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 A179417 a(n) = is binary number (shown here in decimal) constructed from quadratic residues of 65537 in range [(n^2)+1,(n+1)^2], in such way that quadratic residues are mapped to 1-bits, and non-quadratic residues (as well as the multiples of 65537) to 0-bits, with the lower end of range mapped to less significant, and the higher end of range to more significant bits. 3
 1, 5, 24, 104, 279, 2001, 4131, 17453, 88826, 362532, 1655660, 6120642, 25376649, 128526482, 301370205, 1756488602, 8046359747, 30854867177, 73845140753, 488906501177, 2106640948770, 6573967883049, 29711211505300 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS The binary width of terms are 1, 3, 5, 7, 9, ... i.e. the successive odd numbers, as their partial sums give the squares, 1, 4, 9, 16, ... at which points there certainly is always a quadratic residue, which thus gives the most significant bit for each number. LINKS A. Karttunen, Table of n, a(n) for n = 0..256 EXAMPLE In range [(2^2)+1,(2+1)^2] (i.e. [5,9]) we have A165471(5)=A165471(6)=A165471(7)=-1 and A165471(8)=A165471(9)=+1, i.e. there are quadratic non-residues at points 5, 6 and 7, and quadratic residues at 8 and 9, so we construct a binary number 11000, which is 24 in decimal, thus a(2)=24. PROG (MIT Scheme:) (define (A179417 n) (let ((ul (A005408 n))) (let loop ((i (A000290 n)) (j 0) (s 0)) (cond ((= j ul) s) ((= 0 (1+halved (A165471 (1+ i)))) (loop (1+ i) (1+ j) s)) (else (loop (1+ i) (1+ j) (+ s (expt 2 j)))))))) (define (1+halved n) (floor->exact (/ (1+ n) 2))) CROSSREFS Compare to similar bit triangle illustrations given in A080070, A122229, A122232, A122235, A122239, A122242, A122245. Sequence in context: A276139 A078820 A291395 * A181305 A046724 A272578 Adjacent sequences:  A179414 A179415 A179416 * A179418 A179419 A179420 KEYWORD nonn,base AUTHOR Antti Karttunen, Jul 27 2010 STATUS approved

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Last modified September 20 22:20 EDT 2019. Contains 327252 sequences. (Running on oeis4.)