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

0,2

Comments

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.

References

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

Links

Table of n, a(n) for n=0..65.

Formula

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

Example

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).

Maple

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

Mathematica

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 *)

Crossrefs

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

Sequence in context: A050309 A103217 A136273 * A139727 A257468 A278867

Adjacent sequences: A125230 A125231 A125232 * A125234 A125235 A125236

Keyword

nonn,tabl

Author

Gary W. Adamson, Nov 24 2006

Extensions

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

Status

approved