Dhilan Lahoti - cp's OEIS frontend

A261751 Numbers n with property that binary expansion of n^3 begins with the binary expansion of n.

0, 1, 2, 3, 4, 6, 8, 16, 23, 32, 64, 91, 128, 256, 512, 1024, 2048, 4096, 5793, 8192, 16384, 32768, 46341, 65536, 92682, 131072, 185364, 262144, 370728, 524288, 1048576, 2097152, 2965821, 4194304, 5931642, 8388608, 16777216, 33554432, 47453133, 67108864, 94906266

Offset: 1

Dhilan Lahoti, Aug 30 2015

2^k is always a term in this sequence.

It appears that all solutions are either a power of 2 or approximately sqrt(2) * a power of 2. - Andrew Howroyd, Dec 24 2019

			23 is a term of this sequence because its cube written in base 2 (10111110000111) starts with its representation in base 2 (10111).

Andrew Howroyd, Table of n, a(n) for n = 1..1000

Base 2 version of A052210.

Cf. A004539.

SetBeginSet[set1_, set2_] :=
  Do[For[i = 1, i <= Length[set1], i++,If[! set1[[i]] == set2[[i]], Return[False]]];Return[True], {1}];
For[k = 0; set = {}, k <= 100000, k++,If[SetBeginSet[IntegerDigits[k, 2], IntegerDigits[k^3, 2]],Print[k]]]

ok(n)={my(t=n^3); t == 0 || t>>(logint(t,2)-logint(n,2))==n} \\ Andrew Howroyd, Dec 23 2019

\\ for larger values
viable(b,k)={my(p=b^3, q=(b+2^k-1)^3, s=logint(q,2), t=s-logint(b,2)+k); (p>>s)==0 || ((p>>t)<=(b>>k) && (b>>k)<=(q>>t))}
upto(n)={
  local(L=List([0]));
  my(recurse(b,k)=; if(b <= n && viable(b,k), k--; if(k<0, listput(L, b), self()(b,k); self()(b+2^k,k))));
  for(k=0, logint(n,2), recurse(2^k, k));
  Vec(L);
} \\ Andrew Howroyd, Dec 24 2019

Terms a(31) and beyond from Andrew Howroyd, Dec 23 2019

A261749 Numbers k where k^2 is an anagram of (k+2)^2.

Original entry on oeis.org

206, 224, 314, 1799, 2006, 11087, 13364, 15839, 17153, 17324, 20006, 22184, 22706, 24524, 24542, 40031, 40247, 45314, 47069, 48824, 55556, 61694, 64691, 70559, 71351, 89774, 90224, 102374, 108251, 112292, 129824, 132506, 137987, 151757, 154295, 157706, 162089, 167273, 170324, 171557, 175031

Offset: 1

Dhilan Lahoti, Aug 30 2015

Numbers of the form 2*10^k + 6 where k > 1 always appear in this sequence.
Numbers of the form 4*10^k + 31 and 86*10^k + 39 always appear when k > 3.
Similar to A072841 but with (n+2)^2 instead of (n+1)^2.
All numbers in the sequence are of the form 3n + 2.
Multiples of 5 seem to be uncommon.
Another subsequence is numbers of the form 5*(10^(5+9*k)-1)/9 + 1, i.e. 4+9*k 5's followed by a 6: 55556, 55555555555556, 55555555555555555555556, etc. - Robert Israel, Aug 31 2015

			206 is a term in the sequence because 206^2 (42436) and 208^2 (43264) are anagrams.

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

Cf. A072841.

filter:= proc(n) local L1, L2;
  L1:= convert(n^2,base,10);
  L2:= convert((n+2)^2,base,10);
  evalb(sort(L1)=sort(L2));
end proc:
select(filter, [3*i+2 $ i = 1..10^5]); # Robert Israel, Aug 31 2015

Select[Range[10^4], Sort[IntegerDigits[#^2]] == Sort[IntegerDigits[(# + 2)^2]] &] (* Typo fixed by Ivan N. Ianakiev, Sep 02 2015 *)

isok(n) = vecsort(digits(n^2)) == vecsort(digits((n+2)^2)); \\ Michel Marcus, Aug 31 2015

A261749_list = [n for n in range(1,10**6) if sorted(str(n**2)) == sorted(str((n+2)**2))] # Chai Wah Wu, Sep 02 2015

A260477 Numbers n where n^2 is an anagram of (n+1)^2 in base 2.

