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 1-3 of 3 results.

A243025 Fixed points of the transform n = d_(k)*10^(k-1) + d_(k-1)*10^(k-2) + ... + d_(2)*10 + d_(1) -> Sum_{i=1..k-1}{d_(i)^d(i+1)}+d(k)^d(1) (A243023).

Original entry on oeis.org

1, 4155, 4355, 1953504, 1954329, 522169982
Offset: 1

Views

Author

Paolo P. Lava, May 29 2014

Keywords

Comments

Subset of A243023.
This sequence is finite by using the same argument that Armstrong numbers (A005188) are finite. - Robert G. Wilson v, Jun 01 2014

Examples

			1^1 = 1.
5^5 + 5^1 + 1^4 + 4^5 = 4155.
5^5 + 5^3 + 3^4 + 4^5 = 4355.
4^0 + 0^5 + 5^3 + 3^5 + 5^9 + 9^1 + 1^4 = 1953504.
9^2 + 2^3 + 3^4 + 4^5 + 5^9 + 9^1 + 1^9 = 1954329.
		

Crossrefs

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,b,k,ok,n; for n from 10 to q do a:=[]; b:=n;
    while b>0 do a:=[op(a),b mod 10]; b:=trunc(b/10); od; b:=0; ok:=1; for k from 2 to nops(a)
    do if a[k-1]=0 and a[k]=0 then ok:=0; break; else b:=b+a[k-1]^a[k]; fi; od;
    if ok=1 then if n=(b+a[nops(a)]^a[nops(1)]) then print(n);
    fi; fi; od; end: P(10^10);
  • Mathematica
    fQ[n_] := Block[{r = Reverse@ IntegerDigits@ n}, n == Plus @@ (r^RotateLeft@ r)]; k = 1; lst = {}; While[k < 1000000001, If[ fQ@ k, AppendTo[ lst, k]; Print@ k]; k++] (* Robert G. Wilson v, Jun 01 2014 *)

Extensions

Added a(1) as 1 and a(6) by Robert G. Wilson v, Jun 01 2014

A243024 Consider a k-digit number m = d_(k)*10^(k-1) + d_(k-1)*10^(k-2) + ... + d_(2)*10 + d_(1). Sequence lists the numbers m that divide Sum_{i=1..k-1}{d_(i)^d_(i+1)}+d_(k)^d_(1).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 63, 448, 1899, 1942, 4155, 4355, 8503, 28375, 44569, 73687, 1953504, 1954329, 70860598, 522169982
Offset: 1

Views

Author

Paolo P. Lava, May 29 2014

Keywords

Comments

Numbers with two consecutive zeros are not considered, to avoid the case 0^0. Nevertheless, even if we consider 0^0=1 the results do not change (at least up to m=10^8).

Examples

			For 1899 we have: 9^9 + 9^8 + 8^1 + 1^9 = 430467219.
Finally 430467219/1899 = 226681.
For 1954329 we have: 9^2 + 2^3 +3^4 + 4^5 + 5^9 + 9^1 + 1^9 = 1954329.
		

Crossrefs

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,b,k,ok,n; for n from 10 to q do a:=[]; b:=n;
    while b>0 do a:=[op(a),b mod 10]; b:=trunc(b/10); od; b:=0; ok:=1; for k from 2 to nops(a)
    do if a[k-1]=0 and a[k]=0 then ok:=0; break; else b:=b+a[k-1]^a[k]; fi; od;
    if ok=1 then if type((b+a[nops(a)]^a[nops(1)])/n,integer) then print(n);
    fi; fi; od; end: P(10^10);
  • PARI
    isok(n) = d = digits(n); k = #d; (sum(i=1, k-1, j=k-i+1; d[j]^d[(j-1)])+ d[1]^d[k]) % n == 0; \\ Michel Marcus, Sep 29 2014

Extensions

a(1)-a(9) prepended and a(23) from Hiroaki Yamanouchi, Sep 29 2014

A243507 Consider a decimal number, n, with k digits. n = d(k)*10^(k-1) + d(k-1)*10^(k-2) + … + d(2)*10 + d_(1). Sequence lists the numbers n that divide s = Sum_{i=1..k} d(i)^d(i).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 63, 64, 93, 377, 643, 699, 760, 2428, 3435, 13073, 46864, 184405, 208858, 1313290, 2326990, 2868720, 2868741, 18273988, 25265859, 33690905, 87889176, 194123725, 589957694
Offset: 1

Views

Author

Keywords

Comments

Since 0^0 is indeterminate, but for all other Xs, X^0 is 1, we define 0^0 here to be 1. (Since 0 does not divide 1, 0 is not a member.)
For Münchhausen numbers (A046253) the ratio is 1. [Paolo P. Lava, Apr 08 2016]

Examples

			63 is in the sequence because 6^6+3^3 = 46683 and 46683/63 = 741, an integer.
		

Crossrefs

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,b,k,n; for n from 1 to q do a:=[]; b:=n; while b>0 do a:=[op(a),b mod 10]; b:=trunc(b/10); od; b:=0; for k from 1 to nops(a) do if a[k]=0 then b:=b+1; else b:=b+a[k]^a[k]; fi; od; if type(b/n,integer) then print(n); fi; od; end: P(10^10);
  • Mathematica
    fQ[n_] := Block[{id = IntegerDigits@ n /. {0 -> 1}}, Mod[ Total[ id^id], n] == 0]; k = 1; lst = {}; While[k < 10000000001, If[ fQ@ k, AppendTo[ lst, k]; Print@ k]; k++]; lst
Showing 1-3 of 3 results.