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I created this puzzle. Find the next number in the sequence and the logic behind it. $$2, 1536, 5, 1792, 23, 1184, 53, 1168, 113, 1048, 137, 1384, 179, 1576, 311, 1268, 431, 1234, 719, 2026, 1031, 1097, 1499$$

Look closely at each number and its properties. This will help.

Hint 1:

Hint 2:

Yes, previous one was a hint. (And yes, this hint saying that the previous one was a hint is a hint since it will seem redundant after explicitly mentioning the previous one as Hint 1)

My other puzzles can be found here.

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    $\begingroup$ is hint 1 correct? $\endgroup$ – Sagar Chand Jun 5 at 3:27
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    $\begingroup$ Yes. @SagarChand $\endgroup$ – John Brookfields Jun 5 at 5:09
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    $\begingroup$ upvoting the question for Hint 2 XD $\endgroup$ – Sagar Chand Jun 5 at 5:34
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    $\begingroup$ Possibly a rot13-encoded beginning of an answer: vs lbh erirefr rirel bgure ahzore ovgjvfr, lbh trg n frdhrapr bs gjb,guerr,svir,frira,gjragl-guerr,sbegl-bar,svsgl-guerr naq fb ba; zbfg ahzoref ner cevzrf rkprcg guerr uhaqerq friragl-frira, frira uhaqerq naq guerr, naq bar gubhfnaq bar uhaqerq fvkgl-avar. $\endgroup$ – the default. Jun 5 at 11:50
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    $\begingroup$ Vs jr ybbx ng nygreangr ahzoref naq gnxr gurve cevzr snpgbevmngvba, jr frr gung gurl ner fbzr cbjre bs 2 zhygvcyvrq jvgu nabgure cevzr ahzore. Guvf cevzr ahzore frrzf gb or vapernfvat naq gur cbjre 2 frrzf gb or qrpernfvat. Abj vs lbh erzbir nyy gur cbjref bs 2 naq whfg sbphf ba gur cevzr ahzoref, lbh jvyy frr gung gur RAGVER frdhrapr(abg gur nygreangr bar), frrzf gb or va vapernfvat snfuvba. V fgvyy pbhyq abg svther bhg gur tnc orgjrra gur cevzr ahzoref(be jurgure guvf nccebnpu vf pbeerpg, be whfg n pbvapvqrapr) $\endgroup$ – Sagar Chand Jun 5 at 17:09
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The next number is

1997

Why:

Every other entry starting with the first (2) is a prime number. Every other entry starting with the second (1536) is a prime number multiplied by a power of two. Looking at just the prime number parts, these are the four one-digit primes 2, 3, 5, 7, and then all the primes in sequence with the property that if any one digit is deleted from the decimal representation, the remaining decimal string is also a prime number. For example, 431 is in the sequence of primes because 431, 43, 41, and 31 are all prime. 239 is not in the sequence because even though 239, 23, and 29 are prime, 39 is composite. The number in this sequence after 1499 is 1997. (1997, 997, 197, and 199 are all prime.)

The next number in the original sequence is in the subsequence that would be multiplied by a power of two. The exact rule for this factor is tricky to pin down, but: The power of two factors form a non-increasing sequence. A rule of repeatedly multiplying by two until the number is at least $N$ would give the results seen, given any $N$ in $1014 \leq N \leq 1048$. Either way, the prime 1997 must be multiplied by $2^0=1$, so the answer is just 1997.

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  • $\begingroup$ Well done @aschepler. Nice work! $\endgroup$ – John Brookfields Jun 6 at 5:42
  • $\begingroup$ There is a logic behind multiplying with the power of 2. Try to guess it. $\endgroup$ – John Brookfields Jun 6 at 5:43
  • $\begingroup$ I think the logic is simply to pad them with zeroes on the right side until a specific bit length is reached. $\endgroup$ – the default. Jun 7 at 6:28
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All the odd numbers are prime numbers including the the even prime 2. All the even numbers are a product of p*2^n where p a prime and n a positive integer. The next number in the sequence is 1549 because 1031 is a prime of the form 4x+3. 1097 is of the form 4x+1. 1499 is of the form 4x+3. 1549 is of the form 4x+1. I'm sorry I forgot to put the last number of the sequence when I first posted my answer. I've added it now, 1549.

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  • $\begingroup$ Sorry, @VassilisParassidis. That's not the correct logic and that's not the correct number too. rot13(Unq vg orra pubfra gb unir gur sbez bs $4k+r ; r\in\{1,3\}$, gurer zhfg unir orra n qrsvavgr cnggrea. Lbhe ybtvp qbrfa'g svg gur ahzore $2$ naq nyfb, $4k+1, 4k+3$ qb abg nygreangr va gur frdhrapr boivbhfyl (vs lbh unir sbhaq gur pbeerpg frdhrapr). Ohg V ernyyl nccerpvngr lbhe rssbeg ba guvf.) $\endgroup$ – John Brookfields Jun 5 at 20:50

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