A066910 -id:A066910 - cp's OEIS frontend

A073117 a(n+1) = a(n) + a(n) mod n; a(1) = 1.

1, 1, 2, 4, 4, 8, 10, 13, 18, 18, 26, 30, 36, 46, 50, 55, 62, 73, 74, 91, 102, 120, 130, 145, 146, 167, 178, 194, 220, 237, 264, 280, 304, 311, 316, 317, 346, 359, 376, 401, 402, 435, 450, 470, 500, 505, 550, 583, 590, 592, 634, 656, 688, 740, 778

Offset: 1

Reinhard Zumkeller, Aug 19 2002

nonn

Conjecture (seems provable): More generally let a and b(1) be integers. If b(n+1) = b(n) + b(n) (mod(n+a)) there is an integer x(a,b(1)) such that b(n+1) = b(n) + x(a,b(1)) for n sufficiently large. We have x(0,1) = x(1,1) = x(2,1) = 97, x(3,1) = 1, x(4,1) = 3, x(5,1) = 3, x(6,1) = 6, ..., x(97,1) = 43, x(0,11) = 2, etc. - Benoit Cloitre, Aug 20 2002

			a(397) = 38606 = 2*97*199 = (2*199)*97 = 398*97 = (397+1)*97; a(397) mod 397 = (397*97 + 97) mod 397 = 97, a(398) = a(397) + a(397) mod 397 = (397+1)*97 + 97 = (398+1)*97, etc.: a(n+1) = a(n) + 97 for n >= 397.

Rémy Sigrist, Table of n, a(n) for n = 1..1000

Cf. A066910. - Rémy Sigrist, Mar 24 2017

s=1;lst={s};Do[s+=Mod[s, n];AppendTo[lst, s], {n, 1, 6!, 1}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 07 2008 *)

A094405 a(1) = 1; a(n) = (sum of previous terms) mod n.

Original entry on oeis.org

1, 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, 62, 26

Offset: 1

Chuck Seggelin (seqfan(AT)plastereddragon.com), Jun 03 2004

nonn

Theorem. For all values of n>=397, a(n)=97. Proof. Let s(n) denote Sum[a(i), i=1..n-1]. Calculation shows that s(397)=38606=397*97+97. Thus a(397)=397*97+97 mod 397=97. Then s(398)=s(397)+97=398*97+97, giving a(398)=97. A simple inductive argument shows that a(397+k)=97 for all integers k>=0. - John W. Layman, Jun 07 2004
Conjecture: For any seed a(1) the sequence "a(n) = (sum of previous terms) mod n" ends with repeating constant. This is true for a(1) = 1,...,941. - Zak Seidov, Feb 24 2006
Essentially the same as A066910. [From R. J. Mathar, Sep 05 2008]

			a(4) = 0 because the previous terms 1, 1, 2 sum to 4 and 4 mod 4 is 0. a(5) = 4 because the previous terms 1, 1, 2, 0 sum to 4 and 4 mod 5 is 4.

L := [1]; s := 1; p := 2; while (nops(L) < 90) do; if 1>0 then; t := s mod p; L := [op(L),t]; s := s+t; p := p+1; fi; od; L;

A215451 a(0)=1, a(n) = (sum of previous terms) mod (a(n-1)+n).

Original entry on oeis.org

1, 1, 2, 4, 0, 3, 2, 4, 5, 8, 12, 19, 30, 5, 1, 1, 13, 21, 15, 11, 3, 17, 22, 20, 0, 20, 10, 28, 54, 0, 2, 4, 14, 23, 33, 0, 12, 28, 52, 45, 35, 48, 88, 61, 42, 36, 35, 70, 16, 1, 8, 41, 3, 21, 0, 5, 18, 23, 43, 17, 1, 41, 65, 111, 149, 25, 1, 53, 29, 63, 98, 102, 154, 5

Offset: 0

Alex Ratushnyak, Aug 11 2012

nonn

Indices of 0's: 4, 24, 29, 35, 54, 267, 5284, 17827, 43631, 120871, 843813, 1854903, 2226536, 4208775, 5525594, ...
Indices of 1's: 0, 1, 14, 15, 49, 60, 66, 116, 331, 6053, 23760, 288502, 496670, 4281666, ...
Indices such that a(n)=n: 1, 2, 22, 315, 1172, 1441, 1846, 2140, 47376, 593870, 16538298, 111824649, 565597433, 791186876, ...

