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#55 by Michael De Vlieger at Sat Jun 08 00:01:46 EDT 2024
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#54 by Jon E. Schoenfield at Fri Jun 07 19:38:18 EDT 2024
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#53 by Jon E. Schoenfield at Fri Jun 07 19:38:12 EDT 2024
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| COMMENTS
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a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(11;n-1-k,k) with n>= >= 1, a(-1)=10. These are the SW-NE diagonals in P(11;n,k), the (11,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
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| FORMULA
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a(n) = a(n-1)+) + a(n-2), n>= >= 2, a(0)=1, a(1)=11. a(-1)=10.
a(n-1) = ((1+sqrt5)^sqrt(5))^n-( - (1-sqrt5)^sqrt(5))^n)/(2^n*sqrt5)+ sqrt(5)) + 5*((1+sqrt5)^(sqrt(5))^(n-1) -() - (1-sqrt5)^(sqrt(5))^(n-1))/ ())/(2^(n-2)*sqrt5). - sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = A000045(n+4) + A000032(n-4) for n> > 0. - Bruno Berselli, Sep 27 2017
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| STATUS
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approved
editing
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#52 by Joerg Arndt at Sun Feb 18 01:58:56 EST 2024
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#51 by Paolo P. Lava at Sat Feb 17 12:32:06 EST 2024
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| FORMULA
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a(n) = 11*A000045(n) + A000045(n-1). - Paolo P. Lava, May 19 2015
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| MAPLE
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with(numtheory): with(combinat): P:=proc(q) local n;
for n from 0 to q do print(11*fibonacci(n)+fibonacci(n-1));
od; end: P(30); # Paolo P. Lava, May 19 2015
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| STATUS
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approved
editing
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#50 by Charles R Greathouse IV at Thu Sep 08 08:44:45 EDT 2022
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| PROG
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(MAGMAMagma) a0:=1; a1:=11; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
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Discussion
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| Thu Sep 08
| 08:44
| OEIS Server: https://oeis.org/edit/global/2944
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#49 by N. J. A. Sloane at Wed Sep 27 09:12:42 EDT 2017
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#48 by Bruno Berselli at Wed Sep 27 06:30:53 EDT 2017
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Discussion
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| Wed Sep 27
| 07:06
| Michel Marcus: I tend to find the A101220 crossref questionable (but there are over 45 such crossrefs ...)
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#47 by Bruno Berselli at Wed Sep 27 06:30:24 EDT 2017
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#46 by N. J. A. Sloane at Mon Feb 20 12:07:21 EST 2017
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