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First palindrome greater than n+2 in bases n+2 and n.
3

%I #12 Aug 10 2024 01:29:33

%S 5,26,21,24,154,40,121,60,181,84,253,112,337,144,433,180,541,220,661,

%T 264,793,312,937,364,1093,420,1261,480,1441,544,1633,612,1837,684,

%U 2053,760,2281,840,2521,924,2773,1012,3037,1104,3313,1200,3601,1300,3901

%N First palindrome greater than n+2 in bases n+2 and n.

%C a(2), a(3), a(4) and a(6) must be found explicitly.

%H Colin Barker, <a href="/A048269/b048269.txt">Table of n, a(n) for n = 2..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-3,0,1).

%F n even and n >= 8: a(n) = n^2+(n/2+3)*n+1 (which is (1 n/2+3 1) in base n and (1 n/2-2 1) in base n+2).

%F n odd and n >= 5: a(n) = (n+1)*(n+3)/2 (which is ((n+3)/2 (n+3)/2) in base n and ((n+1)/2 (n+1)/2) in base n+2).

%F From _Colin Barker_, Jun 30 2019: (Start)

%F G.f.: x^2*(5 + 26*x + 6*x^2 - 54*x^3 + 106*x^4 + 46*x^5 - 283*x^6 - 14*x^7 + 259*x^8 - 81*x^10) / ((1 - x)^3*(1 + x)^3).

%F a(n) = (5 + (-1)^(1 + n) + 2*(5 + (-1)^n)*n + 2*(2 + (-1)^n)*n^2) / 4 for n>6.

%F a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>10.

%F (End)

%e a(15)= (15+3)/2*15+(15+3)/2=144, which is (99) in base 15 and (88) in base 17.

%t Do[ k = n + 3; While[ RealDigits[ k, n + 2 ][[ 1 ] ] != Reverse[ RealDigits[ k, n + 2 ][[ 1 ] ] ] || RealDigits[ k, n ][[ 1 ] ] != Reverse[ RealDigits[ k, n ][[ 1 ] ] ], k++ ]; Print[ k ], {n, 2, 50} ]

%o (PARI) Vec(x^2*(5 + 26*x + 6*x^2 - 54*x^3 + 106*x^4 + 46*x^5 - 283*x^6 - 14*x^7 + 259*x^8 - 81*x^10) / ((1 - x)^3*(1 + x)^3) + O(x^50)) \\ _Colin Barker_, Jun 30 2019

%Y Cf. A048268.

%K nonn,easy,base

%O 2,1

%A Ulrich Schimke (ulrschimke(AT)aol.com)

%E More terms from _Robert G. Wilson v_, Aug 15 2000