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 A213636 Remainder when n is divided by its least nondivisor. 6
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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) LINKS Antti Karttunen, Table of n, a(n) for n = 1..10000 FORMULA a(n) = n - A213635(n). a(n) = n - m(n)*floor(n/m(n)), where m(n) = A007978(n). EXAMPLE a(10) = 10-3*[10/3] = 1. MAPLE 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 MATHEMATICA 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 *) PROG (Scheme) (define (A213636 n) (modulo n (A007978 n))) ;; Antti Karttunen, Jul 27 2017 (Python) 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 (Python) 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 CROSSREFS Cf. A000961, A003418, A007978, A213633, A213635, A213637, A213638. Sequence in context: A363057 A242481 A228287 * A192393 A184303 A218545 Adjacent sequences: A213633 A213634 A213635 * A213637 A213638 A213639 KEYWORD nonn AUTHOR Clark Kimberling, Jun 16 2012 STATUS approved

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Last modified May 29 12:57 EDT 2024. Contains 372940 sequences. (Running on oeis4.)