|
%I #9 Aug 12 2015 20:58:59
%S 0,5,11,19,31,59,137,337,795,1767,3759,7813,16097,33075,67793,138347,
%T 280677,566041,1136129,2274529,4554047,9143516,18450225,37464726,
%U 76561127,157278265,324136399,668557741,1376893670,2826272837
%N a(0) = 0, a(1) = 5; for n>1, a(n) is determined by the rule that the concatenation of the leading terms of the difference triangle is the same as the concatenation of the digits of the sequence.
%C a(0) = 0, a(1) = 5; binomial transform of sequence gives successive digits of sequence.
%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%e Triangle of successive differences begins:
%e 0...5...11...19....31....59....137...337...795....
%e ..5...6....8....12....28....78....200...458
%e ....1....2....4....16....50....122...258
%e ......1....2....12....34....72....136
%e .........1...10....22....38....64
%e ...........9....12....16....26
%e ..............3.....4....10
%e .................1.....6
%e ....................5
%p revert := proc(n) local Linv,i,L ; L := convert(n,base,10) ; Linv := [] ; for i from 1 to nops(L) do Linv := [op(Linv), op(-i,L)] ; od ; RETURN(Linv) ; end: A125003 := proc(nmax) local ldigs,T,diag,row ; T := array(1..nmax,1..nmax) ; ldigs := [0,5,1,1] ; T[1,1] := ldigs[1] ; T[1,2] := ldigs[2] ; T[2,1] := ldigs[2] ; for diag from 3 to nmax do T[diag,1] := ldigs[diag] ; for row from diag-1 to 1 by -1 do T[row,diag-row+1] := T[row,diag-row]+T[row+1,diag-row] ; od ; if diag > 3 then ldigs := [op(ldigs), op(revert(T[1,diag])) ] ; fi ; od ; RETURN(T) ; end : nmax := 50 : T := A125003(nmax) : for i from 1 to nmax do printf("%d,",T[1,i]) ; od ; # _R. J. Mathar_, Jan 10 2007
%Y Cf. A125588, A125004, A125591.
%K nonn,easy,base
%O 0,2
%A _Eric Angelini_, Jan 06 2007
%E More terms from _R. J. Mathar_, Jan 10 2007
|
