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A125003
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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.
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3
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0, 5, 11, 19, 31, 59, 137, 337, 795, 1767, 3759, 7813, 16097, 33075, 67793, 138347, 280677, 566041, 1136129, 2274529, 4554047, 9143516, 18450225, 37464726, 76561127, 157278265, 324136399, 668557741, 1376893670, 2826272837
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OFFSET
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0,2
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COMMENTS
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a(0) = 0, a(1) = 5; binomial transform of sequence gives successive digits of sequence.
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LINKS
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EXAMPLE
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Triangle of successive differences begins:
0...5...11...19....31....59....137...337...795....
..5...6....8....12....28....78....200...458
....1....2....4....16....50....122...258
......1....2....12....34....72....136
.........1...10....22....38....64
...........9....12....16....26
..............3.....4....10
.................1.....6
....................5
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MAPLE
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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
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CROSSREFS
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KEYWORD
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nonn,easy,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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