A068052 Start from 1, shift one left and sum mod 2 (bitwise-XOR) to get 3 (11 in binary), then shift two steps left and XOR to get 15 (1111 in binary), then three steps and XOR to get 119 (1110111 in binary), then four steps and so on.

1, 3, 15, 119, 1799, 59367, 3743271, 481693095, 123123509927, 62989418816679, 64491023022979239, 132015402419352060071, 540829047855347718631591, 4430403202865824763042320551, 72583450474242118015031400337575, 2378466805556971511916001231449723047

Offset: 0

Author

Antti Karttunen, Feb 13 2002

Keywords

• nonn
• base

Comments

a(n) = each row of A053632 reduced mod 2 and interpreted as a binary number.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..64

Crossrefs

Same sequence shown in binary: A068053.

Cf. A000120, A003987, A028362 (using + instead of XOR), A048675, A053632, A080791, A248663, A285101, A285102, A285103, A285105.

Programs

Maple

with(gfun,seriestolist); [seq(foo(map(`mod`,seriestolist(series(mul(1+(z^i),i=1..n),z,binomial(n+1,2)+1)),2)), n=0..20)];
foo := proc(a) local i; add(a[i]*2^(i-1),i=1..nops(a)); end;
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 1,
      (t-> Bits[Xor](2^n*t, t))(a(n-1)))
    end:
seq(a(n), n=0..16);  # Alois P. Heinz, Mar 07 2024

Mathematica

FoldList[BitXor[#, #*#2]&, 1, 2^Range[20]] (* Paolo Xausa, Mar 07 2024 *)

PARI

a(n) = if(n<1, 1, bitxor(a(n - 1), 2^n*a(n - 1))); \\ Indranil Ghosh, Apr 15 2017, after formula by Antti Karttunen

Formula

a(0) = 1; for n > 0, a(n) = a(n-1) XOR (2^n)*a(n-1), where XOR is bitwise-XOR (A003987).

a(n) = A248663(A285101(n)) = A048675(A285102(n)).

A000120(a(n)) = A285103(n). [Number of ones in binary representation.]

A080791(a(n)) = A285105(n). [Number of nonleading zeros.]

Extensions

Formulas added by Antti Karttunen, Apr 15 2017