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 1-9 of 9 results.

A260519 Concatenate the positions of digits 0, 1,..., 9 in the decimal representation of n, using 1 for the rightmost digit etc., and 0 when the digit does not occur.

Original entry on oeis.org

1000000000, 100000000, 10000000, 1000000, 100000, 10000, 1000, 100, 10, 1, 1200000000, 1200000000, 210000000, 201000000, 200100000, 200010000, 200001000, 200000100, 200000010, 200000001, 1020000000, 120000000, 120000000, 21000000, 20100000, 20010000, 20001000, 20000100, 20000010, 20000001
Offset: 0

Views

Author

M. F. Hasler, Jul 28 2015

Keywords

Examples

			Consider n=1103, the digit '0' is at 2nd position (looking from the right), digit '1' in 3rd and 4th place, digit '2' does not occur, digit '3' at 1st position, digits 4-9 do not occur. Thus a(1103) = 23401000000.

Crossrefs

Cf. A260520 - A260522 for variants; A045918 for the classical "look and say" sequence.

Programs

  • PARI
    a(n)={my(p=vector(10));for(i=1,#n=if(n,Vecrev(digits(n)),[0]),p[n[i]+1]=if(p[n[i]+1],concat(p[n[i]+1],i),i));p=concat(p);p[1]=Str(p[1]);eval(concat(p))}

A260385 Numbers such that n divides A260519(n), where A260519(n) is the concatenation of the positions of the digits 0, 1, ..., 9 in n, positions counted from the right, and 0 if a given digit does not occur.

Original entry on oeis.org

1, 2, 4, 5, 10, 12, 15, 20, 24, 25, 30, 32, 40, 50, 56, 58, 59, 60, 69, 75, 100, 112, 120, 125, 150, 170, 175, 200, 220, 240, 250, 275, 276, 300, 320, 325, 366, 399, 400, 440, 500, 510, 525, 600, 660, 726, 775, 1000, 1250, 2000, 2200, 2360, 2413, 2500, 3142, 3200
Offset: 1

Views

Author

Paolo P. Lava, Jul 24 2015

Keywords

Comments

A260274 is a subset of this sequence

Examples

			Consider number 56. We have no digit 0, 1, 2, 3 and 4, one digit 5 in position 2, one digit 6 in position 1 and no digit 7, 8 and 9. Therefore we get 0000021000 and 21000 / 56 = 375.
Consider 774452318582. We have no digit 0, one digit 1 in position 5, two digits 2 in position 1 and 7, one digit 3 in position 6, two digits 4 in position 9 and 10, two digits 5 in position 3 and 8, no digit 6, two digits 7 in position 11 and 12, two digits 8 in position 2 and 4 and no digit 9. Therefore 774452318582 is transformed in 51769103801112240. But 51769103801112240 / 774452318582 = 66846.08278... Therefore 774452318582 is not part of the sequence.

Links

Crossrefs

Cf. A260274, A260275, A260386, A260519-A260522.

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,b,j,k,ok,n;
for n from 1 to q do a:=convert(n,base,10); b:=0;
for k from 0 to 9 do ok:=0; for j from 1 to nops(a) do
if a[j]=k then ok:=1; b:=b*10^(ilog10(j)+1)+j; fi; od;
if ok=0 then b:=10*b; fi; od; if type(b/n,integer) then print(n);
fi; od; end: P(10^9);
  • PARI
    for(n=1,9e9,A260519(n)%n||print1(n",")) \\ M. F. Hasler, Jul 28 2015

A260386 Numbers n which divide A260521(n), the concatenation of the positions of the digits 9, 8, ..., 0 in the decimal representation of n, where positions are counted from the right, and 0 if a given digit does not occur.

Original entry on oeis.org

1, 2, 4, 5, 8, 12, 15, 19, 21, 24, 25, 48, 68, 75, 96, 123, 228, 231, 275, 312, 321, 375, 451, 484, 712, 726, 768, 868, 1234, 1324, 2143, 2341, 3412, 3421, 4123, 4231, 4312, 4321, 4544, 11425, 12345, 13425, 14235, 14325, 21354, 23451, 24153, 24351, 31524, 32541
Offset: 1

Views

Author

Paolo P. Lava, Jul 24 2015

Keywords

Comments

Given a number n with k digits, label the positions of the digits starting from LSD = 1 to MSD = k. Then concatenate in ascending order the positions of digit 9 in n. Repeat the same process for digits from 8 down to 0. If a digit is not present in n put 0. Sequence lists the numbers that under this transform produce a multiple of the number itself.

