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

0,2

Comments

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

Links

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

Formula

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). - R. J. Mathar, Aug 09 2013

Example

Example:
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

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]

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).

Sequence in context: A256693 A255458 A256135 * A223186 A071340 A162962

Adjacent sequences: A227073 A227074 A227075 * A227077 A227078 A227079

Keyword

nonn,tabl

Author

T. D. Noe, Aug 06 2013

Status

approved