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%I #10 Dec 18 2018 00:47:44
%S 1,2,3,6,7,8,9,14,16,18,19,20,21,22,26,27,33,34,44,55,59,63,67,68,69,
%T 70,74,89,90,91,92,93,94,109,125,126,127,128,129,130,131,132,133,134,
%U 137,138,139,140,141,142,143,144,145,146,147,150,151,152,153,169
%N The lexicographically earliest increasing sequence such that n divides the sum of the first a(n) + 1 terms.
%C Sequence b(n) of the sums of the first a(n)+1 terms = Sum_{k=1..a(n)+1} a(k): 3, 6, 12, 36, 50, 66, 84, 192, 252, 330, 385, 444, 507, 574, 855, 944, 1513, ...
%C Sequence c(n) of quotients when a(n) is calculated = (Sum_{k=1..a(n)+1} a(k) ) / n: 3, 3, 4, 9, 10, 11, 12, 24, 28, 33, 35, 37, 39, 41, 57, 59, 89, ...
%C Is there a lexicographically earliest bijective sequence such that n divides the sum of the first a(n)+1 terms?
%e a(1) = 1.
%e a(2) = 2 because 2 is the smallest number > a(1) and n = 1 divides the sum of the first a(1) + 1 = 2 terms for all any term > 1.
%e a(3) = 3 because 3 is the smallest number > a(2) such that n = 2 divides the sum of the first a(2) + 1 = 3 terms.
%e a(4) = 6 because 6 is the smallest number > a(3) such that n = 3 divides the sum of the first a(3) + 1 = 4 terms.
%e a(5) = 7 and a(6) = 8; a(4) < a(5) < a(6).
%e a(7) = 9 because 9 is the smallest number > a(6) such that n = 4 divides the sum of the first a(4) + 1 = 7 terms.
%e a(8) = 14 because 14 is the smallest number > a(7) such that n = 5 divides the sum of the first a(5) + 1 = 8 terms.
%Y Cf. A316571 (similar sequence for n divides the sum of the first n+1 terms).
%Y Cf. A319736.
%K nonn
%O 1,2
%A _Jaroslav Krizek_, Sep 26 2018
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