Original entry on oeis.org

9, 17, 33, 49, 50, 65, 73, 77, 85, 86, 97, 106, 129, 137, 149, 157, 161, 165, 178, 186, 193, 201, 212, 213, 225, 226, 257, 265, 273, 279, 281, 285, 300, 305, 307, 310, 317, 321, 325, 332, 334, 345, 355, 365, 366, 378, 385, 393, 394, 413, 426, 427, 449, 469

Offset: 1

Dhilan Lahoti, Aug 28 2015

Base 2 equivalent of A072841.

It appears that one of these numbers has a 1/n chance of being divisible by an odd number n, but a smaller than 1/n chance if n is even.

			17 is a term of the sequence because its square base 2 (100100001) and 18's square base 2 (101000100) are anagrams.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

For[i = 1, i <= 10000, i++,
If[Sort[IntegerDigits[i^2, 2]] == Sort[IntegerDigits[(i + 1)^2, 2]],
  Print[i]]]

is(n)=hammingweight(n^2)==hammingweight((n+1)^2) && #binary(n^2)==#binary((n+1)^2) \\ Charles R Greathouse IV, Aug 29 2015

a(27)-a(54) from Charles R Greathouse IV, Aug 29 2015

A247107 a(0) = 0, a(n) = previous term + repunit of length of previous term for n > 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 220, 331, 442, 553, 664, 775, 886, 997, 1108, 2219, 3330, 4441, 5552, 6663, 7774, 8885, 9996, 11107, 22218, 33329, 44440, 55551, 66662, 77773, 88884, 99995, 111106, 222217, 333328, 444439

Offset: 0

Dhilan Lahoti, Nov 30 2014

			98 = 9*10 + 8 -> 10*10 + 9 = 109.
109 = 1*100 + 0*10 + 9*1 -> 2*100 + 1*10 + 10*1 = 220.
a(42) = 44440 + (10^(floor(log_10(44440))+1)-1) / 9 = 44440 + (10^(4+1)-1) / 9 = 44440 + 99999/9 = 44440 + 11111 = 55551.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,10,-20,10).

Similar to A158699, but with simpler rules.

Cf. A002275, A055642.

a[0]=0; a[n_]:=FromDigits[IntegerDigits[a[n-1]]+1]; Array[a,50,0] (* Stefano Spezia, Sep 19 2024 *)

a(0) = 0, a(n) = a(n-1) + A002275(A055642(a(n-1))) for n>0.
From Jon E. Schoenfield, Nov 30 2014: (Start)
For n > 1, a(n) = a(n-1) + (10^(floor(log_10(a(n-1))) + 1) - 1) / 9.
For n > 0, a(n) = ((n-1) mod 9 + 1) * (10^D - 1) / 9 + 1 - D, where D = floor((n-1)/9) + 1. (There are exactly D digits in a(n).) (End)
G.f.: -(10*x^10-10*x^9+1)*x/((10*x^9-1)*(x-1)^2). - Alois P. Heinz, Nov 30 2014

A219735 a(n) = (a(n-2) * a(n-1))^2, with a(0) = 1 and a(1) = 2.

Original entry on oeis.org

1, 2, 4, 64, 65536, 17592186044416, 1329227995784915872903807060280344576

Offset: 0

Dhilan Lahoti, Nov 26 2012

nonn

Every one of these numbers is a power of 2.

Cf. A000079.

RecurrenceTable[{a[0]==1,a[1]==2,a[n]==(a[n-1]*a[n-2])^2},a,{n,7}] (* Harvey P. Dale, Apr 23 2015 *)

a(n) = 2^A002605(n).

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User: Dhilan Lahoti

A261751 Numbers n with property that binary expansion of n^3 begins with the binary expansion of n.

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A261749 Numbers k where k^2 is an anagram of (k+2)^2.

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A260477 Numbers n where n^2 is an anagram of (n+1)^2 in base 2.

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A247107 a(0) = 0, a(n) = previous term + repunit of length of previous term for n > 0.

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A219735 a(n) = (a(n-2) * a(n-1))^2, with a(0) = 1 and a(1) = 2.

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