1, t(n-1,k-1) + t(n-1,k))));
tabl(nn) = {for (n = 1, nn, for (k = 1, n, print1(t(n, k), ", ");); print(););} \\ Michel Marcus, Mar 04 2014

T(n,1) = A000567(n).
T(n,k) = T(n-1,k-1) + T(n-1,k), k>1.
T(n,2) = A002414(n-1).
T(n,3) = A002419(n-2).
T(n,4) = A051843(n-4).
T(n,5) = A027810(n-6).

More terms from Michel Marcus, Mar 04 2014

cp's OEIS Frontend

A125233 Triangle T(n,k) read by rows, the (n-k)-th term of the k times repeated partial sum of the hexagonal numbers, 0 <= k < n, 0 < n.

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A126277 Triangle generated from Eulerian numbers.

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A125234 Triangle T(n,k) read by rows: the k-th column contains the k-fold iterated partial sum of A000566.

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A095264 a(n) = 2^(n+2) - 3*n - 4.

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A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

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A125235 Triangle with the partial column sums of the octagonal numbers.

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