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.

Showing 1-5 of 5 results.

A182654 Floor-sum sequence of r, with r=sqrt(2) and a(1)=1, a(2)=2.

Original entry on oeis.org

1, 2, 4, 7, 8, 11, 12, 14, 15, 16, 18, 19, 21, 22, 24, 25, 26, 28, 29, 31, 32, 33, 35, 36, 38, 39, 41, 42, 43, 45, 46, 48, 49, 50, 52, 53, 55, 56, 57, 59, 60, 62, 63, 65, 66, 67, 69, 70, 72, 73, 74, 76, 77, 79, 80, 82, 83, 84, 86, 87, 89, 90, 91, 93, 94, 96, 97, 98, 100
Offset: 1

Views

Author

Clark Kimberling, Nov 26 2010

Keywords

Comments

Let S be the set generated by these rules: (1) if m and n are in S and m
Let B be the Beatty sequence of r. Then a floor-sum sequence of r is a subsequence of B if and only if a(1) and a(2) are terms of B. For example, 5 is A001951 but not in A182654.

Crossrefs

Cf. A182653, A182655, A182656.

Programs

  • Maple
    A182654 := proc(amax)
        a := {1,2} ;
        r := sqrt(2) ;
        while true do
                anew := {} ;
                for i in a do
                for j in a do
                        if i <> j then S := floor(r*(i+j)) ;                     if is(S <= amax) then anew := anew union { S }; end if;
                        end if;
                end do:
                end do:
                if a union anew = a then
                        return sort(a) ;
                end if;
                a := a union anew ;
        end do:
end proc:
A182654(100) ;

A182655 Floor-sum sequence of r, with r=(3+sqrt(5))/2 and a(1)=1, a(2)=2.

Original entry on oeis.org

1, 2, 7, 20, 23, 54, 57, 62, 65, 70, 78, 112, 143, 146, 151, 154, 159, 164, 167, 172, 175, 180, 185, 188, 193, 201, 206, 209, 214, 222, 230, 235, 243, 256, 264, 290, 295, 298, 303, 311, 319, 324, 332, 345, 353, 366
Offset: 1

Views

Author

Clark Kimberling, Nov 26 2010

Keywords

Comments

Let S be the set generated by these rules: (1) if m and n are in S and m
Let B be the Beatty sequence of r. Then a floor-sum sequence of r is a subsequence of B if and only if a(1) and a(2) are terms of B.

Examples

			7 is in the sequence because floor(r*a(1)+r*a(2))=floor(r+2r)=7
57 is in the sequence because floor(r*a(2)+r*a(4))=floor(r*22)=57
61 is not in the sequence because 23*r=60.21... and 24*r=62.83... so there are no integers x,y with floor(r*(x+y))=61
60 is not in the sequence because floor(r*(x+y))=60 requires x+y=23, and no pair of elements of the sequence sum to 23

Crossrefs

Cf. A182653, A182654, A182656. A001950.

A182656 Floor-sum sequence of r, with r=sqrt(3), and a(1)=1, a(2)=2.

Original entry on oeis.org

1, 2, 5, 10, 12, 19, 20, 22, 24, 25, 29, 34, 36, 38, 39, 41, 43, 45, 46, 50, 51, 53, 55, 58, 60, 62, 64, 65, 67, 69, 71, 72, 74, 76, 77, 79, 81, 83, 84, 86, 88, 90, 91, 93, 95, 96, 98, 100, 102, 103, 105, 107, 109, 110, 112, 114, 116, 117
Offset: 1

Views

Author

Clark Kimberling, Nov 26 2010

Keywords

Comments

Let S be the set generated by these rules: (1) if m and n are in S and m
Let B be the Beatty sequence of r. Then a floor-sum sequence of r is a subsequence of B if and only if a(1) and a(2) are terms of B.

Examples

			a(3)=floor(r+2r)=5.

Crossrefs

Cf. A182653, A182654, A182655, A022838.

A182669 Floor-sum sequence of r, with r = golden ratio = (1+sqrt(5))/2 and a(1)=1, a(2)=3.

