Search: id:A022095 Results 1-1 of 1 results found. %I A022095 %S A022095 1,5,6,11,17,28,45,73,118,191,309,500,809,1309,2118,3427, %T A022095 5545,8972,14517,23489,38006,61495,99501,160996,260497, %U A022095 421493,681990,1103483,1785473,2888956,4674429,7563385 %N A022095 Fibonacci sequence beginning 1 5. %C A022095 a(n-1)=sum(P(5;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=4. These are the sums of the SW-NE diagonals in P(5;n,k), the (5,1) Pascal triangle A093562. Observation by Paul Barry (pbarry(AT)wit.ie, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also sums of the SW-NE diagonals in the (1,4)-Pascal triangle A095666. %H A022095 Index entries for sequences related to linear recurrences with constant coefficients %H A022095 Tanya Khovanova, Recursive Sequences %F A022095 a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=5. a(-1):=4. %F A022095 G.f.: (1+4*x)/(1-x-x^2). %F A022095 Row sums of triangle A131776 starting (1, 5, 6, 11, 17, 28,...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 14 2007 %F A022095 a(n)=4*fibonacci(n-1)+fibonacci(n), n>=1 - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 05 2007 %F A022095 a(n)=((1+sqrt5)^n-(1-sqrt5)^n)/(2^n*sqrt5)+ 2*((1+sqrt5)^(n-1)-(1-sqrt5)^(n-1))/ (2^(n-2)*sqrt5). Offset 1. a(3)=6 [From Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009] %p A022095 a:=n->4*fibonacci(n-1)+fibonacci(n): seq(a(n), n=1..32); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 05 2007 %t A022095 a={};b=1;c=5;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c; AppendTo[a,c],{n,1,9,1}];a (Vladimir Orlovsky, Jul 22 2008) %Y A022095 a(n) = A101220(4, 0, n+1). %Y A022095 a(n) = A109754(4, n+1). %Y A022095 Cf. A131776. %Y A022095 Sequence in context: A136974 A101187 A070373 this_sequence A042531 A042839 A041373 %Y A022095 Adjacent sequences: A022092 A022093 A022094 this_sequence A022096 A022097 A022098 %K A022095 nonn %O A022095 0,2 %A A022095 N. J. A. Sloane (njas(AT)research.att.com). Search completed in 0.001 seconds