OFFSET
1,4
COMMENTS
Compare with A066910.
More generally, define a sequence {a(n, s) : n >= 1} with starting parameter s by a(n, s) = (Sum_{k = 1..n-1} k*a(k, s)) (mod n) with a(1, s) = s. The sequence {a(n, s)} is conjectured to be one of 3 types as illustrated by the following examples for s in [1..100].
1) It is easy to verify that the sequence {a(n, 8)} = {8, 0, 2, 2, 2, 2, ...} becomes constant at n = 3 and the sequence {a(n, 38)} = {38, 0, 2, 0, 4, 4, 4, ...} becomes constant at n = 5.
2) For s in {2, 5, 20, 21, 22, 31, 33, 34, 35, 36, 40, 42, 60, 65, 85, 87, 88, 92, 93, 97, 98, 100} the sequence {a(n, s)} appears to be quasipolynomial in n with 6 constituent polynomials of degree 1.
3) For the remaining values of s <= 100, the sequence {a(n, s)} appears to be an eventually periodic sequence with period 6, so again quasipolynomial in n with 6 constituent polynomials of degree 0. For example, an easy induction argument shows that {a(n, 3)} = {3, 1, 2, 3, 3, 2, 1, 1, 2, 3, 3, 2, 1, 1, 2, 3, 3, 2, 1, ...} has period 6 starting at n = 2.
LINKS
FORMULA
a(n) is quasipolynomial in n (proved by induction): a(6*n) = 3*n for n >= 1, and for n >= 0, a(6*n+1) = 4*n + 1, a(6*n+2) = 3*n + 1, a(6*n+3) = n, a(6*n+4) = 6*n + 3 and a(6*n+5) = n.
G.f.: A(x) = x*(x^8 + 4*x^7 + 4*x^6 + 2*x^5 + x^4 + 2*x^3 + x^2 + 2*x + 1)/((x + 1)^2*(x - 1)^2*(x^2 - x + 1)*(x^2 + x + 1)^2).
MAPLE
a := proc(n, s) option remember; if n = 1 then s else irem(add(k*a(k, s), k = 1 .. n-1), n) end if; end proc:
seq(a(n, 1), n = 1..80);
MATHEMATICA
CoefficientList[Series[x(x^8 + 4*x^7 + 4*x^6 + 2*x^5 + x^4 + 2*x^3 + x^2 + 2*x + 1)/((x + 1)^2*(x - 1)^2*(x^2 - x + 1)*(x^2 + x + 1)^2), {x, 0, 80}], x] (* Stefano Spezia, May 18 2024 *)
PROG
(PARI) lista(nn) = my(v=vector(nn)); v[1]=1; for(n=2, nn, v[n]=sum(k=1, n-1, k*v[k])%n); v; \\ Michel Marcus, May 18 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, May 15 2024
STATUS
approved