Search: id:A005191 Results 1-1 of 1 results found. %I A005191 M3891 %S A005191 1,1,5,19,85,381,1751,8135,38165,180325,856945,4091495,19611175,94309099, %T A005191 454805755,2198649549,10651488789,51698642405,251345549849,1223798004815, %U A005191 5966636799745,29125608152345,142330448514875,696235630761115 %N A005191 Central pentanomial coefficients: largest coefficient of (1+x+...+x^4)^n. %C A005191 Coefficient of x^n in ((1-x^10)/((1-x^5)(1-x^2)(1-x)))^n. - Michael Somos, Sep 24 2003 %C A005191 Note that n divides a(n+1)-a(n). - T. D. Noe (noe(AT)sspectra.com), Mar 16 2005 %C A005191 Terms that are not a multiple of 5 have zero density, namely, there are fewer than n^(log(4)/log(5)) such terms among A005191(1..n). In particular, A005191(5k+2) and A005191(5k+4) are multiples of 5 for every k. - Max Alekseyev (maxale(AT)gmail.com), Apr 25 2005 %D A005191 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). %D A005191 V. E. Hoggatt, Jr. and M. Bicknell, Diagonal sums of generalized Pascal triangles, Fib. Quart., 7 (1969), 341-358, 393. %H A005191 T. D. Noe, Table of n, a(n) for n=0..200 %F A005191 a(n) = sum(k=0..[2n/5], binomial(n, k)*binomial(-n, 2n-5k) ); a(n) = (5^n + sum(j=1..2n-1, (sin(5j*Pi/(2n))/sin(j*Pi/(2n)))^n))/(2n) - 2. - Max Alekseyev (maxale(AT)gmail.com), Mar 04 2005 %o A005191 (PARI) a(n)=if(n<0,0,polcoeff(((1-x^5)/(1-x)+x*O(x^(2*n)))^n,2*n)) %o A005191 (PARI) a(n)=if(n<0,0,polcoeff(((1-x^10)/((1-x^5)*(1-x^2)*(1-x))+x*O(x^n))^n, n)) %o A005191 (PARI) a(n) = sum(k=0,(2*n)\5,binomial(n,k)*binomial(-n,2*n-5*k)) a(n) = round((5^n+sum(j=1,2*n-1,(sin(5*Pi*j/2/n)/sin(Pi*j/2/n))^n))/2/ n)-2 (Alekseyev) %Y A005191 Cf. A001405, A002426, A005190, A018901, A025012, A025013, A025014 %Y A005191 Sequence in context: A149794 A149795 A149796 this_sequence A147091 A149797 A149798 %Y A005191 Adjacent sequences: A005188 A005189 A005190 this_sequence A005192 A005193 A005194 %K A005191 nonn %O A005191 0,3 %A A005191 N. J. A. Sloane (njas(AT)research.att.com). Search completed in 0.002 seconds