0,1

1059831 = A075421(1105 ) is the fifth term of A075421 whose base 4 trajectory provably does not contain a palindrome. A proof along the lines of K. Brockhaus, On the 'Reverse and Add!' algorithm in base 2, can be based on the formula given below.

K. Brockhaus, On the 'Reverse and Add!' algorithm in base 2

Index entries for sequences related to Reverse and Add!

a(0), ..., a(7) as above; for n > 7 and n = 2 (mod 6): a(n) = 5*4^(2*k+9)+3836395*4^k-15 where k = (n+4)/6; n = 3 (mod 6): a(n) = 10*4^(2*k+9)+2450070*4^k-10 where k = (n+3)/6; n = 4 (mod 6): a(n) = 20*4^(2*k+9)-326420*4^k where k = (n+2)/6; n = 5 (mod 6): a(n) = 20*4^(2*k+9)+3544540*4^k-15 where k = (n+1)/6; n = 0 (mod 6): a(n) = 40*4^(2*k+9)+1927800*4^k-10 where k = n/6; n = 1 (mod 6): a(n) = 80*4^(2*k+9)-322580*4^k where k = (n-1)/6. G.f.: -3*(668508000*x^19+444361200*x^18+222142800*x^17-528080680*x^16-356464620*x^15 -125753060*x^14-299532884*x^13-188180432*x^12-143040640*x^11+128992350*x^10+90219415*x^9 +38288125*x^8+28112975*x^7+6666425*x^6+5752375*x^5+424135*x^4+3044705*x^3+2610355*x^2 + 1576104*x+353277)/((x-1)*(x^2+x+1)*(2*x^3-1)*(2*x^3+1)*(4*x^3-1))

1059831 (decimal) = 10002233313 -> 10002233313 + 31333220001 = 102002113320 = 4728312 (decimal).

(PARI) {m=1059831; stop=19; c=0; while(c<stop, print1(k=m, ", "); rev=0; while(k>0, d=divrem(k, 4); k=d[1]; rev=4*rev+d[2]); c++; m=m+rev)}

Cf. A075421, A075153, A075466, A075467, A076247.

Sequence in context: A036098 A043679 A076247 this_sequence A081638 A070042 A109148

Adjacent sequences: A076245 A076246 A076247 this_sequence A076249 A076250 A076251

base,nonn

Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Oct 03 2002