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A094405, A066910, A215452.

nxt[{n_,t_,a_}]:=Module[{c=Mod[t,a+n+1]},{n+1,t+c,c}]; NestList[nxt,{0,1,1},80][[All,3]] (* Harvey P. Dale, Dec 30 2017 *)

s = a = 1
for n in range(1,333):
    print(a, end=', ')
    a = s % (a+n)
    s += a

a(0)=1, a(n) = (a(0)+...+a(n-1)) mod (a(n-1)+n).

A215452 a(1)=1, a(n) = (sum of previous terms) mod (a(n-1) + n).

Original entry on oeis.org

1, 1, 2, 4, 8, 2, 0, 2, 9, 10, 18, 27, 4, 16, 11, 7, 2, 4, 13, 9, 0, 18, 4, 4, 2, 10, 3, 5, 26, 54, 21, 32, 4, 29, 42, 14, 10, 44, 57, 44, 63, 6, 5, 10, 52, 23, 32, 44, 64, 74, 71, 33, 18, 60, 93, 29, 46, 48, 60, 84, 38, 26, 39, 46, 83, 81, 25, 59, 93, 22, 47, 24, 34, 53

Offset: 1

Alex Ratushnyak, Aug 11 2012

nonn

Indices of 0's: 7, 21, 956, 1576, 1964, 55346, 70460, 99845, 399114, 544095, 35321849, 77073660, ...
Indices of 1's: 1, 2, 129, 193, 262, 4495, 99781, 651856, 35351437, ...
Indices such that a(n)=n: 1, 4, 9, 10, 32, 176, 266, 414, 432, 440, 858, 5953, 6030, 15146, 1408096, 3138130, 35659404, 44722350, 109021513, 790542727, ...

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from Harvey P. Dale)

Cf. A094405, A066910, A215451.

s:= proc(n) option remember; `if`(n<1, 0, s(n-1)+a(n)) end:
a:= proc(n) option remember; `if`(n=1, 1, irem(s(n-1), a(n-1)+n)) end:
seq(a(n), n=1..74);  # Alois P. Heinz, Jun 16 2025

nxt[{t_,n_,a_}]:={t+a,n+1,Mod[t+a,a+n+1]}; NestList[nxt,{0,1,1},80][[All,3]] (* Harvey P. Dale, Sep 01 2016 *)

s = a = 1
for n in range(2,333):
    print(a, end=", ")
    a = s % (a+n)
    s += a

a(1)=1, a(n) = (a(0)+...+a(n-1)) mod (a(n-1)+n).

A330249 a(0) = 0; for n > 0, a(n) = n + a((Sum_{k=0..n-1} a(k)) mod n).

Original entry on oeis.org

0, 1, 3, 4, 4, 8, 9, 8, 16, 25, 26, 19, 16, 38, 39, 24, 16, 18, 22, 38, 59, 45, 81, 61, 28, 41, 67, 66, 95, 37, 69, 112, 40, 71, 50, 147, 183, 77, 75, 120, 62, 91, 119, 104, 94, 116, 55, 63, 70, 196, 145, 75, 92, 91, 170, 110, 176, 177, 241, 109

Offset: 0

Samuel B. Reid, Dec 06 2019

nonn

			a(1) = 1 + a(0 mod 1) = 1.
a(2) = 2 + a((0+1) mod 2) = 3.
a(3) = 3 + a((0+1+3) mod 3) = 4.
a(4) = 4 + a((0+1+3+4) mod 4) = 4.

Samuel B. Reid, Table of n, a(n) for n = 0..10000

Cf. A330256, A066910.

a[0] = 0; a[n_] := a[n] = n + a[Mod[Sum[a[k], {k, 0, n-1}], n]]; Array[a, 100, 0] (* Amiram Eldar, Dec 07 2019 *)

lista(nn) = {my(v=vector(nn+1), s=0); v[1]=0; for(n=1, nn, v[n+1]=n+v[s%n+1]; s+=v[n+1]); v; } \\ Jinyuan Wang, Dec 07 2019

A330256 a(0) = 0; for n > 0, a(n) = n - a((Sum_{k=0..n-1} a(k)) mod n).

