A370611 Largest square such that any two consecutive digits of its base-n expansion differ by 1 after arranging the digits in decreasing order.

1, 1, 225, 4, 38025, 751689, 10323369, 355624164, 9814072356, 279740499025, 8706730814089, 86847500601, 11027486960232964, 435408094460869201, 18362780530794065025, 2492638430009890761, 39207739576969100808801, 1972312183619434816475625, 104566626183621314286288961, 13215338757299095309775089

Offset: 2

Author

Jianing Song, Feb 23 2024

Keywords

• nonn
• base
• hard

Comments

By definition, a(n) <= Sum_{i=0..n-1} i*n^i = 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) <= Sum_{i=1..n-1} i*n^(i-1) = A051846(n-1); see A258103.

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

Examples

			See the Example section of A370371.
		

Crossrefs

Cf. A215014, A370370, A370610, A258103 (number of pandigital squares in base n).

The square roots are given by A370371.

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^2)))

Extensions

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

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