A366370 Square array A(n,k) giving the length of the least significant run of 0-bits in binary expansion of A000225(n)^k, or 0 if A000225(n)^k is a binary repunit.
0, 0, 0, 0, 2, 0, 0, 1, 3, 0, 0, 3, 1, 4, 0, 0, 2, 4, 1, 5, 0, 0, 2, 2, 5, 1, 6, 0, 0, 1, 3, 2, 6, 1, 7, 0, 0, 4, 1, 4, 2, 7, 1, 8, 0, 0, 3, 5, 1, 5, 2, 8, 1, 9, 0, 0, 2, 3, 6, 1, 6, 2, 9, 1, 10, 0, 0, 1, 3, 3, 7, 1, 7, 2, 10, 1, 11, 0, 0, 3, 1, 4, 3, 8, 1, 8, 2, 11, 1, 12, 0, 0, 2, 4, 1, 5, 3, 9, 1, 9, 2, 12, 1, 13, 0
Offset: 1
Examples
The top left corner of the square array: n\k| 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 -----+------------------------------------------------------------------- 1 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2 | 0, 2, 1, 3, 2, 2, 1, 4, 3, 2, 1, 3, 2, 2, 1, 5, 4, 3 | 0, 3, 1, 4, 2, 3, 1, 5, 3, 3, 1, 4, 2, 3, 1, 6, 4, 4 | 0, 4, 1, 5, 2, 4, 1, 6, 3, 4, 1, 5, 2, 4, 1, 7, 4, 5 | 0, 5, 1, 6, 2, 5, 1, 7, 3, 5, 1, 6, 2, 5, 1, 8, 4, 6 | 0, 6, 1, 7, 2, 6, 1, 8, 3, 6, 1, 7, 2, 6, 1, 9, 4, 7 | 0, 7, 1, 8, 2, 7, 1, 9, 3, 7, 1, 8, 2, 7, 1, 10, 4, 8 | 0, 8, 1, 9, 2, 8, 1, 10, 3, 8, 1, 9, 2, 8, 1, 11, 4, 9 | 0, 9, 1, 10, 2, 9, 1, 11, 3, 9, 1, 10, 2, 9, 1, 12, 4, 10 | 0, 10, 1, 11, 2, 10, 1, 12, 3, 10, 1, 11, 2, 10, 1, 13, 4, 11 | 0, 11, 1, 12, 2, 11, 1, 13, 3, 11, 1, 12, 2, 11, 1, 14, 4, 12 | 0, 12, 1, 13, 2, 12, 1, 14, 3, 12, 1, 13, 2, 12, 1, 15, 4, 13 | 0, 13, 1, 14, 2, 13, 1, 15, 3, 13, 1, 14, 2, 13, 1, 16, 4, 14 | 0, 14, 1, 15, 2, 14, 1, 16, 3, 14, 1, 15, 2, 14, 1, 17, 4, 15 | 0, 15, 1, 16, 2, 15, 1, 17, 3, 15, 1, 16, 2, 15, 1, 18, 4, 16 | 0, 16, 1, 17, 2, 16, 1, 18, 3, 16, 1, 17, 2, 16, 1, 19, 4, 17 | 0, 17, 1, 18, 2, 17, 1, 19, 3, 17, 1, 18, 2, 17, 1, 20, 4, etc. A000225(4)^4 = ((2^4)-1)^4 = 50625 and A007088(50625) = "1100010111000001", where the rightmost run of 0-bits has length 5, therefore A(4,4) = 5. A000225(3)^5 = ((2^3)-1)^5 = 16807 and A007088(16807) = "100000110100111", where the rightmost run of 0-bits has length 2, therefore A(3,5) = 2. A000225(5)^3 = ((2^5)-1)^3 = 29791 and A007088(29791) = "111010001011111", where the rightmost run of 0-bits is a singleton, therefore A(5,3) = 1.
Links
Programs
-
Mathematica
A285097[n_]:=If[DigitCount[n,2,1]<2,0,IntegerExponent[BitAnd[n-1,n],2]-IntegerExponent[n,2]];A366370[n_,k_]:=A285097[1+(2^n-1)^k]; Table[A366370[k,n-k+1],{n,20},{k,n}] (* Paolo Xausa, Dec 02 2023 *)
-
PARI
up_to = 105; A285097(n) = if(!n || !bitand(n,n-1), 0, valuation((n>>valuation(n,2))-1, 2)); A366370sq(n,k) = A285097(1+(((2^n)-1)^k)); \\ Or more directly as: A366370sq(n,k) = if(1==n||1==k, 0, if(!(k%2), n, 1)+valuation(k>>1,2)); A366370list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A366370sq(col,(a-(col-1))))); (v); }; v366370 = A366370list(up_to); A366370(n) = v366370[n];