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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A307720 Lexicographically earliest sequence of positive integers in which, for all positive k, there are exactly k pairs of consecutive terms whose product is k.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 3, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 2, 4, 2, 4, 2, 4, 2, 4, 3, 4, 3, 4, 3, 4, 3, 4, 3, 4, 3, 5, 1, 5, 1, 5, 1, 7, 1, 7, 1, 7, 1, 7, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 7, 2, 7, 2, 7, 2, 7, 2, 7, 2, 7, 2, 7, 3, 5, 3, 5, 3, 5, 3, 5, 3, 5, 3, 5, 3, 5, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 7, 3, 7, 3, 7, 3, 7, 3, 7, 3, 7, 3, 7, 3, 7, 3, 7, 3, 7, 3, 8, 2, 8
Offset: 1

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Author

Eric Angelini and Jean-Marc Falcoz, Apr 24 2019

Keywords

Comments

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},...
Comment from N. J. A. Sloane, Oct 19 2021: (Start)
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)

Examples

			The sequence starts with 1,1,2,1,3,1,3,2,2,2,2,2,3,...
The product a(n)*a(n+1) = 1 is true exactly once [this is the product a(1) * a(2) = 1 * 1 = 1];
The product a(n)*a(n+1) = 2 is true exactly twice [these are the products a(2) * a(3) = 1 * 2 = 2 and a(3) * a(4) = 2 * 1 = 2];
The product a(n)*a(n+1) = 3 is true exactly three times [these are the products a(4) * a(5) = 1 * 3 = 3 ; a(5) * a(6) = 3 * 1 = 3, and a(6) * a(7) = 1 * 3 = 3];
...
The product a(n)*a(n+1) = 4 is true exactly four times [these are the products a(8) * a(9) = 2 * 2 = 4 ; a(9) * a(10) = 2 * 2 = 4 ; a(10) * a(11) = 2 * 2 = 4 ; a(11) * a(12) = 2 * 2 = 4] ; and so on.
		

Crossrefs

Cf. A307707 (same idea, but with the sum of contiguous terms instead of the product), A307730 (the products), A307630 (when n appears), A307631 (indices of records), A307632 (indices of primes), A348241 and A348242 (bisections), A307633 and A307634 (RUNS transforms of bisections), A348446 (bisection differences), A348458 (partial sums).
See also A307747.

Programs

  • Mathematica
    nmax = 1000; time = {0}; v = 1;
    A307720 = Reap[For[n = 1, n <= nmax, n++, Sow[v]; For[o = 1, True, o++, While[Length[time] < o*v, time = Join[time, Table[0, {Length[time]}]]]; If[time[[o*v]]+1 <= o*v, time[[o*v]]++; v = o; Break[]]]]][[2, 1]] (* Jean-François Alcover, Oct 23 2021, after Rémy Sigrist's PARI program *)
  • PARI
    \\ See Links section.
    
  • Python
    from itertools import islice
    from collections import Counter
    def A307720(): # generator of terms. Greedy algorithm
        yield 1
        c, b = Counter(), 1
        while True:
            k, kb = 1, b
            while c[kb] >= kb:
                k += 1
                kb += b
            c[kb] += 1
            b = k
            yield k
    A307720_list = list(islice(A307720(),100)) # Chai Wah Wu, Oct 21 2021

Extensions

Definition revised slightly by Allan C. Wechsler, Apr 24 2019
Example clarified by Rémy Sigrist, Oct 24 2021