|
Search: id:A117030
|
|
|
| A117030 |
|
a(1) = a(2) = 1; a(n) = a(n - 1)*a(n - 2) - a(n - 3) - a(n - 4) - ... - a(1) for n>2. |
|
+0
2
|
|
| 1, 1, 1, 0, -2, -3, 3, -10, -28, 279, -7803, -2177000, 16987130758, -36980983660158439, -628200804994572838287982201, 23231483704802676028750227275477328286998042, -14594036764575342428539025427350979161630036659925283421091485142638200
(list; graph; listen)
|
|
|
OFFSET
|
1,5
|
|
|
COMMENT
|
Form the product of the previous two terms, and then subtract all other previous terms.
Additionally, with a(1)=1, a(2)=2, this gives: 1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10... cf. A008619
|
|
MATHEMATICA
|
f[s_] := Block[{}, Append[s, s[[ -1]]s[[ -2]] - Plus @@ Drop[s, -2]]]; Nest[f, {1, 1}, 15] (from Robert G. Wilson v (rgwv(at)rgwv.com), May 26 2006)
|
|
PROGRAM
|
(C) #include <stdio.h> #include <inttypes.h> int main (void) { int64_t n1=1; int64_t n2=1; int i; int64_t sum=0, next; printf("%lld, %lld, ", n1, n2); for (i=0; i<12; i++) { next=n1*n2-sum; sum+=n1; n1=n2; n2=next; printf("%lld, ", n2); } }
(PARI) {m=16; a=1; b=1; print1(a=1, ", ", b=1, ", "); v=[]; for(n=3, m, print1(k=a*b-sum(j=1, #v, v[j]), ", "); v=concat(v, a); a=b; b=k)} - (Klaus Brockhaus)
|
|
CROSSREFS
|
Cf. A117157.
Adjacent sequences: A117027 A117028 A117029 this_sequence A117031 A117032 A117033
Sequence in context: A123027 A100652 A094416 this_sequence A009097 A112858 A127003
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
Gabriel Finch (salsaman(AT)xs4all.nl), Apr 16 2006
|
|
EXTENSIONS
|
a(12) corrected; a(15) and a(16) from Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Apr 17 2006
|
|
|
Search completed in 0.002 seconds
|
