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 1-4 of 4 results.

A034087 Numbers divisible by the sum of the squares of their digits.

Original entry on oeis.org

1, 10, 20, 50, 100, 110, 111, 120, 130, 133, 200, 210, 240, 267, 298, 310, 315, 360, 372, 376, 400, 420, 480, 500, 532, 550, 630, 803, 917, 973, 1000, 1010, 1011, 1020, 1030, 1071, 1100, 1101, 1110, 1134, 1148, 1200, 1211, 1222, 1290, 1300, 1302, 1316
Offset: 1

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Examples

			a(100) = 4131 since 4^2+1^2+3^2+1^2=27 divides 4131. - _Carmine Suriano_, May 04 2013
		

Crossrefs

Programs

  • Maple
    isA034087 := proc(n) if n mod A003132(n) = 0 then true ; else false ; end if ; end proc:
    for n from 1 to 1800 do if isA034087(n) then printf("%d ",n) ; end if ; end do ; # R. J. Mathar, Feb 25 2007
  • Mathematica
    Select[Range[1500], Divisible[#, Plus @@ (IntegerDigits[#]^2)] &] (* Amiram Eldar, Jan 31 2021 *)
  • PARI
    isok(m) = !(m % norml2(digits(m))); \\ Michel Marcus, Jan 31 2021
    
  • Python
    def ok(n): return n and n%sum(di**2 for di in map(int, str(n))) == 0
    print([k for k in range(1317) if ok(k)]) # Michael S. Branicky, Jan 10 2025

Formula

A003132[a(n)] | a(n). - R. J. Mathar, Feb 25 2007

A034088 Numbers divisible by the sum of the cubes of their digits.

Original entry on oeis.org

1, 10, 100, 110, 111, 153, 200, 221, 370, 371, 407, 500, 702, 1000, 1010, 1011, 1040, 1100, 1101, 1110, 1120, 1210, 1215, 1232, 1323, 1530, 1728, 2000, 2030, 2080, 2110, 2210, 2240, 2331, 2352, 2376, 2464, 2580, 3212, 3213, 3304, 3456, 3520, 3700, 3710
Offset: 1

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Comments

Ratio between original number and a(n) = 1 for n = 1, 6, 9, 10, 11, ... a(n)= 1, 153, 370, 371, 407,... - Carmine Suriano, May 04 2013

Examples

			12672 is a term since 1^3+2^3+6^3+7^3+2^3 = 576 divides 12672. - _Carmine Suriano_, May 04 2013
		

Crossrefs

Programs

  • Mathematica
    Select[Range[4000], Divisible[#, Plus @@ (IntegerDigits[#]^3)] &] (* Amiram Eldar, Jan 31 2021 *)

Extensions

Offset corrected by Amiram Eldar, Jan 31 2021

A169665 Numbers divisible by the sum of 4th powers of their digits.

Original entry on oeis.org

1, 10, 100, 102, 110, 111, 1000, 1010, 1011, 1020, 1100, 1101, 1110, 1121, 1122, 1634, 2000, 2322, 4104, 5000, 8208, 9474, 10000, 10010, 10011, 10100, 10101, 10110, 10200, 10412, 11000, 11001, 11010, 11100, 11210, 11220, 12502, 12521, 14758
Offset: 1

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Author

Michel Lagneau, Apr 05 2010

Keywords

Examples

			12521 is a term since 1^4 + 2^4 + 5^4 + 2^4 + 1^4 = 659, and 12521 = 19*659;
89295 is a term since 8^4 + 9^4 + 2^4 + 9^4 + 5^4 = 17859, and 89295 = 5*17859.
		

Crossrefs

Programs

  • Maple
    A:= proc(n) add(d^4, d=convert(n, base, 10)) ; end proc: for n from 1 to 200000 do:if irem( n,A(n))=0 then printf(`%d, `,n):else fi:od:
  • Mathematica
    Select[Range[15000], Divisible[#, Plus @@ (IntegerDigits[#]^4)] &] (* Amiram Eldar, Jan 31 2021 *)

Formula

Numbers k such that A055013(k) | k.

A007603 Power-sum numbers: let n = a_1 a_2 ... a_k be a k-digit number; n is a power-sum number if there are exponents e_1 ... e_m such that n = Sum_{i=1..m} Sum_{j=1..k} a_j^e_i.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 23, 24, 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80, 81, 84, 90, 100, 102, 104, 108, 110, 111, 112, 113, 114, 115, 116, 117, 120, 122, 126, 130, 131, 132, 133, 134, 135, 136, 140, 144, 150, 151, 152, 153, 154, 156, 160, 162, 170, 171, 172, 173, 174, 178, 180, 182
Offset: 1

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Examples

			21 = (2+1)+(2^3+1^3)+(2^3+1^3), with e_1, e_2, e_3 = 1, 3, 3.
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Mathematica
    q[n_] := Module[{d = IntegerDigits[n], v = {}, k = 1, s, ans = False}, If[Max[d] == 1, ans = Divisible[n, Total[d]], While[(s = Total[d^k]) <= n, AppendTo[v, s]; If[Length[IntegerPartitions[n, All, v]] > 0, ans = True; Break[]]; k++]]; ans]; Select[Range[200], q] (* Amiram Eldar, Sep 04 2021 *)

Extensions

Corrected and extended by Naohiro Nomoto, Mar 11 2001
Showing 1-4 of 4 results.