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A125769
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a(n) is the least number j such that j*T_k +/- 1 is n-th prime for some k-th triangular number.
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6
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1, 2, 1, 1, 1, 2, 3, 2, 4, 1, 2, 1, 2, 2, 8, 9, 4, 4, 1, 2, 2, 1, 3, 2, 16, 10, 17, 3, 2, 4, 6, 2, 1, 5, 10, 10, 2, 27, 6, 29, 4, 2, 1, 32, 3, 3, 1, 8, 38, 23, 3, 2, 2, 7, 43, 4, 6, 2, 1, 10, 47, 14, 2, 4, 4, 53, 5, 12, 58, 35, 59, 3, 61, 62, 1, 64, 5, 6, 40, 3, 2, 2, 12, 12, 8, 74, 10, 76, 2, 2, 6, 4
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Eventually all primes p appear since (p +/-1) times 1(1+1)/2 equals (p +/- 1).
If we asked for the least number k then k always equals 1 since all primes p appear since (p +/-1) times 1(1+1)/2 equals (p +/- 1).
The k's for the corresponding j's are: round(sqrt(2p/j)).
First occurrence of i is A125770: 1, 2, 7, 9, 34, 31, 54, 15, 16, 26, 148, 68, 398, 62, 193, 25, 27, 140, 550, 397, 107, 113, ...,.
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EXAMPLE
| a(1) = 1 because 1*1+1 = 2 which is the first prime,
a(2) = 2 because 2*1+1 = 3 which is the second prime,
a(3) = 4 because 1*6-1 = 5 which is the third prime,
a(8) = 3 because 2*10-1 = 19 which is the eighth prime, ...
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MATHEMATICA
| triQ[n_] := IntegerQ@ Sqrt[8n + 1]; f[n_] := Block[{j = 1, p = Prime@n}, While[ !triQ[(p - 1)/j] && !triQ[(p + 1)/j], j++ ]; j]; Array[f, 92]
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CROSSREFS
| Cf. A000217, A125765, A125766, A125767, A125768, A125770.
Sequence in context: A178526 A039958 A029344 * A003023 A156070 A114731
Adjacent sequences: A125766 A125767 A125768 * A125770 A125771 A125772
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KEYWORD
| nonn
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AUTHOR
| Jonathan Vos Post (jvospost3(AT)gmail.com) & Robert G. Wilson v (rgwv(AT)rgwv.com), Dec 01 2006
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