Original entry on oeis.org

0, 1, 1, 2, 4, 3, 3, 7, 5, 4, 10, 4, 7, 6, 13, 5, 12, 16, 12, 18, 14, 21, 9, 11, 10, 14, 22, 15, 14, 28, 9, 10, 23, 31, 24, 33, 22, 15, 37, 24, 16, 40, 14, 34, 37, 36, 43, 23, 34, 42, 13, 18, 37, 50, 17, 18, 32, 40, 40, 19, 46, 57, 39, 59, 30, 15, 32, 21, 11, 32, 40, 65, 32, 62, 41, 58, 63, 60

Offset: 0

Samuel B. Reid, Dec 07 2019

nonn

			a(1) = 1 - a(0 mod 1) = 1.
a(2) = 2 - a((0+1) mod 2) = 1.
a(3) = 3 - a((0+1+1) mod 3) = 2.
a(4) = 4 - a((0+1+1+2) mod 4) = 4.

Samuel B. Reid, Table of n, a(n) for n = 0..10000
Samuel B. Reid, Colored plot of one billion terms. This plot is normalized by column. Within each column, density corresponds, in a linear fashion, to this spectrum.
Samuel B. Reid, Distribution, between 0 and n, of the first billion terms

Cf. A330249, A066910.

a[0] = 0; a[n_] := a[n] = n - a[Mod[Sum[a[k], {k, 0, n - 1}], n]]; Array[a, 100, 0] (* Amiram Eldar, Dec 07 2019 *)

s=0; for (n=1, #(a=vector(78)), print1 (a[n]=if (n==1, 0, (n-1)-a[1+(s%(n-1))])", "); s+=a[n]) \\ Rémy Sigrist, Dec 08 2019

A337307 a(1) = 1; a(n) = 1 + Product_{k=1..n-1} a(k) (mod n-1).

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 1, 3, 1, 1, 7, 6, 1, 12, 1, 10, 1, 12, 1, 3, 1, 1, 21, 12, 1, 6, 21, 1, 1, 15, 1, 20, 1, 31, 15, 1, 1, 32, 13, 1, 1, 18, 1, 7, 25, 1, 17, 38, 1, 1, 1, 1, 1, 26, 1, 6, 1, 1, 29, 47, 1, 42, 1, 1, 1, 1, 61, 26, 1, 25, 1, 21, 1, 64, 21, 1, 1, 29, 1, 18

Offset: 1

Matt Donahoe, Aug 22 2020

nonn

Note that the running product for each a(n) is incrementally computed mod n-1.

Inspired by A066910.

Cf. A129871 (without the mod operation).

a[1] = 1; a[n_] := a[n] = 1 + Mod[Product[a[k], {k, 1, n - 1}], n - 1]; Array[a, 100] (* Amiram Eldar, Aug 22 2020 *)

lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = 1 + prod(k=1, n-1, va[k]) % (n-1);); va;} \\ Michel Marcus, Aug 23 2020

def f(n):
    if n == 1: return 1
    a = 1
    for k in range(1, n):
        a = a * f(k) % (n - 1)
    return a + 1

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

A073117 a(n+1) = a(n) + a(n) mod n; a(1) = 1.

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A094405 a(1) = 1; a(n) = (sum of previous terms) mod n.

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A215451 a(0)=1, a(n) = (sum of previous terms) mod (a(n-1)+n).

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A215452 a(1)=1, a(n) = (sum of previous terms) mod (a(n-1) + n).

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A330249 a(0) = 0; for n > 0, a(n) = n + a((Sum_{k=0..n-1} a(k)) mod n).

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A330256 a(0) = 0; for n > 0, a(n) = n - a((Sum_{k=0..n-1} a(k)) mod n).

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A337307 a(1) = 1; a(n) = 1 + Product_{k=1..n-1} a(k) (mod n-1).

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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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