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A372865
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a(n) = (Sum_{k = 1..n-1} k*a(k)) (mod n) with a(1) = 1.
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0
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1, 1, 0, 3, 0, 3, 5, 4, 1, 9, 1, 6, 9, 7, 2, 15, 2, 9, 13, 10, 3, 21, 3, 12, 17, 13, 4, 27, 4, 15, 21, 16, 5, 33, 5, 18, 25, 19, 6, 39, 6, 21, 29, 22, 7, 45, 7, 24, 33, 25, 8, 51, 8, 27, 37, 28, 9, 57, 9, 30, 41, 31, 10, 63, 10, 33, 45, 34, 11, 69, 11, 36, 49, 37, 12, 75, 12, 39, 53, 40
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OFFSET
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1,4
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COMMENTS
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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.
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LINKS
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FORMULA
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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).
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MAPLE
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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);
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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