A073117 -id:A073117 - cp's OEIS frontend

A066910 a(1) = 1; a(n+1) = (sum{k=1 to n} a(k) ) (mod n).

1, 0, 1, 2, 0, 4, 2, 3, 5, 0, 8, 4, 6, 10, 4, 5, 7, 11, 1, 17, 11, 18, 10, 15, 1, 21, 11, 16, 26, 17, 27, 16, 24, 7, 5, 1, 29, 13, 17, 25, 1, 33, 15, 20, 30, 5, 45, 33, 7, 2, 42, 22, 32, 52, 38, 8, 2, 47, 23, 32, 50, 25, 35, 55, 31, 46, 10, 3, 57, 29, 41, 65, 41, 64, 36, 53, 11, 2

Offset: 1

Leroy Quet, Jan 22 2002

Steven Taschuk and Phil Carmody posted to sci.math (http://www.mathforum.com/epigone/sci.math/sazhazhi ) that a(k) = 97 for k >= 398.

Apart from the initial term, this is the first differences of A073117. - Rémy Sigrist, Mar 22 2017

			a(7) = (1 + 0 + 1 + 2 + 0 + 4) (mod 6) = 8 (mod 6) = 2.

Ivan Neretin, Table of n, a(n) for n = 1..501

Cf. A073117.

Fold[Append[#1, Mod[Total@#1, #2]] &, {1}, Range@78] (* Ivan Neretin, Nov 22 2015 *)

first(m)=my(v=vector(m));v[1]=1;for(i=2,m,v[i]=sum(k=1,i-1,v[k])%(i-1));v \\ Anders Hellström, Nov 22 2015

A074482 Consider the recursion b(1,n) = 1, b(k+1,n) = b(k,n) + (b(k,n) reduced mod(k+n)); then there is a number x such that b(k,n) - b(k-1,n) is a constant x depending only on n, for k > y = A074483(n). Sequence gives values of x.

Original entry on oeis.org

97, 97, 97, 1, 3, 3, 6, 6, 8, 4, 1, 8, 8, 3, 2, 5, 17143, 5, 3, 4, 5, 316, 22, 41, 28, 1, 41, 41, 3, 74, 39, 5, 316, 37, 37, 37, 12178, 12178, 67382, 67382, 73191, 52, 25, 51, 50, 67382, 6001, 25, 6001, 51, 22, 17, 2, 5, 23, 50, 1, 50, 50, 14, 50, 492, 20, 50, 20, 52, 17, 17143

Offset: 0

Reinhard Zumkeller and Benoit Cloitre, Aug 23 2002

nonn

Conjecture: a(n) is defined for all n (as well as A074483);
A074484(n) = a(n)*(A074483(n)+ n + 1);
b(k, n) = a(n)*(k + n + 1) for k > A074483(n).

			a(0) = A073117(A074483(0)) mod A074483(0) = A073117(397) mod 397 = 38606 mod 397 = 97.

David W. Wilson, Table of n, a(n) for n = 0..10000

Cf. A073117.

A074483 Consider the recursion b(1,n) = 1, b(k+1,n) = b(k,n) + (b(k,n) reduced mod(k+n)); then there is a number y such that b(k,n)-b(k-1,n) is a constant (= A074482(n)) for k > y. Sequence gives values of y.

Original entry on oeis.org

397, 396, 395, 4, 11, 10, 25, 24, 29, 14, 5, 26, 25, 10, 7, 16, 68265, 14, 13, 12, 17, 1220, 67, 136, 93, 6, 133, 132, 9, 272, 129, 14, 1209, 126, 125, 124, 48605, 48604, 269393, 269392, 292695, 180, 77, 178, 177, 269386, 24017, 72, 24015, 172, 67, 44, 11, 16, 65

Offset: 0

Reinhard Zumkeller and Benoit Cloitre, Aug 23 2002

nonn

Conjecture: a(n) is defined for all n (as well as A074482);
A074484(n) = A074482(n)*(a(n)+ n + 1);
b(k, n) = A074482(n)*(k + n + 1) for k > a(n).

			A074482(0) = A073117(a(0)) mod a(0) = A073117(397) mod 397 = 38606 mod 397 = 97.

