A049343 -id:A049343 - cp's OEIS frontend

A260702 Numbers n such that 3*n and n^2 have the same digit sum.

0, 3, 6, 9, 12, 15, 18, 21, 30, 33, 39, 45, 48, 51, 60, 66, 90, 96, 99, 102, 105, 111, 120, 123, 129, 132, 150, 153, 156, 159, 162, 165, 180, 189, 195, 198, 201, 210, 225, 231, 246, 252, 255, 261, 285, 300, 330, 333, 348, 351, 390, 399, 429, 450, 453, 459, 462

Offset: 1

Vincenzo Librandi, Nov 17 2015

All terms are multiple of 3.

If n is in the sequence, then so is 10*n. - Robert Israel, Apr 05 2020

			159 is in the sequence because 159^2 = 25281 and 3*159 = 477 have the same digit sum: 18.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000290, A007953, A008585, A049343, A058369.

Contains A133384 and A199682.

[n: n in [0..500] | &+Intseq(3*n) eq &+Intseq(n^2)];

select(n -> convert(convert(3*n,base,10),`+`)=convert(convert(n^2,base,10),`+`), [seq(i,i=0..1000,3)]); # Robert Israel, Apr 05 2020

Select[Range[0, 500], Total[IntegerDigits[3 #]] == Total[IntegerDigits[#^2]] &]

isok(n) = sumdigits(3*n) == sumdigits(n^2); \\ Michel Marcus, Nov 17 2015

[n for n in (0..500) if sum((3*n).digits())==sum((n^2).digits())] # Bruno Berselli, Nov 17 2015

A007953(A008585(a(n))) = A007953(A000290(a(n))).

A260906 Numbers n such that 3*n and n^3 have the same digit sum.

Original entry on oeis.org

0, 3, 6, 30, 60, 63, 126, 171, 252, 300, 324, 543, 585, 600, 630, 1260, 1281, 1710, 2520, 2925, 3000, 3240, 5430, 5850, 5946, 6000, 6300, 12600, 12606, 12633, 12810, 14631, 16263, 17100, 21618, 22308, 22971, 24663, 25200, 27633, 28845, 28887, 28965, 29241

Offset: 1

Vincenzo Librandi, Nov 18 2015

All terms are multiples of 3.

n is in the sequence iff 10*n is. Are there infinitely many terms not divisible by 10? - Robert Israel, Nov 20 2015

			126 is in the sequence because 126^3 = 2000376 and 3*126 = 378 have the same digit sum: 18.

Robert Israel, Table of n, a(n) for n = 1..320

Cf. A000578, A007953, A008585, A049343, A260702.

[n: n in [0..50000] | &+Intseq(3*n) eq &+Intseq(n^3)];

select(n -> convert(convert(n^3,base,10),`+`)=convert(convert(3*n,base,10),`+`), 3*[$0..10^5]); # Robert Israel, Nov 20 2015

Select[Range[0, 50000], Total[IntegerDigits[3 #]] == Total[IntegerDigits[#^3]] &]

for(n=0, 1e5, if(sumdigits(n^3)==sumdigits(3*n), print1(n, ", "))) \\ Altug Alkan, Nov 20 2015

A007953(A008585(a(n))) = A007953(A000578(a(n))).

cp's OEIS Frontend

A260702 Numbers n such that 3*n and n^2 have the same digit sum.

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