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a(n) is the least number m such that Sum_{k=0..(n-1)}{d(m + k)} | Sum_{k=0..(n-1)}{m+k}, where d(x) is the number of divisors of x.
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%I #17 Jun 22 2016 23:41:43

%S 1,1,19,4,4,8,65,2,36,6,30,5,39,27,13,105,8,114,11,22,68,191,130,51,

%T 38,70,31,117,163,69,286,313,86,159,15,145,90,574,244,45,100,62,105,

%U 457,61,9,1,139,7,7,60,231,347,144,344,3,36,489,103,185,292,682,19

%N a(n) is the least number m such that Sum_{k=0..(n-1)}{d(m + k)} | Sum_{k=0..(n-1)}{m+k}, where d(x) is the number of divisors of x.

%H Paolo P. Lava, <a href="/A274249/b274249.txt">Table of n, a(n) for n = 1..2000</a>

%F a( A160922(n)) = 1. - _Michel Marcus_, Jun 22 2016

%e a(3) = 19 because it is the least number such that (19 + 20 + 21) / (d(19) + d(20) + d(21)) = 60 / (2 + 6 + 4) = 60 / 12 = 5 is integer.

%p with(numtheory): P:=proc(q) local i,k,n;

%p for i from 1 to q do for n from 1 to q do

%p if type(i*(2*n+i-1)/(2*add(tau(n+k),k=0..i-1)),integer)

%p then print(n); break; fi; od; od; end: P(10^6);

%t Table[m = 1; While[! Divisible[Sum[m + k, {k, 0, n - 1}], Sum[ DivisorSigma[0, m + k], {k, 0, n - 1}]], m++]; m, {n, 63}] (* _Michael De Vlieger_, Jun 22 2016 *)

%o (PARI) a(n) = {my(m = 1); while (sum(k=0, n-1, m+k) % sum(k=0, n-1, numdiv(m+k)), m++); m;} \\ _Michel Marcus_, Jun 20 2016

%Y Cf. A000005, A160922.

%K nonn,easy

%O 1,3

%A _Paolo P. Lava_, Jun 16 2016