A064434 -id:A064434 - cp's OEIS frontend

A064456 A064434(n) = 0.

1, 3, 79, 235, 431, 1503, 2943, 6059, 6619, 18911, 54223, 302467995, 1772665631, 2148845167, 5145362667, 129465909327, 212089391807

Offset: 1

Jonathan Ayres (JonathanAyres(AT)btinternet.com), Oct 01 2001

a(18) > 4*10^12. - Donovan Johnson, Jan 19 2011

Also indices n where A079878(n)=1. - R. J. Mathar, Nov 13 2011

Cf. A064433.

: a := 0; for n := 1 to 3000000000 do am := a; a := (am*2 + 1) mod n; if a = 0 then write(n," "); end; end;

More terms from Klaus Brockhaus, Oct 04 2001

Offset corrected and a(15)-a(17) from Donovan Johnson, Jan 19 2011

A079878 a(1)=1, then a(n)=2a(n-1) if 2a(n-1)<=n, a(n)=2*a(n-1)-n otherwise.

Original entry on oeis.org

1, 2, 1, 2, 4, 2, 4, 8, 7, 4, 8, 4, 8, 2, 4, 8, 16, 14, 9, 18, 15, 8, 16, 8, 16, 6, 12, 24, 19, 8, 16, 32, 31, 28, 21, 6, 12, 24, 9, 18, 36, 30, 17, 34, 23, 46, 45, 42, 35, 20, 40, 28, 3, 6, 12, 24, 48, 38, 17, 34, 7, 14, 28, 56, 47, 28, 56, 44, 19, 38, 5, 10, 20, 40, 5, 10, 20, 40, 1, 2

Offset: 1

Benoit Cloitre, Feb 20 2003

nonn

a(A200087(n)) = n and a(m) <> n for m < A200087(n). [Reinhard Zumkeller, Nov 13 2011]

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A200063 (fixed points), A064456 (positions of 1).

a079878 n = a079878_list !! (n-1)
a079878_list = 1 : zipWith (\x n -> if x <= n then x else x - n)
                           (map (* 2) a079878_list) [2..]
-- Reinhard Zumkeller, Nov 13 2011

nxt[{n_,a_}]:={n+1,If[2a<=n+1,2a,2a-n-1]}; Transpose[NestList[nxt,{1,1},80]][[2]] (* Harvey P. Dale, Jul 20 2015 *)

a=1; for(n=2,100,b=if(sign(2*a-n)-1,2*a,2*a-n); a=b; print1(b,","))

a(n) = A064434(n)+1.

It seems that sum(k=1, n, a(k))/n^2 ->1/4

A330405 a(1) = 0; thereafter a(n) = (a(n-1)^2+1) mod n.

Original entry on oeis.org

0, 1, 2, 1, 2, 5, 5, 2, 5, 6, 4, 5, 0, 1, 2, 5, 9, 10, 6, 17, 17, 4, 17, 2, 5, 0, 1, 2, 5, 26, 26, 5, 26, 31, 17, 2, 5, 26, 14, 37, 17, 38, 26, 17, 20, 33, 9, 34, 30, 1, 2, 5, 26, 29, 17, 10, 44, 23, 58, 5, 26, 57, 37, 26, 27, 4, 17, 18, 49, 22, 59, 26, 20, 31, 62, 45, 24

Offset: 1

Matthew Ryan, Dec 12 2019

Does the value 0 appear infinitely many times? - Rémy Sigrist, Dec 16 2019
From Michael De Vlieger, Jan 26 2020: (Start)
Observations based on a(n) for 1 <= n <= 300000:
The value 0 appears at indices n = {1, 13, 26, 89, 205, 530, 2041, 276205, ...}.
The value 1 appears at indices n = {2, 4, 14, 27, 50, 90, 99, 175, 188, 206, 531, 2042, 5445, 6845, 7200, 18225, 24389, 25215, 37538, 46875, 48672, 53066, 79527, 93900, 147875, 176267, 186576, 196025, 254457, 276206, ...}. Let M be the indices in a(n) where 1 appears.
The subsequence {1, 2, 5, 26} appears with the first term at index n = 27, and apparently for all subsequent indices listed n M.
The subsequence {1, 2, 5, 26, 677} appears with the first term at index n = 2042, and apparently for all subsequent indices listed n M.
A stable next term in the subsequence S = {1, 2, 5, 26, 677} is not yet apparent, given 300000 terms of a(n). (End)

			a(1) = 0; a(2) = (0^2+1) mod 2 = 1; a(3) = (1^2+1) mod 2 = 2.

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Cf. A003095, A064434.

Nest[Append[#1, Mod[#1[[#2 - 1]]^2 + 1, #2]] & @@ {#, Length@ # + 1} &, {0}, 76] (* Michael De Vlieger, Dec 16 2019 *)

v=0; for (n=1, 77, print1 (v=(v^2+1)%n", ")) \\ Rémy Sigrist, Dec 16 2019

a(1) = 0; a(n) = (a(n-1)^2+1) mod n.

A357261 a(n) is the number of blocks in the bottom row after adding n blocks to the preceding structure of rows. See Comments and Example sections for more details.

