It needs to find an exponential generating function for the next sequence: $(2^n-1)B_n$. Where $B_n$ is the n-th number of Bernoulli. I found that exponential generating function for sequence of $B_n$ is $B(t)=t/(e^t-1)$. But I do not know what to do with $2^nB_n$. Thank you in advance.

  • $\begingroup$ Hint: what happens if you replace $t$ by $2t$ ? $\endgroup$ – Raymond Manzoni Nov 9 '14 at 12:00
  • $\begingroup$ I think about it, but then it seemed very easy to me and I think that there is not right way of solution $\endgroup$ – Lex Nov 9 '14 at 12:01
  • $\begingroup$ The right answer is $2t/(e^{2t}-1)$? $\endgroup$ – Lex Nov 9 '14 at 12:02
  • $\begingroup$ the expansion in powers of $t$ will be replaced by the expansion if powers of $(2t)$ so.... $\endgroup$ – Raymond Manzoni Nov 9 '14 at 12:02
  • $\begingroup$ so $2t/(e^{2t}-1)$? $\endgroup$ – Lex Nov 9 '14 at 12:05

Hint 1: what happens if you replace $t$ by $2t$ ?
Hint 2: subtract the two generating functions.


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