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<DIV>Seqfans, </DIV>
<DIV> Consider David Wilson's nice 10-adic
constant A025016: <BR>x = Sum_{n>=0} n! (10-adic)
<BR>=...92247479478684738621107994804323593105039052556442336528920420940314</DIV>
<DIV> <BR>Investigating the Bell number
analogue:<BR> <BR> B(k) = Sum_{n>=0} n^k*n!
(10-adic)<BR> <BR>I was quite surprised to find that
<BR> <BR> B(1) = -1 = ...99999999999 (10-adic).<BR> <BR>I
went further to find the remarkable relation </DIV>
<DIV>(see examples at bottom of message):<BR> <BR> B(n) =
A014182(n)*B(0) + A014619(n)<BR> <BR>But, wait - this is
a base-independent formula ! </DIV>
<DIV><BR>Does the same formula hold for p-adic bases other than 10-adic?
<DIV>I have not had time to check, but I think it does. </DIV>
<DIV> </DIV>
<DIV>This constant A025016 has a b-adic expression for all base b.
<BR>I believe that the constant A025016 is base-independent,
</DIV></DIV>
<DIV>but the digits recorded in A025016 are base-10. </DIV>
<DIV> <BR>Any comments?</DIV>
<DIV>Thanks,</DIV>
<DIV> Paul<BR> <BR>Conjectured
Formula: </DIV>
<DIV>B(n) = A014182(n)*B(0) + A014619(n)</DIV>
<DIV>Examples: <BR>B(1) = 0*B(0) - 1 <BR>B(2) = -1*B(0) + 1<BR>B(3) =
1*B(0) + 1<BR>B(4) = 2*B(0) - 5<BR>B(5) = -9*B(0) + 5<BR>B(6) = 9*B(0) +
21<BR>B(7) = 50*B(0) - 105<BR>B(8) = -267*B(0) + 141<BR>B(9) = 413*B(0) +
777<BR>B(10) = 2180*B(0) - 5513<BR>B(11) = -17731*B(0) + 13209<BR>B(12) =
50533*B(0) + 39821<BR>B(13) = 110176*B(0) - 527525<BR>B(14) = -1966797*B(0) +
2257425<BR>B(15) = 9938669*B(0) -
41511<BR> <BR> <BR>A014182 <BR>Expansion of
exp(1-x-exp(-x)). <BR>1, 0, -1, 1, 2, -9, 9, 50, -267, 413, 2180, -17731,
50533, 110176, <BR>-1966797, 9938669, -8638718, -278475061, 2540956509,
-9816860358, <BR> <BR>A014619 <BR>Exponential generating
function is -f(x) * int(exp(exp(-t)-1),t,0,x) <BR>where f(x) = exp(1-x-exp(-x))
is an exponential generating function for A014182. <BR>-1, 1, 1, -5, 5,
21, -105, 141, 777, -5513, 13209, 39821, -527525, 2257425, <BR>
<BR>END.</DIV></BODY></HTML>