A332542 - OEIS

%I #52 Aug 27 2021 11:19:53

%S 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,

%T 7,86,27,30,959,62,16,34,8,73,35,17,24,163,39,42,127,9,22,46,92,62,19,

%U 23,15,158,51,10,20,75,28,58,116,121,59,29,127,254,11

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

%C 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

%C 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

%H Seiichi Manyama, <a href="/A332542/b332542.txt">Table of n, a(n) for n = 3..10000</a>

%H J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, <a href="http://arxiv.org/abs/2004.14000">Three Cousins of Recaman's Sequence</a>, arXiv:2004:14000  [math.NT], 2020-2021.

%e 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.

%p grow2 := proc(n,M) local p,q,k; # searches out to a limit of M

%p # returns n, k (A332542(n)), n+k+1 (A332543(n)), p (A332544(n)), and q (which appears to match A082183(n-1))

%p for k from 1 to M do

%p    if ((k+1)*n + k*(k+1)/2) mod (n+k+1) = 0 then

%p    p := (k+1)*n+k*(k+1)/2;

%p    q := p/(n+k+1); return([n,k,n+k+1,p,q]);

%p    fi;

%p od:

%p # if no success, return -1's

%p [n,-1,-1,-1,-1]; end; # _N. J. A. Sloane_, Feb 18 2020

%t a[n_] := NestWhile[#1+1&,0,!IntegerQ[Divide[(#+1)*n+#*(#+1)/2,n+#+1]]&]

%t a/@Range[3,100] (* _Bradley Klee_, Apr 30 2020 *)

%o (Ruby)

%o def A(n)

%o   s = n

%o   t = n + 1

%o   while s % t > 0

%o     s += t

%o     t += 1

%o   end

%o   t - n - 1

%o end

%o def A332542(n)

%o   (3..n).map{|i| A(i)}

%o end

%o p A332542(100) # _Seiichi Manyama_, Feb 19 2020

%o (PARI) a(n) = my(k=1); while (sum(i=0, k, n+i) % (n+k+1), k++); k; \\ _Michel Marcus_, Aug 26 2021

%o (Python)

%o def a(n):

%o     k, s = 1, 2*n+1

%o     while s%(n+k+1) != 0: k += 1; s += n+k

%o     return k

%o print([a(n) for n in range(3, 67)]) # _Michael S. Branicky_, Aug 26 2021

%Y Cf. A332543, A332544, A082183.

%Y See A332558-A332561 for a multiplicative analog.

%K nonn

%O 3,1

%A _Scott R. Shannon_, Feb 18 2020