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%I #20 Jan 30 2013 02:50:50
%S 1,1,-3,1,-5,5,1,-7,25,-5,1,-9,23,-35,49,1,-11,73,-27,112,-49,1,-13,
%T 53,-77,629,-91,58,1,-15,145,-130,1399,-451,753,-58,1,-17,95,-135,
%U 2699,-2301,8573,-869,341,1,-19,241
%N Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.
%C In A190339 we saw that (the second Bernoulli numbers) A164555/A027642 is an eigensequence (its inverse binomial transform is the sequence signed) of the second kind, see A192456/A191302. We consider this array preceded by 1 for the second row, by 1, -3/2, for the third one; 1 is chosen and is followed by the differences of successive rows.
%C Hence
%C 1 1/2 1/6 0
%C 1 -1/2 -1/3 -1/6 -1/30
%C 1 -3/2 1/6 1/6 2/15 1/15
%C 1 -5/2 5/3 0 -1/30 -1/15 -8/105.
%C The second row is A051716/A051717.
%C The (reduced) triangle before the square array (T(n,m) in A190339) is a(n)/b(n)=
%C B(0)= 1 = 1 Redbernou1li
%C B(1)= -1/2 = 1 -3/2
%C B(2)= 1/6 = 1 -5/2 5/3
%C B(3)= 0 = 1 -7/2 25/6 -5/3
%C B(4)=-1/30 = 1 -9/2 23/3 -35/6 49/30
%C B(5)= 0 = 1 -11/2 73/6 -27/2 112/15 -49/30.
%C For the main diagonal, see A165142.
%C Denominator b(n) will be submitted.
%C This transform is valuable for every eigensequence of the second kind. For instance Leibniz's 1/n (A003506).
%C With increasing exponents for coefficients, polynomials CB(n,x) create Redbernou1li. See the formula.
%C Triangle Bernou1li for A027641/A027642 with the same denominator A080326 for every column is
%C 1
%C 1 -3/2
%C 1 -5/2 10/6
%C 1 -7/2 25/6 -10/6
%C 1 -9/2 46/6 -35/6 49/30
%C 1 -11/2 73/6 -81/6 224/30 -49/30.
%C For numerator by columns,see A000012, -A144396, A100536, Q(n)=n*(2*n^2+9*n+9)/2 , new.
%C Triangle Checkbernou1 with the same denominator A080326 for every row is
%C 1/1
%C (2 -3)/2
%C (6 -15 +10)/6
%C (6 -21 +25 -10)/6
%C (30 -135 +230 -175 +49)/30
%C (30 -165 +365 -405 +224 -49)/30;
%C Hence for numerator: 1, 2-3, 16-15, 31-31, 309-310, 619-619, 8171-8166.
%C Absolute sum: 1, 5, 31, 62, 619, 1238, 17337. Reduced division by A080326:
%C 1, 5/2, 31/6, 31/3, 619/30, 619/15, 5779/70, = A172030(n+1)/A172031(n+1).
%F CB(0,x) = 1,
%F CB(1,x) = 1 - 3*x/2,
%F CB(n,x) = (1-x)*CB(n-1,x) + B(n)*x^n , n > 1.
%Y Cf. A028246 (Worpitzky), A085737/A085738 (Conway-Sloane), A051714/A051715 (Akiyama-Tanigawa), A192456/A191302 for other triangles that lead to the Bernoulli numbers.
%K sign,frac,tabl
%O 0,3
%A _Paul Curtz_, Apr 27 2012
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