David W. Wilson, Table of n, a(n) for n = 0..10000

Cf. A073117.

A074484 a(n) = b(A074483(n), n), where b(k) is the recursion: b(1,n)=1, b(k+1,n)=b(k,n) + (b(k,n) reduced mod(k+n)) (cf. A074482).

Original entry on oeis.org

38606, 38606, 38606, 8, 48, 48, 192, 192, 304, 96, 16, 304, 304, 72, 44, 160, 1170558326, 160, 96, 128, 190, 392472, 1980, 6560, 3304, 32, 6560, 6560, 114, 22348, 6240, 230, 392472, 5920, 5920, 5920, 592362276, 592362276, 18154867024, 18154867024

Offset: 0

Reinhard Zumkeller and Benoit Cloitre, Aug 23 2002

nonn

a(n) = A074482(n)*(A074483(n)+ n + 1).

			a(0) = A073117(A074483(0)) = A073117(397) = 38606.

David W. Wilson, Table of n, a(n) for n = 0..10000

Cf. A074482, A074483, A073117.

A117846 Given n, define a(n) as follows: let a sequence b(k) be defined by b(k+1)=b(k)+b(k)mod k; b(1)=2n-1. (Here b(k)mod k denotes the least nonnegative residue of b(k) modulo k). Let a(n) be the common value of b(k+1)-b(k) for all large k if such exists; otherwise let a(n) be 0.

Original entry on oeis.org

97, 1, 2, 2, 316, 2, 3, 3, 3, 4, 12, 4, 4, 12, 11, 11, 316, 11, 316, 316, 6, 316, 316, 316, 316, 97, 316, 316, 13, 316, 13, 13, 13, 13, 8, 13, 13, 12, 13, 13, 13, 13, 13, 13, 14, 14, 316, 14, 316, 316, 316, 97, 9, 97, 97, 13, 10, 10, 11, 10, 14, 11, 12, 12, 97, 12, 97, 132

Offset: 1

Alex Abercrombie, Mar 22 2007

nonn

Putting b(1)=2n gives essentially the same sequence as putting b(1)=2n-1. It is a plausible conjecture or at least an interesting open problem that a(n) is never zero; that is all the sequences b(k) are arithmetic progressions from some point on. Sequence A073117 is the sequence b(k) with b(1)=1. Do the values a(n) include all positive numbers?

			n=4: b(1)=7 and the sequence b(k) continues 7,8,10,12,14...with b(k+1)-b(k)=2 for all k>3, so a(4)=2.

Cf. A073117.

A372865 a(n) = (Sum_{k = 1..n-1} k*a(k)) (mod n) with a(1) = 1.

Original entry on oeis.org

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

Offset: 1

Peter Bala, May 15 2024

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.

MathOverflow, Mod sequences that seem to become constant

Cf. A066910, A073117, A094405, A074482, A117846, A253387.

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);

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 *)

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

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).

cp's OEIS Frontend

A066910 a(1) = 1; a(n+1) = (sum{k=1 to n} a(k) ) (mod n).

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A074482 Consider the recursion b(1,n) = 1, b(k+1,n) = b(k,n) + (b(k,n) reduced mod(k+n)); then there is a number x such that b(k,n) - b(k-1,n) is a constant x depending only on n, for k > y = A074483(n). Sequence gives values of x.

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A074483 Consider the recursion b(1,n) = 1, b(k+1,n) = b(k,n) + (b(k,n) reduced mod(k+n)); then there is a number y such that b(k,n)-b(k-1,n) is a constant (= A074482(n)) for k > y. Sequence gives values of y.

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A074484 a(n) = b(A074483(n), n), where b(k) is the recursion: b(1,n)=1, b(k+1,n)=b(k,n) + (b(k,n) reduced mod(k+n)) (cf. A074482).

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A117846 Given n, define a(n) as follows: let a sequence b(k) be defined by b(k+1)=b(k)+b(k)mod k; b(1)=2n-1. (Here b(k)mod k denotes the least nonnegative residue of b(k) modulo k). Let a(n) be the common value of b(k+1)-b(k) for all large k if such exists; otherwise let a(n) be 0.

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A372865 a(n) = (Sum_{k = 1..n-1} k*a(k)) (mod n) with a(1) = 1.

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