A213633 -id:A213633 - cp's OEIS frontend

A213636 Remainder when n is divided by its least nondivisor.

1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 4, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 4, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 4, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1

Offset: 1

Clark Kimberling, Jun 16 2012

nonn

Experimentation suggests that every positive integer occurs in this sequence and that
2 occurs only in even numbered positions,
3 occurs in only in positions that are multiples of 12,
4 occurs only in positions that are multiples of 12,
5 occurs only in positions that are multiples of 60,
6 occurs only in positions that are multiples of 60,
7 occurs only in positions that are multiples of 2520, etc.
See A213637 for positions of 1 and A213638 for positions of 2.
From Robert Israel, Jul 28 2017: (Start)
Given any positive number m, let q be a prime > m and r = A003418(q-1).  Then a(n) = m if n == m (mod q) and n == 0 (mod r).  By the Chinese Remainder Theorem, such n exists.
On the other hand, if a(n) = m, we must have A007978(n) > m, and then n must be divisible by A003418(q-1) where q = A007978(n) is a member of A000961 greater than m.  Moreover, if q=p^j with j>1, n is divisible by p^(j-1) so m must be divisible by p^(j-1).  Thus:
For m=2, A003418(2)=2.
For m=3, A007978(n) can't be 4 because m is odd, so A007978(n)>= 5 and n must be divisible by A003418(4)=12.
For m=4, A003418(4)=12.
For m=5 or 6, A003418(6)=60.
For m=7, A007978(n) can't be 8 because m is odd, and can't be 9 because m is not divisible by 3, so n must be divisible by A003418(10)=2520. (End)

			a(10) = 10-3*[10/3] = 1.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000961, A003418, A007978, A213633, A213635, A213637, A213638.

f:= proc(n) local k;
  for k from 2 do if n mod k <> 0 then return n mod k fi od
end proc:
map(f, [$1..100]); # Robert Israel, Jul 27 2017

y=120; z=2000;
t = Table[k := 1; While[Mod[n, k] == 0, k++];
   k, {n, 1, z}]  (*A007978*)
Table[Floor[n/t[[n]]], {n, 1, y}] (*A213633*)
Table[n - Floor[n/t[[n]]], {n, 1, y}] (*A213634*)
Table[t[[n]]*Floor[n/t[[n]]], {n, 1, y}] (*A213635*)
t1 = Table[n - t[[n]]*Floor[n/t[[n]]],
   {n, 1, z}] (* A213636 *)
Flatten[Position[t1, 1]] (* A213637 *)
Flatten[Position[t1, 2]] (* A213638 *)
rem[n_]:=Module[{lnd=First[Complement[Range[n],Divisors[n]]]},Mod[n,lnd]]; Join[{1,2},Array[rem,100,3]] (* Harvey P. Dale, Mar 26 2013 *)
Table[Mod[n, SelectFirst[Range[n + 1], ! Divisible[n, #] &]], {n, 105}] (* Michael De Vlieger, Jul 29 2017 *)

def a(n):
    k=2
    while n%k==0: k+=1
    return n%k
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jul 28 2017

def A213636(n): return next(filter(None, (n%d for d in range(2,n)))) if n>2 else n # Chai Wah Wu, Feb 22 2023

(define (A213636 n) (modulo n (A007978 n))) ;; Antti Karttunen, Jul 27 2017

a(n) = n - A213635(n).

a(n) = n - m(n)*floor(n/m(n)), where m(n) = A007978(n).

A213634 n-[n/m], where m is the least nondivisor of n (as in A007978) and [ ] = floor.

