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%I #31 Nov 17 2019 15:54:53
%S 574,1185,1240,1269,1376,1906,1910,2572,2689,2980,3238,3384,3801,3904,
%T 4121,4570,4691,4789,5236,5862,5902,6227,6332,6402,6438,6568,7234,
%U 7900,8113,8434,8543,8557,8566,9232,9611,9670,9824,9891,9898,10564,10587,10754,11230,11247,11535,11691,11896,12562,12965,13019,13228,13246,13355,13484,13894,14560,14714,14957,15176,15226,15346,15892,16558,16668,16944,17035,17224,17387,17890,18379,18406,18534,18556,18761,19222,19598,19888,20090
%N Numbers n such that numerator(Bernoulli(2*n)/(2*n)) is different from numerator(Bernoulli(2*n)/(2*n*(2*n-1))).
%C _Michael Somos_ (Feb 01 2004) discovered the remarkable fact that A001067 is different from A046968, even though they agree for the first 573 terms.
%C Numbers n such that A001067 is different from A046968, or alternatively, those n such that gcd(A001067(n),2n-1) is > 1.
%C If gcd(A000367(n), A000367(n+2)) <>1 then n = A090495(n) - (3*A090496(n) + 1)/2. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Feb 08 2004
%C So far, all terms correspond to irregular primes. Notice that these numbers are generated by n=((2k+1)p+1)/2 where p is an irregular prime and k is some integer = 1,2,... . In the Excel spreadsheet provided at the link, you will notice that much larger firstborn irregular primes p tend to produce smaller values of k. E.g., p = 691, 683, 653, k = 5, 15, 23. So by some guessing we could test a given large irregular prime for the first few values of k. I found ip's 257, 293, 311 this way, but not the index. Also the spreadsheet shows the corresponding irregular primes where the Bacher forecast fails for firstborn irregular prime. - _Cino Hilliard_, Feb 15 2004
%H Robert G. Wilson v, <a href="/A090495/b090495.txt">Table of n, a(n) for n = 1..200 </a>
%H Cino Hilliard, <a href="/A090495/a090495.pdf">Bernoulli ratios</a> [posted on Yahoo group B2LCC, Feb 04 2004]
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/StirlingsSeries.html">Stirling's Series</a>
%p a := n->numer(bernoulli(2*n)/(2*n)): b := n->numer(bernoulli(2*n)/(2*n*(2*n-1))): for n from 1 to 2000 do if a(n)<>b(n) then print(n,a(n)/b(n)); fi; od:
%t a[n_] := Numerator[BernoulliB[2n]/(2n)] (* A001067 *); b[n_] := Numerator[BernoulliB[2n]/(2n(2n-1))] (* A046968 *); For[n=1, n <= 580, n++, If[ a[n] != b[n], Print[n, " ", a[n]/b[n]] ] ]
%t k = 1; lst = {}; While[k < 38001, b = BernoulliB[2 k]; If [Numerator[b/(2 k)] != Numerator[b/(2 k (2 k - 1))], AppendTo[lst, k]; Print[{k}]]; k++ ]; lst (* _Robert G. Wilson v_, Aug 19 2010 *)
%o (PARI) bern2(c,m1,m2) = { for(n=m1,m2, n2=n+n; a = numerator(bernfrac(n2)/(n2)); \ A001067 b = numerator(a/(n2-1)); if(a <> b,print("A("c") = "n","a/b);c++) ) } \\ _Cino Hilliard_
%Y Cf. A090496, A001067, A046968, A092291. A274297.
%K nonn,nice
%O 1,1
%A _N. J. A. Sloane_, Feb 03 2004
%E a(1)-a(7) from _Michael Somos_ and _W. Edwin Clark_, Feb 03 2004
%E a(8)-a(9) from _Robert G. Wilson v_, Feb 03 2004
%E a(10)-a(12) from _Eric W. Weisstein_, Feb 03 2004
%E a(13)-a(39) from _Cino Hilliard_, Feb 03 2004
%E a(40) from _Eric W. Weisstein_, Feb 04 2004
%E Many further terms from _Cino Hilliard_, Feb 15 2004
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