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 1-6 of 6 results.

A169662 Numbers divisible by the sum of their digits, and by the sum of their digits squared, by the sum of their digits cubed and by the sum of 4th powers of their digits.

Original entry on oeis.org

1, 10, 100, 110, 111, 1000, 1010, 1011, 1100, 1101, 1110, 2000, 5000, 10000, 10010, 10011, 10100, 10101, 10110, 11000, 11001, 11010, 11100, 20000, 50000, 55000, 100000, 100010, 100011, 100100, 100101, 100110, 101000, 101001, 101010, 101100
Offset: 1

Views

Author

Michel Lagneau, Apr 05 2010

Keywords

Comments

The numbers such that all digits are nonzero are rare (see the subsequence A176194).

Examples

			1121211 is a term since 1^4 + 1^4 + 2^4 + 1^4 + 2^4 + 1^4 + 1^4 = 37 and 1121211 = 37*30303 ; 1^3 + 1^3 + 2^3 + 1^3 + 2^3 + 1^3 + 1^3 = 21 and 1121211 = 21*53391 ; 1^2 + 1^2 + 2^2 + 1^2 + 2^2 + 1^2 + 1^2 = 13 and 1121211 = 13* 86247 ; 1 + 1 + 2 + 1 + 2 + 1 + 1 = 9 and 1121211 = 9*124579.
		

Crossrefs

Intersection of A005349, A034087, A034088 and A169665.

Programs

  • Maple
    isA169662 := proc(n)
            dgs := convert(n,base,10) ;
            if (n mod ( add(d,d=dgs) ) = 0)  and (n mod (add(d^2,d=dgs) )) =0 and (n mod (add(d^3,d=dgs))) =0 and (n mod (add(d^4,d=dgs))) = 0 then
                    true;
            else
                    false;
            end if;
    end proc:
    for i from 1 to 110000 do
            if isA169662(i) then
                    printf("%d,",i) ;
            end if;
    end do: # R. J. Mathar, Nov 07 2011
  • Mathematica
    q[n_] := And @@ Divisible[n, Plus @@@ Transpose @ Map[#^Range[4] &, IntegerDigits[n]]]; Select[Range[10^5], q] (* Amiram Eldar, Jan 31 2021 *)

Formula

{n : A007953(n)|n and A003132(n)|n and A055012(n)| n and A055013(n)| n}.

A176194 Numbers with no zero digits divisible by the sum of the k-th powers of their digits, for each k = 1,2,3,4.

Original entry on oeis.org

1, 111, 1121211, 11243232, 12132432, 12413232, 22331232, 23111352, 23411232, 24113232, 41223312, 42131232, 44662464, 111111111, 112452144, 114251424, 135964224, 211412544, 246134592, 313212312, 332131212, 382941675, 416283624, 442114512, 523173456, 671635575, 979652772
Offset: 1

Views

Author

Michel Lagneau, Apr 11 2010

Keywords

Comments

For the numbers divisible by the sum of k-th powers of digits including 0, see A169662. The numbers such that the digits are > 0 are rare.

Examples

			For n = 246134592 we obtain :
2^4 + 4^4 + 6^4 + 1^4 + 3^4 + 4^4 + 5^4 + 9^4 + 2^4 = 9108, and 246134592 = 9108*27024 ;
2^3 + 4^3 + 6^3 + 1^3 + 3^3 + 4^3 + 5^3 + 9^3 + 2^3 = 1242, and 246134592 = 1242*198176 ;
2^2 + 4^2 + 6^2 + 1^2 + 3^2 + 4^2 + 5^2 + 9^2 + 2^2 = 192, and 246134592 = 192*1281951 ;
2 + 4 + 6 + 1 + 3 + 4 + 5 + 9 + 2 = 36, and 246134592 = 36*6837072.
		

Crossrefs

Programs

  • Maple
    with(numtheory):for n from 2 to 500000000 do:l:=evalf(floor(ilog10(n))+1):n0:=n:s1:=0:s2:=0:s3:=0:s4:=0:p:=1:for m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :s1:=s1+u:p:=p*u:s2:=s2+u^2:s3:=s3+u^3:s4:=s4+u^4: od:if irem(n,s1)=0 and irem(n,s2)=0 and irem(n,s3)=0 and irem(n,s4)=0 and p<>0 then print(n):else fi:od:

Formula

A007953 (n)|n and A003132(n)|n and A055012 (n)| n and A055013 (n)| n and all digits < > 0.

Extensions

a(1)-a(2) and more terms add by Amiram Eldar, Apr 20 2023

A169663 Numbers k divisible by the sum of the digits and the sum of the squares of digits of k (in base 10).

Original entry on oeis.org

1, 10, 20, 50, 100, 110, 111, 120, 133, 200, 210, 240, 315, 360, 372, 400, 420, 480, 500, 550, 630, 803, 1000, 1010, 1011, 1020, 1071, 1100, 1101, 1110, 1134, 1148, 1200, 1300, 1302, 1330, 1344, 1431, 1547, 2000, 2010, 2023, 2040, 2100, 2196, 2200, 2220
Offset: 1

Views

Author

Michel Lagneau, Apr 05 2010

Keywords

Examples

			For k = 2196, 2^2 + 1^2 + 9^2 + 6^2 = 122, 2 + 1 + 9 + 6 = 18, and 2196 = 18*122 so it is divisible by both 18 and 122.
		

Crossrefs

Intersection of A005349 and A034087.

