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%I #25 Mar 04 2018 19:47:21
%S 0,1,1,2,2,1,3,3,3,2,4,4,5,5,3,5,5,7,8,8,5,6,6,9,11,13,13,8,7,7,11,14,
%T 18,21,21,13,8,8,13,17,23,29,34,34,21,9,9,15,20,28,37,47,55,55,34,10,
%U 10,17,23,33,45,60,76,89,89,55,11,11,19,26,38,53,73,97,123,144,144
%N Table read by ascending antidiagonals: T(n,k) = A000045(k+1)*n + A000045(k).
%C T(n,k) is entry k in the Fibonacci-like sequence which starts n, n+1, i.e., T(n,0) = n, T(n,1) = n+1, T(n,k) = T(n,k-1) + T(n,k-2).
%C Therefore, T(0,k), T(1,k), and T(2,k) are equivalent to A000045(k), A000045(k+2), and A000045(k+3) respectively.
%C Any two rows T(x,.) and T(y,.) contain infinitely many differing terms for any values of x and y, provided that max(x,y) >= 3.
%C Column k is the list of all positive integers congruent to A000045(k) mod A000045(k+1), with the exception of row 0 which is the same as row 1 with 0 included.
%H Ely Golden, <a href="/A292030/b292030.txt">Table of n, a(n) for n = 0..10000</a>(contains 140 antidiagonals, flattened).
%H Ely Golden, <a href="/A292030/a292030.txt">List of ordered pairs (j,k) such that T(j,k)=n for n = 0..10000</a>
%e T(4,2) = 9 because 9 is element 2 in the Fibonacci-like sequence starting with 4 and 5.
%e The array begins:
%e 0, 1, 1, 2, 3, 5, 8, ...
%e 1, 2, 3, 5, 8, 13, 21, ...
%e 2, 3, 5, 8, 13, 21, 34, ...
%e 3, 4, 7, 11, 18, 29, 47, ...
%e 4, 5, 9, 14, 23, 37, 60, ...
%e 5, 6, 11, 17, 28, 45, 73, ...
%e 6, 7, 13, 20, 33, 53, 86, ...
%o (Python)
%o def nextAntidiagonal(d):
%o if(d[0]==0): return [d[1]+1,0]
%o return [d[0]-1,d[1]+1]
%o def fibonacciM(m,n):
%o f0,f1=0,1
%o if(n<0):
%o for _ in range(-n): f1,f0=f0,f1-f0
%o else:
%o for _ in range(n): f0,f1=f1,f0+f1
%o return f1*m+f0
%o d=[0,0]
%o for i in range(10001):
%o print(str(i)+" "+str(fibonacciM(d[0],d[1])))
%o d=nextAntidiagonal(d)
%o (PARI) T(n, k) = fibonacci(k+1)*n + fibonacci(k);
%o tabl(nn) = matrix(nn+1,nn+1, i,j, T(i-1,j-1)); \\ _Michel Marcus_, Sep 27 2017
%Y Cf. A292031, A292032.
%K nonn,tabl
%O 0,4
%A _Ely Golden_, Sep 07 2017
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