A072081 -id:A072081 - cp's OEIS frontend

A072083 Numbers divisible by the 4th power of the sum of their digits in base 10.

1, 10, 100, 1000, 2000, 2401, 5000, 10000, 13122, 20000, 24010, 50000, 100000, 110000, 131220, 140000, 190000, 200000, 230000, 234256, 240100, 280000, 320000, 370000, 390625, 400221, 410000, 460000, 500000, 512000, 550000, 614656, 640000

Offset: 1

Labos Elemer, Jun 14 2002

If k is a term, then 10*k is a term. There are an infinite number of terms that are not divisible by 10. The numbers m = 24 * 10^(294*k - 292) + 1, k = 7*a - 6, a >= 1, are divisible by 7^4 = digsum(m)^4. Also, the numbers s = 491 * 10^(4624*k - 4623) + 3, k = 17*u - 11, u >= 1, are divisible by 17^4 = digsum(s)^4. - Marius A. Burtea, Mar 19 2020

The numbers 2^A095412(n), n >= 6, are terms. - Marius A. Burtea, Apr 02 2020

			k=614656: sumdigits(614656)=28, q=1, since k=28*28*28*28.

Donovan Johnson, Table of n, a(n) for n = 1..1000

Cf. A005349, A003634, A072081, A072082.

[k:k in [1..640000]| k mod &+Intseq(k)^4 eq 0]; // Marius A. Burtea, Mar 19 2020

sud[x_] := Apply[Plus, IntegerDigits[x]] Do[s=sud[n]^4; If[IntegerQ[n/s], Print[n]], {n, 1, 10000}]
Select[Range[700000],Divisible[#,Total[IntegerDigits[ #]]^4]&] (* Harvey P. Dale, Jun 28 2011 *)

isok(m) = (m % sumdigits(m)^4) == 0; \\ Michel Marcus, Apr 02 2020

A072408 Least number > 1 which equals n-th power of the sum of its digits in decimal base.

Original entry on oeis.org

2, 81, 512, 2401, 17210368, 34012224, 612220032, 20047612231936, 3904305912313344, 13744803133596058624, 8007313507497959524352, 2518170116818978404827136, 81920000000000000, 2670419511272061205254504361, 2759031540715333904109053133443

Offset: 1

Labos Elemer, Jun 17 2002

			a(1) = 2 > 1;
a(9) = 3904305912313344 = 54^9 = (3+9+0+4+3+0+5+9+1+2+3+1+3+3+4+4)^9.

Michael De Vlieger, Table of n, a(n) for n = 1..274
Leo Moser, Digits of Powers, Recreational Mathematics, Curiosa 145. Scripta Mathematica Vol. XIII, March-June 1947, Nos. 1-2, page 117.

Cf. A007953, A046017, A072081-A072083.

Array[Block[{k = 2}, While[k^# != Total[IntegerDigits[k^#]]^#, k++]; k^#] &, 13] (* Michael De Vlieger, Nov 05 2020 *)

a(n) = Min{ x; x=SumDigit(x)^n} = Min{x; x=A007953(x)^n}

a(n) = A046017(n)^n. - Michael De Vlieger, Nov 05 2020

A379980 Numbers that are divisible by the square of the sum of the squares of their digits.

Original entry on oeis.org

1, 10, 100, 1000, 1100, 1200, 1300, 2000, 2023, 2100, 2400, 3100, 4332, 5000, 10000, 10100, 10200, 10300, 11000, 12000, 13000, 20000, 20100, 20230, 20400, 21000, 24000, 30100, 30324, 31000, 31311, 42000, 43011, 43320, 50000, 52022, 52215, 55000, 71824, 100000

Offset: 1

Amiram Eldar, Jan 07 2025

Called "Second-order Harshad numbers" by Pal and Gopalan (2023).

If k is a term, then 10*k is also a term.

			10 is a term since 10 is divisible by (1^2 + 0^2)^2 = 1.
1100 is a term since 1100 is divisible by (1^2 + 1^2 + 0^2 + 0^2)^2 = 4.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Pradip Kumar Pal and Kaushik Gopalan, Second Order Harshad Number, International Journal of Mathematical Education, Vol. 13, No. 1 (2023), pp. 25-26.

Cf. A003132, A005349, A072081, A180490 (binary analog).
Subsequence of A034087.
Subsequences: A379981, A379982.

Select[Range[10^5], Divisible[#, (Plus @@ (IntegerDigits[#]^2))^2] &]

isok(k) = !(k % vecsum(apply(x -> x^2, digits(k)))^2);

def ok(n): return n and n%sum(di**2 for di in map(int, str(n)))**2 == 0
print([k for k in range(100001) if ok(k)]) # Michael S. Branicky, Jan 10 2025

A072082 Numbers divisible by the cube of the sum of their digits in base 10.

