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.

User: Shmelev Aleksei

Shmelev Aleksei's wiki page.

Shmelev Aleksei has authored 4 sequences.

A282839 Numbers that are equal to the sum of descending numbers raised to their digits' powers.

Original entry on oeis.org

1, 65, 6796
Offset: 1

Author

Shmelev Aleksei, Feb 22 2017

Comments

Sequence is complete.

Examples

			1 = 1^1;
65 = 2^6 + 1^5;
6796 = 4^6 + 3^7 + 2^9 + 1^6.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[10^5], # == Total[ Reverse[ Range@ IntegerLength@ #]^ IntegerDigits@ #] &] (* Giovanni Resta, Feb 23 2017 *)
  • PARI
    isok(n) = my(d=digits(n)); sum(k=1, #d, (#d-k+1)^d[k]) == n; \\ Michel Marcus, Feb 24 2017
  • VBA
    sub calcul()
    sheets("Result").select
    range("A1").select
    for i=1 to 10^13
    sum=0
    for k=1 to len(i)
    sum=sum+(len(i)-k+1)^mid(i,k,1)
    next
    if i=sum then
    activecell.value=i
    activesheet.offset(1,0).select
    end if
    next
    end sub
    

A283034 Numbers k such that k = (sum of digits of k)^(last digit of k).

Original entry on oeis.org

1, 4913, 19683, 52521875, 24794911296, 68719476736, 271818611107, 1174711139837
Offset: 1

Author

Shmelev Aleksei, Feb 27 2017

Keywords

Comments

The check must be done up to 10^22 (then for 23 digits in a number max result can be (23*10)^9 = 4, 6*10^20 < 10^22).

Examples

			1 = 1^1,
4913 = (4+9+1+3)^3,
19683 = (1+9+6+8+3)^3,
52521875 = (5+2+5+2+1+8+7+5)^5.
		

Crossrefs

Programs

  • Mathematica
    Union[Reap[nd=1; Sow[1]; While[Ceiling[(10^(nd-1))^(1/9)] <= 9 nd, Do[ Do[v = s^e; If[Mod[v, 10] == e && Plus @@ IntegerDigits@ v == s, Sow[v]], {s, Ceiling[ (10^(nd-1))^(1/e)], Min[ Floor[10^(nd/e)], 9 nd]}], {e, 2, 9}]; nd++]][[2, 1]]] (* all terms, Giovanni Resta, Feb 27 2017 *)
  • VBA
    Sub calcul()
       Sheets("Result").Select
       Range("A1").Select
       For i = 1 To 10000000
          Sum = 0
          For k = 1 To Len(i)
             Sum = Sum + Mid(i, k, 1)
          Next
          If Sum ^Mid(i, len(i), 1)= i Then
             ActiveCell.Value = i
             ActiveCell.Offset(1, 0).Select
          End If
       Next
    End Sub

Extensions

a(5)-a(8) from Giovanni Resta, Feb 27 2017

A282693 Numbers k such that k = (sum of digits of k)*((sum of digits of k) + 1).

Original entry on oeis.org

0, 12, 42, 90, 156
Offset: 1

Author

Shmelev Aleksei, Feb 24 2017

Examples

			    0 = 0*(0+1).
   12 = (1+2)*(1+2+1).
   42 = (4+2)*(4+2+1).
   90 = (9+0)*(9+0+1).
  156 = (1+5+6)*(1+5+6+1).
		

Programs

  • Mathematica
    fQ[n_] := Block[{a = Plus @@ IntegerDigits@ n}, a*(a +1) == n]; Select[ Range[0, 1000], fQ] (* Robert G. Wilson v, Feb 24 2017 *)
  • VBA
    Sub calcul()
       Sheets("Result").Select
       Range("A1").Select
       For i = 1 To 10000000
          Sum = 0
          For k = 1 To Len(i)
             Sum = Sum + Mid(i, k, 1)
          Next
          res = Sum * (Sum + 1)
          If res = i Then
             ActiveCell.Value = i
             ActiveCell.Offset(1, 0).Select
          End If
       Next
    End Sub

A282782 Numbers that are equal to a product of powers of digits where the exponents from left to right decrease with 1 and the exponent for the units digit is 1.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1715
Offset: 1

Author

Shmelev Aleksei, Feb 21 2017

Keywords

Comments

Up to 10^9 no other number matches the rule.
There are no other terms up to 10^200; this list is probably complete. - Charles R Greathouse IV, Feb 22 2017

Examples

			1 = 1^1, 2 = 2^1, ..., 1715 = (1^4)*(7^3)*(1^2)*(5^1).
These numbers do not match the rule:
46: (4^2)*(6^1) = 96 <> 46.
234: (2^3)*(3^2)*(4^1) = 288 <> 234.
4342: (4^4)*(3^3)*(4^2)*(2^1) = 221184 <> 4342.
46914: (4^5)*(6^4)*(9^3)*(1^2)*(4^1) = 3869835264 <> 46914.
		

Crossrefs

Subsequence of A002473; apart from the first term, a subsequence of A238985.

Programs

  • Mathematica
    mx = 10^50; test[n_] := n == Times @@ (IntegerDigits[n] ^Reverse[Range@ IntegerLength@ n]); Union@Reap[Do[n = 2^i 3^j 7^k; If[test@n, Sow@n], {i, 0, Log2[mx]}, {j, 0, Log[3, mx/2^i]}, {k, 0, Log[7, mx/2^i/3^j]}]; Do[n = 5  3^j 7^k; If[test@n, Sow@n], {j, 0, Log[3, mx/5]}, {k, 0, Log[7, mx/ 5/ 3^j]}]][[2, 1]] (* Search up to 10^50, Giovanni Resta, Feb 22 2017 *)
    Select[Range[0, 2000], Times @@ MapIndexed[#1^First[#2] &, Reverse@ IntegerDigits@ #] == # &] (* Michael De Vlieger, Feb 22 2017 *)
  • PARI
    is(n)=my(d=digits(n)); prod(i=1,#d,d[#d+1-i]^i)==n || !n \\ Charles R Greathouse IV, Feb 22 2017
    
  • PARI
    list(lim)=my(v=List([0]),t7,t37,t); for(a=0,logint(lim\1,7), t7=7^a; for(b=0,logint(lim\t7,3), t=t37=t7*3^b; while(t<=lim, if(is(t), listput(v,t)); t<<=1); t=t37; while(t<=lim, if(is(t), listput(v,t)); t*=5))); Set(v) \\ Charles R Greathouse IV, Feb 22 2017
  • VBA
    ' For example for 5-figure numbers:
    Dim zahl As String
    For i = 10000 To 99999
    zahl = i
    If i = CInt(Left(zahl, 1)) ^ 5 * CInt(Right(Left(zahl, 2), 1)) ^ 4 * CInt(Right(Left(zahl, 3), 1)) ^ 3 * CInt(Right(Left(zahl, 4), 1)) ^ 2 * CInt(Right(zahl, 1)) ^ 1 Then
    MsgBox (i)
    End If
    Next i
    

Extensions

Leading 0 prepended by David A. Corneth, Feb 22 2017