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a(n) = 2*A000111(n+1) + A000111(n).
1

%I #17 Jun 15 2022 01:49:48

%S 3,3,5,12,37,138,605,3042,17257,108978,758105,5759322,47439277,

%T 421090218,4006875605,40686781602,439122198097,5019624693858,

%U 60582649901105,769831261587882,10273367294485717,143649246839399898,2100196647406842605,32044492213621026162,509357494543973054137

%N a(n) = 2*A000111(n+1) + A000111(n).

%C It is clear from the Berry et al. article that they intended to consider 2*A000111(n+1) - A000111(n) (which is A104854), not 2*A000111(n+1) + A000111(n).

%H D. Berry, J. Broom, D. Dixon, and A. Flaherty, <a href="https://www.math.lsu.edu/system/files/DeAngelisProject2.pdf">Umbral Calculus and the Boustrophedon Transform</a>, 2013.

%F E.g.f.: 1 + (sec(x) + tan(x) + 1)*(sec(x) + tan(x)). - _Sergei N. Gladkovskii_, Jun 11 2015

%o (Python)

%o from itertools import accumulate, islice

%o def A231895_gen(): # generator of terms

%o yield 3

%o blist = (0,1)

%o while True:

%o yield blist[-1]+2*(blist := tuple(accumulate(reversed(blist),initial=0)))[-1]

%o A231895_list = list(islice(A231895_gen(),40)) # _Chai Wah Wu_, Jun 14 2022

%Y Cf. A000111, A104854.

%K nonn

%O 0,1

%A _N. J. A. Sloane_, Nov 18 2013