Original entry on oeis.org

1, 3, 3, 3, 4, 1, 3, 1, 5, 4, 3, 3, 4, 6, 1, 3, 6, 3, 1, 7, 5, 3, 2, 2, 3, 5, 8, 1, 3, 6, 1, 6, 3, 1, 9, 6, 3, 1, 10, 7, 4, 2, 1, 1, 2, 4, 7, 11, 1, 3, 6, 10, 3, 9, 4, 12, 5, 11, 5, 13, 5, 11, 4, 12, 7, 3, 14, 8, 2, 12, 8, 5, 3, 2, 2, 3, 5

Offset: 1

John Tyler Rascoe, Oct 08 2022

A structure of rows is built up successively where each n blocks are added onto the preceding structure. The first row has an initial width of 3. After n = 1, n blocks are first added filling in the last row where n - 1 left off.
Once a row is filled a new row is started below it. After adding n blocks, if the final row reached is filled exactly, then the width of the next row is increased by one. Otherwise the width of the next row is the same as the previous.
Assuming the rows are built according to the given algorithm, a(n) is the number of blocks tagged 'n' in the last row that includes a block tagged 'n'." - Peter Luschny, Dec 20 2022
Will this sequence ever reach a point after which a(n) grows linearly? This is the case using an initial width of 2 instead of 3.

			After blocks 1 and 2, the initial row width 3 is exactly filled and hence the next row (of 3's and 4) is 1 longer.
  |1|2|2|       initial row
  |3|3|3|4|
  |4|4|4|5|
  |5|5|5|5|
  |6|6|6|6|6|
  |6|_|_|_|_|   last row after n=6
For n=6, the structure after blocks 1 through 6 have been added is as shown above and its final row has just one block (one 6) so that a(6) = 1.

John Tyler Rascoe, Table of n, a(n) for n = 1..10000

Cf. A002024, A057176, A064434, A096535, A104647, A275204.

A357261_list := proc(maxn) local A, g, c, n, r;
A := []; g := 3; c := 0;
for n from 1 to maxn do
    r := irem(n + c, g);
    c := r;
    if r = 0 then
        r := g; g := g + 1;
    fi;
    A := [op(A), r];
od; return A end:
A357261_list(77); # Peter Luschny, Dec 21 2022

lista(nn) = my(nbc=3, col=0, list=List()); for (n=1, nn, col = lift(Mod(col+n, nbc)); listput(list, if (col, col, nbc)); if (!col, nbc++);); Vec(list); \\ Michel Marcus, Oct 17 2022

def A357261_list(maxn):
    """Returns a list of the first maxn terms"""
    A = []
    g = 3
    c = 0
    for n in range(1,maxn+1):
        if (n + c)%g == 0:
            A.append(g)
            g += 1
            c = 0
        else:
            A.append((n + c)%g)
            c = A[-1]
    return A

A358073 a(n) is the row position of the n-th number n after adding the number n, n times to the preceding triangle. A variant of A357261, see Comments and Examples for more details.

Original entry on oeis.org

1, 2, 3, 3, 4, 6, 4, 3, 3, 4, 6, 9, 13, 6, 21, 16, 33, 15, 34, 18, 3, 25, 12, 36, 25, 51, 18, 46, 15, 45, 16, 48, 21, 55, 30, 6, 43, 21, 60, 40, 81, 24, 67, 12, 57, 4, 51, 99, 49, 99, 3, 55, 108, 15, 70, 126, 36, 94, 6, 66, 127, 42, 105, 22, 87, 6, 73, 141, 63

Offset: 1

John Tyler Rascoe, Oct 29 2022

A triangle is built up successively where n appears n times within the triangle. Each row has a set width before n is added, and the first row begins with a width of 1.
Numbers n are added to the first open position within the triangle or where the previous n left off so that no gaps are left in the rows of the triangle. If the row position of the n-th number n placed is the rightmost position within that row, then the width of the next row is increased by n. Otherwise, the width of the next row stays the same as the previous one.
The next row's width can only increase after a given n is added all n times. So when a row is filled after adding fewer than n n's, the next row, by definition, will have the same width.

			After 5 is added 5 times, the fifth 5 falls in the rightmost row position. So the width of the next row is increased by 5.
  |1|       initial row
  |2|2|
  |3|3|3|4|
  |4|4|4|5|
  |5|5|5|5|
  |6|6|6|6|6|6|7|7|7|
  |7|7|7|7|_|_|_|_|_|
a(7) = 4 because the row position of the seventh 7 added is 4.

John Tyler Rascoe, Table of n, a(n) for n = 1..10000
John Tyler Rascoe, Scatterplot of a(n) for n = 1...50000

Cf. A002024, A057176, A064434, A096535, A104647, A275204, A357261.

A358073_list := proc(maxn)  local A, g, c, n, r;
A := []; g := 1; c := 0;
for n from 1 to maxn do
    r := irem(n + c, g);
    c := r;
    if r = 0 then
        r := g;
        g := g + n;
    fi;
    A := [op(A), r];
od; return A end:
A358073_list(69); # Peter Luschny, Dec 21 2022

def A358073_list(maxn):
    """Returns a list of the first maxn terms"""
    A = []
    g = 1
    c = 0
    for n in range(1,maxn+1):
        if (n + c)%g ==0:
            A.append(g)
            g += n
            c = 0
        else:
            A.append((n + c)%g)
            c = A[-1]
    return A

cp's OEIS Frontend

A064456 A064434(n) = 0.

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A079878 a(1)=1, then a(n)=2*a(n-1) if 2*a(n-1)<=n, a(n)=2*a(n-1)-n otherwise.

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A330405 a(1) = 0; thereafter a(n) = (a(n-1)^2+1) mod n.

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Formula

A357261 a(n) is the number of blocks in the bottom row after adding n blocks to the preceding structure of rows. See Comments and Example sections for more details.

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A358073 a(n) is the row position of the n-th number n after adding the number n, n times to the preceding triangle. A variant of A357261, see Comments and Examples for more details.

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Maple

Python

A079878 a(1)=1, then a(n)=2a(n-1) if 2a(n-1)<=n, a(n)=2*a(n-1)-n otherwise.