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A379525 a(1) = 1. For n > 1, if a(n-1) is a novel term, a(n) = A003132(a(n-1)). If a(n-1) has been seen k (>1) times already, a(n) = k*a(n-1).

%I A379525 #15 Dec 26 2024 20:08:09
%S A379525 1,1,2,4,16,37,58,89,145,42,20,4,8,64,52,29,85,89,178,114,18,65,61,37,
%T A379525 74,65,130,10,1,3,9,81,65,195,107,50,25,29,58,116,38,73,58,174,66,72,
%U A379525 53,34,25,50,100,1,4,12,5,25,75,74,148,81,162,41,17,50,150
%N A379525 a(1) = 1. For n > 1, if a(n-1) is a novel term, a(n) = A003132(a(n-1)). If a(n-1) has been seen k (>1) times already, a(n) = k*a(n-1).
%C A379525 2 and 3 appear only once since the only way they can be expressed as the sum of squares is as 2 = 1^2 + 1^2 and 3 = 1^2 + 1^2 + 1^2 (see Example). Maximum number of times prime(k) can appear is A379551(k).
%C A379525 Conjecture: Every positive integer appears in the sequence at least once.
%H A379525 Michael De Vlieger, <a href="/A379525/b379525.txt">Table of n, a(n) for n = 1..10000</a>
%H A379525 Michael De Vlieger, <a href="/A379525/a379525.png">Log log scatterplot of a(n)</a>, n = 1..2^20.
%e A379525 a(2) = 1^2 = 1 so a(3) = 2 since 1 has occurred twice. This is the only occasion of a(n) = 2 in the sequence. Because a(3) = 2 is a novel term, a(4) = 2^2 = 4, then a(5) = 4^2 = 16.
%e A379525 a(8) = a(18) = 89, so a(19) = 2*89 = 178.
%e A379525 a(28) = 10, a novel term, so a(29) = 1, the 3rd occurrence of 1, so a(30) = 3 (the only occasion of 3 in the sequence).
%t A379525 nn = 120; c[_] := 0; a[1] = j = 1;
%t A379525 Do[If[c[j] == 0,
%t A379525     k = Total[IntegerDigits[j]^2]; c[j]++,
%t A379525     k = ++c[j]*j ];
%t A379525   Set[{a[n], j}, {k, k}], {n, 2, nn}];
%t A379525 Array[a, nn] (* _Michael De Vlieger_, Dec 25 2024 *)
%Y A379525 Cf. A003132, A379551.
%K A379525 nonn,base
%O A379525 1,3
%A A379525 _David James Sycamore_, Dec 24 2024