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

2,3

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

Links

Table of n, a(n) for n=2..21.

Example

See the Example section of A370371.

Prog

(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)))

Crossrefs

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

The square roots are given by A370371.

Cf. A062813, A051846.

Sequence in context: A204704 A206649 A340501 * A265432 A247885 A345552

Adjacent sequences: A370608 A370609 A370610 * A370636 A370637 A370638

Keyword

nonn,base,hard

Author

Jianing Song, Feb 23 2024

Extensions

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

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

Status

approved