Original entry on oeis.org

1, 2, 2, 3, 3, 5, 4, 6, 5, 7, 6, 10, 7, 10, 8, 11, 9, 14, 10, 14, 11, 15, 12, 20, 13, 18, 14, 19, 15, 23, 16, 22, 17, 23, 18, 29, 19, 26, 20, 27, 21, 32, 22, 30, 23, 31, 24, 39, 25, 34, 26, 35, 27, 41, 28, 38, 29, 39, 30, 52, 31, 42, 32, 43, 33, 50, 34, 46, 35, 47, 36

Offset: 1

Clark Kimberling, Jun 16 2012

If n is odd, a(n) = (n+1)/2. - Robert Israel, Oct 09 2016

			a(10) = 10 - [10/3] = 7.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A007978, A213633.

f:= proc(n) local m;
       for m from 2 do if n mod m <> 0 then return n - iquo(n,m) fi od
end proc:
map(f, [$1..100]); # Robert Israel, Oct 09 2016

Mathematica
```
(See A213633.)
```

A213635 m*[n/m], where m is the least nondivisor of n (as in A007978) and [ ] = floor.

Original entry on oeis.org

0, 0, 2, 3, 4, 4, 6, 6, 8, 9, 10, 10, 12, 12, 14, 15, 16, 16, 18, 18, 20, 21, 22, 20, 24, 24, 26, 27, 28, 28, 30, 30, 32, 33, 34, 35, 36, 36, 38, 39, 40, 40, 42, 42, 44, 45, 46, 45, 48, 48, 50, 51, 52, 52, 54, 54, 56, 57, 58, 56, 60, 60, 62, 63, 64, 64, 66, 66, 68

Offset: 1

Clark Kimberling, Jun 16 2012

Cf. A213633.

Mathematica
```
(See A213633.)
```

def A213635(n): return n-next(filter(None, (n%d for d in range(2,n)))) if n>2 else 0 # Chai Wah Wu, Feb 22 2023

A331902 T(n, k) = floor(n/m) where m is the least positive integer such that floor(n/m) = floor(k/m). Square array read by antidiagonals, for n >= 0 and k >= 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 4, 1, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 5, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 6, 1, 0

Offset: 0

Rémy Sigrist, Jan 31 2020

For any n > 0, the n-th row has A001651(n) nonzero terms.

			Array T(n, k) begins (with dots instead of 0's for readability):
   n\k|   0   1   2   3   4   5   6   7   8   9  10  11  12
   ---+----------------------------------------------------
     0|   .   .   .   .   .   .   .   .   .   .   .   .   .
     1|   .   1   .   .   .   .   .   .   .   .   .   .   .
     2|   .   .   2   1   .   .   .   .   .   .   .   .   .
     3|   .   .   1   3   1   1   .   .   .   .   .   .   .
     4|   .   .   .   1   4   2   1   1   .   .   .   .   .
     5|   .   .   .   1   2   5   1   1   1   1   .   .   .
     6|   .   .   .   .   1   1   6   3   2   1   1   1   .
     7|   .   .   .   .   1   1   3   7   2   1   1   1   1
     8|   .   .   .   .   .   1   2   2   8   4   2   2   1
     9|   .   .   .   .   .   1   1   1   4   9   3   3   1
    10|   .   .   .   .   .   .   1   1   2   3  10   5   2
    11|   .   .   .   .   .   .   1   1   2   3   5  11   2
    12|   .   .   .   .   .   .   .   1   1   1   2   2  12

Rémy Sigrist, Table of n, a(n) for n = 0..10010 (antidiagonals 0..140)
Rémy Sigrist, Colored representation of T(n, k) for n, k = 0..1000 (where the hue is function of T(n, k), red pixels correspond to 0's)

Cf. A001651, A213633, A331886.

T(n,k) = for (x=1, oo, if (n\x==k\x, return (n\x)))

T(n, k) = floor(n/A331886(n, k)) = floor(k/A331886(n, k)).
T(n, k) = T(k, n).
T(n, k) = 0 iff max(n, k) >= 2*min(n, k).
T(n, n+1) = A213633(n+1).

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A213636 Remainder when n is divided by its least nondivisor.

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A213634 n-[n/m], where m is the least nondivisor of n (as in A007978) and [ ] = floor.

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A213635 m*[n/m], where m is the least nondivisor of n (as in A007978) and [ ] = floor.

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Mathematica

Python

A331902 T(n, k) = floor(n/m) where m is the least positive integer such that floor(n/m) = floor(k/m). Square array read by antidiagonals, for n >= 0 and k >= 0.

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