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COMMENTS
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All natural integers will appear sooner or later in the sequence (from the definition) - but mostly "later"! Indeed, the sequence increases very slowly: after 100000 terms the smallest term not yet present is 32.
Here is, in the same range, a sample of the count {term, occurrences} so far:
{1,192},{2,396},{3,618},{4,796},{5,1160},{6,1296},{7,2294},{8,2080},{9,2489},{10,2826},{11,3487},{12,1596},{13,2295},{14,1960},{15,2370},{16,2640},{17,4097},{18,2214},{19,4598},{20,2770},{21,3759},{22,4477},{23,5612},{24,4884},{25,5825},{26,6006},{27,6359},{28,4676},{29,5481},{30,3060},{31,1411},{32,0},{33,182},{34,0},{35,315},{36,0},{37,1221},{38,0},{39,214},{40,0},{41,1353},{42,0},{43,1183},{44,0},{45,0},{46,0},{47,1058},{48,0},{49,172},{50,0},{51,0},{52,0},{53,580},...
After 100000 terms, the first products that are not yet present are (the primes): 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, ... and (the composites) 118, 122, 134, ...
Here is again a sample so far (100000 terms computed) of {product, number of occurrences of the product}:
{1,1},{2,2},{3,3},{4,4},{5,5},{6,6},{7,7},{8,8},{9,9},{10,10},{11,11},{12,12},{13,13},{14,14},{15,15},{16,16},{17,17},{18,18},{19,19},{20,20},{21,21},{22,22},{23,23},{24,24},{25,25},{26,26},{27,27},{28,28},{29,29},{30,30},{31,31},{32,32},{33,33},{34,34},{35,35},{36,36},{37,37},{38,38},{39,39},{40,40},{41,41},{42,42},{43,43},{44,44},{45,45},{46,46},{47,47},{48,48},{49,49},{50,50},{51,51},{52,52},{53,53},{54,54},{55,55},{56,56},{57,57},{58,58},{59,0},{60,60},{61,0},{62,62},{63,63},{64,64},{65,65},{66,66},{67,0},{68,68},{69,69},{70,70},{71,0},{72,72},{73,0},{74,74},{75,75},{76,76},{77,77},{78,78},{79,0},{80,80},{81,81},{82,82},{83,0},{84,84},{85,85},{86,86},{87,87},{88,88},{89,0},{90,90},{91,91},{92,92},{93,93},{94,94},{95,95},{96,96},{97,0},{98,98},{99,99},{100,100},{101,0},{102,102},{103,0},{104,104},{105,105},{106,106},{107,0},{108,108},{109,0},{110,110},{111,111},{112,112},{113,0},{114,114},{115,115},{116,116},{117,117},{118,0},{119,119},{120,120},{121,121},{122,0},{123,123},{124,124},{125,125},{126,126},{127,0},{128,128},{129,129},{130,130},{131,0},{132,132},{133,133},{134,0},{135,135},{136,136},{137,0},{138,138},{139,0},{140,140},{141,141},{142,0},...
Theorem. This sequence can also be defined by a greedy algorithm. That is, let b(1)=1, and for n >= 1, let b(n+1) be the smallest positive integer k such that m = k*b(n) has appeared at most n-1 times in the list [b(i)*b(i+1): i=1..n-1]. Then b(n) = a(n) for all n >= 1.
(Note that for n=1 the list is empty, and so we take k = b(1) = 1.)
Remark: The theorem is not obvious and requires a proof, given in a link below. "Lexicographically earliest" sequences often require some backtracking, but the point of the theorem is that no backtracking is needed here.
The proof also shows that there are infinitely many 1's in the sequence, and that each k appears k times in the sequence of products a(i)*a(i+1). (End)
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