Search: id:A042964 Results 1-1 of 1 results found. %I A042964 %S A042964 2,3,6,7,10,11,14,15,18,19,22,23,26,27,30,31,34,35,38,39,42,43,46,47, %T A042964 50,51,54,55,58,59,62,63,66,67,70,71,74,75,78,79,82,83,86,87,90,91,94, %U A042964 95,98,99,102,103,106,107,110,111,114,115,118,119,122,123,126,127 %N A042964 Numbers congruent to 2 or 3 mod 4. %C A042964 Also numbers m such that binomial(m+2,m) mod 2 = 0. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Oct 20 2007 %C A042964 Also numbers m such that floor(1+(m/2)) mod 2 = 0. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Oct 20 2007 %C A042964 Partial sums of the sequence 2,1,3,1,3,1,3,1,3,1, ... which has period 2. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Oct 20 2007 %C A042964 In groups of four add and divide by two the odd and even numbers - George E. Antoniou (george.antoniou(AT)montclair.edu), Dec 12 2001. %C A042964 Comments from Jeremy Gardiner (jeremy.gardiner(AT)btinternet.com) on the "mystery calculator". There are 6 cards. %C A042964 Card 0 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, ... (A005408 sequence) %C A042964 Card 1 2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, ... (this sequence) %C A042964 Card 2 4, 5, 6, 7, 12, 13, 14, 15, 20, 21, 22, 23, 28, 29, 30, 31, 36, 37, 38, 39, ... ( A047566) %C A042964 Card 3 8, 9, 10, 11, 12, 13, 14, 15, 24, 25, 26, 27, 28, 29, 30, 31, 40, 41, 42, ... (A115419) %C A042964 Card 4 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 48, 49, 50, ... (A115420) %C A042964 Card 5 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ... (A115421) %C A042964 The trick: You secretly select a number between 1 and 63 from one of the cards. You indicate to me the cards on which that number appears; I tell you the number you selected! %C A042964 The solution: I add together the first term from each of the indicated cards. The total equals the selected number. The numbers in each sequence all have a "1" in the same position in their binary expansion. Example: You indicate cards 1, 3 and 5. Your selected number is 2+8+32 = 42. %C A042964 Numbers having a 1 in position 1 of their binary expansion. One of the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. - Jeremy Gardiner (jeremy.gardiner(AT)btinternet.com), Jan 22 2006 %C A042964 A133872(a(n)) = 0; complement of A042948. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 03 2008] %C A042964 Also the 2nd Witt transform of A040000 [Moree]. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 08 2008] %H A042964 Maths Magic, Mystery Calculator. %H A042964 Pieter Moree, The formal series Witt transform, Discr. Math. no. 295 vol. 1-3 (2005) 143-160. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 08 2008] %F A042964 G.f.: (2+x+x^2)/((1-x)*(1-x^2)). a(n)=a(n-1)+2+(-1)^n - Michael Somos, Jan 12 2000. %F A042964 a(n) = 2n if n is odd else n = 2n-1. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Oct 16 2003 %F A042964 a(n)=[2+(-1)^n+(-1)^(n+1)]*n-[1+(-1)^n]/2, n>=1 - Paolo P. Lava (ppl(AT)spl.at), Feb 15 2008 %F A042964 a(n)=(3+(-1)^n)/2 + 2n = 2n+2-(n mod 2). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Oct 20 2007 %F A042964 a(n)=4*n-a(n-1)-3 (with a(1)=2) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 22 2009] %e A042964 For n=2, a(2)=4*2-2-3=3; n=3, a(3)=4*3-3-3=6; n=4, a(4)=4*4-6-3=7 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 22 2009] %o A042964 (PARI) a(n)=2*n+2-n%2 %Y A042964 a(n) = A047406(n)/2. %Y A042964 Cf. A000040, A133620, A133621, A133622, A133630, A133635. %Y A042964 Cf. A133872, A133882, A133890, A133900, A133910. %Y A042964 Sequence in context: A073439 A107998 A053438 this_sequence A062837 A073170 A014689 %Y A042964 Adjacent sequences: A042961 A042962 A042963 this_sequence A042965 A042966 A042967 %K A042964 nonn,new %O A042964 1,1 %A A042964 N. J. A. Sloane (njas(AT)research.att.com). %E A042964 Edited by N. J. A. Sloane (njas(AT)research.att.com), Jun 30 2008 at the suggestion of R. J. Mathar Search completed in 0.002 seconds