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 A162363 Binary Keith numbers, excluding positive powers of 2. 2
 1, 3, 143, 285, 569, 683, 1138, 1366, 2276, 154203, 308405, 616810, 678491, 1356981, 1480343, 2713962, 2960686, 2212558911, 4425117821, 8850235641, 17700471281 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS This sequence uses the binary expansion of n rather than the decimal expansion used in the usual Keith numbers, A007629. A number n (having a t-bit binary representation) is in this sequence if n is a term in the t-step Fibonacci-like series beginning with the t bits of n. See the example below. Let the bits of n be b(i) for i=1 to t. Then b(t+1) = sum_{i=1..t} b(i). Subsequent terms are b(t+k+1) = 2*b(t+k) - b(k) for k=1,2,3,.... (This is equivalent to, but faster than, the usual method of adding the previous t terms to find the next term.) Due to the growth rate of the numbers in the series, the term equal to n occurs on or before position 2t in the series. Terms in this sequence fall into families having the same number of 1 bits. For instance, 143, 285, 569, 1138, and 2276 all have 5 bits set. Numbers in each family are either 2x or 2x-1, where x is the previous number in the family. The binary expansion of each number in family f begins with f-1 (in binary). This sequence is infinite because for any odd prime (or base-2 pseudoprime, A001567) p=2k+1, we can create a family of numbers with 2^(2k)+1 bits set. The first number in that family is 2^c + c(2^c-2)/(4^p-1) + 1, where c=2^p-1. In binary, this number is a 1 followed by a repeating pattern of p zeros and p ones and terminated by 1, for a total of 2^p bits. For example, 2212558911 is 10000011111000001111100000111111 in binary. LINKS EXAMPLE In binary, 143 = (1,0,0,0,1,1,1,1). Subsequent terms are 5,9,18,36,72,143. MAPLE isA162363 := proc(n)     local L, t, a ;     if numtheory[factorset](n) = {2} then         return false;     end if;     L := ListTools[Reverse](convert(n, base, 2)) ;     t :=  nops(L) ;     while true do         a := add(op(-i, L), i=1..t) ;         L := [op(L), a] ;         if a > n then             return false;         elif a = n then             return true;         end if;     end do: end proc: for n from 1 do     if isA162363(n) then         printf("%d, \n", n);     end if; end do: # R. J. Mathar, Jan 12 2016 MATHEMATICA IsKeith2[n_Integer] := Module[{b, s}, b=IntegerDigits[n, 2]; s=Total[b]; If[s<=1, n==1, k=1; While[s=2*s-b[[k]]; s

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Last modified February 27 17:20 EST 2020. Contains 332307 sequences. (Running on oeis4.)