A332542 a(n) is the smallest k such that n+(n+1)+(n+2)+...+(n+k) is divisible by n+k+1.

2, 7, 14, 3, 6, 47, 14, 4, 10, 20, 25, 11, 5, 31, 254, 15, 18, 55, 6, 10, 22, 44, 14, 23, 11, 7, 86, 27, 30, 959, 62, 16, 34, 8, 73, 35, 17, 24, 163, 39, 42, 127, 9, 22, 46, 92, 62, 19, 23, 15, 158, 51, 10, 20, 75, 28, 58, 116, 121, 59, 29, 127, 254, 11

Offset

3,1

Comments

Note that (n+(n+1)+(n+2)+...+(n+k))/(n+k+1) = A332544(n)/(n+k+1) = A082183(n-1). See the Myers et al. link for proof. - N. J. A. Sloane, Apr 30 2020

We can always take k = n^2-2*n-1, for then the sum in the definition becomes (n+1)*n*(n-1)*(n-2)/2, which is an integral multiple of n+k+1 = n*(n-1). So a(n) always exists. - N. J. A. Sloane, Feb 20 2020

Links

Seiichi Manyama, Table of n, a(n) for n = 3..10000

J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, Three Cousins of Recaman's Sequence, arXiv:2004:14000 [math.NT], 2020-2021.

Example

n=4: we get 4 -> 4+5=9 -> 9+6=15 -> 15+7=22 -> 22+8=30 -> 30+9=39 -> 39+10=49 -> 49+11=60, which is divisible by 12, and took k=7 steps, so a(4) = 7. Also A332543(4) = 12, A332544(4) = 60, and A082183(3) = 60/12 = 5.

Maple

grow2 := proc(n, M) local p, q, k; # searches out to a limit of M
# returns n, k (A332542(n)), n+k+1 (A332543(n)), p (A332544(n)), and q (which appears to match A082183(n-1))
for k from 1 to M do
   if ((k+1)*n + k*(k+1)/2) mod (n+k+1) = 0 then
   p := (k+1)*n+k*(k+1)/2;
   q := p/(n+k+1); return([n, k, n+k+1, p, q]);
   fi;
od:
# if no success, return -1's
[n, -1, -1, -1, -1]; end; # N. J. A. Sloane, Feb 18 2020

Mathematica

a[n_] := NestWhile[#1+1&, 0, !IntegerQ[Divide[(#+1)*n+#*(#+1)/2, n+#+1]]&]
a/@Range[3, 100] (* Bradley Klee, Apr 30 2020 *)

Prog

(Ruby)
def A(n)
  s = n
  t = n + 1
  while s % t > 0
    s += t
    t += 1
  end
  t - n - 1
end
def A332542(n)
  (3..n).map{|i| A(i)}
end
p A332542(100) # Seiichi Manyama, Feb 19 2020
(PARI) a(n) = my(k=1); while (sum(i=0, k, n+i) % (n+k+1), k++); k; \\ Michel Marcus, Aug 26 2021
(Python)
def a(n):
    k, s = 1, 2*n+1
    while s%(n+k+1) != 0: k += 1; s += n+k
    return k
print([a(n) for n in range(3, 67)]) # Michael S. Branicky, Aug 26 2021

Crossrefs

Cf. A332543, A332544, A082183.

See A332558-A332561 for a multiplicative analog.

Sequence in context: A102155 A102154 A102153 * A368895 A331703 A032537

Adjacent sequences: A332539 A332540 A332541 * A332543 A332544 A332545

Keyword

nonn

Author

Scott R. Shannon, Feb 18 2020

Status

approved