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A380494 a(1) = 1. For n > 1 a(n) is the smallest positive integer not yet in the sequence which is divisible by A007953(a(n-1)) + 1.

%I A380494 #10 Jan 29 2025 12:47:50
%S A380494 1,2,3,4,5,6,7,8,9,10,12,16,24,14,18,20,15,21,28,11,27,30,32,36,40,25,
%T A380494 48,13,35,45,50,42,49,56,60,63,70,64,22,55,33,77,75,26,54,80,72,90,
%U A380494 100,34,88,17,81,110,39,52,96,112,65,84,78,128,108,120,44,99
%N A380494 a(1) = 1. For n > 1 a(n) is the smallest positive integer not yet in the sequence which is divisible by A007953(a(n-1)) + 1.
%C A380494 For n > 1 a(n) is the smallest novel multiple of digsum(a(n-1)) + 1. The "+1" is intended to avoid an infinite number of consecutive multiples of 9, which would otherwise occur after a(13). The only way a(n) = prime p can occur is if A007953(a(n-1)) is the first occasion of p-1. Sequence is conjectured to be a permutation of the natural numbers, with primes in order.
%H A380494 Michael De Vlieger, <a href="/A380494/b380494.txt">Table of n, a(n) for n = 1..10000</a>
%H A380494 Michael De Vlieger, <a href="/A380494/a380494.png">Log log scatterplot of a(n)</a>, n = 1..10^6.
%e A380494 a(1) = 1 has digsum = 1, therefore a(2) = 2, the smallest novel multiple of 1+1 = 2.
%e A380494 Likewise, a(n) = n for all n <= 10. a(11) = 12 because 1 + digsum(10) = 2 and 12 is the smallest number not already a term which is divisible by 2.
%e A380494 a(20) = 28, and digsum(28) + 1 = 11. Therefore a(21) = 11 because 28 is the first occasion of digsum(a(n-1)) = 10.
%t A380494 nn = 10^4; j = 1; u = 2; c[_] := False; m[_] := 1;
%t A380494 {j}~Join~Reap[Do[
%t A380494   k = u; d = DigitSum[j] + 1;
%t A380494   While[c[Set[k, m[d]*d]], m[d]++];
%t A380494   Set[{j, c[k]}, {k, True}]; Sow[k];
%t A380494   If[k == u, While[c[u], u++]],
%t A380494   {n, 2, nn}] ][[-1, 1]]
%Y A380494 Cf. A007953, A051855.
%K A380494 nonn,base,easy
%O A380494 1,2
%A A380494 _David James Sycamore_ and _Ali Sada_, Jan 25 2025
%E A380494 More terms from _Michael De Vlieger_, Jan 25 2025.