A283960 a(n) = (Sum_{j=1..h-1} a(n-j) + a(n-1)*a(n-h+1))/a(n-h) with a(1), ..., a(h)=1, where h = 6.

1, 1, 1, 1, 1, 1, 6, 16, 41, 106, 276, 2101, 6026, 15976, 41901, 109726, 835906, 2397991, 6358066, 16676206, 43670551, 332688201, 954394051, 2530493951, 6637087801, 17380769451, 132409067806, 379846433966, 1007130234091, 2641544268306, 6917502570826, 52698476298301

Offset

1,7

Links

Seiichi Manyama, Table of n, a(n) for n = 1..1929

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,399,0,0,0,0,-399,0,0,0,0,1).

Formula

a(5*k-2) = 3*a(5*k-3) - a(5*k-4) - 1,

a(5*k-1) = 3*a(5*k-2) - a(5*k-3) - 1,

a(5*k) = 3*a(5*k-1) - a(5*k-2) - 1,

a(5*k+1) = 3*a(5*k) - a(5*k-1) - 1,

a(5*k+2) = 8*a(5*k+1) - a(5*k) - 1.

From Colin Barker, Nov 03 2020: (Start)

G.f.: x*(1 + x + x^2 + x^3 + x^4 - 398*x^5 - 393*x^6 - 383*x^7 - 358*x^8 - 293*x^9 + 276*x^10 + 106*x^11 + 41*x^12 + 16*x^13 + 6*x^14) / ((1 - x)*(1 + x + x^2 + x^3 + x^4)*(1 - 398*x^5 + x^10)).

a(n) = 399*a(n-5) - 399*a(n-10) + a(n-15) for n>15.

(End)

Mathematica

a[n_]:= If[n<7, 1, (Sum[a[n-j] , {j, 5}] +  a[n - 1] a[n - 5])/a[n - 6]]; Table[a[n], {n, 30}] (* Indranil Ghosh, Mar 18 2017 *)

Prog

(PARI) a(n) = if(n<7, 1, (sum(j=1, 5, a(n - j)) + a(n - 1)*a(n - 5))/a(n - 6));
for(n=1, 30, print1(a(n), ", ")) \\ Indranil Ghosh, Mar 18 2017

Crossrefs

Cf. A072881 (h=3), A283958 (h=4), A283959 (h=5), this sequence (h=6).

Sequence in context: A261819 A347642 A073570 * A283330 A263325 A107614

Adjacent sequences: A283957 A283958 A283959 * A283961 A283962 A283963

Keyword

nonn

Author

Seiichi Manyama, Mar 18 2017

Status

approved