cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A114834 Each term is previous term plus floor of root mean square of two previous terms.

Original entry on oeis.org

1, 2, 3, 5, 9, 16, 28, 50, 90, 162, 293, 529, 956, 1728, 3124, 5648, 10211, 18462, 33380, 60352, 109119, 197293, 356716, 644961, 1166123, 2108412, 3812120, 6892514, 12462029, 22532007, 40739059, 73658371, 133178227, 240793271, 435366958, 787166465
Offset: 1

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Author

Jonathan Vos Post, Feb 19 2006

Keywords

Comments

What is this sequence and the ratio of adjacent terms, asymptotically? Primes in this sequence include 2, 3, 5, 293. Squares in this sequence include 9, 16, 529.

Examples

			a(3) = 2 + floor[sqrt[(1^2 + 2^2)/2]] = 2 + floor[Sqrt[5/2]] = 2 + 1 = 3.
a(4) = 3 + floor[sqrt[(2^2 + 3^2)/2]] = 4 + floor[Sqrt[13/2]] = 3 + 2 = 5.
a(5) = 5 + floor[sqrt[(3^2 + 5^2)/2]] = 8 + floor[Sqrt[34/2]] = 5 + 4 = 9.
a(6) = 9 + floor[sqrt[(5^2 + 9^2)/2]] = 15 + floor[Sqrt[106/2]] = 9 + 7 = 16.
a(7) = 16 + floor[sqrt[(9^2 + 16^2)/2]] = 15 + floor[Sqrt[337/2]] = 16 + 12 = 28.
a(8) = 28 + floor[sqrt[(16^2 + 28^2)/2]] = 15 + floor[Sqrt[1040/2]] = 28 + 22 = 50.
a(9) = 50 + floor[sqrt[(28^2 + 50^2)/2]] = 50 + floor[Sqrt[3284/2]] = 50 + 40 = 90.
a(10) = 90 + floor[sqrt[(50^2 + 90^2)/2]] = 50 + floor[Sqrt[10600/2]] = 90 + 72 = 162.
a(11) = 162 + floor[sqrt[(90^2 + 162^2)/2]] = 50 + floor[Sqrt[34344/2]] = 162 + 131 = 293.
a(12) = 293 + floor[sqrt[(162^2 + 293^2)/2]] = 293 + floor[Sqrt[112093/2]] = 293 + 236 = 529.
		

Crossrefs

Programs

  • Maple
    rms := proc(a,b)
        sqrt((a^2+b^2)/2) ;
    end proc:
    A114834 := proc(n)
        option remember;
        if n<= 2 then
            n;
        else
            procname(n-1)+floor(rms(procname(n-1),procname(n-2))) ;
        end if;
    end proc: # R. J. Mathar, Jun 23 2014

Formula

a(1) = 1, a(2) = 2, for n>2: a(n+1) = a(n) + floor(RMS[a(n),a(n-1)]). a(n+1) = a(n) + floor[Sqrt[[a(n)^2]+[a(n-1)^2]/2]].