Programs

  • Maple
    with(numtheory):for n from 1 to 1000000 do:l:=evalf(floor(ilog10(n))+1):n0:=n:s1:=0:s2:=0:for m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :s1:=s1+u:s2:=s2+u^2:od:if irem(n,s1)=0 and irem(n,s2)=0 then print(n):else fi:od:
  • Mathematica
    Select[Range[2220], Divisible[#, Plus @@ (d = IntegerDigits[#])] && Divisible[#, Plus @@ (d^2)] &] (* Amiram Eldar, Mar 04 2023 *)
  • PARI
    sd2(n) = my(d=digits(n)); sum(i=1, #d, d[i]^2);
    isok(n) = !(n % sumdigits(n)) && !(n % sd2(n)); \\ Michel Marcus, Dec 21 2014

Formula

A007953(k)|k and A003132(k)|k.

A169664 Numbers k divisible respectively by the sum of digits, the sum of the squares and the sum of the cubes of digits in base 10 of k.

Original entry on oeis.org

1, 10, 100, 110, 111, 200, 500, 1000, 1010, 1011, 1100, 1101, 1110, 2000, 2352, 5000, 5500, 10000, 10010, 10011, 10100, 10101, 10110, 11000, 11001, 11010, 11100, 11112, 20000, 22000, 22200, 23520, 25032, 25110, 30100, 40000, 41013, 44160, 50000
Offset: 1

Views

Author

Michel Lagneau, Apr 05 2010

Keywords

Examples

			For k = 174192, 1^3 + 7^3 + 4^3 + 1^3 + 9^3 + 2^3 = 1146, and 174192 = 152*1146; 1^2 + 7^2 + 4^2 + 1^2 + 9^2 + 2^2 = 152, and 174192 = 152*1146; 1 + 7 + 4 + 1 + 9 + 2 = 24, and 174192 = 24*7258.
		

Crossrefs

Programs

  • Maple
    with(numtheory):for n from 1 to 200000 do:l:=evalf(floor(ilog10(n))+1) : n0:=n:s1:=0:s2:=0: s3:=0:for m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :s1:=s1+u:s2:=s2+u^2:s3:=s3+u^3:od:if irem(n,s1)=0 and irem(n,s2)=0 and irem(n,s3)=0 then print(n):else fi:od:
  • Mathematica
    dsQ[n_]:=Module[{idn=IntegerDigits[n]}, Divisible[n,Total[idn]] && Divisible[n,Total[idn^2]] && Divisible[n,Total[idn^3]]]; Select[Range[50000],dsQ]  (* Harvey P. Dale, Feb 24 2011 *)

Formula

A007953(k)|k and A003132(k)|k and A055012(k)| k.

A244857 Numbers divisible by both the sum of the squares of their digits and the product of their digits.

Original entry on oeis.org

1, 111, 315, 1344, 3312, 4416, 6624, 11112, 12312, 31311, 114192, 121716, 134112, 134136, 141312, 231336, 282624, 313416, 314112, 411648, 431136, 613116, 628224, 1112232, 1112832, 1121232, 1122112, 1122312, 1122912, 1143216, 1211232, 1212112, 1212192, 1212312
Offset: 1

Views

Author

Michel Lagneau, Jul 07 2014

Keywords

Comments

Subsequence of A034087.
The property "numbers divisible by the sum of the squares and product of their digits" leads to the Diophantine equation t*x1*x2*...*xr=s*(x1^2+x2^2+...+xr^2), where t and s are divisors of n; xi is from [1...9].
Intersection of A034087 and A007602. - Jens Kruse Andersen, Jul 13 2014

Examples

			315 is in the sequence because 3^2+1^2+5^2 = 35 divides 315 and 3*1*5 = 15 divides 315.
		

Crossrefs

Programs

  • Mathematica
    dspQ[n_]:=Module[{idn=IntegerDigits[n], t}, t=Times@@idn; t!=0 && Divisible[n, Total[idn^2]] && Divisible[n, t]]; Select[Range[2*10^6], dspQ]
  • PARI
    isok(n) = (d = digits(n)) && (prd = prod(i=1, #d, d[i])) && !(n % prd) && !(n % sum(i=1, #d, d[i]^2)); \\ Michel Marcus, Jul 07 2014

A118122 Least prime of level 2n-1 (cf. A117563).

Original entry on oeis.org

5, 11, 17, 509, 29, 83, 41, 79, 887, 59, 109, 71, 331, 193, 383, 190717, 101, 107, 787, 277, 1129, 911, 137, 1181, 149, 463, 1013, 839, 1087, 179, 433, 191, 197, 4093, 349, 503, 2423, 227, 701, 239, 5378731, 587, 601, 439, 269, 6491, 281, 1621, 877, 499
Offset: 1

Views

Author

Keywords

Examples

			The first occurrence of 1 in A117563 is a(3) which implies the third prime which is 5.
The first occurrence of 3 in A117562 is a(5) which implies the fifth prime which is 11.
The first occurrence of 5 in A117562 is a(7) which implies the seventh prime which is 17, etc.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := If[n == 1, 0, Block[{p = Prime@n, np = Prime[n + 1]}, (2p - np)/Min@Select[Divisors[2p - np], # >= np - p &]]]; t = Table[0, {100}]; Do[a = (f@n + 1)/2; If[a < 101 && t[[a]] == 0, t[[a]] = Prime@n; Print[{a, n, Prime@n}]], {n, 10^6}]

Formula

Levels of primes are defined in A117563. Conjecture: there are an infinite number of prime members at each level.
Showing 1-6 of 6 results.