A330904 -id:A330904 - cp's OEIS frontend

A037301 Numbers whose base-2 and base-3 expansions have the same digit sum.

0, 1, 6, 7, 10, 11, 12, 13, 18, 19, 21, 36, 37, 46, 47, 58, 59, 60, 61, 86, 92, 102, 103, 114, 115, 120, 121, 166, 167, 172, 173, 180, 181, 198, 199, 216, 217, 222, 223, 261, 273, 282, 283, 285, 298, 299, 300, 301, 306, 307, 309, 318

Offset: 1

Clark Kimberling

If Sum_{i=0..k} (binomial(k,i) mod 2) == Sum_{i=0..k} (binomial(k,i) mod 3) then k is in the sequence. (The converse does not hold.) - Benoit Cloitre, Nov 16 2003
Problem: To prove that the sequence is infinite. A generalization: Let s_m(k) denote the sum of digits of k in base m; does the Diophantine equation s_p(k) = s_q(k), where p,q are fixed distinct primes, have infinitely many solutions? - Vladimir Shevelev, Jul 30 2009
Also, numbers k such that the exponent of the largest power of 2 dividing k! is exactly twice the exponent of the largest power of 3 dividing k!. - Ivan Neretin, Mar 08 2015
a(5) = 10, a(6) = 11, a(7) = 12 and a(8) = 13 is the first time that four consecutive terms appear in this sequence. Conjecture: There is no occurrence of five or more consecutive terms of a(n). Tested by exhaustive search up to a(n) = 3^29. - Thomas König, Aug 15 2020

Stanislav Sykora, Table of n, a(n) for n = 1..10000
Vladimir Shevelev, Compact integers and factorials, Acta Arith. 126 (2007), no. 3, 195-236.
Vladimir Shevelev, Binomial predictors, arXiv:0907.3302 [math.NT], 2009.
Lukas Spiegelhofer, Collisions of the binary and ternary sum-of-digits functions, arXiv:2105.11173 [math.NT], 2021. Proves that sequence is infinite.

Cf. A000120, A053735, A180017.

Cf. A001316, A051638, A212222, A330904, A334765.

Select[ Range@ 320, Total@ IntegerDigits[#, 2] == Total@ IntegerDigits[#, 3] &] (* Robert G. Wilson v, Oct 24 2014 *)

is(n)=sumdigits(n,3)==hammingweight(n) \\ Charles R Greathouse IV, May 21 2015

A053735(a(n)) = A000120(a(n)); A180017(a(n)) = 0. - Reinhard Zumkeller, Aug 06 2010

Zero prepended by Zak Seidov, May 31 2010

A334765 Numbers m such that the numbers of 1's in the binary expansion of m equals the negative sum of balanced ternary trits of m.

Original entry on oeis.org

0, 41, 68, 131, 132, 368, 384, 528, 1095, 1098, 1100, 1106, 1112, 1122, 1124, 1152, 1176, 1346, 1824, 2561, 3282, 3284, 3336, 3344, 3392, 3524, 4098, 4101, 4104, 4112, 4118, 4128, 4172, 4352, 4496, 4739, 4740, 5504, 6224, 9856, 9857, 9869, 9896, 9923, 9924

Offset: 1

Thomas König, May 10 2020

a(116) = 32770, a(117) = 32771 and a(118) = 32772 is the first time that three consecutive numbers appear in this sequence. Conjecture: There is no occurrence of four or more consecutive numbers. Tested by exhaustive search up to a(n) = 3^26. - Thomas König, Jul 19 2020

			41_10 = 1TTTT_bt = 101001_2, the sum of the digits is 1-1-1-1-1 = -3 for balanced ternary and 1+1+1 = 3 for base 2, so 41 is a term.

Thomas König, Table of n, a(n) for n = 1..30000

Aside from the first term, subsequence of A174657.

Cf. A000120, A037301, A065363, A330904.

Select[Range[0, 10^4], -Total@ If[First@ # == 0, Rest@ #, #] &[Prepend[IntegerDigits[#, 3], 0] //. {x___, y_, k_ /; k > 1, z___} :> {x, y + 1, -1, z}] == DigitCount[#, 2, 1] &] (* Michael De Vlieger, Jul 08 2020 *)

bt(n)= if (n==0, return (0)); my(d=digits(n, 3), c=1); while(c, if(d[1]==2, d=concat(0, d)); c=0; for(i=2, #d, if(d[i]==2, d[i]=-1; d[i-1]+=1; c=1))); vecsum(d); \\ A065363
isok(m) = bt(m) + hammingweight(m) == 0; \\ Michel Marcus, Jun 07 2020

Integers m such that -A065363(m) = A000120(m).

cp's OEIS Frontend

A037301 Numbers whose base-2 and base-3 expansions have the same digit sum.

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