Search: id:A129557 Results 1-1 of 1 results found. %I A129557 %S A129557 1,4,34,151,1291,5734,49024,217741,1861621,8268424,70692574,313982371, %T A129557 2684456191,11923061674,101938642684,452762361241,3870983965801, %U A129557 17193046665484,146995452057754,652883010927151,5581956194228851 %N A129557 Numbers k>0 such that k^2 is a centered pentagonal number. %C A129557 Corresponding numbers n such that centered pentagonal number A005891(n) = (5n^2+5n+2)/2 is a perfect square are listed in A129556(n) = {0, 2, 21, 95, 816, 3626, 31005, ...}. %H A129557 Eric Weisstein, Link to a section of The World of Mathematics, Centered Pentagonal Number. %F A129557 a(n) = Sqrt[ (5*A129556(n)^2 + 5*A129556(n) + 2)/2 ]. %F A129557 For n>=5, a(n) = 38*a(n-2) - a(n-4). [From Max Alekseyev (maxale(AT)gmail.com), May 08 2009] %t A129557 Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[ Sqrt[f] ] ], {n, 1,40000} ] %t A129557 q=5;s=0;lst={};Do[s+=n;If[Sqrt[q*s+1]==Floor[Sqrt[q*s+1]],AppendTo[lst, Sqrt[q*s+1]]],{n,0,8!}];lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Apr 02 2009] %o A129557 (PARI) A129557()={ for(n=1,1000000000, f=(5*n^2+5*n+2)/2 ; if(issquare(f), print(round(sqrt(f))) ; ); ) ; } A129557() ; - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 11 2007 %Y A129557 Cf. A005891 = Centered pentagonal numbers: (5n^2+5n+2)/2. Cf. A129556 = numbers n such that centered pentagonal number A005891(n) = (5n^2+5n+2)/ 2 is a perfect square. %Y A129557 Sequence in context: A053902 A054464 A002101 this_sequence A085695 A049293 A116430 %Y A129557 Adjacent sequences: A129554 A129555 A129556 this_sequence A129558 A129559 A129560 %K A129557 nonn %O A129557 1,2 %A A129557 Alexander Adamchuk (alex(AT)kolmogorov.com), Apr 20 2007 %E A129557 More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 11 2007 %E A129557 Formula and further terms from Max Alekseyev (maxale(AT)gmail.com), May 08 2009 Search completed in 0.001 seconds