Original entry on oeis.org

1, 3, 6, 11, 14, 19, 22, 24, 27, 32, 35, 37, 40, 43, 45, 48, 53, 56, 58, 61, 64, 66, 69, 71, 74, 77, 79, 82, 87, 90, 92, 95, 98, 100, 103, 105, 108, 111, 113, 116, 119, 121, 124, 126, 129, 132, 134, 137, 139, 142, 145, 147, 150, 153, 155, 158, 160, 163, 166, 168, 171, 173, 174, 176, 179, 181, 184
Offset: 1

Views

Author

Clark Kimberling, Nov 27 2010

Keywords

Comments

Let S be the set generated by these rules: (1) if m and n are in S and m
Let B be the Beatty sequence of r. Then a floor-sum sequence of r is a subsequence of B if and only if a(1) and a(2) are terms of B. Thus, A182669 is a subsequence of the lower Wythoff sequence, A000201.

Examples

			a(3)=floor(r+3r)=6.

Links

Crossrefs

Cf. A000201, A182653, A182570.

Programs

  • PARI
    lista(nn) = my(S=[1, 3], r=(1+sqrt(5))/2, new, k); while(1, new=[]; for(m=1, #S, for(n=m+1, #S, k=floor(r*(S[m]+S[n])); if(k<=nn, new=setunion(new,[k])))); if(S==setunion(S,new), return(S)); S=setunion(S,new)) \\ Iain Fox, Apr 25 2019
  • QBasic
    r=(1+5^(1/2))/2: s(1)=1: s(2)=3: s(5)=6
For h=2 to 200: c(h)=h+c(h-1): next h
For h=1 to 100: c=c(h): d=0
  For i=1 to h+1: d=d+1: s(c+d)=int(s(i)+s(h+2)*r)
  Next i
Next h
For i=1 to 1000: for j=i+1 to 1001
if s(i)>=s(j) then swap s(i),s(j)
next j,i
For i=1 to 120: if s(i+1)<>s(i) then print s(i);
next i

Extensions

139 (generated by m=22, n=64) added by R. J. Mathar, Nov 28 2010

A182670 Floor-sum sequence of r, where r = golden ratio = (1+sqrt(5))/2 and a(1)=2, a(2)=3.

Original entry on oeis.org

2, 3, 8, 16, 17, 29, 30, 32, 38, 40, 50, 51, 53, 55, 56, 59, 61, 64, 66, 67, 69, 72, 74, 76, 77, 79, 84, 85, 87, 88, 90, 92, 93, 95, 98, 100, 101, 103, 106, 108, 110, 111, 113, 114, 116, 118, 119, 121, 122, 124, 126, 127, 129, 131, 132, 134, 135, 137, 139, 140
Offset: 1

Views

Author

Clark Kimberling, Nov 27 2010

Keywords

Comments

Let S be the set generated by these rules: (1) if m and n are in S and m
Let B be the Beatty sequence of r. Then a floor-sum sequence of r is a subsequence of B if and only if a(1) and a(2) are terms of B. Thus, A182670 is not a subsequence of the lower Wythoff sequence, A000201.

Examples

			a(3) = floor(2r+3r) = 8.

Links

Crossrefs

Cf. A000201, A182653, A182669.

Programs

  • Maple
    A182670 := proc(amax)
        a := {2,3} ;
        r := (1+sqrt(5))/2 ;
        while true do
                anew := {} ;
                for i in a do for j in a do
                        if i <> j then  S := floor(r*(i+j)) ;                     if is(S <= amax) then anew := anew union { S }; end if;
                        end if;
                end do:
                end do:
                if a union anew = a then
                        return sort(a) ;
                end if;
                a := a union anew ;
        end do:
end proc:
A182670(140) ;
  • PARI
    lista(nn) = my(S=[2, 3], r=(1+sqrt(5))/2, new, k); while(1, new=[]; for(m=1, #S, for(n=m+1, #S, k=floor(r*(S[m]+S[n])); if(k<=nn, new=setunion(new, [k])))); if(S==setunion(S, new), return(S)); S=setunion(S, new)) \\ Iain Fox, Apr 25 2019
Showing 1-5 of 5 results.

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