A234541 Least k such that floor(n/k) + (n mod k) is a prime, or 0 if no such k exists.

0, 1, 1, 2, 1, 2, 1, 4, 2, 2, 1, 4, 1, 2, 3, 8, 1, 6, 1, 4, 2, 2, 1, 8, 2, 2, 5, 4, 1, 6, 1, 6, 2, 2, 3, 12, 1, 2, 3, 8, 1, 6, 1, 4, 2, 2, 1, 16, 3, 10, 3, 4, 1, 18, 3, 6, 2, 2, 1, 8, 1, 2, 6, 18, 3, 6, 1, 4, 3, 4, 1, 14, 1, 2, 9, 4, 5, 6, 1, 16, 2, 2, 1, 12, 2, 2

Offset

1,4

Comments

a(n) = 1 only if n is a prime.

a(2m) <= m, because with k=m, floor(2m/m)+(2m mod m) = 2.

a(2m+1) <= 2m: floor((2m+1)/2m) + ((2m+1) mod 2m) = 1 + 1 = 2.

Links

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

Prog

(Python)
primes = [2, 3]
primFlg = [0]*100000
primFlg[2] = primFlg[3] = 1
def appPrime(k):
  for p in primes:
    if k%p==0:  return
    if p*p > k:  break
  primes.append(k)
  primFlg[k] = 1
for n in range(5, 100000, 6):
  appPrime(n)
  appPrime(n+2)
for n in range(1, 100000):
  a = 0
  for k in range(1, n):
    c = n//k + n%k
    if primFlg[c]:  # if c in primes:
      a = k
      break
  print(str(a), end=', ')
(Scheme)
;; MIT/GNU Scheme, with Aubrey Jaffer's SLIB Scheme library and function A234575bi as defined in A234575
(require 'factor) ;; For predicate prime? from SLIB-library.
(define (A234541 n) (let loop ((k 1)) (cond ((prime? (A234575bi n k)) k) ((> k n) 0) (else (loop (+ 1 k))))))
;; Antti Karttunen, Dec 29 2013

Crossrefs

Cf. A000040, A234575.

Sequence in context: A296131 A345344 A319004 * A066389 A077191 A317545

Adjacent sequences: A234538 A234539 A234540 * A234542 A234543 A234544

Keyword

nonn,easy

Author

Alex Ratushnyak, Dec 27 2013

Status

approved