A125233 -id:A125233 - cp's OEIS frontend

A126277 Triangle generated from Eulerian numbers.

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

A126284 a(n) = 52^n - 4n - 5.

Original entry on oeis.org

1, 7, 23, 59, 135, 291, 607, 1243, 2519, 5075, 10191, 20427, 40903, 81859, 163775, 327611, 655287, 1310643, 2621359, 5242795, 10485671, 20971427, 41942943, 83885979, 167772055, 335544211, 671088527, 1342177163, 2684354439

Offset: 1

Gary W. Adamson, Dec 24 2006

Row sums of A125233.
A triangle with left and right borders being the odd numbers 1,3,5,7,... will give the same partial sums for the sum of its rows. - J. M. Bergot, Sep 29 2012
The triangle in the above comment is constructed the same way as Pascal's triangle, i.e., C(n, k) = C(n-1, k) + C(n-1, k-1). - Michael B. Porter, Oct 03 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A000384, A125233, A126277.

List([1..30],n->5*2^n-4*n-5); # Muniru A Asiru, Oct 24 2018

[5*2^n - 4*n - 5: n in [1..30]]; // G. C. Greubel, Oct 23 2018

A126284:=n->5*2^n-4*n-5; seq(A126284(n), n=1..50); # Wesley Ivan Hurt, Mar 27 2014

CoefficientList[Series[(1 + 3 x)/(1 - 4 x + 5 x^2 - 2 x^3), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 28 2014 *)

a(n)=5<Charles R Greathouse IV, Oct 03 2012

a(1) = 1; a(2) = 7; a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3), n > 2.
The 6th diagonal from the right of A126277.
G.f.: x*(1+3*x)/(1-4*x+5*x^2-2*x^3). - Colin Barker, Feb 12 2012
E.g.f.: 5*exp(2*x) - (5+4*x)*exp(x). - G. C. Greubel, Oct 23 2018

More terms from Vladimir Joseph Stephan Orlovsky, Oct 18 2008

New definition from R. J. Mathar, Sep 29 2012

A135856 A007318 * a bidiagonal matrix with all 1's in the main diagonal and all 4's in the subdiagonal.

Original entry on oeis.org

1, 5, 1, 9, 6, 1, 13, 15, 7, 1, 17, 28, 22, 8, 1, 21, 45, 50, 30, 9, 1, 25, 66, 95, 80, 39, 10, 1, 29, 91, 161, 175, 119, 49, 11, 1, 33, 120, 252, 336, 294, 168, 60, 12, 1, 37, 153, 372, 588, 630, 462, 228, 72, 13, 1

Offset: 1

Gary W. Adamson, Dec 01 2007

Row sums = A048487.

When the first column is removed from this triangle, the result is A125233. - Georg Fischer, Jul 26 2023

			First few rows of the triangle:
   1;
   5,  1;
   9,  6,  1;
  13, 15,  7,  1;
  17, 28, 22,  8,  1;
  21, 45, 50, 30,  9,  1;
  25, 66, 95, 80, 39, 10,  1;
  ...

Cf. A048487, A125233.

Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal and all 4's in the subdiagonal (by columns, (1, 4, 0, 0, 0, ...) in every column.

cp's OEIS Frontend

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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Maple

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A126284 a(n) = 5*2^n - 4*n - 5.

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GAP

Magma

Maple

Mathematica

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A135856 A007318 * a bidiagonal matrix with all 1's in the main diagonal and all 4's in the subdiagonal.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A126284 a(n) = 52^n - 4n - 5.