Examples

			Consider number 75. We have no digit 9 and 8, one digit 7 in position 2, no digit 6, one digit 5 in position 1, no digit 4, 3, 2, 1 and 0. Therefore we get 002100000 and 2100000 / 75 = 28000.
Consider 774452318582. We have no digit 9, two digits 8 in positions 2 and 4, two digits 7 in positions 11 and 12, no digit 6, two digit 5 in positions 3 and 8, two digits 4 in positions 9 and 10, one digit 3 in position 6, two digits 2 in positions 1 and 7, one digit 1 in position 5 and no digit 0. Therefore 774452318582 is transformed in 024111203891061750. But 24111203891061750 / 774452318582 = 31133.2322... Therefore 774452318582 is not part of the sequence.

Links

Crossrefs

A260275 is a subset of this sequence.
Cf. A260274, A260275, A260385, A260519-A260522.

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,b,j,k,ok,n;
for n from 1 to q do a:=convert(n,base,10); b:=0;
for k from 9 by -1 to 0 do ok:=0; for j from 1 to nops(a) do
if a[j]=k then ok:=1; b:=b*10^(ilog10(j)+1)+j; fi; od;
if ok=0 then b:=10*b; fi; od; if type(b/n,integer) then print(n);
fi; od; end: P(10^9);
  • PARI
    for(n=1,9e9,A260521(n)%n||print1(n",")) \\ M. F. Hasler, Jul 28 2015

A260521 Concatenate the positions of digits 9, 8,..., 0 in the decimal representation of n, using 1 for the rightmost digit etc., and 0 when the digit does not occur.

Original entry on oeis.org

1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000, 21, 120, 120, 1020, 10020, 100020, 1000020, 10000020, 100000020, 1000000020, 201, 210, 1200, 1200, 10200, 100200, 1000200, 10000200, 100000200, 1000000200
Offset: 0

Views

Author

M. F. Hasler, Jul 28 2015

Keywords

Examples

			Consider n=1103, digits '9' through '4' do not occur, digit '3' at 1st position (looking from the right), digit '2' does not occur, digit '1' in 3rd and 4th place, and digit '0' is at 2nd position. Thus a(1103) = 00000010342 = 10342.

Crossrefs

Cf. A045918 (the classical "look and say" sequence).
Cf. A260519 - A260522 for variants (counting from the right or starting with digit 0).

Programs

  • PARI
    A260521=a(n)={my(p=vector(10));for(i=1,#n=if(n,Vecrev(digits(n)),[0]),p[n[i]+1]=if(p[n[i]+1],concat(p[n[i]+1],i),i));p=concat(Vecrev(p));p[1]=Str(p[1]);eval(concat(p))}

A260529 Concatenate the positions of digits 9, 8,..., 0 in the decimal representation of n, using 1 for the rightmost digit etc., skip it if a digit does not occur.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 21, 12, 12, 12, 12, 12, 12, 12, 12, 12, 21, 21, 12, 12, 12, 12, 12, 12, 12, 12, 21, 21, 21, 12, 12, 12, 12, 12, 12, 12, 21, 21, 21, 21, 12, 12, 12, 12, 12, 12, 21, 21, 21, 21, 21, 12, 12, 12, 12, 12, 21, 21, 21, 21, 21, 21, 12
Offset: 0

Views

Author

M. F. Hasler, Jul 28 2015

Keywords

Comments

Sequence A260275 lists fixed points of this function.

Examples

			Consider n=1103, digits 4-9 do not occur, the digit '3' is at the 1st position (looking from the right), digit '2' does not occur, digit '1' is in 3rd and 4th place, digit '0' at 2nd position. Thus a(1103) = 1342.

Crossrefs

Cf. A260526 - A260528 and A260519 - A260522 for variants; A045918 for the classical "look and say" sequence.

Programs

  • PARI
    a(n)={my(p=vector(10));for(i=1,#n=if(n,Vecrev(digits(n)),[0]),p[n[i]+1]=if(p[n[i]+1],concat(p[n[i]+1],i),i));if(1<#p=concat(Vecrev(select(t->#t,p))),p[1]=Str(p[1]);eval(concat(p)),p)}

A260526 Concatenate the positions of digits 0, 1,..., 9 in the decimal representation of n, using 1 for the rightmost digit etc., skip it if a digit does not occur.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 12, 21, 21, 21, 21, 21, 21, 21, 21, 12, 12, 12, 21, 21, 21, 21, 21, 21, 21, 12, 12, 12, 12, 21, 21, 21, 21, 21, 21, 12, 12, 12, 12, 12, 21, 21, 21, 21, 21, 12, 12, 12, 12, 12, 12, 21, 21, 21, 21, 12, 12, 12, 12, 12, 12, 12, 21, 21
Offset: 0

Views

Author

M. F. Hasler, Jul 28 2015

Keywords

Comments

Sequence A260274 lists fixed points of this function.
Sequence A260519 is the same except for concatenating a 0 for digits which do not occur.