Original entry on oeis.org

1, 10, 100, 200, 500, 512, 1000, 2000, 2401, 4913, 5000, 5103, 5120, 5832, 10000, 10206, 11000, 11200, 11664, 13122, 14000, 17576, 19000, 19683, 20000, 20412, 21141, 23000, 23328, 24010, 28000, 29160, 32000, 37000, 39366, 40000, 40824

Offset: 1

Labos Elemer, Jun 14 2002

If k is a term, then 10 * k is a term. There are an infinite number of terms that are not divisible by 10. The numbers m = 24 * 10^(294 * k - 292) +1 are divisible by 7^3 = digsum(m)^3. Also, the numbers s = 491 * 10^(4624 * k - 4623) + 3, k >= 1, are divisible by 17^3 = digsum(s)^3. - Marius A. Burtea, Mar 18 2020

			k=98415: sumdigits(98415)=27, q=98415=5*27*27*27.

Donovan Johnson, Table of n, a(n) for n = 1..1000

Cf. A055012, A005349, A003634, A072081, A072083.

[k:k in [1..41000]| k mod &+Intseq(k)^3 eq 0]; // Marius A. Burtea, Mar 18 2020

sud[x_] := Apply[Plus, IntegerDigits[x]] Do[s=sud[n]^3; If[IntegerQ[n/s], Print[n]], {n, 1, 10000}]
Select[Range[50000],Divisible[#,Total[IntegerDigits[#]]^3]&] (* Harvey P. Dale, Mar 22 2016 *)

is(n)=n%sumdigits(n)^3==0 \\ Charles R Greathouse IV, Mar 19 2020

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

Michel Lagneau, Apr 05 2010

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

			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.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Intersection of A005349, A034087, A034088 and A169665.

Cf. A072081, A117562, A176194.

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

q[n_] := And @@ Divisible[n, Plus @@@ Transpose @ Map[#^Range[4] &, IntegerDigits[n]]]; Select[Range[10^5], q] (* Amiram Eldar, Jan 31 2021 *)

{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

Michel Lagneau, Apr 11 2010

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.

			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.

Amiram Eldar, Table of n, a(n) for n = 1..464

Intersection of A052382 and A169662.

Cf. A034087, A072081, A117562, A034087, A072081, A117562, A034088, A034088.

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:

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

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

Michel Lagneau, Apr 05 2010

			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.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Digit.

Cf. A003132, A007953, A034087, A072081, A117562.

Intersection of A005349 and A034087.

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:

Select[Range[2220], Divisible[#, Plus @@ (d = IntegerDigits[#])] && Divisible[#, Plus @@ (d^2)] &] (* Amiram Eldar, Mar 04 2023 *)

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

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

Michel Lagneau, Apr 05 2010

			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.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Digit.

Cf. A003132, A007953, A034087, A055012, A072081, A117562.

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:

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 *)

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

A243008 Triangular numbers divisible by the square of the sum of their digits.

Original entry on oeis.org

1, 10, 3240, 3321, 13041, 13203, 15400, 65341, 80200, 90100, 161028, 210276, 260281, 265356, 266085, 300700, 346528, 500500, 937765, 947376, 1043290, 1228528, 1313010, 1628110, 2049300, 2390391, 2421100, 3357936, 3746953, 4020030, 5250420, 6641190, 6857956, 6939675

Offset: 1

K. D. Bajpai, Aug 20 2014

Intersection of A000217 and A072081.

			a(3) = 3240 = 80 * (80 + 1)/2 is a triangular number. Since 3240 is divisible by (3 + 2 + 4 + 0)^2 = 81, it appears in the sequence.
a(3) = 3321 = 81 * (81 + 1)/2 is a triangular number. Since 3321 is divisible by (3 + 3 + 2 + 1)^2 = 81, it appears in the sequence.

K. D. Bajpai, Table of n, a(n) for n = 1..12445

Cf. A000217, A072081, A076713, A118693.

Select[Table[n*(n + 1)/2, {n, 10000}], Divisible[#, Plus @@ IntegerDigits[#]^2] &]

for(n=1,10^4,s=n*(n+1)/2;if(s%(sumdigits(s)^2)==0,print1(s,", "))) \\ Derek Orr, Aug 23 2014

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A072083 Numbers divisible by the 4th power of the sum of their digits in base 10.

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A072408 Least number > 1 which equals n-th power of the sum of its digits in decimal base.

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A379980 Numbers that are divisible by the square of the sum of the squares of their digits.

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A072082 Numbers divisible by the cube of the sum of their digits in base 10.

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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.

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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.

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

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