[seqfan] Re: Fw: Closed form?

Prof. Dr. Alois Heinz heinz at hs-heilbronn.de
Thu Apr 30 17:00:22 CEST 2009

```I have no other "closed form" (yet) but your observation is
very interesting.  It seems that there is a constant c=0.9233..
with a(n)/f(n) -> c for n -> oo.

Digits:=70:
f:= proc(x) option remember;
evalf (sum (x^k/k!/2^binomial(k,2), k=0..infinity)) end:
g:= proc(b, n) option remember; local t;
if b<0 then 0 elif b=0 or n=0 then 1
elif b>=n then add (g(b-t, n) *binomial (n+1, t) *(-1)^(t+1), t=1..n+1)
else g(b-1, n) +g(2*b, n-1) fi end:
a:= proc(n) local t; t:= nops (convert (n, base, 2)); g(n /2^(t-1), t) end:
for p from 25 to 30 do m:= 10^p: print (a(m)/f(m)) od:

.9233049132867869282854328449063186689783879094200987655866432714032601
.9233055492034789360128094855433414334325830219927073909182819254619767
.9233101656306179062025549798530255741041873743718112821787767740848994
.9233086902130197332283162644294024178639791138718439995819446220587922
.9233045005693638679516587374502832667457864006480990249814586610979267
.9233069322864489830441904416767875050659436261404406133336182581267821

m:= 10^30;     1000000000000000000000000000000

evalf (a(m));

.4231336582427852099242942356652580530952967820713964111848326519868935*10^1359

evalf (f(m)*.923307);

.4231336892745922423499436238080564317955152484472213702015459481593758*10^1359

a(m);

42313365824278520992429423566525805309529678207139641118483265198689345959
64001498167167356162114676737033632820724092651564984491621353774573247441
17790927298365767017684969367859091298253645336041872105238091343372306822
94623001508975382388339463360770725583084128391152252697794817712257231571
76488503389110577499142214340931183921986321495089914066792977324090497161
17990051150784185718361670651376732838840128542211522321488813474518955709
67765040976545024341662818890382963109707530684608201077993658545710924773
00122686813233536547602509753531560891116878392164106017595251070471997875
07508895754597191073147240314556836995396699699872117200670473542995511615
12576627967466902892691509662131791519277857223010879474021158610522938478
88184581491020427189764706517371783708752078369443081374533604831642789703
38811101596056096812931270255663587973451415184920132961637401695473206487
07561137676816567475936611143912497666013420676158427595003672683929597324
02498560155938188237212338723139810006700147003249652985656567565353228299
90553069535098565561577658106713984610497397089587512357735103138220610758
16468674304481286484436640458511803226356815114002223707691750302012254291
59807609191188841625495786370819957175960100297613697220739486315969978988
10143508269051951861961688261109117245862189596342621203700512812621232068
802130794670107610191711594

David Wilson schrieb:

>
>
>A recurrence for A000123 is a(n) = a(n-1) + a([n/2]).  This gives
>a(n)-a(n-1) = a([n/2]). One might hope to approximate a with real function f
>satisfying f'(n) = f(n/2). Setting f(0) = a(0) = 1, power series analysis of
>f'(n) = f(n/2) gives the power series below.
>
>f is encouragingly close to A000123 for small values of f (<= 10000) but
>drops away for larger values. Maybe someone with better math software can
>fudge the equations. Maybe f'(n-1/2) = f(n/2) is closer.
>
>
>

```