Examples

			Consider n=1103, the digit '0' is at 2nd position (looking from the right), digit '1' in 3rd and 4th place, digit '3' at 1st position, digits 2 and 4-9 do not occur. Thus a(1103) = 2341.

Crossrefs

Cf. A260527 - A260529 and A260519 - A260522 for variants; A045918 for the classical "look and say" sequence.

Programs

  • PARI
    a(n)={my(p=vector(10));for(i=1,#n=if(n,Vecrev(digits(n)),[0]),p[n[i]+1]=if(p[n[i]+1],concat(p[n[i]+1],i),i));if(1<#p=concat(select(t->#t,p)),p[1]=Str(p[1]);eval(concat(p)),p)}

A260520 Concatenate the positions of digits 0, 1, ..., 9 in the decimal representation of n, using 1 for the leftmost digit etc., and 0 when the digit does not occur.

Original entry on oeis.org

1000000000, 100000000, 10000000, 1000000, 100000, 10000, 1000, 100, 10, 1, 2100000000, 1200000000, 120000000, 102000000, 100200000, 100020000, 100002000, 100000200, 100000020, 100000002, 2010000000, 210000000, 120000000, 12000000, 10200000, 10020000, 10002000, 10000200, 10000020, 10000002
Offset: 0

Views

Author

M. F. Hasler, Jul 28 2015

Keywords

Examples

			Consider n=1103: the digit '0' is at the 3rd position, digit '1' in the 1st and 2nd places, digit '2' does not occur, digit '3' at the 4th position, digits 4-9 do not occur. Thus a(1103) = 31204000000.

Crossrefs

Cf. A260519, A260521 and A260522 for variants.
Cf. A045918 for the classical "look and say" sequence.

Programs

  • PARI
    a(n)={my(p=vector(10));for(i=1,#n=if(n,digits(n),[0]),p[n[i]+1]=if(p[n[i]+1],concat(p[n[i]+1],i),i));p=concat(p);p[1]=Str(p[1]);eval(concat(p))}

A260527 Concatenate the positions of digits 0, 1,..., 9 in the decimal representation of n, using 1 for the leftmost digit etc., skip digits which don't occur.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 21, 12, 12, 12, 12, 12, 12, 12, 12, 12, 21, 21, 12, 12, 12, 12, 12, 12, 12, 12, 21, 21, 21, 12, 12, 12, 12, 12, 12, 12, 21, 21, 21, 21, 12, 12, 12, 12, 12, 12
Offset: 0

Views

Author

M. F. Hasler, Jul 28 2015

Keywords

Examples

			Consider n=1103, the digit '0' is at 3rd position (looking from the left), digit '1' in 1st and 2nd place, digit '3' at 4th position, digits 2 and 4-9 do not occur. Thus a(1103) = 3124.

Crossrefs

Cf. A260526 - A260529 and A260519 - A260522 for variants; A045918 for the classical "look and say" sequence.

Programs

  • PARI
    a(n)={my(p=vector(10));for(i=1,#n=if(n,digits(n),[0]),p[n[i]+1]=if(p[n[i]+1],concat(p[n[i]+1],i),i));if(1<#p=concat(select(t->#t,p)),p[1]=Str(p[1]);eval(concat(p)),p)}

A260528 Concatenate the positions of digits 9, 8,..., 0 in the decimal representation of n, using 1 for the leftmost digit etc., skip digits which don't occur.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 12, 21, 21, 21, 21, 21, 21, 21, 21, 12, 12, 12, 21, 21, 21, 21, 21, 21, 21, 12, 12, 12, 12, 21, 21, 21, 21, 21, 21, 12, 12, 12, 12, 12, 21, 21, 21, 21, 21, 12, 12, 12, 12, 12, 12, 21, 21, 21, 21, 12, 12, 12, 12, 12
Offset: 0

Views

Author

M. F. Hasler, Jul 28 2015

Keywords

Examples

			Consider n=1103: the digit '3' is at the 4th position (looking from the left), digit '1' is in 1st and 2nd place, digit '0' is at 3rd position, digits 2 and 4-9 do not occur. Thus a(1103) = 4123.

Crossrefs

Cf. A260526 - A260529 and A260519 - A260522 for variants; A045918 for the classical "look and say" sequence.

Programs

  • PARI
    a(n)={my(p=vector(10));for(i=1,#n=if(n,digits(n),[0]),p[n[i]+1]=if(p[n[i]+1],concat(p[n[i]+1],i),i));if(1<#p=concat(Vecrev(select(t->#t,p))),p[1]=Str(p[1]);eval(concat(p)),p)}
Showing 1-9 of 9 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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