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A090825
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Nonprimes n such that (3/2)*(1/n)*(2*n+1)*(3^n+1)*B(2*n) is an integer, where B(k) denotes the k-th Bernoulli number.
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0
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1, 49, 91, 119, 133, 169, 217, 221, 247, 259, 289, 301, 323, 329, 335, 343, 361, 403, 413, 427, 469, 481, 497, 511, 517, 527, 553, 559, 589, 611, 629, 637, 679, 703, 707, 721, 731, 749
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Conjecture: composite numbers with all prime factors in A053176 are in the sequence. For p prime (3/2)*(1/p)*(2*p+1)*(3^p+1)*B(2*p) == 1 (mod p). There are few terms n with (3/2)*(1/n)*(2*n+1)*(3^n+1)*B(2*n) == 1 (mod n): 91,247,....Is this subsequence finite?
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LINKS
| F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and A. R. Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), #07.1.2.
F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and A. R. Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence [pdf, ps].
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PROG
| (PARI) for(n=1, 750, if(frac( (3/2)*(1/n)*(2*n+1)*(3^n+1)*bernfrac(2*n))==0, if(isprime(n)==0, print1(n, ", "))))
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CROSSREFS
| Sequence in context: A006832 A093894 A158725 * A157342 A118886 A198773
Adjacent sequences: A090822 A090823 A090824 * A090826 A090827 A090828
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KEYWORD
| nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Feb 11 2004
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