

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
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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 N1 do
S:= [seq(A[m]+m*i, i=1..floor((NA[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, n1, (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



