login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A269427 a(1) = 1, a(n) counts m < n for which n == a(m) (mod m). 3
1, 1, 2, 1, 4, 1, 3, 2, 4, 3, 3, 1, 7, 4, 2, 1, 7, 3, 4, 3, 4, 2, 6, 5, 7, 3, 2, 1, 10, 1, 6, 5, 6, 3, 3, 2, 8, 5, 6, 2, 5, 4, 6, 3, 6, 7, 6, 1, 10, 3, 3, 3, 9, 3, 5, 3, 7, 5, 8, 3, 7, 4, 6, 3, 5, 4, 7, 6, 7, 3, 4, 3, 9, 8, 7, 3, 6, 1, 6, 5, 6 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

I conjecture that this sequence is unbounded. Consider the first k terms of this sequence, and let L be the floor of log(k). If we count the times that each number 1,2,...,2L appears among the first k terms of this sequence, it appears that these sums form a normal distribution centered at L, so that L appears approximately k/10 times among the first k terms of this sequence. (For instance, in the first k = 10000 terms of the sequence, L = log(10000) = 9 appears 1174 times, a maximal count among any value that appears at all.) Thus the sequence appears to be unbounded.

The sequence is unbounded. For any k, consider k pairwise coprime integers m_1, ..., m_k. By the Chinese Remainder Theorem, there are infinitely many n such that n == a(m_j) (mod m_j) for each j, and thus a(n) >= k. - Robert Israel, Mar 21 2016

LINKS

Peter Kagey, Table of n, a(n) for n = 1..10000

EXAMPLE

a(1) = 1;

a(2) = 1 because 2 == a(1) (mod 1);

a(3) = 2 because 3 == a(1) (mod 1) and 3 == a(2) (mod 2);

a(4) = 1 because 4 == a(1) (mod 1);

a(5) = 4 because 5 == a(1) (mod 1), 5 == a(2) (mod 2), 5 == a(3) (mod 3), and 5 == a(4) (mod 4).

MAPLE

N:= 200: # to get a(1) to a(N)

A:= Vector(N, 1):

for m from 2 to N-1 do

S:= [seq(A[m]+m*i, i=1..floor((N-A[m])/m))];

A[S]:= map(`+`, A[S], 1);

od:

convert(A, list); # Robert Israel, Mar 21 2016

MATHEMATICA

a[1] = 1; a[n_] := a[n] = Count[Range[n - 1], m_ /; Mod[a[m], m] == Mod[n, m]]; Table[a@ n, {n, 81}] (* Michael De Vlieger, Mar 21 2016 *)

PROG

(Java)

int[] terms = new int[10000];

terms[0] = 1;

for (int i = 1; i < 10000; i++) {

int count = 0;

for (int j = 0; j < i; j++) {

if (((i+1) - terms[j]) % (j+1) == 0) {

count++;

}

}

terms[i] = count;

}

(PARI) lista(nn) = {va = vector(nn); print1(va[1] = 1, ", "); for (n=2, nn, va[n] = sum(m=1, n-1, (Mod(va[m], m) == Mod(n, m))); print1(va[n], ", "); ); } \\ Michel Marcus, Feb 26 2016

CROSSREFS

Cf. A269423.

Sequence in context: A353379 A263653 A330328 * A349391 A077808 A021471

Adjacent sequences: A269424 A269425 A269426 * A269428 A269429 A269430

KEYWORD

easy,nonn

AUTHOR

Alec Jones, Feb 25 2016

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified February 4 13:47 EST 2023. Contains 360055 sequences. (Running on oeis4.)