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%I #17 Jul 05 2019 03:01:05
%S 0,0,1,2,7,8,13,18,27,35,50,61,75,79,96,113,120,150,173,180,204,227,
%T 245,274,295,318,346,363,398,438,448,484,524,537,584,625,648,707,749,
%U 771,830,882,914,983,1041,1073,1143,1207,1238,1307,1372,1405,1480,1544,1573,1645
%N a(n) is the number of palindromic numbers with 7 digits in base n which are also palindromic in base n+1.
%C If an integer m is palindromic in both bases n and n+1, then m has an odd number of digits in base n (see also A048268).
%C If m has 1, 3 or 5 digits in base n, the number of integers that are palindromic in bases n and n+1 is of order O(n) (see also A048268).
%C If m has at least 7 digits in base n, it seems that a(n) is of order O(n^2*log(n)).
%o (Python)
%o def nextpal(n,base): # m is the first palindrome successor of n in base base
%o m, pl = n+1, 0
%o while m > 0:
%o m, pl = m//base, pl+1
%o if n+1 == base**pl:
%o pl = pl+1
%o n = n//(base**(pl//2))+1
%o m, n = n, n//(base**(pl%2))
%o while n > 0:
%o m, n = m*base+n%base, n//base
%o return m
%o def ispal(n,b):
%o if n%b == 0:
%o return 0
%o else:
%o nn, m = n, 0
%o while n > 0:
%o n, m = n//b, m*b+n%b
%o return m == nn
%o n, d = 1, 7
%o while n < 20000:
%o n = n+1
%o p = n**(d-1)-1
%o a = 0
%o while p < n**d:
%o p = nextpal(p,n+1)
%o if ispal(p,n):
%o a = a+1
%o print(n,a)
%o (PARI) nextpal(n, b) = {my(m=n+1, p = 0); while (m > 0, m = m\b; p++;); if (n+1 == b^p, p++); n = n\(b^(p\2))+1; m = n; n = n\(b^(p%2)); while (n > 0, m = m*b + n%b; n = n\b;); m;} \\ after Python
%o ispal(n, b) = my(d=digits(n, b)); Vecrev(d) == d;
%o a(n) = {my(d=7, p = n^(d-1)-1, nb = 0); while (p < n^d, p = nextpal(p, n+1); if (ispal(p, n), nb++);); nb;} \\ _Michel Marcus_, Jul 04 2019
%Y Cf. A048268.
%K nonn,base
%O 2,4
%A _A.H.M. Smeets_, Jun 30 2019
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