A308918 a(n) is the number of palindromic numbers with 7 digits in base n which are also palindromic in base n+1.

0, 0, 1, 2, 7, 8, 13, 18, 27, 35, 50, 61, 75, 79, 96, 113, 120, 150, 173, 180, 204, 227, 245, 274, 295, 318, 346, 363, 398, 438, 448, 484, 524, 537, 584, 625, 648, 707, 749, 771, 830, 882, 914, 983, 1041, 1073, 1143, 1207, 1238, 1307, 1372, 1405, 1480, 1544, 1573, 1645

Offset: 2

Author

A.H.M. Smeets, Jun 30 2019

Keywords

• nonn
• base

Comments

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).

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).

If m has at least 7 digits in base n, it seems that a(n) is of order O(n^2*log(n)).

Crossrefs

Cf. A048268.

Programs

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
ispal(n, b) = my(d=digits(n, b)); Vecrev(d) == d;
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

Python

def nextpal(n,base): # m is the first palindrome successor of n in base base
    m, pl = n+1, 0
    while m > 0:
        m, pl = m//base, pl+1
    if n+1 == base**pl:
        pl = pl+1
    n = n//(base**(pl//2))+1
    m, n = n, n//(base**(pl%2))
    while n > 0:
        m, n = m*base+n%base, n//base
    return m
def ispal(n,b):
    if n%b == 0:
        return 0
    else:
        nn, m = n, 0
        while n > 0:
            n, m = n//b, m*b+n%b
        return m == nn
n, d = 1, 7
while n < 20000:
    n = n+1
    p = n**(d-1)-1
    a = 0
    while p < n**d:
        p = nextpal(p,n+1)
        if ispal(p,n):
            a = a+1
    print(n,a)