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Digitsum of a(n) + digitsum of a(n+1) divides a(n+2). This is the lexicographically earliest sequence of distinct positive terms with this property.
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%I #29 Dec 20 2023 08:04:20

%S 1,2,3,5,8,13,12,7,10,16,24,26,14,39,17,20,30,15,9,45,18,36,54,72,90,

%T 108,126,144,162,180,198,27,81,216,234,252,270,288,135,189,243,297,

%U 324,351,306,342,360,378,405,432,396,459,468,504,486,513,540,414,450,522,558,567,576,612,594,621,648,675,684

%N Digitsum of a(n) + digitsum of a(n+1) divides a(n+2). This is the lexicographically earliest sequence of distinct positive terms with this property.

%H Michael De Vlieger, <a href="/A364120/b364120.txt">Table of n, a(n) for n = 1..10000</a>

%H Eric Angelini, <a href="http://cinquantesignes.blogspot.com/2023/07/supersums-superproducts.html">SuperSums, SuperProducts</a>, personal blog.

%H Michael De Vlieger, <a href="/A364120/a364120.png">Log log scatterplot of a(n)</a>, n = 1..2^20.

%e digitsum a(1) + digitsum a(2) = 1 + 2 = 3 and 3 divides exactly a(3) = 3;

%e digitsum a(2) + digitsum a(3) = 2 + 3 = 5 and 5 divides exactly a(4) = 5;

%e digitsum a(3) + digitsum a(4) = 3 + 5 = 8 and 8 divides exactly a(5) = 8;

%e digitsum a(4) + digitsum a(5) = 5 + 8 = 13 and 13 divides exactly a(6) = 13;

%e digitsum a(5) + digitsum a(6) = 8 + 1 + 3 = 12 and 12 divides exactly a(7) = 12; etc.

%t nn = 80; c[_] := False; m[_] := 1; Array[Set[{a[#], c[#]}, {#, True}] &, 2]; d = i = 1; dd = j = 2; Do[k = d + dd; While[c[k m[k]], m[k]++]; k *= m[k]; Set[{a[n], c[k], i, j, d, dd}, {k, True, j, k, dd, Total@ IntegerDigits[k]}], {n, 3, nn}]; Array[a, nn] (* _Michael De Vlieger_, Jul 12 2023 *)

%o (PARI) {a=List(u=[1,2]); for(n=2,99, s=sumdigits(a[n-1])+sumdigits(a[n]);

%o forstep(k=s,oo,s, set search (u,k)&& next; listput(a,k); u=setunion(u,[k]);

%o break));a}

%Y Cf. A007953, A364187, A364188.

%K base,nonn

%O 1,2

%A _Eric Angelini_ and _M. F. Hasler_, Jul 12 2023

%E Edited definition. - _N. J. A. Sloane_, Aug 26 2023