A094568 - OEIS

%I #7 May 04 2016 10:36:25

%S 2,3,5,8,10,13,21,24,26,34,55,63,65,68,89,144,165,168,170,178,233,377,

%T 432,440,442,445,466,610,987,1131,1152,1155,1157,1165,1220,1597,2584,

%U 2961,3016,3024,3026,3029,3050,3194,4181,6765,7752,7896,7917,7920,7922

%N Triangle of binary products of Fibonacci numbers.

%C Start with the triangle in A094566: starting with row 2, expel from each row the term that is a square of a Fibonacci number (A007598). The remaining triangle is this sequence.

%C In each row, the difference between neighboring terms is a Fibonacci number. For n>1, row n consists of n numbers, first F(2n) and last F(2n+1).

%C Central numbers: (2,10,65,442,...), essentially A064170.

%C Alternating row sums: 2,2,11,11,78,78,...; the sequence b=(2,11,78,...) is A094569.

%H Clark Kimberling, <a href="http://www.fq.math.ca/Papers1/42-1/quartkimberling01_2004.pdf">Orderings of products of Fibonacci numbers</a>, Fibonacci Quarterly 42:1 (2004), pp. 28-35.

%e First four rows:

%e 2

%e 3 5

%e 8 10 13

%e 21 24 26 34

%o (PARI) pef(k, n) = fibonacci(2*k)*fibonacci(2*n-2*k);

%o pof(k, n) = fibonacci(2*n-2*k+1)*fibonacci(2*k-1);

%o isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8)); \\ from A010056

%o isfib2(x) = issquare(x) && isfib(sqrtint(x));

%o tabl(nn) = {for (n=2, nn, if (n % 2 == 0, for (k=1, n/2, if (! isfib2(x = pef(k,n)), print1(x, ", "));); forstep (k=n/2, 1, -1, if (! isfib2(x = pof(k,n)), print1(x, ", "));), for (k=1, n\2, if (! isfib2(x = pef(k,n)), print1(x, ", "));); forstep (k=n\2+1, 1, -1, if (! isfib2(x = pof(k,n)), print1(x, ", ")););); print(););} \\ _Michel Marcus_, May 04 2016

%Y Cf. A000045, A007598, A094565, A094566, A094569.

%K nonn,tabl

%O 1,1

%A _Clark Kimberling_, May 12 2004