A227076 A triangle formed like Pascal's triangle, but with 5^n on the borders instead of 1.

1, 5, 5, 25, 10, 25, 125, 35, 35, 125, 625, 160, 70, 160, 625, 3125, 785, 230, 230, 785, 3125, 15625, 3910, 1015, 460, 1015, 3910, 15625, 78125, 19535, 4925, 1475, 1475, 4925, 19535, 78125, 390625, 97660, 24460, 6400, 2950, 6400, 24460, 97660, 390625

Offset: 0

Author

T. D. Noe, Aug 06 2013

Keywords

• nonn
• tabl

Comments

All rows except the zeroth are divisible by 5. Is there a closed-form formula for these numbers, as there is for binomial coefficients?

Examples

			Triangle begins as:
       1;
       5,     5;
      25,    10,    25;
     125,    35,    35,  125;
     625,   160,    70,  160,  625;
    3125,   785,   230,  230,  785, 3125;
   15625,  3910,  1015,  460, 1015, 3910, 15625;
   78125, 19535,  4925, 1475, 1475, 4925, 19535, 78125;
  390625, 97660, 24460, 6400, 2950, 6400, 24460, 97660, 390625;
		

Links

• T. D. Noe, Rows n = 0..50 of triangle, flattened

Crossrefs

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).

Cf. A051601 (n on the borders), A137688 (2^n on borders).

Cf. A083585 (row sums: (8*5^n - 5*2^n)/3), A227074 (4^n edges), A227075 (3^n edges).

Cf. A000351.

Programs

Magma

function T(n,k) // T = A227076
  if k eq 0 or k eq n then return 5^n;
  else return T(n-1,k) + T(n-1,k-1);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 10 2025

Maple

A227076 := proc(n,k)
    if k = 0 or k = n then
        5^n ;
    elif k < 0 or k > n then
        0;
    else
        procname(n-1,k)+procname(n-1,k-1) ;
    end if;
end proc: # R. J. Mathar, Aug 09 2013

Mathematica

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 5^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

Python

from sage.all import *
@CachedFunction
def T(n,k): # T = A227076
    if k==0 or k==n: return pow(5,n)
    else: return T(n-1,k) + T(n-1,k-1)
print(flatten([[T(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 10 2025

Formula

From R. J. Mathar, Aug 09 2013: (Start)

T(n,0) = 5^n.

T(n,1) = 5*A047850(n-1).

T(n,2) = 5*(5^n/80 + 3*n/4 + 51/16).

T(n,3) = 5*(5^n/320 + 45*n/16 + 3*n^2/8 + 819/64). (End)

Sum_{k=0..n} (-1)^k*T(n, k) = 20*(1+(-1)^n)*A009969(floor((n-1)/2)) - (3/5)*[n = 0]. - G. C. Greubel, Jan 10 2025