A370371 Largest m such that any two consecutive digits of the base-n expansion of m^2 differ by 1 after arranging the digits in decreasing order.

1, 1, 15, 2, 195, 867, 3213, 18858, 99066, 528905, 2950717, 294699, 105011842, 659854601, 4285181505, 1578809181, 198009443151, 1404390324525, 10225782424031, 3635290739033, 583655347579584, 4564790605900107, 36485812146621733, 297764406866494254, 2479167155959358950

Offset: 2

Author

Jianing Song, Feb 16 2024

Keywords

• nonn
• base
• hard

Comments

By definition, a(n) <= sqrt(Sum_{i=0..n-1} i*n^i) = sqrt(A062813(n)). If n is odd and n-1 has an even number of 2s as prime factors, then there are no pandigital squares in base n, so a(n) <= sqrt(Sum_{i=1..n-1} i*n^(i-1)) = sqrt(A051846(n-1)); see A258103.

If n is odd and n-1 has an even 2-adic valuation, then a(n) <= sqrt(Sum_{i=2..n-1} i*n^(i-2)); see A258103. - Chai Wah Wu, Feb 25 2024

Examples

			Base 4: 15^2 = 225 = 3201_4;
Base 6: 195^2 = 38025 = 452013_6;
Base 7: 867^2 = 751689 = 6250341_7;
Base 8: 3213^2 = 10323369 = 47302651_8;
Base 9: 18858^2 = 355624164 = 823146570_9;
Base 10: 99066^2 = 9814072356;
Base 11: 528905^2 = 279740499025 = A8701245369_11;
Base 12: 2950717^2 = 8706730814089 = B8750A649321_12;
Base 13: 294699^2 = 86847500601 = 8260975314_13.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 2..28

Crossrefs

Cf. A215014, A370362, A370370, A258103 (number of pandigital squares in base n).

The actual squares are given by A370611.

Cf. A062813, A051846.

Programs

PARI

isconsecutive(m,n)=my(v=vecsort(digits(m,n))); for(i=2, #v, if(v[i]!=1+v[i-1], return(0))); 1 \\ isconsecutive(k,n) == 1 if and only if any two consecutive digits of the base-n expansion of m differ by 1 after arranging the digits in decreasing order
a(n) = forstep(m=sqrtint(if(n%2==1 && valuation(n-1, 2)%2==0, n^(n-1) - (n^(n-1)-1)/(n-1)^2, n^n - (n^n-n)/(n-1)^2)), 0, -1, if(isconsecutive(m^2,n), return(m)))

Python

from math import isqrt
from sympy import multiplicity
from sympy.ntheory import digits
def a(n):
    ub = isqrt(sum(i*n**i for i in range(n)))
    if n%2 == 1 and multiplicity(2, n-1)%2 == 0:
        ub = isqrt(sum(i*n**(i-2) for i in range(2, n)))
    return(next(i for i in range(ub, -1, -1) if len(d:=sorted(digits(i*i, n)[1:])) == d[-1]-d[0]+1 == len(set(d))))
print([a(n) for n in range(2, 13)]) # Michael S. Branicky, Feb 23 2024

Extensions

a(17)-a(20) and a(22)-a(26) from Michael S. Branicky, Feb 23 2024

a(21) from Chai Wah Wu, Feb 25 2024